# stacks/stacks-project

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 \input{preamble} % OK, start here. % \begin{document} \title{Morphisms of Algebraic Stacks} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent In this chapter we introduce some types of morphisms of algebraic stacks. A reference in the case of quasi-separated algebraic stacks with representable diagonal is \cite{LM-B}. \medskip\noindent The goal is to extend the definition of each of the types of morphisms of algebraic spaces to morphisms of algebraic stacks. Each case is slightly different and it seems best to treat them all separately. \medskip\noindent For morphisms of algebraic stacks which are representable by algebraic spaces we have already defined a large number of types of morphisms, see Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. For each corresponding case in this chapter we have to make sure the definition in the general case is compatible with the definition given there. \section{Conventions and abuse of language} \label{section-conventions} \noindent We continue to use the conventions and the abuse of language introduced in Properties of Stacks, Section \ref{stacks-properties-section-conventions}. \section{Properties of diagonals} \label{section-diagonals} \noindent The diagonal of an algebraic stack is closely related to the $\mathit{Isom}$-sheaves, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-diagonal}. By the second defining property of an algebraic stack these $\mathit{Isom}$-sheaves are always algebraic spaces. \begin{lemma} \label{lemma-isom-locally-finite-type} Let $\mathcal{X}$ be an algebraic stack. Let $T$ be a scheme and let $x, y$ be objects of the fibre category of $\mathcal{X}$ over $T$. Then the morphism $\mathit{Isom}_\mathcal{X}(x, y) \to T$ is locally of finite type. \end{lemma} \begin{proof} By Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation} we may assume that $\mathcal{X} = [U/R]$ for some smooth groupoid in algebraic spaces. By Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-locally-finite-type} it suffices to check the property fppf locally on $T$. Thus we may assume that $x, y$ come from morphisms $x', y' : T \to U$. By Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-morphisms} we see that in this case $\mathit{Isom}_\mathcal{X}(x, y) = T \times_{(y', x'), U \times_S U} R$. Hence it suffices to prove that $R \to U \times_S U$ is locally of finite type. This follows from the fact that the composition $s : R \to U \times_S U \to U$ is smooth (hence locally of finite type, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation} and \ref{spaces-morphisms-lemma-finite-presentation-finite-type}) and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. \end{proof} \begin{lemma} \label{lemma-isom-pseudo-torsor-aut} Let $\mathcal{X}$ be an algebraic stack. Let $T$ be a scheme and let $x, y$ be objects of the fibre category of $\mathcal{X}$ over $T$. Then \begin{enumerate} \item $\mathit{Isom}_\mathcal{X}(y, y)$ is a group algebraic space over $T$, and \item $\mathit{Isom}_\mathcal{X}(x, y)$ is a pseudo torsor for $\mathit{Isom}_\mathcal{X}(y, y)$ over $T$. \end{enumerate} \end{lemma} \begin{proof} See Groupoids in Spaces, Definitions \ref{spaces-groupoids-definition-group-space} and \ref{spaces-groupoids-definition-pseudo-torsor}. The lemma follows immediately from the fact that $\mathcal{X}$ is a stack in groupoids. \end{proof} \noindent Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The {\it diagonal of $f$} is the morphism $$\Delta_f : \mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Y} \mathcal{X}$$ Here are two properties that every diagonal morphism has. \begin{lemma} \label{lemma-properties-diagonal} \begin{slogan} Diagonals of morphisms of algebraic stacks are representable by algebraic spaces and locally of finite type. \end{slogan} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then \begin{enumerate} \item $\Delta_f$ is representable by algebraic spaces, and \item $\Delta_f$ is locally of finite type. \end{enumerate} \end{lemma} \begin{proof} Let $T$ be a scheme and let $a : T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ be a morphism. By definition of the fibre product and the $2$-Yoneda lemma the morphism $a$ is given by a triple $a = (x, x', \alpha)$ where $x, x'$ are objects of $\mathcal{X}$ over $T$, and $\alpha : f(x) \to f(x')$ is a morphism in the fibre category of $\mathcal{Y}$ over $T$. By definition of an algebraic stack the sheaves $\mathit{Isom}_\mathcal{X}(x, x')$ and $\mathit{Isom}_\mathcal{Y}(f(x), f(x'))$ are algebraic spaces over $T$. In this language $\alpha$ defines a section of the morphism $\mathit{Isom}_\mathcal{Y}(f(x), f(x')) \to T$. A $T'$-valued point of $\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T$ for $T' \to T$ a scheme over $T$ is the same thing as an isomorphism $x|_{T'} \to x'|_{T'}$ whose image under $f$ is $\alpha|_{T'}$. Thus we see that \begin{equation} \label{equation-diagonal} \vcenter{ \xymatrix{ \mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T \ar[d] \ar[r] & \mathit{Isom}_\mathcal{X}(x, x') \ar[d] \\ T\ar[r]^-\alpha & \mathit{Isom}_\mathcal{Y}(f(x), f(x')) } } \end{equation} is a fibre square of sheaves over $T$. In particular we see that $\mathcal{X} \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}, a} T$ is an algebraic space which proves part (1) of the lemma. \medskip\noindent To prove the second statement we have to show that the left vertical arrow of Diagram (\ref{equation-diagonal}) is locally of finite type. By Lemma \ref{lemma-isom-locally-finite-type} the algebraic space $\mathit{Isom}_\mathcal{X}(x, x')$ and is locally of finite type over $T$. Hence the right vertical arrow of Diagram (\ref{equation-diagonal}) is locally of finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. We conclude by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type}. \end{proof} \begin{lemma} \label{lemma-properties-diagonal-representable} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. Then \begin{enumerate} \item $\Delta_f$ is representable (by schemes), \item $\Delta_f$ is locally of finite type, \item $\Delta_f$ is a monomorphism, \item $\Delta_f$ is separated, and \item $\Delta_f$ is locally quasi-finite. \end{enumerate} \end{lemma} \begin{proof} We have already seen in Lemma \ref{lemma-properties-diagonal} that $\Delta_f$ is representable by algebraic spaces. Hence the statements (2) -- (5) make sense, see Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Also Lemma \ref{lemma-properties-diagonal} guarantees (2) holds. Let $T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ be a morphism and contemplate Diagram (\ref{equation-diagonal}). By Algebraic Stacks, Lemma \ref{algebraic-lemma-criterion-map-representable-spaces-fibred-in-groupoids} the right vertical arrow is injective as a map of sheaves, i.e., a monomorphism of algebraic spaces. Hence also the morphism $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T$ is a monomorphism. Thus (3) holds. We already know that $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T$ is locally of finite type. Thus Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-loc-finite-type-loc-quasi-finite} allows us to conclude that $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X} \to T$ is locally quasi-finite and separated. This proves (4) and (5). Finally, Morphisms of Spaces, Proposition \ref{spaces-morphisms-proposition-locally-quasi-finite-separated-over-scheme} implies that $T \times_{\mathcal{X} \times_\mathcal{Y} \mathcal{X}} \mathcal{X}$ is a scheme which proves (1). \end{proof} \begin{lemma} \label{lemma-representable-separated-diagonal-closed} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent \begin{enumerate} \item $f$ is separated, \item $\Delta_f$ is a closed immersion, \item $\Delta_f$ is proper, or \item $\Delta_f$ is universally closed. \end{enumerate} \end{lemma} \begin{proof} The statements $f$ is separated'', $\Delta_f$ is a closed immersion'', $\Delta_f$ is universally closed'', and $\Delta_f$ is proper'' refer to the notions defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times_\mathcal{Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma \ref{categories-lemma-base-change-diagonal} and Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering} we see that for any property $P$ (as in that lemma) we have: $\Delta_f$ has $P$ if and only if $\Delta_{U/V} : U \to U \times_V U$ has $P$. Hence the equivalence of (2), (3) and (4) follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-separated-diagonal-proper} applied to $U \to V$. Moreover, if (1) holds, then $U \to V$ is separated and we see that $\Delta_{U/V}$ is a closed immersion, i.e., (2) holds. Finally, assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times_\mathcal{Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is separated. Using Categories, Lemma \ref{categories-lemma-base-change-diagonal} once more we see that $\Delta_{T'/T}$ is the base change of $\Delta_f$. Hence our assumption (2) implies that $\Delta_{T'/T}$ is a closed immersion, hence $T' \to T$ is separated as desired. \end{proof} \begin{lemma} \label{lemma-representable-quasi-separated-diagonal-quasi-compact} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent \begin{enumerate} \item $f$ is quasi-separated, \item $\Delta_f$ is quasi-compact, or \item $\Delta_f$ is of finite type. \end{enumerate} \end{lemma} \begin{proof} The statements $f$ is quasi-separated'', $\Delta_f$ is quasi-compact'', and $\Delta_f$ is of finite type'' refer to the notions defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Note that (2) and (3) are equivalent in view of the fact that $\Delta_f$ is locally of finite type by Lemma \ref{lemma-properties-diagonal-representable} (and Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-transformations-property-implication}). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = \mathcal{X} \times_\mathcal{Y} V$ which is an algebraic space by assumption, and the morphism $U \to \mathcal{X}$ is surjective and smooth. By Categories, Lemma \ref{categories-lemma-base-change-diagonal} and Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering} we see that we have: $\Delta_f$ is quasi-compact if and only if $\Delta_{U/V} : U \to U \times_V U$ is quasi-compact. If (1) holds, then $U \to V$ is quasi-separated and we see that $\Delta_{U/V}$ is quasi-compact, i.e., (2) holds. Assume (2) holds. Let $T$ be a scheme, and $a : T \to \mathcal{Y}$ a morphism. Set $T' = \mathcal{X} \times_\mathcal{Y} T$. To prove (1) we have to show that the morphism of algebraic spaces $T' \to T$ is quasi-separated. Using Categories, Lemma \ref{categories-lemma-base-change-diagonal} once more we see that $\Delta_{T'/T}$ is the base change of $\Delta_f$. Hence our assumption (2) implies that $\Delta_{T'/T}$ is quasi-compact, hence $T' \to T$ is quasi-separated as desired. \end{proof} \begin{lemma} \label{lemma-representable-locally-separated-diagonal-immersion} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then the following are equivalent \begin{enumerate} \item $f$ is locally separated, and \item $\Delta_f$ is an immersion. \end{enumerate} \end{lemma} \begin{proof} The statements $f$ is locally separated'', and $\Delta_f$ is an immersion'' refer to the notions defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. Proof omitted. Hint: Argue as in the proofs of Lemmas \ref{lemma-representable-separated-diagonal-closed} and \ref{lemma-representable-quasi-separated-diagonal-quasi-compact}. \end{proof} \section{Separation axioms} \label{section-separated} \noindent Let $\mathcal{X} = [U/R]$ be a presentation of an algebraic stack. Then the properties of the diagonal of $\mathcal{X}$ over $S$, are the properties of the morphism $j : R \to U \times_S U$. For example, if $\mathcal{X} = [S/G]$ for some smooth group $G$ in algebraic spaces over $S$ then $j$ is the structure morphism $G \to S$. Hence the diagonal is not automatically separated itself (contrary to what happens in the case of schemes and algebraic spaces). To say that $[S/G]$ is quasi-separated over $S$ should certainly imply that $G \to S$ is quasi-compact, but we hesitate to say that $[S/G]$ is quasi-separated over $S$ without also requiring the morphism $G \to S$ to be quasi-separated. In other words, requiring the diagonal morphism to be quasi-compact does not really agree with our intuition for a quasi-separated algebraic stack'', and we should also require the diagonal itself to be quasi-separated. \medskip\noindent What about separated algebraic stacks''? We have seen in Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-separated-diagonal-proper} that an algebraic space is separated if and only if the diagonal is proper. This is the condition that is usually used to define separated algebraic stacks too. In the example $[S/G] \to S$ above this means that $G \to S$ is a proper group scheme. This means algebraic stacks of the form $[\Spec(k)/E]$ are proper over $k$ where $E$ is an elliptic curve over $k$ (insert future reference here). In certain situations it may be more natural to assume the diagonal is finite. \begin{definition} \label{definition-separated} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item We say $f$ is {\it DM} if $\Delta_f$ is unramified\footnote{The letters DM stand for Deligne-Mumford. If $f$ is DM then given any scheme $T$ and any morphism $T \to \mathcal{Y}$ the fibre product $\mathcal{X}_T = \mathcal{X} \times_\mathcal{Y} T$ is an algebraic stack over $T$ whose diagonal is unramified, i.e., $\mathcal{X}_T$ is DM. This implies $\mathcal{X}_T$ is a Deligne-Mumford stack, see Theorem \ref{theorem-DM}. In other words a DM morphism is one whose fibres'' are Deligne-Mumford stacks. This hopefully at least motivates the terminology.}. \item We say $f$ is {\it quasi-DM} if $\Delta_f$ is locally quasi-finite\footnote{If $f$ is quasi-DM, then the fibres'' $\mathcal{X}_T$ of $\mathcal{X} \to \mathcal{Y}$ are quasi-DM. An algebraic stack $\mathcal{X}$ is quasi-DM exactly if there exists a scheme $U$ and a surjective flat morphism $U \to \mathcal{X}$ of finite presentation which is locally quasi-finite, see Theorem \ref{theorem-quasi-DM}. Note the similarity to being Deligne-Mumford, which is defined in terms of having an \'etale covering by a scheme.}. \item We say $f$ is {\it separated} if $\Delta_f$ is proper. \item We say $f$ is {\it quasi-separated} if $\Delta_f$ is quasi-compact and quasi-separated. \end{enumerate} \end{definition} \noindent In this definition we are using that $\Delta_f$ is representable by algebraic spaces and we are using Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} to make sense out of imposing conditions on $\Delta_f$. We note that these definitions do not conflict with the already existing notions if $f$ is representable by algebraic spaces, see Lemmas \ref{lemma-representable-quasi-separated-diagonal-quasi-compact} and \ref{lemma-representable-separated-diagonal-closed}. There is an interesting way to characterize these conditions by looking at higher diagonals, see Lemma \ref{lemma-definition-separated}. \begin{definition} \label{definition-absolute-separated} Let $\mathcal{X}$ be an algebraic stack over the base scheme $S$. Denote $p : \mathcal{X} \to S$ the structure morphism. \begin{enumerate} \item We say $\mathcal{X}$ is {\it DM over $S$} if $p : \mathcal{X} \to S$ is DM. \item We say $\mathcal{X}$ is {\it quasi-DM over $S$} if $p : \mathcal{X} \to S$ is quasi-DM. \item We say $\mathcal{X}$ is {\it separated over $S$} if $p : \mathcal{X} \to S$ is separated. \item We say $\mathcal{X}$ is {\it quasi-separated over $S$} if $p : \mathcal{X} \to S$ is quasi-separated. \item We say $\mathcal{X}$ is {\it DM} if $\mathcal{X}$ is DM\footnote{Theorem \ref{theorem-DM} shows that this is equivalent to $\mathcal{X}$ being a Deligne-Mumford stack.} over $\Spec(\mathbf{Z})$. \item We say $\mathcal{X}$ is {\it quasi-DM} if $\mathcal{X}$ is quasi-DM over $\Spec(\mathbf{Z})$. \item We say $\mathcal{X}$ is {\it separated} if $\mathcal{X}$ is separated over $\Spec(\mathbf{Z})$. \item We say $\mathcal{X}$ is {\it quasi-separated} if $\mathcal{X}$ is quasi-separated over $\Spec(\mathbf{Z})$. \end{enumerate} In the last 4 definitions we view $\mathcal{X}$ as an algebraic stack over $\Spec(\mathbf{Z})$ via Algebraic Stacks, Definition \ref{algebraic-definition-viewed-as}. \end{definition} \noindent Thus in each case we have an absolute notion and a notion relative to our given base scheme (mention of which is usually suppressed by our abuse of notation introduced in Properties of Stacks, Section \ref{stacks-properties-section-conventions}). We will see that (1) $\Leftrightarrow$ (5) and (2) $\Leftrightarrow$ (6) in Lemma \ref{lemma-separated-implies-morphism-separated}. We spend some time proving some standard results on these notions. \begin{lemma} \label{lemma-trivial-implications} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item If $f$ is separated, then $f$ is quasi-separated. \item If $f$ is DM, then $f$ is quasi-DM. \item If $f$ is representable by algebraic spaces, then $f$ is DM. \end{enumerate} \end{lemma} \begin{proof} To see (1) note that a proper morphism of algebraic spaces is quasi-compact and quasi-separated, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-proper}. To see (2) note that an unramified morphism of algebraic spaces is locally quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-quasi-finite}. Finally (3) follows from Lemma \ref{lemma-properties-diagonal-representable}. \end{proof} \begin{lemma} \label{lemma-base-change-separated} All of the separation axioms listed in Definition \ref{definition-separated} are stable under base change. \end{lemma} \begin{proof} Let $f : \mathcal{X} \to \mathcal{Y}$ and $\mathcal{Y}' \to \mathcal{Y}$ be morphisms of algebraic stacks. Let $f' : \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$ be the base change of $f$ by $\mathcal{Y}' \to \mathcal{Y}$. Then $\Delta_{f'}$ is the base change of $\Delta_f$ by the morphism $\mathcal{X}' \times_{\mathcal{Y}'} \mathcal{X}' \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$, see Categories, Lemma \ref{categories-lemma-base-change-diagonal}. By the results of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} each of the properties of the diagonal used in Definition \ref{definition-separated} is stable under base change. Hence the lemma is true. \end{proof} \begin{lemma} \label{lemma-check-separated-covering} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be a surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times_\mathcal{Y} \mathcal{X} \to W$ has one of the separation properties of Definition \ref{definition-separated} then so does $f$. \end{lemma} \begin{proof} Denote $g : W \times_\mathcal{Y} \mathcal{X} \to W$ the base change. Then $\Delta_g$ is the base change of $\Delta_f$ by the morphism $q : W \times_\mathcal{Y} (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$. Since $q$ is the base change of $W \to \mathcal{Y}$ we see that $q$ is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Hence the result follows from Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-weak-covering}. \end{proof} \begin{lemma} \label{lemma-change-of-base-separated} Let $S$ be a scheme. The property of being quasi-DM over $S$, quasi-separated over $S$, or separated over $S$ (see Definition \ref{definition-absolute-separated}) is stable under change of base scheme, see Algebraic Stacks, Definition \ref{algebraic-definition-change-of-base}. \end{lemma} \begin{proof} Follows immediately from Lemma \ref{lemma-base-change-separated}. \end{proof} \begin{lemma} \label{lemma-fibre-product-after-map} Let $f : \mathcal{X} \to \mathcal{Z}$, $g : \mathcal{Y} \to \mathcal{Z}$ and $\mathcal{Z} \to \mathcal{T}$ be morphisms of algebraic stacks. Consider the induced morphism $i : \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{X} \times_\mathcal{T} \mathcal{Y}$. Then \begin{enumerate} \item $i$ is representable by algebraic spaces and locally of finite type, \item if $\Delta_{\mathcal{Z}/\mathcal{T}}$ is quasi-separated, then $i$ is quasi-separated, \item if $\Delta_{\mathcal{Z}/\mathcal{T}}$ is separated, then $i$ is separated, \item if $\mathcal{Z} \to \mathcal{T}$ is DM, then $i$ is unramified, \item if $\mathcal{Z} \to \mathcal{T}$ is quasi-DM, then $i$ is locally quasi-finite, \item if $\mathcal{Z} \to \mathcal{T}$ is separated, then $i$ is proper, and \item if $\mathcal{Z} \to \mathcal{T}$ is quasi-separated, then $i$ is quasi-compact and quasi-separated. \end{enumerate} \end{lemma} \begin{proof} The following diagram $$\xymatrix{ \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \ar[r]_i \ar[d] & \mathcal{X} \times_\mathcal{T} \mathcal{Y} \ar[d] \\ \mathcal{Z} \ar[r]^-{\Delta_{\mathcal{Z}/\mathcal{T}}} \ar[r] & \mathcal{Z} \times_\mathcal{T} \mathcal{Z} }$$ is a $2$-fibre product diagram, see Categories, Lemma \ref{categories-lemma-fibre-product-after-map}. Hence $i$ is the base change of the diagonal morphism $\Delta_{\mathcal{Z}/\mathcal{T}}$. Thus the lemma follows from Lemma \ref{lemma-properties-diagonal}, and the material in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \end{proof} \begin{lemma} \label{lemma-semi-diagonal} Let $\mathcal{T}$ be an algebraic stack. Let $g : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks over $\mathcal{T}$. Consider the graph $i : \mathcal{X} \to \mathcal{X} \times_\mathcal{T} \mathcal{Y}$ of $g$. Then \begin{enumerate} \item $i$ is representable by algebraic spaces and locally of finite type, \item if $\mathcal{Y} \to \mathcal{T}$ is DM, then $i$ is unramified, \item if $\mathcal{Y} \to \mathcal{T}$ is quasi-DM, then $i$ is locally quasi-finite, \item if $\mathcal{Y} \to \mathcal{T}$ is separated, then $i$ is proper, and \item if $\mathcal{Y} \to \mathcal{T}$ is quasi-separated, then $i$ is quasi-compact and quasi-separated. \end{enumerate} \end{lemma} \begin{proof} This is a special case of Lemma \ref{lemma-fibre-product-after-map} applied to the morphism $\mathcal{X} = \mathcal{X} \times_\mathcal{Y} \mathcal{Y} \to \mathcal{X} \times_\mathcal{T} \mathcal{Y}$. \end{proof} \begin{lemma} \label{lemma-section-immersion} Let $f : \mathcal{X} \to \mathcal{T}$ be a morphism of algebraic stacks. Let $s : \mathcal{T} \to \mathcal{X}$ be a morphism such that $f \circ s$ is $2$-isomorphic to $\text{id}_\mathcal{T}$. Then \begin{enumerate} \item $s$ is representable by algebraic spaces and locally of finite type, \item if $f$ is DM, then $s$ is unramified, \item if $f$ is quasi-DM, then $s$ is locally quasi-finite, \item if $f$ is separated, then $s$ is proper, and \item if $f$ is quasi-separated, then $s$ is quasi-compact and quasi-separated. \end{enumerate} \end{lemma} \begin{proof} This is a special case of Lemma \ref{lemma-semi-diagonal} applied to $g = s$ and $\mathcal{Y} = \mathcal{T}$ in which case $i : \mathcal{T} \to \mathcal{T} \times_\mathcal{T} \mathcal{X}$ is $2$-isomorphic to $s$. \end{proof} \begin{lemma} \label{lemma-composition-separated} All of the separation axioms listed in Definition \ref{definition-separated} are stable under composition of morphisms. \end{lemma} \begin{proof} Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks to which the axiom in question applies. The diagonal $\Delta_{\mathcal{X}/\mathcal{Z}}$ is the composition $$\mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Y} \mathcal{X} \longrightarrow \mathcal{X} \times_\mathcal{Z} \mathcal{X}.$$ Our separation axiom is defined by requiring the diagonal to have some property $\mathcal{P}$. By Lemma \ref{lemma-fibre-product-after-map} above we see that the second arrow also has this property. Hence the lemma follows since the composition of morphisms which are representable by algebraic spaces with property $\mathcal{P}$ also is a morphism with property $\mathcal{P}$, see our general discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} and Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-unramified}, \ref{spaces-morphisms-lemma-composition-quasi-finite}, \ref{spaces-morphisms-lemma-composition-proper}, \ref{spaces-morphisms-lemma-composition-quasi-compact}, and \ref{spaces-morphisms-lemma-composition-separated}. \end{proof} \begin{lemma} \label{lemma-separated-over-separated} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks over the base scheme $S$. \begin{enumerate} \item If $\mathcal{Y}$ is DM over $S$ and $f$ is DM, then $\mathcal{X}$ is DM over $S$. \item If $\mathcal{Y}$ is quasi-DM over $S$ and $f$ is quasi-DM, then $\mathcal{X}$ is quasi-DM over $S$. \item If $\mathcal{Y}$ is separated over $S$ and $f$ is separated, then $\mathcal{X}$ is separated over $S$. \item If $\mathcal{Y}$ is quasi-separated over $S$ and $f$ is quasi-separated, then $\mathcal{X}$ is quasi-separated over $S$. \item If $\mathcal{Y}$ is DM and $f$ is DM, then $\mathcal{X}$ is DM. \item If $\mathcal{Y}$ is quasi-DM and $f$ is quasi-DM, then $\mathcal{X}$ is quasi-DM. \item If $\mathcal{Y}$ is separated and $f$ is separated, then $\mathcal{X}$ is separated. \item If $\mathcal{Y}$ is quasi-separated and $f$ is quasi-separated, then $\mathcal{X}$ is quasi-separated. \end{enumerate} \end{lemma} \begin{proof} Parts (1), (2), (3), and (4) follow immediately from Lemma \ref{lemma-composition-separated} and Definition \ref{definition-absolute-separated}. For (5), (6), (7), and (8) think of $\mathcal{X}$ and $\mathcal{Y}$ as algebraic stacks over $\Spec(\mathbf{Z})$ and apply Lemma \ref{lemma-composition-separated}. Details omitted. \end{proof} \noindent The following lemma is a bit different to the analogue for algebraic spaces. To compare take a look at Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-compose-after-separated}. \begin{lemma} \label{lemma-compose-after-separated} Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. \begin{enumerate} \item If $g \circ f$ is DM then so is $f$. \item If $g \circ f$ is quasi-DM then so is $f$. \item If $g \circ f$ is separated and $\Delta_g$ is separated, then $f$ is separated. \item If $g \circ f$ is quasi-separated and $\Delta_g$ is quasi-separated, then $f$ is quasi-separated. \end{enumerate} \end{lemma} \begin{proof} Consider the factorization $$\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}$$ of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces, see Lemmas \ref{lemma-properties-diagonal} and \ref{lemma-fibre-product-after-map}. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ we get morphisms of algebraic spaces $$A = \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow T.$$ If $g \circ f$ is DM (resp.\ quasi-DM), then the composition $A \to T$ is unramified (resp.\ locally quasi-finite). Hence (1) (resp.\ (2)) follows on applying Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-unramified} (resp. Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-quasi-finite}). This proves (1) and (2). \medskip\noindent Proof of (4). Assume $g \circ f$ is quasi-separated and $\Delta_g$ is quasi-separated. Consider the factorization $$\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}$$ of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces and the second one is quasi-separated, see Lemmas \ref{lemma-properties-diagonal} and \ref{lemma-fibre-product-after-map}. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ we get morphisms of algebraic spaces $$A = \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow T$$ such that $B \to T$ is quasi-separated. The composition $A \to T$ is quasi-compact and quasi-separated as we have assumed that $g \circ f$ is quasi-separated. Hence $A \to B$ is quasi-separated by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-compose-after-separated}. And $A \to B$ is quasi-compact by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-permanence}. Thus $f$ is quasi-separated. \medskip\noindent Proof of (3). Assume $g \circ f$ is separated and $\Delta_g$ is separated. Consider the factorization $$\mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{X}$$ of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces and the second one is separated, see Lemmas \ref{lemma-properties-diagonal} and \ref{lemma-fibre-product-after-map}. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ we get morphisms of algebraic spaces $$A = \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times_\mathcal{Y} \mathcal{X}) \times_{(\mathcal{X} \times_\mathcal{Z} \mathcal{X})} T \longrightarrow T$$ such that $B \to T$ is separated. The composition $A \to T$ is proper as we have assumed that $g \circ f$ is quasi-separated. Hence $A \to B$ is proper by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-closed-permanence} which means that $f$ is separated. \end{proof} \begin{lemma} \label{lemma-separated-implies-morphism-separated} Let $\mathcal{X}$ be an algebraic stack over the base scheme $S$. \begin{enumerate} \item $\mathcal{X}$ is DM $\Leftrightarrow$ $\mathcal{X}$ is DM over $S$. \item $\mathcal{X}$ is quasi-DM $\Leftrightarrow$ $\mathcal{X}$ is quasi-DM over $S$. \item If $\mathcal{X}$ is separated, then $\mathcal{X}$ is separated over $S$. \item If $\mathcal{X}$ is quasi-separated, then $\mathcal{X}$ is quasi-separated over $S$. \end{enumerate} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks over the base scheme $S$. \begin{enumerate} \item[(5)] If $\mathcal{X}$ is DM over $S$, then $f$ is DM. \item[(6)] If $\mathcal{X}$ is quasi-DM over $S$, then $f$ is quasi-DM. \item[(7)] If $\mathcal{X}$ is separated over $S$ and $\Delta_{\mathcal{Y}/S}$ is separated, then $f$ is separated. \item[(8)] If $\mathcal{X}$ is quasi-separated over $S$ and $\Delta_{\mathcal{Y}/S}$ is quasi-separated, then $f$ is quasi-separated. \end{enumerate} \end{lemma} \begin{proof} Parts (5), (6), (7), and (8) follow immediately from Lemma \ref{lemma-compose-after-separated} and Spaces, Definition \ref{spaces-definition-separated}. To prove (3) and (4) think of $X$ and $Y$ as algebraic stacks over $\Spec(\mathbf{Z})$ and apply Lemma \ref{lemma-compose-after-separated}. Similarly, to prove (1) and (2), think of $\mathcal{X}$ as an algebraic stack over $\Spec(\mathbf{Z})$ consider the morphisms $$\mathcal{X} \longrightarrow \mathcal{X} \times_S \mathcal{X} \longrightarrow \mathcal{X} \times_{\Spec(\mathbf{Z})} \mathcal{X}$$ Both arrows are representable by algebraic spaces. The second arrow is unramified and locally quasi-finite as the base change of the immersion $\Delta_{S/\mathbf{Z}}$. Hence the composition is unramified (resp.\ locally quasi-finite) if and only if the first arrow is unramified (resp.\ locally quasi-finite), see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-unramified} and \ref{spaces-morphisms-lemma-permanence-unramified} (resp.\ Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-quasi-finite} and \ref{spaces-morphisms-lemma-permanence-quasi-finite}). \end{proof} \begin{lemma} \label{lemma-properties-covering-imply-diagonal} Let $\mathcal{X}$ be an algebraic stack. Let $W$ be an algebraic space, and let $f : W \to \mathcal{X}$ be a surjective, flat, locally finitely presented morphism. \begin{enumerate} \item If $f$ is unramified (i.e., \'etale, i.e., $\mathcal{X}$ is Deligne-Mumford), then $\mathcal{X}$ is DM. \item If $f$ is locally quasi-finite, then $\mathcal{X}$ is quasi-DM. \end{enumerate} \end{lemma} \begin{proof} Note that if $f$ is unramified, then it is \'etale by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-flat-lfp-etale}. This explains the parenthetical remark in (1). Assume $f$ is unramified (resp.\ locally quasi-finite). We have to show that $\Delta_\mathcal{X} : \mathcal{X} \to \mathcal{X} \times \mathcal{X}$ is unramified (resp.\ locally quasi-finite). Note that $W \times W \to \mathcal{X} \times \mathcal{X}$ is also surjective, flat, and locally of finite presentation. Hence it suffices to show that $$W \times_{\mathcal{X} \times \mathcal{X}, \Delta_\mathcal{X}} \mathcal{X} = W \times_\mathcal{X} W \longrightarrow W \times W$$ is unramified (resp.\ locally quasi-finite), see Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-covering}. By assumption the morphism $\text{pr}_i : W \times_\mathcal{X} W \to W$ is unramified (resp.\ locally quasi-finite). Hence the displayed arrow is unramified (resp.\ locally quasi-finite) by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-unramified} (resp.\ Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-quasi-finite}). \end{proof} \begin{lemma} \label{lemma-monomorphism-separated} A monomorphism of algebraic stacks is separated and DM. The same is true for immersions of algebraic stacks. \end{lemma} \begin{proof} If $f : \mathcal{X} \to \mathcal{Y}$ is a monomorphism of algebraic stacks, then $\Delta_f$ is an isomorphism, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism}. Since an isomorphism of algebraic spaces is proper and unramified we see that $f$ is separated and DM. The second assertion follows from the first as an immersion is a monomorphism, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-immersion-monomorphism}. \end{proof} \begin{lemma} \label{lemma-separation-properties-residual-gerbe} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Assume the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ at $x$ exists. If $\mathcal{X}$ is DM, resp.\ quasi-DM, resp.\ separated, resp.\ quasi-separated, then so is $\mathcal{Z}_x$. \end{lemma} \begin{proof} This is true because $\mathcal{Z}_x \to \mathcal{X}$ is a monomorphism hence DM and separated by Lemma \ref{lemma-monomorphism-separated}. Apply Lemma \ref{lemma-separated-over-separated} to conclude. \end{proof} \section{Inertia stacks} \label{section-inertia} \noindent The (relative) inertia stack of a stack in groupoids is defined in Stacks, Section \ref{stacks-section-the-inertia-stack}. The actual construction, in the setting of fibred categories, and some of its properties is in Categories, Section \ref{categories-section-inertia}. \begin{lemma} \label{lemma-inertia} Let $\mathcal{X}$ be an algebraic stack. Then the inertia stack $\mathcal{I}_\mathcal{X}$ is an algebraic stack as well. The morphism $$\mathcal{I}_\mathcal{X} \longrightarrow \mathcal{X}$$ is representable by algebraic spaces and locally of finite type. More generally, let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then the relative inertia $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic stack and the morphism $$\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \longrightarrow \mathcal{X}$$ is representable by algebraic spaces and locally of finite type. \end{lemma} \begin{proof} By Categories, Lemma \ref{categories-lemma-inertia-fibred-category} there are equivalences $$\mathcal{I}_\mathcal{X} \to \mathcal{X} \times_{\Delta, \mathcal{X} \times_S \mathcal{X}, \Delta} \mathcal{X} \quad\text{and}\quad \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X} \times_{\Delta, \mathcal{X} \times_\mathcal{Y} \mathcal{X}, \Delta} \mathcal{X}$$ which shows that the inertia stacks are algebraic stacks. Let $T \to \mathcal{X}$ be a morphism given by the object $x$ of the fibre category of $\mathcal{X}$ over $T$. Then we get a $2$-fibre product square $$\xymatrix{ \mathit{Isom}_\mathcal{X}(x, x) \ar[d] \ar[r] & \mathcal{I}_\mathcal{X} \ar[d] \\ T \ar[r]^x & \mathcal{X} }$$ This follows immediately from the definition of $\mathcal{I}_\mathcal{X}$. Since $\mathit{Isom}_\mathcal{X}(x, x)$ is always an algebraic space locally of finite type over $T$ (see Lemma \ref{lemma-isom-locally-finite-type}) we conclude that $\mathcal{I}_\mathcal{X} \to \mathcal{X}$ is representable by algebraic spaces and locally of finite type. Finally, for the relative inertia we get $$\vcenter{ \xymatrix{ \mathit{Isom}_\mathcal{X}(x, x) \ar[d] & K \ar[l] \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ \mathit{Isom}_\mathcal{Y}(f(x), f(x)) & T \ar[l]_-e \ar[r]^x & \mathcal{X} } }$$ with both squares $2$-fibre products. This follows from Categories, Lemma \ref{categories-lemma-relative-inertia-as-fibre-product}. The left vertical arrow is a morphism of algebraic spaces locally of finite type over $T$, and hence is locally of finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-permanence-finite-type}. Thus $K$ is an algebraic space and $K \to T$ is locally of finite type. This proves the assertion on the relative inertia. \end{proof} \begin{remark} \label{remark-inertia-is-group-in-spaces} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. In Properties of Stacks, Remark \ref{stacks-properties-remark-representable-over} we have seen that the $2$-category of morphisms $\mathcal{Z} \to \mathcal{X}$ representable by algebraic spaces with target $\mathcal{X}$ forms a category. In this category the inertia stack of $\mathcal{X}/\mathcal{Y}$ is a {\it group object}. Recall that an object of $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is just a pair $(x, \alpha)$ where $x$ is an object of $\mathcal{X}$ and $\alpha$ is an automorphism of $x$ in the fibre category of $\mathcal{X}$ that $x$ lives in with $f(\alpha) = \text{id}$. The composition $$c : \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \longrightarrow \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$$ is given by the rule on objects $$((x, \alpha), (x', \alpha'), \beta) \mapsto (x, \alpha \circ \beta^{-1} \circ \alpha' \circ \beta)$$ which makes sense as $\beta : x \to x'$ is an isomorphism in the fibre category by our definition of fibre products. The neutral element $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is given by the functor $x \mapsto (x, \text{id}_x)$. We omit the proof that the axioms of a group object hold. \end{remark} \noindent Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks and let $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ be its inertia stack. Let $T$ be a scheme and let $x$ be an object of $\mathcal{X}$ over $T$. Set $y = f(x)$. We have seen in the proof of Lemma \ref{lemma-inertia} that for any scheme $T$ and object $x$ of $\mathcal{X}$ over $T$ there is an exact sequence of sheaves of groups \begin{equation} \label{equation-exact-sequence-isom} 0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \to \mathit{Isom}_\mathcal{X}(x, x) \to \mathit{Isom}_\mathcal{Y}(y, y) \end{equation} The group structure on the second and third term is the one defined in Lemma \ref{lemma-isom-pseudo-torsor-aut} and the sequence gives a meaning to the first term. Also, there is a canonical cartesian square $$\xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) \ar[d] \ar[r] & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[d] \\ T \ar[r]^x & \mathcal{X} }$$ In fact, the group structure on $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ discussed in Remark \ref{remark-inertia-is-group-in-spaces} induces the group structure on $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$. This allows us to define the sheaf $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$ also for morphisms from algebraic spaces to $\mathcal{X}$. We formalize this in the following definition. \begin{definition} \label{definition-isom} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $Z$ be an algebraic space. \begin{enumerate} \item Let $x : Z \to \mathcal{X}$ be a morphism. We set $$\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x) = Z \times_{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$$ We endow it with the structure of a group algebraic space over $Z$ by pulling back the composition law discussed in Remark \ref{remark-inertia-is-group-in-spaces}. We will sometimes refer to $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x, x)$ as the {\it relative sheaf of automorphisms of $x$}. \item Let $x_1, x_2 : Z \to \mathcal{X}$ be morphisms. Set $y_i = f \circ x_i$. Let $\alpha : y_1 \to y_2$ be a $2$-morphism. Then $\alpha$ determines a morphism $\Delta^\alpha : Z \to Z \times_{y_1, \mathcal{Y}, y_2} Z$ and we set $$\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) = (Z \times_{x_1, \mathcal{X}, x_2} Z) \times_{Z \times_{y_1, \mathcal{Y}, y_2} Z, \Delta^\alpha} Z.$$ We will sometimes refer to $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)$ as the {\it relative sheaf of isomorphisms from $x_1$ to $x_2$}. \end{enumerate} If $\mathcal{Y} = \Spec(\mathbf{Z})$ or more generally when $\mathcal{Y}$ is an algebraic space, then we use the notation $\mathit{Isom}_\mathcal{X}(x, x)$ and $\mathit{Isom}_\mathcal{X}(x_1, x_2)$ and we use the terminology {\it sheaf of automorphisms of $x$} and {\it sheaf of isomorphisms from $x_1$ to $x_2$}. \end{definition} \begin{lemma} \label{lemma-isom-pseudo-torsor-aut-over-space} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $Z$ be an algebraic space and let $x_i : Z \to \mathcal{X}$, $i = 1, 2$ be morphisms. Then \begin{enumerate} \item $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ is a group algebraic space over $Z$, \item there is an exact sequence of groups $$0 \to \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2) \to \mathit{Isom}_\mathcal{X}(x_2, x_2) \to \mathit{Isom}_\mathcal{Y}(f \circ x_2, f \circ x_2)$$ \item there is a map of algebraic spaces $\mathit{Isom}_\mathcal{X}(x_1, x_2) \to \mathit{Isom}_\mathcal{Y}(f \circ x_1, f \circ x_2)$ such that for any $2$-morphism $\alpha : f \circ x_1 \to f \circ x_2$ we obtain a cartesian diagram $$\xymatrix{ \mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2) \ar[d] \ar[r] & Z \ar[d]^\alpha \\ \mathit{Isom}_\mathcal{X}(x_1, x_2) \ar[r] & \mathit{Isom}_\mathcal{Y}(f \circ x_1, f \circ x_2) }$$ \item for any $2$-morphism $\alpha : f \circ x_1 \to f \circ x_2$ the algebraic space $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}^\alpha(x_1, x_2)$ is a pseudo torsor for $\mathit{Isom}_{\mathcal{X}/\mathcal{Y}}(x_2, x_2)$ over $Z$. \end{enumerate} \end{lemma} \begin{proof} Part (1) follows from Definition \ref{definition-isom}. Part (2) comes from the exact sequence (\ref{equation-exact-sequence-isom}) \'etale locally on $Z$. Part (3) can be seen by unwinding the definitions. Locally on $Z$ in the \'etale topology part (4) reduces to part (2) of Lemma \ref{lemma-isom-pseudo-torsor-aut}. \end{proof} \begin{lemma} \label{lemma-cartesian-square-inertia} Let $\pi : \mathcal{X} \to \mathcal{Y}$ and $f : \mathcal{Y}' \to \mathcal{Y}$ be morphisms of algebraic stacks. Set $\mathcal{X}' = \mathcal{X} \times_\mathcal{Y} \mathcal{Y}'$. Then both squares in the diagram $$\xymatrix{ \mathcal{I}_{\mathcal{X}'/\mathcal{Y}'} \ar[r] \ar[d]_{ \text{Categories, Equation}\ (\ref{categories-equation-functorial}) } & \mathcal{X}' \ar[r]_{\pi'} \ar[d] & \mathcal{Y}' \ar[d]^f \\ \mathcal{I}_{\mathcal{X}/\mathcal{Y}} \ar[r] & \mathcal{X} \ar[r]^\pi & \mathcal{Y} }$$ are fibre product squares. \end{lemma} \begin{proof} The inertia stack $\mathcal{I}_{\mathcal{X}'/\mathcal{Y}'}$ is defined as the category of pairs $(x', \alpha')$ where $x'$ is an object of $\mathcal{X}'$ and $\alpha'$ is an automorphism of $x'$ with $\pi'(\alpha') = \text{id}$, see Categories, Section \ref{categories-section-inertia}. Suppose that $x'$ lies over the scheme $U$ and maps to the object $x$ of $\mathcal{X}$. By the construction of the $2$-fibre product in Categories, Lemma \ref{categories-lemma-2-product-categories-over-C} we see that $x' = (U, x, y', \beta)$ where $y'$ is an object of $\mathcal{Y}'$ over $U$ and $\beta$ is an isomorphism $\beta : \pi(x) \to f(y')$ in the fibre category of $\mathcal{Y}$ over $U$. By the very construction of the $2$-fibre product the automorphism $\alpha'$ is a pair $(\alpha, \gamma)$ where $\alpha$ is an automorphism of $x$ over $U$ and $\gamma$ is an automorphism of $y'$ over $U$ such that $\alpha$ and $\gamma$ are compatible via $\beta$. The condition $\pi'(\alpha') = \text{id}$ signifies that $\gamma = \text{id}$ whereupon the condition that $\alpha, \beta, \gamma$ are compatible is exactly the condition $\pi(\alpha) = \text{id}$, i.e., means exactly that $(x, \alpha)$ is an object of $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$. In this way we see that the left square is a fibre product square (some details omitted). \end{proof} \begin{lemma} \label{lemma-monomorphism-cartesian-square-inertia} Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Then the diagram $$\xymatrix{ \mathcal{I}_\mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{I}_\mathcal{Y} \ar[r] & \mathcal{Y} }$$ is a fibre product square. \end{lemma} \begin{proof} This follows immediately from the fact that $f$ is fully faithful (see Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism}) and the definition of the inertia in Categories, Section \ref{categories-section-inertia}. Namely, an object of $\mathcal{I}_\mathcal{X}$ over a scheme $T$ is the same thing as a pair $(x, \alpha)$ consisting of an object $x$ of $\mathcal{X}$ over $T$ and a morphism $\alpha : x \to x$ in the fibre category of $\mathcal{X}$ over $T$. As $f$ is fully faithful we see that $\alpha$ is the same thing as a morphism $\beta : f(x) \to f(x)$ in the fibre category of $\mathcal{Y}$ over $T$. Hence we can think of objects of $\mathcal{I}_\mathcal{X}$ over $T$ as triples $((y, \beta), x, \gamma)$ where $y$ is an object of $\mathcal{Y}$ over $T$, $\beta : y \to y$ in $\mathcal{Y}_T$ and $\gamma : y \to f(x)$ is an isomorphism over $T$, i.e., an object of $\mathcal{I}_\mathcal{Y} \times_\mathcal{Y} \mathcal{X}$ over $T$. \end{proof} \begin{lemma} \label{lemma-presentation-inertia} Let $\mathcal{X}$ be an algebraic stack. Let $[U/R] \to \mathcal{X}$ be a presentation. Let $G/U$ be the stabilizer group algebraic space associated to the groupoid $(U, R, s, t, c)$. Then $$\xymatrix{ G \ar[d] \ar[r] & U \ar[d] \\ \mathcal{I}_\mathcal{X} \ar[r] & \mathcal{X} }$$ is a fibre product diagram. \end{lemma} \begin{proof} Immediate from Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-2-cartesian-inertia}. \end{proof} \section{Higher diagonals} \label{section-higher-diagonals} \noindent Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. In this situation it makes sense to consider not only the diagonal $$\Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$$ but also the diagonal of the diagonal, i.e., the morphism $$\Delta_{\Delta_f} : \mathcal{X} \longrightarrow \mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Y} \mathcal{X})} \mathcal{X}$$ Because of this we sometimes use the following terminology. We denote $\Delta_{f, 0} = f$ the {\it zeroth diagonal}, we denote $\Delta_{f, 1} = \Delta_f$ the {\it first diagonal}, and we denote $\Delta_{f, 2} = \Delta_{\Delta_f}$ the {\it second diagonal}. Note that $\Delta_{f, 1}$ is representable by algebraic spaces and locally of finite type, see Lemma \ref{lemma-properties-diagonal}. Hence $\Delta_{f, 2}$ is representable, a monomorphism, locally of finite type, separated, and locally quasi-finite, see Lemma \ref{lemma-properties-diagonal-representable}. \medskip\noindent We can describe the second diagonal using the relative inertia stack. Namely, the fibre product $\mathcal{X} \times_{(\mathcal{X} \times_\mathcal{Y} \mathcal{X})} \mathcal{X}$ is equivalent to the relative inertia stack $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by Categories, Lemma \ref{categories-lemma-inertia-fibred-category}. Moreover, via this identification the second diagonal becomes the {\it neutral section} $$\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$$ of the relative inertia stack. By analogy with what happens for groupoids in algebraic spaces (Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-diagonal}) we have the following equivalences. \begin{lemma} \label{lemma-diagonal-diagonal} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item The following are equivalent \begin{enumerate} \item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is separated, \item $\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is separated, and \item $\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion. \end{enumerate} \item The following are equivalent \begin{enumerate} \item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is quasi-separated, \item $\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is quasi-separated, and \item $\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a quasi-compact. \end{enumerate} \item The following are equivalent \begin{enumerate} \item $\mathcal{I}_{\mathcal{X}/\mathcal{Y}} \to \mathcal{X}$ is locally separated, \item $\Delta_{f, 1} = \Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is locally separated, and \item $\Delta_{f, 2} = e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an immersion. \end{enumerate} \end{enumerate} \end{lemma} \begin{proof} Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then $G = U \times_\mathcal{X} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is an algebraic space over $U$ (Lemma \ref{lemma-inertia}). In fact, $G$ is a group algebraic space over $U$ by the group law on relative inertia constructed in Remark \ref{remark-inertia-is-group-in-spaces}. Moreover, $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is surjective and smooth as a base change of $U \to \mathcal{X}$. Finally, the base change of $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ by $G \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is the identity $U \to G$ of $G/U$. Thus the equivalence of (a) and (c) follows from Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-group-scheme-separated}. Since $\Delta_{f, 2}$ is the diagonal of $\Delta_f$ we have (b) $\Leftrightarrow$ (c) by definition. \end{proof} \begin{lemma} \label{lemma-second-diagonal} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent: \begin{enumerate} \item the morphism $f$ is representable by algebraic spaces, \item the second diagonal of $f$ is an isomorphism, \item the group stack $\mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is trivial over $\mathcal X$, and \item for a scheme $T$ and a morphism $x : T \to \mathcal{X}$ the kernel of $\mathit{Isom}_\mathcal{X}(x, x) \to \mathit{Isom}_\mathcal{Y}(f(x), f(x))$ is trivial. \end{enumerate} \end{lemma} \begin{proof} We first prove the equivalence of (1) and (2). Namely, $f$ is representable by algebraic spaces if and only if $f$ is faithful, see Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}. On the other hand, $f$ is faithful if and only if for every object $x$ of $\mathcal{X}$ over a scheme $T$ the functor $f$ induces an injection $\mathit{Isom}_\mathcal{X}(x, x) \to \mathit{Isom}_\mathcal{Y}(f(x), f(x))$, which happens if and only if the kernel $K$ is trivial, which happens if and only if $e : T \to K$ is an isomorphism for every $x : T \to \mathcal{X}$. Since $K = T \times_{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ as discussed above, this proves the equivalence of (1) and (2). To prove the equivalence of (2) and (3), by the discussion above, it suffices to note that a group stack is trivial if and only if its identity section is an isomorphism. Finally, the equivalence of (3) and (4) follows from the definitions: in the proof of Lemma \ref{lemma-inertia} we have seen that the kernel in (4) corresponds to the fibre product $T \times_{x, \mathcal{X}} \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ over $T$. \end{proof} \noindent This lemma leads to the following hierarchy for morphisms of algebraic stacks. \begin{lemma} \label{lemma-hierarchy} A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is \begin{enumerate} \item a monomorphism if and only if $\Delta_{f, 1}$ is an isomorphism, and \item representable by algebraic spaces if and only if $\Delta_{f, 1}$ is a monomorphism. \end{enumerate} Moreover, the second diagonal $\Delta_{f, 2}$ is always a monomorphism. \end{lemma} \begin{proof} Recall from Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism} that a morphism of algebraic stacks is a monomorphism if and only if its diagonal is an isomorphism of stacks. Thus Lemma \ref{lemma-second-diagonal} can be rephrased as saying that a morphism is representable by algebraic spaces if the diagonal is a monomorphism. In particular, it shows that condition (3) of Lemma \ref{lemma-properties-diagonal-representable} is actually an if and only if, i.e., a morphism of algebraic stacks is representable by algebraic spaces if and only if its diagonal is a monomorphism. \end{proof} \begin{lemma} \label{lemma-first-diagonal-separated-second-diagonal-closed} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then \begin{enumerate} \item $\Delta_{f, 1}$ separated $\Leftrightarrow$ $\Delta_{f, 2}$ closed immersion $\Leftrightarrow$ $\Delta_{f, 2}$ proper $\Leftrightarrow$ $\Delta_{f, 2}$ universally closed, \item $\Delta_{f, 1}$ quasi-separated $\Leftrightarrow$ $\Delta_{f, 2}$ finite type $\Leftrightarrow$ $\Delta_{f, 2}$ quasi-compact, and \item $\Delta_{f, 1}$ locally separated $\Leftrightarrow$ $\Delta_{f, 2}$ immersion. \end{enumerate} \end{lemma} \begin{proof} Follows from Lemmas \ref{lemma-representable-separated-diagonal-closed}, \ref{lemma-representable-quasi-separated-diagonal-quasi-compact}, and \ref{lemma-representable-locally-separated-diagonal-immersion} applied to $\Delta_{f, 1}$. \end{proof} \noindent The following lemma is kind of cute and it may suggest a generalization of these conditions to higher algebraic stacks. \begin{lemma} \label{lemma-definition-separated} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then \begin{enumerate} \item $f$ is separated if and only if $\Delta_{f, 1}$ and $\Delta_{f, 2}$ are universally closed, and \item $f$ is quasi-separated if and only if $\Delta_{f, 1}$ and $\Delta_{f, 2}$ are quasi-compact. \item $f$ is quasi-DM if and only if $\Delta_{f, 1}$ and $\Delta_{f, 2}$ are locally quasi-finite. \item $f$ is DM if and only if $\Delta_{f, 1}$ and $\Delta_{f, 2}$ are unramified. \end{enumerate} \end{lemma} \begin{proof} Proof of (1). Assume that $\Delta_{f, 2}$ and $\Delta_{f, 1}$ are universally closed. Then $\Delta_{f, 1}$ is separated and universally closed by Lemma \ref{lemma-first-diagonal-separated-second-diagonal-closed}. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-universally-closed-quasi-compact} and Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-transformations-property-implication} we see that $\Delta_{f, 1}$ is quasi-compact. Hence it is quasi-compact, separated, universally closed and locally of finite type (by Lemma \ref{lemma-properties-diagonal}) so proper. This proves $\Leftarrow$'' of (1). The proof of the implication in the other direction is omitted. \medskip\noindent Proof of (2). This follows immediately from Lemma \ref{lemma-first-diagonal-separated-second-diagonal-closed}. \medskip\noindent Proof of (3). This follows from the fact that $\Delta_{f, 2}$ is always locally quasi-finite by Lemma \ref{lemma-properties-diagonal-representable} applied to $\Delta_f = \Delta_{f, 1}$. \medskip\noindent Proof of (4). This follows from the fact that $\Delta_{f, 2}$ is always unramified as Lemma \ref{lemma-properties-diagonal-representable} applied to $\Delta_f = \Delta_{f, 1}$ shows that $\Delta_{f, 2}$ is locally of finite type and a monomorphism. See More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-universally-injective-unramified}. \end{proof} \begin{lemma} \label{lemma-separated-implies-isom} Let $f : \mathcal{X} \to \mathcal{Y}$ be a separated (resp.\ quasi-separated, resp.\ quasi-DM, resp.\ DM) morphism of algebraic stacks. Then \begin{enumerate} \item given algebraic spaces $T_i$, $i = 1, 2$ and morphisms $x_i : T_i \to \mathcal{X}$, with $y_i = f \circ x_i$ the morphism $$T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \longrightarrow T_1 \times_{y_1, \mathcal{Y}, y_2} T_2$$ is proper (resp.\ quasi-compact and quasi-separated, resp.\ locally quasi-finite, resp.\ unramified), \item given an algebraic space $T$ and morphisms $x_i : T \to \mathcal{X}$, $i = 1, 2$, with $y_i = f \circ x_i$ the morphism $$\mathit{Isom}_\mathcal{X}(x_1, x_2) \longrightarrow \mathit{Isom}_\mathcal{Y}(y_1, y_2)$$ is proper (resp.\ quasi-compact and quasi-separated, resp.\ locally quasi-finite, resp.\ unramified). \end{enumerate} \end{lemma} \begin{proof} Proof of (1). Observe that the diagram $$\xymatrix{ T_1 \times_{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times_{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} }$$ is cartesian. Hence this follows from the fact that $f$ is separated (resp.\ quasi-separated, resp.\ quasi-DM, resp.\ DM) if and only if the diagonal is proper (resp.\ quasi-compact and quasi-separated, resp.\ locally quasi-finite, resp.\ unramified). \medskip\noindent Proof of (2). This is true because $$\mathit{Isom}_\mathcal{X}(x_1, x_2) = (T \times_{x_1, \mathcal{X}, x_2} T) \times_{T \times T, \Delta_T} T$$ hence the morphism in (2) is a base change of the morphism in (1). \end{proof} \section{Quasi-compact morphisms} \label{section-quasi-compact} \noindent Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} we have defined what it means for $f$ to be quasi-compact. Here is another characterization. \begin{lemma} \label{lemma-characterize-representable-quasi-compact} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent: \begin{enumerate} \item $f$ is quasi-compact (as in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}), and \item for every quasi-compact algebraic stack $\mathcal{Z}$ and any morphism $\mathcal{Z} \to \mathcal{Y}$ the algebraic stack $\mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ is quasi-compact. \end{enumerate} \end{lemma} \begin{proof} Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks with $\mathcal{Z}$ quasi-compact. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack} there exists a quasi-compact scheme $U$ and a surjective smooth morphism $U \to \mathcal{Z}$. Since $f$ is representable by algebraic spaces and quasi-compact we see by definition that $U \times_\mathcal{Y} \mathcal{X}$ is an algebraic space, and that $U \times_\mathcal{Y} \mathcal{X} \to U$ is quasi-compact. Hence $U \times_Y X$ is a quasi-compact algebraic space. The morphism $U \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ is smooth and surjective (as the base change of the smooth and surjective morphism $U \to \mathcal{Z}$). Hence $\mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ is quasi-compact by another application of Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack} \medskip\noindent Assume (2). Let $Z \to \mathcal{Y}$ be a morphism, where $Z$ is a scheme. We have to show that the morphism of algebraic spaces $p : Z \times_\mathcal{Y} \mathcal{X} \to Z$ is quasi-compact. Let $U \subset Z$ be affine open. Then $p^{-1}(U) = U \times_\mathcal{Y} \mathcal{Z}$ and the algebraic space $U \times_\mathcal{Y} \mathcal{Z}$ is quasi-compact by assumption (2). Hence $p$ is quasi-compact, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-compact-local}. \end{proof} \noindent This motivates the following definition. \begin{definition} \label{definition-quasi-compact} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is {\it quasi-compact} if for every quasi-compact algebraic stack $\mathcal{Z}$ and morphism $\mathcal{Z} \to \mathcal{Y}$ the fibre product $\mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ is quasi-compact. \end{definition} \noindent By Lemma \ref{lemma-characterize-representable-quasi-compact} above this agrees with the already existing notion for morphisms of algebraic stacks representable by algebraic spaces. In particular this notion agrees with the notions already defined for morphisms between algebraic stacks and schemes. \begin{lemma} \label{lemma-base-change-quasi-compact} The base change of a quasi-compact morphism of algebraic stacks by any morphism of algebraic stacks is quasi-compact. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-composition-quasi-compact} The composition of a pair of quasi-compact morphisms of algebraic stacks is quasi-compact. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-closed-immersion-quasi-compact} A closed immersion of algebraic stacks is quasi-compact. \end{lemma} \begin{proof} This follows from the fact that immersions are always representable and the corresponding fact for closed immersion of algebraic spaces. \end{proof} \begin{lemma} \label{lemma-surjection-from-quasi-compact} Let $$\xymatrix{ \mathcal{X} \ar[rr]_f \ar[rd]_p & & \mathcal{Y} \ar[dl]^q \\ & \mathcal{Z} }$$ be a $2$-commutative diagram of morphisms of algebraic stacks. If $f$ is surjective and $p$ is quasi-compact, then $q$ is quasi-compact. \end{lemma} \begin{proof} Let $\mathcal{T}$ be a quasi-compact algebraic stack, and let $\mathcal{T} \to \mathcal{Z}$ be a morphism. By Properties of Stacks, Lemma \ref{stacks-properties-lemma-base-change-surjective} the morphism $\mathcal{T} \times_\mathcal{Z} \mathcal{X} \to \mathcal{T} \times_\mathcal{Z} \mathcal{Y}$ is surjective and by assumption $\mathcal{T} \times_\mathcal{Z} \mathcal{X}$ is quasi-compact. Hence $\mathcal{T} \times_\mathcal{Z} \mathcal{Y}$ is quasi-compact by Properties of Stacks, Lemma \ref{stacks-properties-lemma-quasi-compact-stack}. \end{proof} \begin{lemma} \label{lemma-quasi-compact-permanence} Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $g \circ f$ is quasi-compact and $g$ is quasi-separated then $f$ is quasi-compact. \end{lemma} \begin{proof} This is true because $f$ equals the composition $(1, f) : \mathcal{X} \to \mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{Y}$. The first map is quasi-compact by Lemma \ref{lemma-section-immersion} because it is a section of the quasi-separated morphism $\mathcal{X} \times_\mathcal{Z} \mathcal{Y} \to \mathcal{X}$ (a base change of $g$, see Lemma \ref{lemma-base-change-separated}). The second map is quasi-compact as it is the base change of $f$, see Lemma \ref{lemma-base-change-quasi-compact}. And compositions of quasi-compact morphisms are quasi-compact, see Lemma \ref{lemma-composition-quasi-compact}. \end{proof} \begin{lemma} \label{lemma-quasi-compact-quasi-separated-permanence} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item If $\mathcal{X}$ is quasi-compact and $\mathcal{Y}$ is quasi-separated, then $f$ is quasi-compact. \item If $\mathcal{X}$ is quasi-compact and quasi-separated and $\mathcal{Y}$ is quasi-separated, then $f$ is quasi-compact and quasi-separated. \item A fibre product of quasi-compact and quasi-separated algebraic stacks is quasi-compact and quasi-separated. \end{enumerate} \end{lemma} \begin{proof} Part (1) follows from Lemma \ref{lemma-quasi-compact-permanence}. Part (2) follows from (1) and Lemma \ref{lemma-compose-after-separated}. For (3) let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Z} \to \mathcal{Y}$ be morphisms of quasi-compact and quasi-separated algebraic stacks. Then $\mathcal{X} \times_\mathcal{Y} \mathcal{Z} \to \mathcal{Z}$ is quasi-compact and quasi-separated as a base change of $\mathcal{X} \to \mathcal{Y}$ using (2) and Lemmas \ref{lemma-base-change-quasi-compact} and \ref{lemma-base-change-separated}. Hence $\mathcal{X} \times_\mathcal{Y} \mathcal{Z}$ is quasi-compact and quasi-separated as an algebraic stack quasi-compact and quasi-separated over $\mathcal{Z}$, see Lemmas \ref{lemma-separated-over-separated} and \ref{lemma-composition-quasi-compact}. \end{proof} \begin{lemma} \label{lemma-reach-points-scheme-theoretic-image} Let $f : \mathcal{X} \to \mathcal{Y}$ be a quasi-compact morphism of algebraic stacks. Let $y \in |\mathcal{Y}|$ be a point in the closure of the image of $|f|$. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram $$\xymatrix{ \Spec(K) \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \Spec(A) \ar[r] & \mathcal{Y} }$$ such that the closed point of $\Spec(A)$ maps to $y$. \end{lemma} \begin{proof} Choose an affine scheme $V$ and a point $v \in V$ and a smooth morphism $V \to \mathcal{Y}$ sending $v$ to $y$. Consider the base change diagram $$\xymatrix{ V \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d]_g & \mathcal{X} \ar[d]^f \\ V \ar[r] & \mathcal{Y} }$$ Recall that $|V \times_\mathcal{Y} \mathcal{X}| \to |V| \times_{|\mathcal{Y}|} |\mathcal{X}|$ is surjective (Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian}). Because $|V| \to |\mathcal{Y}|$ is open (Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}) we conclude that $v$ is in the closure of the image of $|g|$. Thus it suffices to prove the lemma for the quasi-compact morphism $g$ (Lemma \ref{lemma-base-change-quasi-compact}) which we do in the next paragraph. \medskip\noindent Assume $\mathcal{Y} = Y$ is an affine scheme. Then $\mathcal{X}$ is quasi-compact as $f$ is quasi-compact (Definition \ref{definition-quasi-compact}). Choose an affine scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Then the image of $|f|$ is the image of $W \to Y$. By Morphisms, Lemma \ref{morphisms-lemma-reach-points-scheme-theoretic-image} we can choose a diagram $$\xymatrix{ \Spec(K) \ar[r] \ar[d] & W \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \Spec(A) \ar[r] & Y \ar[r] & Y }$$ such that the closed point of $\Spec(A)$ maps to $y$. Composing with $W \to \mathcal{X}$ we obtain a solution. \end{proof} \section{Noetherian algebraic stacks} \label{section-noetherian} \noindent We have already defined locally Noetherian algebraic stacks in Properties of Stacks, Section \ref{stacks-properties-section-types-properties}. \begin{definition} \label{definition-noetherian} Let $\mathcal{X}$ be an algebraic stack. We say $\mathcal{X}$ is {\it Noetherian} if $\mathcal{X}$ is quasi-compact, quasi-separated and locally Noetherian. \end{definition} \noindent Note that a Noetherian algebraic stack $\mathcal{X}$ is not just quasi-compact and locally Noetherian, but also quasi-separated. In the language of Section \ref{section-higher-diagonals} if we denote $p : \mathcal{X} \to \Spec(\mathbf{Z})$ the absolute'' structure morphism (i.e., the structure morphism of $\mathcal{X}$ viewed as an algebraic stack over $\mathbf{Z}$), then $$\mathcal{X}\text{ Noetherian} \Leftrightarrow \mathcal{X}\text{ locally Noetherian and } \Delta_{p, 0}, \Delta_{p, 1}, \Delta_{p, 2} \text{ quasi-compact}.$$ This will later mean that an algebraic stack of finite type over a Noetherian algebraic stack is not automatically Noetherian. \begin{lemma} \label{lemma-locally-closed-in-noetherian} Let $\mathcal{X} \to \mathcal{Y}$ be an immersion of algebraic stacks. If $\mathcal{Y}$ is Noetherian, then $\mathcal{X}$ is Noetherian. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Affine morphisms} \label{section-affine} \noindent Affine morphisms of algebraic stacks are defined as follows. \begin{definition} \label{definition-affine} A morphism of algebraic stacks is said to be {\it affine} if it is representable and affine in the sense of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \end{definition} \noindent For us it is a little bit more convenient to think of an affine morphism of algebraic stacks as a morphism of algebraic stacks which is representable by algebraic spaces and affine in the sense of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. (Recall that the default for representable'' in the Stacks project is representable by schemes.) Since this is clearly equivalent to the notion just defined we shall use this characterization without further mention. We prove a few simple lemmas about this notion. \begin{lemma} \label{lemma-base-change-affine} Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be an affine morphism of algebraic stacks. Then $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}$ is an affine morphism of algebraic stacks. \end{lemma} \begin{proof} This follows from the discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \end{proof} \begin{lemma} \label{lemma-composition-affine} Compositions of affine morphisms of algebraic stacks are affine. \end{lemma} \begin{proof} This follows from the discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-affine}. \end{proof} \section{Integral and finite morphisms} \label{section-integral} \noindent Integral and finite morphisms of algebraic stacks are defined as follows. \begin{definition} \label{definition-integral} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item We say $f$ is {\it integral} if $f$ is representable and integral in the sense of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \item We say $f$ is {\it finite} if $f$ is representable and integral in the sense of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \end{enumerate} \end{definition} \noindent For us it is a little bit more convenient to think of an integral, resp.\ finite morphism of algebraic stacks as a morphism of algebraic stacks which is representable by algebraic spaces and integral, resp.\ finite in the sense of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. (Recall that the default for representable'' in the Stacks project is representable by schemes.) Since this is clearly equivalent to the notion just defined we shall use this characterization without further mention. We prove a few simple lemmas about this notion. \begin{lemma} \label{lemma-base-change-integral} Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be an integral (or finite) morphism of algebraic stacks. Then $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}$ is an integral (or finite) morphism of algebraic stacks. \end{lemma} \begin{proof} This follows from the discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \end{proof} \begin{lemma} \label{lemma-composition-integral} Compositions of integral, resp.\ finite morphisms of algebraic stacks are integral, resp.\ finite. \end{lemma} \begin{proof} This follows from the discussion in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-integral}. \end{proof} \section{Open morphisms} \label{section-open} \noindent Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} we have defined what it means for $f$ to be universally open. Here is another characterization. \begin{lemma} \label{lemma-characterize-representable-universally-open} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent \begin{enumerate} \item $f$ is universally open (as in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}), and \item for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is open. \end{enumerate} \end{lemma} \begin{proof} Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times_\mathcal{Y} \mathcal{X} \to V$ of algebraic spaces is universally open, in particular the map $|V \times_\mathcal{Y} \mathcal{X}| \to |V|$ is open. By Properties of Stacks, Section \ref{stacks-properties-section-points} in the commutative diagram $$\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }$$ the horizontal arrows are open and surjective, and moreover $$|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|$$ is surjective. Hence as the left vertical arrow is open it follows that the right vertical arrow is open. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. \end{proof} \noindent Thus we may use the following natural definition. \begin{definition} \label{definition-open} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item We say $f$ is {\it open} if the map of topological spaces $|\mathcal{X}| \to |\mathcal{Y}|$ is open. \item We say $f$ is {\it universally open} if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $$|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$$ is open, i.e., the base change $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}$ is open. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-base-change-universally-open} The base change of a universally open morphism of algebraic stacks by any morphism of algebraic stacks is universally open. \end{lemma} \begin{proof} This is immediate from the definition. \end{proof} \begin{lemma} \label{lemma-composition-universally-open} The composition of a pair of (universally) open morphisms of algebraic stacks is (universally) open. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Submersive morphisms} \label{section-submersive} \noindent Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} we have defined what it means for $f$ to be universally submersive. Here is another characterization. \begin{lemma} \label{lemma-characterize-representable-universally-submersive} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent \begin{enumerate} \item $f$ is universally submersive (as in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}), and \item for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is submersive. \end{enumerate} \end{lemma} \begin{proof} Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times_\mathcal{Y} \mathcal{X} \to V$ of algebraic spaces is universally submersive, in particular the map $|V \times_\mathcal{Y} \mathcal{X}| \to |V|$ is submersive. By Properties of Stacks, Section \ref{stacks-properties-section-points} in the commutative diagram $$\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }$$ the horizontal arrows are open and surjective, and moreover $$|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|$$ is surjective. Hence as the left vertical arrow is submersive it follows that the right vertical arrow is submersive. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. \end{proof} \noindent Thus we may use the following natural definition. \begin{definition} \label{definition-submersive} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item We say $f$ is {\it submersive}\footnote{This is very different from the notion of a submersion of differential manifolds.} if the continuous map $|\mathcal{X}| \to |\mathcal{Y}|$ is submersive, see Topology, Definition \ref{topology-definition-submersive}. \item We say $f$ is {\it universally submersive} if for every morphism of algebraic stacks $\mathcal{Y}' \to \mathcal{Y}$ the base change $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$ is submersive. \end{enumerate} \end{definition} \noindent We note that a submersive morphism is in particular surjective. \begin{lemma} \label{lemma-base-change-universally-submersive} The base change of a universally submersive morphism of algebraic stacks by any morphism of algebraic stacks is universally submersive. \end{lemma} \begin{proof} This is immediate from the definition. \end{proof} \begin{lemma} \label{lemma-composition-universally-submersive} The composition of a pair of (universally) submersive morphisms of algebraic stacks is (universally) submersive. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Universally closed morphisms} \label{section-universally-closed} \noindent Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} we have defined what it means for $f$ to be universally closed. Here is another characterization. \begin{lemma} \label{lemma-characterize-representable-universally-closed} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent \begin{enumerate} \item $f$ is universally closed (as in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}), and \item for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is closed. \end{enumerate} \end{lemma} \begin{proof} Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times_\mathcal{Y} \mathcal{X} \to V$ of algebraic spaces is universally closed, in particular the map $|V \times_\mathcal{Y} \mathcal{X}| \to |V|$ is closed. By Properties of Stacks, Section \ref{stacks-properties-section-points} in the commutative diagram $$\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }$$ the horizontal arrows are open and surjective, and moreover $$|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|$$ is surjective. Hence as the left vertical arrow is closed it follows that the right vertical arrow is closed. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. \end{proof} \noindent Thus we may use the following natural definition. \begin{definition} \label{definition-closed} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item We say $f$ is {\it closed} if the map of topological spaces $|\mathcal{X}| \to |\mathcal{Y}|$ is closed. \item We say $f$ is {\it universally closed} if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the morphism of topological spaces $$|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$$ is closed, i.e., the base change $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{Z}$ is closed. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-base-change-universally-closed} The base change of a universally closed morphism of algebraic stacks by any morphism of algebraic stacks is universally closed. \end{lemma} \begin{proof} This is immediate from the definition. \end{proof} \begin{lemma} \label{lemma-composition-universally-closed} The composition of a pair of (universally) closed morphisms of algebraic stacks is (universally) closed. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-universally-closed-local} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent \begin{enumerate} \item $f$ is universally closed, \item for every scheme $Z$ and every morphism $Z \to \mathcal{Y}$ the projection $|Z \times_\mathcal{Y} \mathcal{X}| \to |Z|$ is closed, \item for every affine scheme $Z$ and every morphism $Z \to \mathcal{Y}$ the projection $|Z \times_\mathcal{Y} \mathcal{X}| \to |Z|$ is closed, and \item there exists an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$ such that $V \times_\mathcal{Y} \mathcal{X} \to V$ is a universally closed morphism of algebraic stacks. \end{enumerate} \end{lemma} \begin{proof} We omit the proof that (1) implies (2), and that (2) implies (3). \medskip\noindent Assume (3). Choose a surjective smooth morphism $V \to \mathcal{Y}$. We are going to show that $V \times_\mathcal{Y} \mathcal{X} \to V$ is a universally closed morphism of algebraic stacks. Let $\mathcal{Z} \to V$ be a morphism from an algebraic stack to $V$. Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W = \coprod W_i$ is a disjoint union of affine schemes. Then we have the following commutative diagram $$\xymatrix{ \coprod_i |W_i \times_\mathcal{Y} \mathcal{X}| \ar@{=}[r] \ar[d] & |W \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \ar@{=}[r] & |\mathcal{Z} \times_V (V \times_\mathcal{Y} \mathcal{X})| \ar[ld] \\ \coprod |W_i| \ar@{=}[r] & |W| \ar[r] & |\mathcal{Z}| }$$ We have to show the south-east arrow is closed. The middle horizontal arrows are surjective and open (Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}). By assumption (3), and the fact that $W_i$ is affine we see that the left vertical arrows are closed. Hence it follows that the right vertical arrow is closed. \medskip\noindent Assume (4). We will show that $f$ is universally closed. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider the diagram $$\xymatrix{ |(V \times_\mathcal{Y} \mathcal{Z}) \times_V (V \times_\mathcal{Y} \mathcal{X})| \ar@{=}[r] \ar[rd] & |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |Z \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ & |V \times_\mathcal{Y} \mathcal{Z}| \ar[r] & |\mathcal{Z}| }$$ The south-west arrow is closed by assumption. The horizontal arrows are surjective and open because the corresponding morphisms of algebraic stacks are surjective and smooth (see reference above). It follows that the right vertical arrow is closed. \end{proof} \section{Universally injective morphisms} \label{section-universally-injective} \noindent Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} we have defined what it means for $f$ to be universally injective. Here is another characterization. \begin{lemma} \label{lemma-characterize-representable-universally-injective} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent \begin{enumerate} \item $f$ is universally injective (as in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}), and \item for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is injective. \end{enumerate} \end{lemma} \begin{proof} Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times_\mathcal{Y} \mathcal{X} \to V$ of algebraic spaces is universally injective, in particular the map $|V \times_\mathcal{Y} \mathcal{X}| \to |V|$ is injective. By Properties of Stacks, Section \ref{stacks-properties-section-points} in the commutative diagram $$\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }$$ the horizontal arrows are open and surjective, and moreover $$|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|$$ is surjective. Hence as the left vertical arrow is injective it follows that the right vertical arrow is injective. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. \end{proof} \noindent Thus we may use the following natural definition. \begin{definition} \label{definition-universally-injective} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is {\it universally injective} if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map $$|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$$ is injective. \end{definition} \begin{lemma} \label{lemma-base-change-universally-injective} The base change of a universally injective morphism of algebraic stacks by any morphism of algebraic stacks is universally injective. \end{lemma} \begin{proof} This is immediate from the definition. \end{proof} \begin{lemma} \label{lemma-composition-universally-injective} The composition of a pair of universally injective morphisms of algebraic stacks is universally injective. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-universally-injective} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent \begin{enumerate} \item $f$ is universally injective, and \item $\Delta : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is surjective. \end{enumerate} \end{lemma} \begin{proof} If $f$ is universally injective, then the first projection $|\mathcal{X} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{X}|$ is injective, which implies that $|\Delta|$ is surjective. Conversely, assume $\Delta$ is surjective. Then any base change of $\Delta$ is surjective (see Properties of Stacks, Section \ref{stacks-properties-section-surjective}). Since the diagonal of a base change of $f$ is a base change of $\Delta$, we see that it suffices to show that $|\mathcal{X}| \to |\mathcal{Y}|$ is injective. If not, then by Properties of Stacks, Lemma \ref{stacks-properties-lemma-points-cartesian} we find that the first projection $|\mathcal{X} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{X}|$ is not injective. Of course this means that $|\Delta|$ is not surjective. \end{proof} \section{Universal homeomorphisms} \label{section-universal-homeomorphisms} \noindent Let $f$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} we have defined what it means for $f$ to be a universal homeomorphism. Here is another characterization. \begin{lemma} \label{lemma-characterize-representable-universal-homeomorphism} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. The following are equivalent \begin{enumerate} \item $f$ is a universal homeomorphism (Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}), and \item for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map of topological spaces $|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$ is a homeomorphism. \end{enumerate} \end{lemma} \begin{proof} Assume (1), and let $\mathcal{Z} \to \mathcal{Y}$ be as in (2). Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Z}$. By assumption the morphism $V \times_\mathcal{Y} \mathcal{X} \to V$ of algebraic spaces is a universal homeomorphism, in particular the map $|V \times_\mathcal{Y} \mathcal{X}| \to |V|$ is a homeomorphism. By Properties of Stacks, Section \ref{stacks-properties-section-points} in the commutative diagram $$\xymatrix{ |V \times_\mathcal{Y} \mathcal{X}| \ar[r] \ar[d] & |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \ar[d] \\ |V| \ar[r] & |\mathcal{Z}| }$$ the horizontal arrows are open and surjective, and moreover $$|V \times_\mathcal{Y} \mathcal{X}| \longrightarrow |V| \times_{|\mathcal{Z}|} |\mathcal{Z} \times_\mathcal{Y} \mathcal{X}|$$ is surjective. Hence as the left vertical arrow is a homeomorphism it follows that the right vertical arrow is a homeomorphism. This proves (2). The implication (2) $\Rightarrow$ (1) follows from the definitions. \end{proof} \noindent Thus we may use the following natural definition. \begin{definition} \label{definition-universal-homeomorphism} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is a {\it universal homeomorphism} if for every morphism of algebraic stacks $\mathcal{Z} \to \mathcal{Y}$ the map of topological spaces $$|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \to |\mathcal{Z}|$$ is a homeomorphism. \end{definition} \begin{lemma} \label{lemma-base-change-universal-homeomorphism} The base change of a universal homeomorphism of algebraic stacks by any morphism of algebraic stacks is a universal homeomorphism. \end{lemma} \begin{proof} This is immediate from the definition. \end{proof} \begin{lemma} \label{lemma-composition-universal-homeomorphism} The composition of a pair of universal homeomorphisms of algebraic stacks is a universal homeomorphism. \end{lemma} \begin{proof} Omitted. \end{proof} \section{Types of morphisms smooth local on source-and-target} \label{section-local-source-target} \noindent Given a property of morphisms of algebraic spaces which is {\it smooth local on the source-and-target}, see Descent on Spaces, Definition \ref{spaces-descent-definition-local-source-target} we may use it to define a corresponding property of morphisms of algebraic stacks, namely by imposing either of the equivalent conditions of the lemma below. \begin{lemma} \label{lemma-local-source-target} Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is smooth local on the source-and-target. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider commutative diagrams $$\xymatrix{ U \ar[d]_a \ar[r]_h & V \ar[d]^b \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }$$ where $U$ and $V$ are algebraic spaces and the vertical arrows are smooth. The following are equivalent \begin{enumerate} \item for any diagram as above such that in addition $U \to \mathcal{X} \times_\mathcal{Y} V$ is smooth the morphism $h$ has property $\mathcal{P}$, and \item for some diagram as above with $a : U \to \mathcal{X}$ surjective the morphism $h$ has property $\mathcal{P}$. \end{enumerate} If $\mathcal{X}$ and $\mathcal{Y}$ are representable by algebraic spaces, then this is also equivalent to $f$ (as a morphism of algebraic spaces) having property $\mathcal{P}$. If $\mathcal{P}$ is also preserved under any base change, and fppf local on the base, then for morphisms $f$ which are representable by algebraic spaces this is also equivalent to $f$ having property $\mathcal{P}$ in the sense of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms}. \end{lemma} \begin{proof} Let us prove the implication (1) $\Rightarrow$ (2). Pick an algebraic space $V$ and a surjective and smooth morphism $V \to \mathcal{Y}$. Pick an algebraic space $U$ and a surjective and smooth morphism $U \to \mathcal{X} \times_\mathcal{Y} V$. Note that $U \to \mathcal{X}$ is surjective and smooth as well, as a composition of the base change $\mathcal{X} \times_\mathcal{Y} V \to \mathcal{X}$ and the chosen map $U \to \mathcal{X} \times_\mathcal{Y} V$. Hence we obtain a diagram as in (1). Thus if (1) holds, then $h : U \to V$ has property $\mathcal{P}$, which means that (2) holds as $U \to \mathcal{X}$ is surjective. \medskip\noindent Conversely, assume (2) holds and let $U, V, a, b, h$ be as in (2). Next, let $U', V', a', b', h'$ be any diagram as in (1). Picture $$\xymatrix{ U \ar[d] \ar[r]_h & V \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} } \quad\quad \xymatrix{ U' \ar[d] \ar[r]_{h'} & V' \ar[d] \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }$$ To show that (2) implies (1) we have to prove that $h'$ has $\mathcal{P}$. To do this consider the commutative diagram $$\xymatrix{ U \ar[dd]^h & U \times_\mathcal{X} U' \ar[d] \ar[l] \ar@/^6ex/[dd]^{(h, h')} \ar[r] & U' \ar[dd]^{h'} \\ & U \times_\mathcal{Y} V' \ar[lu] \ar[d] & \\ V & V \times_\mathcal{Y} V' \ar[l] \ar[r] & V' }$$ of algebraic spaces. Note that the horizontal arrows are smooth as base changes of the smooth morphisms $V \to \mathcal{Y}$, $V' \to \mathcal{Y}$, $U \to \mathcal{X}$, and $U' \to \mathcal{X}$. Note that $$\xymatrix{ U \times_\mathcal{X} U' \ar[d] \ar[r] & U' \ar[d] \\ U \times_\mathcal{Y} V' \ar[r] & \mathcal{X} \times_\mathcal{Y} V' }$$ is cartesian, hence the left vertical arrow is smooth as $U', V', a', b', h'$ is as in (1). Since $\mathcal{P}$ is smooth local on the target by Descent on Spaces, Lemma \ref{spaces-descent-lemma-local-source-target-implies} part (2) we see that the base change $U \times_\mathcal{Y} V' \to V \times_\mathcal{Y} V'$ has $\mathcal{P}$. Since $\mathcal{P}$ is smooth local on the source by Descent on Spaces, Lemma \ref{spaces-descent-lemma-local-source-target-implies} part (1) we can precompose by the smooth morphism $U \times_\mathcal{X} U' \to U \times_\mathcal{Y} V'$ and conclude $(h, h')$ has $\mathcal{P}$. Since $V \times_\mathcal{Y} V' \to V'$ is smooth we conclude $U \times_\mathcal{X} U' \to V'$ has $\mathcal{P}$ by Descent on Spaces, Lemma \ref{spaces-descent-lemma-local-source-target-implies} part (3). Finally, since $U \times_X U' \to U'$ is surjective and smooth and $\mathcal{P}$ is smooth local on the source (same lemma) we conclude that $h'$ has $\mathcal{P}$. This finishes the proof of the equivalence of (1) and (2). \medskip\noindent If $\mathcal{X}$ and $\mathcal{Y}$ are representable, then Descent on Spaces, Lemma \ref{spaces-descent-lemma-local-source-target-characterize} applies which shows that (1) and (2) are equivalent to $f$ having $\mathcal{P}$. \medskip\noindent Finally, suppose $f$ is representable, and $U, V, a, b, h$ are as in part (2) of the lemma, and that $\mathcal{P}$ is preserved under arbitrary base change. We have to show that for any scheme $Z$ and morphism $Z \to \mathcal{X}$ the base change $Z \times_\mathcal{Y} \mathcal{X} \to Z$ has property $\mathcal{P}$. Consider the diagram $$\xymatrix{ Z \times_\mathcal{Y} U \ar[d] \ar[r] & Z \times_\mathcal{Y} V \ar[d] \\ Z \times_\mathcal{Y} \mathcal{X} \ar[r] & Z }$$ Note that the top horizontal arrow is a base change of $h$ and hence has property $\mathcal{P}$. The left vertical arrow is smooth and surjective and the right vertical arrow is smooth. Thus Descent on Spaces, Lemma \ref{spaces-descent-lemma-local-source-target-characterize} kicks in and shows that $Z \times_\mathcal{Y} \mathcal{X} \to Z$ has property $\mathcal{P}$. \end{proof} \begin{definition} \label{definition-P} Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is smooth local on the source-and-target. We say a morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks {\it has property $\mathcal{P}$} if the equivalent conditions of Lemma \ref{lemma-local-source-target} hold. \end{definition} \begin{remark} \label{remark-composition} Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is smooth local on the source-and-target and stable under composition. Then the property of morphisms of algebraic stacks defined in Definition \ref{definition-P} is stable under composition. Namely, let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks having property $\mathcal{P}$. Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y} \times_\mathcal{Z} W$. Finally, choose an algebraic space $U$ and a surjective and smooth morphism $U \to \mathcal{X} \times_\mathcal{Y} V$. Then the morphisms $V \to W$ and $U \to V$ have property $\mathcal{P}$ by definition. Whence $U \to W$ has property $\mathcal{P}$ as we assumed that $\mathcal{P}$ is stable under composition. Thus, by definition again, we see that $g \circ f : \mathcal{X} \to \mathcal{Z}$ has property $\mathcal{P}$. \end{remark} \begin{remark} \label{remark-base-change} Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is smooth local on the source-and-target and stable under base change. Then the property of morphisms of algebraic stacks defined in Definition \ref{definition-P} is stable under base change. Namely, let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y}' \to \mathcal{Y}$ be morphisms of algebraic stacks and assume $f$ has property $\mathcal{P}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to \mathcal{X} \times_\mathcal{Y} V$. Finally, choose an algebraic space $V'$ and a surjective and smooth morphism $V' \to \mathcal{Y}' \times_\mathcal{Y} V$. Then the morphism $U \to V$ has property $\mathcal{P}$ by definition. Whence $V' \times_V U \to V'$ has property $\mathcal{P}$ as we assumed that $\mathcal{P}$ is stable under base change. Considering the diagram $$\xymatrix{ V' \times_V U \ar[r] \ar[d] & \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ V' \ar[r] & \mathcal{Y}' \ar[r] & \mathcal{Y} }$$ we see that the left top horizontal arrow is smooth and surjective, whence by definition we see that the projection $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$ has property $\mathcal{P}$. \end{remark} \begin{remark} \label{remark-implication} Let $\mathcal{P}, \mathcal{P}'$ be properties of morphisms of algebraic spaces which are smooth local on the source-and-target. Suppose that we have $\mathcal{P} \Rightarrow \mathcal{P}'$ for morphisms of algebraic spaces. Then we also have $\mathcal{P} \Rightarrow \mathcal{P}'$ for the properties of morphisms of algebraic stacks defined in Definition \ref{definition-P} using $\mathcal{P}$ and $\mathcal{P}'$. This is clear from the definition. \end{remark} \section{Morphisms of finite type} \label{section-finite-type} \noindent The property locally of finite type'' of morphisms of algebraic spaces is smooth local on the source-and-target, see Descent on Spaces, Remark \ref{spaces-descent-remark-list-local-source-target}. It is also stable under base change and fpqc local on the target, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-locally-finite-type}. Hence, by Lemma \ref{lemma-local-source-target} above, we may define what it means for a morphism of algebraic spaces to be locally of finite type as follows and it agrees with the already existing notion defined in Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} when the morphism is representable by algebraic spaces. \begin{definition} \label{definition-locally-finite-type} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. \begin{enumerate} \item We say $f$ {\it locally of finite type} if the equivalent conditions of Lemma \ref{lemma-local-source-target} hold with $\mathcal{P} = \text{locally of finite type}$. \item We say $f$ is {\it of finite type} if it is locally of finite type and quasi-compact. \end{enumerate} \end{definition} \begin{lemma} \label{lemma-composition-finite-type} The composition of finite type morphisms is of finite type. The same holds for locally of finite type. \end{lemma} \begin{proof} Combine Remark \ref{remark-composition} with Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-finite-type}. \end{proof} \begin{lemma} \label{lemma-base-change-finite-type} A base change of a finite type morphism is finite type. The same holds for locally of finite type. \end{lemma} \begin{proof} Combine Remark \ref{remark-base-change} with Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type}. \end{proof} \begin{lemma} \label{lemma-immersion-locally-finite-type} An immersion is locally of finite type. \end{lemma} \begin{proof} Combine Remark \ref{remark-implication} with Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-immersion-locally-finite-type}. \end{proof} \begin{lemma} \label{lemma-locally-finite-type-locally-noetherian} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type and $\mathcal{Y}$ is locally Noetherian, then $\mathcal{X}$ is locally Noetherian. \end{lemma} \begin{proof} Let $$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} }$$ be a commutative diagram where $U$, $V$ are schemes, $V \to \mathcal{Y}$ is surjective and smooth, and $U \to V \times_\mathcal{Y} \mathcal{X}$ is surjective and smooth. Then $U \to V$ is locally of finite type. If $\mathcal{Y}$ is locally Noetherian, then $V$ is locally Noetherian. By Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian} we see that $U$ is locally Noetherian, which means that $\mathcal{X}$ is locally Noetherian. \end{proof} \noindent The following two lemmas will be improved on later (after we have discussed morphisms of algebraic stacks which are locally of finite presentation). \begin{lemma} \label{lemma-check-finite-type-covering} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W \to \mathcal{Y}$ be a surjective, flat, and locally of finite presentation where $W$ is an algebraic space. If the base change $W \times_\mathcal{Y} \mathcal{X} \to W$ is locally of finite type, then $f$ is locally of finite type. \end{lemma} \begin{proof} Choose an algebraic space $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Choose an algebraic space $U$ and a surjective smooth morphism $U \to V \times_\mathcal{Y} \mathcal{X}$. We have to show that $U \to V$ is locally of finite presentation. Now we base change everything by $W \to \mathcal{Y}$: Set $U' = W \times_\mathcal{Y} U$, $V' = W \times_\mathcal{Y} V$, $\mathcal{X}' = W \times_\mathcal{Y} \mathcal{X}$, and $\mathcal{Y}' = W \times_\mathcal{Y} \mathcal{Y} = W$. Then it is still true that $U' \to V' \times_{\mathcal{Y}'} \mathcal{X}'$ is smooth by base change. Hence by our definition of locally finite type morphisms of algebraic stacks and the assumption that $\mathcal{X}' \to \mathcal{Y}'$ is locally of finite type, we see that $U' \to V'$ is locally of finite type. Then, since $V' \to V$ is surjective, flat, and locally of finite presentation as a base change of $W \to \mathcal{Y}$ we see that $U \to V$ is locally of finite type by Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-locally-finite-type} and we win. \end{proof} \begin{lemma} \label{lemma-check-finite-type-precompose} Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. Assume $\mathcal{X} \to \mathcal{Z}$ is locally of finite type and that $\mathcal{X} \to \mathcal{Y}$ is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Then $\mathcal{Y} \to \mathcal{Z}$ is locally of finite type. \end{lemma} \begin{proof} Choose an algebraic space $W$ and a surjective smooth morphism $W \to \mathcal{Z}$. Choose an algebraic space $V$ and a surjective smooth morphism $V \to W \times_\mathcal{Z} \mathcal{Y}$. Set $U = V \times_\mathcal{Y} \mathcal{X}$ which is an algebraic space. We know that $U \to V$ is surjective, flat, and locally of finite presentation and that $U \to W$ is locally of finite type. Hence the lemma reduces to the case of morphisms of algebraic spaces. The case of morphisms of algebraic spaces is Descent on Spaces, Lemma \ref{spaces-descent-lemma-locally-finite-type-fppf-local-source}. \end{proof} \begin{lemma} \label{lemma-finite-type-permanence} Let $f : \mathcal{X} \to \mathcal{Y}$, $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $g \circ f : \mathcal{X} \to \mathcal{Z}$ is locally of finite type, then $f : \mathcal{X} \to \mathcal{Y}$ is locally of finite type. \end{lemma} \begin{proof} We can find a diagram $$\xymatrix{ U \ar[r] \ar[d] & V \ar[r] \ar[d] & W \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{Y} \ar[r] & \mathcal{Z} }$$ where $U$, $V$, $W$ are schemes, the vertical arrow $W \to \mathcal{Z}$ is surjective and smooth, the arrow $V \to \mathcal{Y} \times_\mathcal{Z} W$ is surjective and smooth, and the arrow $U \to \mathcal{X} \times_\mathcal{Y} V$ is surjective and smooth. Then also $U \to \mathcal{X} \times_\mathcal{Z} V$ is surjective and smooth (as a composition of a surjective and smooth morphism with a base change of such). By definition we see that $U \to W$ is locally of finite type. Hence $U \to V$ is locally of finite type by Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type} which in turn means (by definition) that $\mathcal{X} \to \mathcal{Y}$ is locally of finite type. \end{proof} \section{Points of finite type} \label{section-points-finite-type} \noindent Let $\mathcal{X}$ be an algebraic stack. A finite type point $x \in |\mathcal{X}|$ is a point which can be represented by a morphism $\Spec(k) \to \mathcal{X}$ which is locally of finite type. Finite type points are a suitable replacement of closed points for algebraic spaces and algebraic stacks. There are always enough of them'' for example. \begin{lemma} \label{lemma-point-finite-type} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent: \begin{enumerate} \item There exists a morphism $\Spec(k) \to \mathcal{X}$ which is locally of finite type and represents $x$. \item There exists a scheme $U$, a closed point $u \in U$, and a smooth morphism $\varphi : U \to \mathcal{X}$ such that $\varphi(u) = x$. \end{enumerate} \end{lemma} \begin{proof} Let $u \in U$ and $U \to \mathcal{X}$ be as in (2). Then $\Spec(\kappa(u)) \to U$ is of finite type, and $U \to \mathcal{X}$ is representable and locally of finite type (by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-locally-finite-presentation} and \ref{spaces-morphisms-lemma-finite-presentation-finite-type}). Hence we see (1) holds by Lemma \ref{lemma-composition-finite-type}. \medskip\noindent Conversely, assume $\Spec(k) \to \mathcal{X}$ is locally of finite type and represents $x$. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. By assumption $U \times_\mathcal{X} \Spec(k) \to U$ is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point $v$ of $U \times_\mathcal{X} \Spec(k)$ (there exists at least one, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-identify-finite-type-points}). By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-type-points-morphism} the image $u \in U$ of $v$ is a finite type point of $U$. Hence by Morphisms, Lemma \ref{morphisms-lemma-identify-finite-type-points} after shrinking $U$ we may assume that $u$ is a closed point of $U$, i.e., (2) holds. \end{proof} \begin{definition} \label{definition-finite-type-point} Let $\mathcal{X}$ be an algebraic stack. We say a point $x \in |\mathcal{X}|$ is a {\it finite type point}\footnote{This is a slight abuse of language as it would perhaps be more correct to say locally finite type point''.} if the equivalent conditions of Lemma \ref{lemma-point-finite-type} are satisfied. We denote $\mathcal{X}_{\text{ft-pts}}$ the set of finite type points of $\mathcal{X}$. \end{definition} \noindent We can describe the set of finite type points as follows. \begin{lemma} \label{lemma-identify-finite-type-points} Let $\mathcal{X}$ be an algebraic stack. We have $$\mathcal{X}_{\text{ft-pts}} = \bigcup\nolimits_{\varphi : U \to X\text{ smooth}} |\varphi|(U_0)$$ where $U_0$ is the set of closed points of $U$. Here we may let $U$ range over all schemes smooth over $\mathcal{X}$ or over all affine schemes smooth over $\mathcal{X}$. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-point-finite-type}. \end{proof} \begin{lemma} \label{lemma-finite-type-points-morphism} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type, then $f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}$. \end{lemma} \begin{proof} Take $x \in \mathcal{X}_{\text{ft-pts}}$. Represent $x$ by a locally finite type morphism $x : \Spec(k) \to \mathcal{X}$. Then $f \circ x$ is locally of finite type by Lemma \ref{lemma-composition-finite-type}. Hence $f(x) \in \mathcal{Y}_{\text{ft-pts}}$. \end{proof} \begin{lemma} \label{lemma-finite-type-points-surjective-morphism} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $f$ is locally of finite type and surjective, then $f(\mathcal{X}_{\text{ft-pts}}) = \mathcal{Y}_{\text{ft-pts}}$. \end{lemma} \begin{proof} We have $f(\mathcal{X}_{\text{ft-pts}}) \subset \mathcal{Y}_{\text{ft-pts}}$ by Lemma \ref{lemma-finite-type-points-morphism}. Let $y \in |\mathcal{Y}|$ be a finite type point. Represent $y$ by a morphism $\Spec(k) \to \mathcal{Y}$ which is locally of finite type. As $f$ is surjective the algebraic stack $\mathcal{X}_k = \Spec(k) \times_\mathcal{Y} \mathcal{X}$ is nonempty, therefore has a finite type point $x \in |\mathcal{X}_k|$ by Lemma \ref{lemma-identify-finite-type-points}. Now $\mathcal{X}_k \to \mathcal{X}$ is a morphism which is locally of finite type as a base change of $\Spec(k) \to \mathcal{Y}$ (Lemma \ref{lemma-base-change-finite-type}). Hence the image of $x$ in $\mathcal{X}$ is a finite type point by Lemma \ref{lemma-finite-type-points-morphism} which maps to $y$ by construction. \end{proof} \begin{lemma} \label{lemma-enough-finite-type-points} Let $\mathcal{X}$ be an algebraic stack. For any locally closed subset $T \subset |\mathcal{X}|$ we have $$T \not = \emptyset \Rightarrow T \cap \mathcal{X}_{\text{ft-pts}} \not = \emptyset.$$ In particular, for any closed subset $T \subset |\mathcal{X}|$ we see that $T \cap \mathcal{X}_{\text{ft-pts}}$ is dense in $T$. \end{lemma} \begin{proof} Let $i : \mathcal{Z} \to \mathcal{X}$ be the reduced induced substack structure on $T$, see Properties of Stacks, Remark \ref{stacks-properties-remark-stack-structure-locally-closed-subset}. An immersion is locally of finite type, see Lemma \ref{lemma-immersion-locally-finite-type}. Hence by Lemma \ref{lemma-finite-type-points-morphism} we see $\mathcal{Z}_{\text{ft-pts}} \subset \mathcal{X}_{\text{ft-pts}} \cap T$. Finally, any nonempty affine scheme $U$ with a smooth morphism towards $\mathcal{Z}$ has at least one closed point, hence $\mathcal{Z}$ has at least one finite type point by Lemma \ref{lemma-identify-finite-type-points}. The lemma follows. \end{proof} \noindent Here is another, more technical, characterization of a finite type point on an algebraic stack. It tells us in particular that the residual gerbe of $\mathcal{X}$ at $x$ exists whenever $x$ is a finite type point! \begin{lemma} \label{lemma-point-finite-type-monomorphism} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent: \begin{enumerate} \item $x$ is a finite type point, \item there exists an algebraic stack $\mathcal{Z}$ whose underlying topological space $|\mathcal{Z}|$ is a singleton, and a morphism $f : \mathcal{Z} \to \mathcal{X}$ which is locally of finite type such that $\{x\} = |f|(|\mathcal{Z}|)$, and \item the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ at $x$ exists and the inclusion morphism $\mathcal{Z}_x \to \mathcal{X}$ is locally of finite type. \end{enumerate} \end{lemma} \begin{proof} (All of the morphisms occurring in this paragraph are representable by algebraic spaces, hence the conventions and results of Properties of Stacks, Section \ref{stacks-properties-section-properties-morphisms} are applicable.) Assume $x$ is a finite type point. Choose an affine scheme $U$, a closed point $u \in U$, and a smooth morphism $\varphi : U \to \mathcal{X}$ with $\varphi(u) = x$, see Lemma \ref{lemma-identify-finite-type-points}. Set $u = \Spec(\kappa(u))$ as usual. Set $R = u \times_\mathcal{X} u$ so that we obtain a groupoid in algebraic spaces $(u, R, s, t, c)$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}. The projection morphisms $R \to u$ are the compositions $$R = u \times_\mathcal{X} u \to u \times_\mathcal{X} U \to u \times_\mathcal{X} X = u$$ where the first arrow is of finite type (a base change of the closed immersion of schemes $u \to U$) and the second arrow is smooth (a base change of the smooth morphism $U \to \mathcal{X}$). Hence $s, t : R \to u$ are locally of finite type (as compositions, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-finite-type}). Since $u$ is the spectrum of a field, it follows that $s, t$ are flat and locally of finite presentation (by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-noetherian-finite-type-finite-presentation}). We see that $\mathcal{Z} = [u/R]$ is an algebraic stack by Criteria for Representability, Theorem \ref{criteria-theorem-flat-groupoid-gives-algebraic-stack}. By Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack} we obtain a canonical morphism $$f : \mathcal{Z} \longrightarrow \mathcal{X}$$ which is fully faithful. Hence this morphism is representable by algebraic spaces, see Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces} and a monomorphism, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-monomorphism}. It follows that the residual gerbe $\mathcal{Z}_x \subset \mathcal{X}$ of $\mathcal{X}$ at $x$ exists and that $f$ factors through an equivalence $\mathcal{Z} \to \mathcal{Z}_x$, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-residual-gerbe-unique}. By construction the diagram $$\xymatrix{ u \ar[d] \ar[r] & U \ar[d] \\ \mathcal{Z} \ar[r]^f & \mathcal{X} }$$ is commutative. By Criteria for Representability, Lemma \ref{criteria-lemma-flat-quotient-flat-presentation} the left vertical arrow is surjective, flat, and locally of finite presentation. Consider $$\xymatrix{ u \times_\mathcal{X} U \ar[d] \ar[r] & \mathcal{Z} \times_\mathcal{X} U \ar[r] \ar[d] & U \ar[d] \\ u \ar[r] & \mathcal{Z} \ar[r]^f & \mathcal{X} }$$ As $u \to \mathcal{X}$ is locally of finite type, we see that the base change $u \times_\mathcal{X} U \to U$ is locally of finite type. Moreover, $u \times_\mathcal{X} U \to \mathcal{Z} \times_\mathcal{X} U$ is surjective, flat, and locally of finite presentation as a base change of $u \to \mathcal{Z}$. Thus $\{u \times_\mathcal{X} U \to \mathcal{Z} \times_\mathcal{X} U\}$ is an fppf covering of algebraic spaces, and we conclude that $\mathcal{Z} \times_\mathcal{X} U \to U$ is locally of finite type by Descent on Spaces, Lemma \ref{spaces-descent-lemma-locally-finite-presentation-fppf-local-source}. By definition this means that $f$ is locally of finite type (because the vertical arrow $\mathcal{Z} \times_\mathcal{X} U \to \mathcal{Z}$ is smooth as a base change of $U \to \mathcal{X}$ and surjective as $\mathcal{Z}$ has only one point). Since $\mathcal{Z} = \mathcal{Z}_x$ we see that (3) holds. \medskip\noindent It is clear that (3) implies (2). If (2) holds then $x$ is a finite type point of $\mathcal{X}$ by Lemma \ref{lemma-finite-type-points-morphism} and Lemma \ref{lemma-enough-finite-type-points} to see that $\mathcal{Z}_{\text{ft-pts}}$ is nonempty, i.e., the unique point of $\mathcal{Z}$ is a finite type point of $\mathcal{Z}$. \end{proof} \section{Special presentations of algebraic stacks} \label{section-presentations} \noindent The following lemma gives a criterion for when a slice'' of a presentation is still flat over the algebraic stack. \begin{lemma} \label{lemma-slice} Let $\mathcal{X}$ be an algebraic stack. Consider a cartesian diagram $$\xymatrix{ U \ar[d] & F \ar[l]^p \ar[d] \\ \mathcal{X} & \Spec(k) \ar[l] }$$ where $U$ is an algebraic space, $k$ is a field, and $U \to \mathcal{X}$ is flat and locally of finite presentation. Let $f_1, \ldots, f_r \in \Gamma(U, \mathcal{O}_U)$ and $z \in |F|$ such that $f_1, \ldots, f_r$ map to a regular sequence in the local ring $\mathcal{O}_{F, \overline{z}}$. Then, after replacing $U$ by an open subspace containing $p(z)$, the morphism $$V(f_1, \ldots, f_r) \longrightarrow \mathcal{X}$$ is flat and locally of finite presentation. \end{lemma} \begin{proof} Choose a scheme $W$ and a surjective smooth morphism $W \to \mathcal{X}$. Choose an extension of fields $k \subset k'$ and a morphism $w : \Spec(k') \to W$ such that $\Spec(k') \to W \to \mathcal{X}$ is $2$-isomorphic to $\Spec(k') \to \Spec(k) \to \mathcal{X}$. This is possible as $W \to \mathcal{X}$ is surjective. Consider the commutative diagram $$\xymatrix{ U \ar[d] & U \times_\mathcal{X} W \ar[l]^-{\text{pr}_0} \ar[d] & F' \ar[l]^-{p'} \ar[d] \\ \mathcal{X} & W \ar[l] & \Spec(k') \ar[l] }$$ both of whose squares are cartesian. By our choice of $w$ we see that $F' = F \times_{\Spec(k)} \Spec(k')$. Thus $F' \to F$ is surjective and we can choose a point $z' \in |F'|$ mapping to $z$. Since $F' \to F$ is flat we see that $\mathcal{O}_{F, \overline{z}} \to \mathcal{O}_{F', \overline{z}'}$ is flat, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-at-point-etale-local-rings}. Hence $f_1, \ldots, f_r$ map to a regular sequence in $\mathcal{O}_{F', \overline{z}'}$, see Algebra, Lemma \ref{algebra-lemma-flat-increases-depth}. Note that $U \times_\mathcal{X} W \to W$ is a morphism of algebraic spaces which is flat and locally of finite presentation. Hence by More on Morphisms of Spaces, Lemma \ref{spaces-more-morphisms-lemma-slice} we see that there exists an open subspace $U'$ of $U \times_\mathcal{X} W$ containing $p(z')$ such that the intersection $U' \cap (V(f_1, \ldots, f_r) \times_\mathcal{X} W)$ is flat and locally of finite presentation over $W$. Note that $\text{pr}_0(U')$ is an open subspace of $U$ containing $p(z)$ as $\text{pr}_0$ is smooth hence open. Now we see that $U' \cap (V(f_1, \ldots, f_r) \times_\mathcal{X} W) \to \mathcal{X}$ is flat and locally of finite presentation as the composition $$U' \cap (V(f_1, \ldots, f_r) \times_\mathcal{X} W) \to W \to \mathcal{X}.$$ Hence Properties of Stacks, Lemma \ref{stacks-properties-lemma-check-property-after-precomposing} implies $\text{pr}_0(U') \cap V(f_1, \ldots, f_r) \to \mathcal{X}$ is flat and locally of finite presentation as desired. \end{proof} \begin{lemma} \label{lemma-quasi-finite-at-point} Let $\mathcal{X}$ be an algebraic stack. Consider a cartesian diagram $$\xymatrix{ U \ar[d] & F \ar[l]^p \ar[d] \\ \mathcal{X} & \Spec(k) \ar[l] }$$ where $U$ is an algebraic space, $k$ is a field, and $U \to \mathcal{X}$ is locally of finite type. Let $z \in |F|$ be such that $\dim_z(F) = 0$. Then, after replacing $U$ by an open subspace containing $p(z)$, the morphism $$U \longrightarrow \mathcal{X}$$ is locally quasi-finite. \end{lemma} \begin{proof} Since $f : U \to \mathcal{X}$ is locally of finite type there exists a maximal open $W(f) \subset U$ such that the restriction $f|_{W(f)} : W(f) \to \mathcal{X}$ is locally quasi-finite, see Properties of Stacks, Remark \ref{stacks-properties-remark-local-source-apply} (\ref{stacks-properties-item-loc-quasi-finite}). Hence all we need to do is prove that $p(z)$ is a point of $W(f)$. Moreover, the remark referenced above also shows the formation of $W(f)$ commutes with arbitrary base change by a morphism which is representable by algebraic spaces. Hence it suffices to show that the morphism $F \to \Spec(k)$ is locally quasi-finite at $z$. This follows immediately from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-quasi-finite-rel-dimension-0}. \end{proof} \noindent A quasi-DM stack has a locally quasi-finite covering'' by a scheme. \begin{theorem} \label{theorem-quasi-DM} Let $\mathcal{X}$ be an algebraic stack. The following are equivalent \begin{enumerate} \item $\mathcal{X}$ is quasi-DM, and \item there exists a scheme $W$ and a surjective, flat, locally finitely presented, locally quasi-finite morphism $W \to \mathcal{X}$. \end{enumerate} \end{theorem} \begin{proof} The implication (2) $\Rightarrow$ (1) is Lemma \ref{lemma-properties-covering-imply-diagonal}. Assume (1). Let $x \in |\mathcal{X}|$ be a finite type point. We will produce a scheme over $\mathcal{X}$ which works'' in a neighbourhood of $x$. At the end of the proof we will take the disjoint union of all of these to conclude. \medskip\noindent Let $U$ be an affine scheme, $U \to \mathcal{X}$ a smooth morphism, and $u \in U$ a closed point which maps to $x$, see Lemma \ref{lemma-point-finite-type}. Denote $u = \Spec(\kappa(u))$ as usual. Consider the following commutative diagram $$\xymatrix{ u \ar[d] & R \ar[l] \ar[d] \\ U \ar[d] & F \ar[d] \ar[l]^p \\ \mathcal{X} & u \ar[l] }$$ with both squares fibre product squares, in particular $R = u \times_\mathcal{X} u$. In the proof of Lemma \ref{lemma-point-finite-type-monomorphism} we have seen that $(u, R, s, t, c)$ is a groupoid in algebraic spaces with $s, t$ locally of finite type. Let $G \to u$ be the stabilizer group algebraic space (see Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-stabilizer-groupoid}). Note that $$G = R \times_{(u \times u)} u = (u \times_\mathcal{X} u) \times_{(u \times u)} u = \mathcal{X} \times_{\mathcal{X} \times \mathcal{X}} u.$$ As $\mathcal{X}$ is quasi-DM we see that $G$ is locally quasi-finite over $u$. By More on Groupoids in Spaces, Lemma \ref{spaces-more-groupoids-lemma-groupoid-on-field-dimension-equal-stabilizer} we have $\dim(R) = 0$. \medskip\noindent Let $e : u \to R$ be the identity of the groupoid. Thus both compositions $u \to R \to u$ are equal to the identity morphism of $u$. Note that $R \subset F$ is a closed subspace as $u \subset U$ is a closed subscheme. Hence we can also think of $e$ as a point of $F$. Consider the maps of \'etale local rings $$\mathcal{O}_{U, u} \xrightarrow{p^\sharp} \mathcal{O}_{F, \overline{e}} \longrightarrow \mathcal{O}_{R, \overline{e}}$$ Note that $\mathcal{O}_{R, \overline{e}}$ has dimension $0$ by the result of the first paragraph. On the other hand, the kernel of the second arrow is $p^\sharp(\mathfrak m_u)\mathcal{O}_{F, \overline{e}}$ as $R$ is cut out in $F$ by $\mathfrak m_u$. Thus we see that $$\mathfrak m_{\overline{z}} = \sqrt{p^\sharp(\mathfrak m_u)\mathcal{O}_{F, \overline{e}}}$$ On the other hand, as the morphism $U \to \mathcal{X}$ is smooth we see that $F \to u$ is a smooth morphism of algebraic spaces. This means that $F$ is a regular algebraic space (Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular}). Hence $\mathcal{O}_{F, \overline{e}}$ is a regular local ring (Properties of Spaces, Lemma \ref{spaces-properties-lemma-regular}). Note that a regular local ring is Cohen-Macaulay (Algebra, Lemma \ref{algebra-lemma-regular-ring-CM}). Let $d = \dim(\mathcal{O}_{F, \overline{e}})$. By Algebra, Lemma \ref{algebra-lemma-find-sequence-image-regular} we can find $f_1, \ldots, f_d \in \mathcal{O}_{U, u}$ whose images $\varphi(f_1), \ldots, \varphi(f_d)$ form a regular sequence in $\mathcal{O}_{F, \overline{z}}$. By Lemma \ref{lemma-slice} after shrinking $U$ we may assume that $Z = V(f_1, \ldots, f_d) \to \mathcal{X}$ is flat and locally of finite presentation. Note that by construction $F_Z = Z \times_\mathcal{X} u$ is a closed subspace of $F = U \times_\mathcal{X} u$, that $e$ is a point of this closed subspace, and that $$\dim(\mathcal{O}_{F_Z, \overline{e}}) = 0.$$ By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-dimension-fibre-at-a-point} it follows that $\dim_e(F_Z) = 0$ because the transcendence degree of $e$ relative to $u$ is zero. Hence it follows from Lemma \ref{lemma-quasi-finite-at-point} that after possibly shrinking $U$ the morphism $Z \to \mathcal{X}$ is locally quasi-finite. \medskip\noindent We conclude that for every finite type point $x$ of $\mathcal{X}$ there exists a locally quasi-finite, flat, locally finitely presented morphism $f_x : Z_x \to \mathcal{X}$ with $x$ in the image of $|f_x|$. Set $W = \coprod_x Z_x$ and $f = \coprod f_x$. Then $f$ is flat, locally of finite presentation, and locally quasi-finite. In particular the image of $|f|$ is open, see Properties of Stacks, Lemma \ref{stacks-properties-lemma-topology-points}. By construction the image contains all finite type points of $\mathcal{X}$, hence $f$ is surjective by Lemma \ref{lemma-enough-finite-type-points} (and Properties of Stacks, Lemma \ref{stacks-properties-lemma-characterize-surjective}). \end{proof} \begin{lemma} \label{lemma-DM-residual-gerbe} Let $\mathcal{Z}$ be a DM, locally Noetherian, reduced algebraic stack with $|\mathcal{Z}|$ a singleton. Then there exists a field $k$ and a surjective \'etale morphism $\Spec(k) \to \mathcal{Z}$. \end{lemma} \begin{proof} By Properties of Stacks, Lemma \ref{stacks-properties-lemma-unique-point-better} there exists a field $k$ and a surjective, flat, locally finitely presented morphism $\Spec(k) \to \mathcal{Z}$. Set $U = \Spec(k)$ and $R = U \times_\mathcal{Z} U$ so we obtain a groupoid in algebraic spaces