# stacks/stacks-project

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 \input{preamble} % OK, start here. % \begin{document} \title{Properties of Algebraic Stacks} \maketitle \phantomsection \label{section-phantom} \tableofcontents \section{Introduction} \label{section-introduction} \noindent Please see Algebraic Stacks, Section \ref{algebraic-section-introduction} for a brief introduction to algebraic stacks, and please read some of that chapter for our foundations of algebraic stacks. The intent is that in that chapter we are careful to distinguish between schemes, algebraic spaces, algebraic stacks, and starting with this chapter we employ the customary abuse of language where all of these concepts are used interchangeably. \medskip\noindent The goal of this chapter is to introduce some basic notions and properties of algebraic stacks. A fundamental reference for the case of quasi-separated algebraic stacks with representable diagonal is \cite{LM-B}. \section{Conventions and abuse of language} \label{section-conventions} \noindent We choose a big fppf site $\Sch_{fppf}$. All schemes are contained in $\Sch_{fppf}$. And all rings $A$ considered have the property that $\Spec(A)$ is (isomorphic) to an object of this big site. \medskip\noindent We also fix a base scheme $S$, by the conventions above an element of $\Sch_{fppf}$. The reader who is only interested in the absolute case can take $S = \Spec(\mathbf{Z})$. \medskip\noindent Here are our conventions regarding algebraic stacks: \begin{enumerate} \item When we say {\it algebraic stack} we will mean an algebraic stacks over $S$, i.e., a category fibred in groupoids $p : \mathcal{X} \to (\Sch/S)_{fppf}$ which satisfies the conditions of Algebraic Stacks, Definition \ref{algebraic-definition-algebraic-stack}. \item We will say $f : \mathcal{X} \to \mathcal{Y}$ is a {\it morphism of algebraic stacks} to indicate a $1$-morphism of algebraic stacks over $S$, i.e., a $1$-morphism of categories fibred in groupoids over $(\Sch/S)_{fppf}$, see Algebraic Stacks, Definition \ref{algebraic-definition-morphism-algebraic-stacks}. \item A {\it $2$-morphism} $\alpha : f \to g$ will indicate a $2$-morphism in the $2$-category of algebraic stacks over $S$, see Algebraic Stacks, Definition \ref{algebraic-definition-morphism-algebraic-stacks}. \item Given morphisms $\mathcal{X} \to \mathcal{Z}$ and $\mathcal{Y} \to \mathcal{Z}$ of algebraic stacks we abusively call the $2$-fibre product $\mathcal{X} \times_\mathcal{Z} \mathcal{Y}$ the {\it fibre product}. \item We will write $\mathcal{X} \times_S \mathcal{Y}$ for the product of the algebraic stacks $\mathcal{X}$, $\mathcal{Y}$. \item We will often abuse notation and say two algebraic stacks $\mathcal{X}$ and $\mathcal{Y}$ are {\it isomorphic} if they are equivalent in this $2$-category. \end{enumerate} Here are our conventions regarding algebraic spaces. \begin{enumerate} \item If we say $X$ is an {\it algebraic space} then we mean that $X$ is an algebraic space over $S$, i.e., $X$ is a presheaf on $(\Sch/S)_{fppf}$ which satisfies the conditions of Spaces, Definition \ref{spaces-definition-algebraic-space}. \item A {\it morphism of algebraic spaces} $f :X \to Y$ is a morphism of algebraic spaces over $S$ as defined in Spaces, Definition \ref{spaces-definition-morphism-algebraic-spaces}. \item We will {\bf not} distinguish between an algebraic space $X$ and the algebraic stack $\mathcal{S}_X \to (\Sch/S)_{fppf}$ it gives rise to, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-algebraic}. \item In particular, a {\it morphism} $f : X \to \mathcal{Y}$ from $X$ to an algebraic stack $\mathcal{Y}$ means a morphism $f : \mathcal{S}_X \to \mathcal{Y}$ of algebraic stacks. Similarly for morphisms $\mathcal{Y} \to X$. \item Moreover, given an algebraic stack $\mathcal{X}$ we say {\it $\mathcal{X}$ is an algebraic space} to indicate that $\mathcal{X}$ is representable by an algebraic space, see Algebraic Stacks, Definition \ref{algebraic-definition-representable-by-algebraic-space}. \item We will use the following notational convention: If we indicate an algebraic stack by a roman capital (such as $X, Y, Z, A, B, \ldots$) then it will be the case that its inertia stack is trivial, and hence it is an algebraic space, see Algebraic Stacks, Proposition \ref{algebraic-proposition-algebraic-stack-no-automorphisms}. \end{enumerate} Here are our conventions regarding schemes. \begin{enumerate} \item If we say $X$ is a {\it scheme} then we mean that $X$ is a scheme over $S$, i.e., $X$ is an object of $(\Sch/S)_{fppf}$. \item By a {\it morphism of schemes} we mean a morphism of schemes over $S$. \item We will {\bf not} distinguish between a scheme $X$ and the algebraic stack $\mathcal{S}_X \to (\Sch/S)_{fppf}$ it gives rise to, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-algebraic}. \item In particular, a {\it morphism} $f : X \to \mathcal{Y}$ from a scheme $X$ to an algebraic stack $\mathcal{Y}$ means a morphism $f : \mathcal{S}_X \to \mathcal{Y}$ of algebraic stacks. Similarly for morphisms $\mathcal{Y} \to X$. \item Moreover, given an algebraic stack $\mathcal{X}$ we say {\it $\mathcal{X}$ is a scheme} to indicate that $\mathcal{X}$ is representable, see Algebraic Stacks, Section \ref{algebraic-section-representable}. \end{enumerate} Here are our conventions regarding morphisms of algebraic stacks: \begin{enumerate} \item A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is {\it representable}, or {\it representable by schemes} if for every scheme $T$ and morphism $T \to \mathcal{Y}$ the fibre product $T \times_\mathcal{Y} \mathcal{X}$ is a scheme. See Algebraic Stacks, Section \ref{algebraic-section-representable-morphism}. \item A morphism $f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks is {\it representable by algebraic spaces} if for every scheme $T$ and morphism $T \to \mathcal{Y}$ the fibre product $T \times_\mathcal{Y} \mathcal{X}$ is an algebraic space. See Algebraic Stacks, Definition \ref{algebraic-definition-representable-by-algebraic-spaces}. In this case $Z \times_\mathcal{Y} \mathcal{X}$ is an algebraic space whenever $Z \to \mathcal{Y}$ is a morphism whose source is an algebraic space, see Algebraic Stacks, Lemma \ref{algebraic-lemma-base-change-by-space-representable-by-space}. \end{enumerate} Note that every morphism $X \to \mathcal{Y}$ from an algebraic space to an algebraic stack is representable by algebraic spaces, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-diagonal}. We will use this basic result without further mention. \section{Properties of morphisms representable by algebraic spaces} \label{section-properties-morphisms} \noindent We will study properties of (arbitrary) morphisms of algebraic stacks in its own chapter. For morphisms representable by algebraic spaces we know what it means to be surjective, smooth, or \'etale, etc. This applies in particular to morphisms $X \to \mathcal{Y}$ from algebraic spaces to algebraic stacks. In this section, we recall how this works, we list the properties to which this applies, and we prove a few easy lemmas. \medskip\noindent Our first lemma says a morphism is representable by algebraic spaces if it is so after a base change by a flat, locally finitely presented, surjective morphism. \begin{lemma} \label{lemma-check-representable-covering} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $W$ be an algebraic space and let $W \to \mathcal{Y}$ be surjective, locally of finite presentation, and flat. The following are equivalent \begin{enumerate} \item $f$ is representable by algebraic spaces, and \item $W \times_\mathcal{Y} \mathcal{X}$ is an algebraic space. \end{enumerate} \end{lemma} \begin{proof} The implication (1) $\Rightarrow$ (2) is Algebraic Stacks, Lemma \ref{algebraic-lemma-base-change-by-space-representable-by-space}. Conversely, let $W \to \mathcal{Y}$ be as in (2). To prove (1) it suffices to show that $f$ is faithful on fibre categories, see Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}. Assumption (2) implies in particular that $W \times_\mathcal{Y} \mathcal{X} \to W$ is faithful. Hence the faithfulness of $f$ follows from Stacks, Lemma \ref{stacks-lemma-faithful-descent}. \end{proof} \noindent Let $P$ be a property of morphisms of algebraic spaces which is fppf local on the target and preserved by arbitrary base change. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Then we say {\it $f$ has property $P$} if and only if for every scheme $T$ and morphism $T \to \mathcal{Y}$ the morphism of algebraic spaces $T \times_\mathcal{Y} \mathcal{X} \to T$ has property $P$, see Algebraic Stacks, Definition \ref{algebraic-definition-relative-representable-property}. \medskip\noindent It turns out that if $f : \mathcal{X} \to \mathcal{Y}$ is representable by algebraic spaces and has property $P$, then for any morphism of algebraic stacks $\mathcal{Y}' \to \mathcal{Y}$ the base change $\mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$ has property $P$, see Algebraic Stacks, Lemmas \ref{algebraic-lemma-base-change-representable-by-spaces} and \ref{algebraic-lemma-base-change-representable-transformations-property}. If the property $P$ is preserved under compositions, then this holds also in the setting of morphisms of algebraic stacks representable by algebraic spaces, see Algebraic Stacks, Lemmas \ref{algebraic-lemma-composition-representable-by-spaces} and \ref{algebraic-lemma-composition-representable-transformations-property}. Moreover, in this case products $\mathcal{X}_1 \times \mathcal{X}_2 \to \mathcal{Y}_1 \times \mathcal{Y}_2$ of morphisms representable by algebraic spaces having property $\mathcal{P}$ have property $\mathcal{P}$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-product-representable-transformations-property}. \medskip\noindent Finally, if we have two properties $P, P'$ of morphisms of algebraic spaces which are fppf local on the target and preserved by arbitrary base change and if $P(f) \Rightarrow P'(f)$ for every morphism $f$, then the same implication holds for the corresponding property of morphisms of algebraic stacks representable by algebraic spaces, see Algebraic Stacks, Lemma \ref{algebraic-lemma-representable-transformations-property-implication}. We will use this {\it without further mention} in the following and in the following chapters. \medskip\noindent The discussion above applies to each of the following properties of morphisms of algebraic spaces \begin{enumerate} \item quasi-compact, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-compact} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-quasi-compact}, \item quasi-separated, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-separated} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-quasi-separated}, \item universally closed, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-universally-closed} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-universally-closed}, \item universally open, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-universally-open} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-universally-open}, \item universally submersive, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-universally-submersive} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-universally-submersive}, \item universal homeomorphism, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-universal-homeomorphism} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-universal-homeomorphism}, \item surjective, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-surjective} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-surjective}, \item universally injective, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-universally-injective} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-universally-injective}, \item locally of finite type, 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}, \item locally of finite presentation, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-presentation} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-locally-finite-presentation}, \item finite type, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-type} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-finite-type}, \item finite presentation, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-presentation} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-finite-presentation}, \item flat, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-flat} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-flat}, \item open immersion, see Morphisms of Spaces, Section \ref{spaces-morphisms-section-immersions} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-open-immersion}, \item isomorphism, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-isomorphism}, \item affine, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-affine} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-affine}, \item closed immersion, see Morphisms of Spaces, Section \ref{spaces-morphisms-section-immersions} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-closed-immersion}, \item separated, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-separated} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-separated}, \item proper, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-proper} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-proper}, \item quasi-affine, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-affine} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-quasi-affine}, \item integral, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-integral} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-integral}, \item finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-integral} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-finite}, \item (locally) quasi-finite, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-quasi-finite} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-quasi-finite}, \item syntomic, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-syntomic} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-syntomic}, \item smooth, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-smooth} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-smooth}, \item unramified, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-unramified} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-unramified}, \item \'etale, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-etale} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-etale}, \item finite locally free, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-finite-locally-free} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-finite-locally-free}, \item monomorphism, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-monomorphism} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-property-monomorphism}, \item immersion, see Morphisms of Spaces, Section \ref{spaces-morphisms-section-immersions} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-fppf-property-immersion}, \item locally separated, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-separated} and Descent on Spaces, Lemma \ref{spaces-descent-lemma-descending-fppf-property-locally-separated}, \end{enumerate} \begin{lemma} \label{lemma-property-spaces-too} Let $P$ be a property of morphisms of algebraic spaces as above. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. The following are equivalent: \begin{enumerate} \item $f$ has $P$, \item for every algebraic space $Z$ and morphism $Z \to \mathcal{Y}$ the morphism $Z \times_\mathcal{Y} \mathcal{X} \to Z$ has $P$. \end{enumerate} \end{lemma} \begin{proof} The implication (2) $\Rightarrow$ (1) is immediate. Assume (1). Let $Z \to \mathcal{Y}$ be as in (2). Choose a scheme $U$ and a surjective \'etale morphism $U \to Z$. By assumption the morphism $U \times_\mathcal{Y} \mathcal{X} \to U$ has $P$. But the diagram $$\xymatrix{ U \times_\mathcal{Y} \mathcal{X} \ar[d] \ar[r] & Z \times_\mathcal{Y} \mathcal{X} \ar[d] \\ U \ar[r] & Z }$$ is cartesian, hence the right vertical arrow has $P$ as $\{U \to Z\}$ is an fppf covering. \end{proof} \noindent The following lemma tells us it suffices to check $P$ after a base change by a surjective, flat, locally finitely presented morphism. \begin{lemma} \label{lemma-check-property-covering} Let $P$ be a property of morphisms of algebraic spaces as above. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Let $W$ be an algebraic space and let $W \to \mathcal{Y}$ be surjective, locally of finite presentation, and flat. Set $V = W \times_\mathcal{Y} \mathcal{X}$. Then $$(f\text{ has }P) \Leftrightarrow (\text{the projection }V \to W\text{ has }P).$$ \end{lemma} \begin{proof} The implication from left to right follows from Lemma \ref{lemma-property-spaces-too}. Assume $V \to W$ has $P$. Let $T$ be a scheme, and let $T \to \mathcal{Y}$ be a morphism. Consider the commutative diagram $$\xymatrix{ T \times_\mathcal{Y} \mathcal{X} \ar[d] & T \times_\mathcal{Y} W \ar[d] \ar[l] \ar[r] & W \ar[d] \\ T & T \times_\mathcal{Y} V \ar[l] \ar[r] & V }$$ of algebraic spaces. The squares are cartesian. The bottom left morphism is a surjective, flat morphism which is locally of finite presentation, hence $\{T \times_\mathcal{Y} V \to T\}$ is an fppf covering. Hence the fact that the right vertical arrow has property $P$ implies that the left vertical arrow has property $P$. \end{proof} \begin{lemma} \label{lemma-check-property-weak-covering} Let $P$ be a property of morphisms of algebraic spaces as above. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Let $\mathcal{Z} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces, surjective, flat, and locally of finite presentation. Set $\mathcal{W} = \mathcal{Z} \times_\mathcal{Y} \mathcal{X}$. Then $$(f\text{ has }P) \Leftrightarrow (\text{the projection }\mathcal{W} \to \mathcal{Z}\text{ has }P).$$ \end{lemma} \begin{proof} Choose an algebraic space $W$ and a morphism $W \to \mathcal{Z}$ which is surjective, flat, and locally of finite presentation. By the discussion above the composition $W \to \mathcal{Y}$ is also surjective, flat, and locally of finite presentation. Denote $V = W \times_\mathcal{Z} \mathcal{W} = V \times_\mathcal{Y} \mathcal{X}$. By Lemma \ref{lemma-check-property-covering} we see that $f$ has $\mathcal{P}$ if and only if $V \to W$ does and that $\mathcal{W} \to \mathcal{Z}$ has $\mathcal{P}$ if and only if $V \to W$ does. The lemma follows. \end{proof} \begin{lemma} \label{lemma-check-property-after-precomposing} Let $P$ be a property of morphisms of algebraic spaces as above. Let $\tau \in \{\etale, smooth, syntomic, fppf\}$. Let $\mathcal{X} \to \mathcal{Y}$ and $\mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks representable by algebraic spaces. Assume \begin{enumerate} \item $\mathcal{X} \to \mathcal{Y}$ is surjective and \'etale, smooth, syntomic, or flat and locally of finite presentation, \item the composition has $P$, and \item $P$ is local on the source in the $\tau$ topology. \end{enumerate} Then $\mathcal{Y} \to \mathcal{Z}$ has property $P$. \end{lemma} \begin{proof} Let $Z$ be a scheme and let $Z \to \mathcal{Z}$ be a morphism. Set $X = \mathcal{X} \times_\mathcal{Z} Z$, $Y = \mathcal{Y} \times_\mathcal{Z} Z$. By (1) $\{X \to Y\}$ is a $\tau$ covering of algebraic spaces and by (2) $X \to Z$ has property $P$. By (3) this implies that $Y \to Z$ has property $P$ and we win. \end{proof} \begin{lemma} \label{lemma-representable-in-terms-presentations} Let $g : \mathcal{X}' \to \mathcal{X}$ be a morphism of algebraic stacks which is representable by algebraic spaces. Let $[U/R] \to \mathcal{X}$ be a presentation. Set $U' = U \times_\mathcal{X} \mathcal{X}'$, and $R' = R \times_\mathcal{X} \mathcal{X}'$. Then there exists a groupoid in algebraic spaces of the form $(U', R', s', t', c')$, a presentation $[U'/R'] \to \mathcal{X}'$, and the diagram $$\xymatrix{ [U'/R'] \ar[d]_{[\text{pr}]} \ar[r] & \mathcal{X}' \ar[d]^g \\ [U/R] \ar[r] & \mathcal{X} }$$ is $2$-commutative where the morphism $[\text{pr}]$ comes from a morphism of groupoids $\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)$. \end{lemma} \begin{proof} Since $U \to \mathcal{Y}$ is surjective and smooth, see Algebraic Stacks, Lemma \ref{algebraic-lemma-smooth-quotient-smooth-presentation} the base change $U' \to \mathcal{X}'$ is also surjective and smooth. Hence, by Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation} it suffices to show that $R' = U' \times_{\mathcal{X}'} U'$ in order to get a smooth groupoid $(U', R', s', t', c')$ and a presentation $[U'/R'] \to \mathcal{X}'$. Using that $R = V \times_\mathcal{Y} V$ (see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}) this follows from $$R' = U \times_\mathcal{X} U \times_\mathcal{X} \mathcal{X}' = (U \times_\mathcal{X} \mathcal{X}') \times_{\mathcal{X}'} (U \times_\mathcal{X} \mathcal{X}')$$ see Categories, Lemmas \ref{categories-lemma-associativity-2-fibre-product} and \ref{categories-lemma-2-fibre-product-erase-factor}. Clearly the projection morphisms $U' \to U$ and $R' \to R$ give the desired morphism of groupoids $\text{pr} : (U', R', s', t', c') \to (U, R, s, t, c)$. Hence the morphism $[\text{pr}]$ of quotient stacks by Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-functorial}. \medskip\noindent We still have to show that the diagram $2$-commutes. It is clear that the diagram $$\xymatrix{ U' \ar[d]_{\text{pr}_U} \ar[r]_{f'} & \mathcal{X}' \ar[d]^g \\ U \ar[r]^f & \mathcal{X} }$$ $2$-commutes where $\text{pr}_U : U' \to U$ is the projection. There is a canonical $2$-arrow $\tau : f \circ t \to f \circ s$ in $\Mor(R, \mathcal{X})$ coming from $R = U \times_\mathcal{X} U$, $t = \text{pr}_0$, and $s = \text{pr}_1$. Using the isomorphism $R' \to U' \times_{\mathcal{X}'} U'$ we get similarly an isomorphism $\tau' : f' \circ t' \to f' \circ s'$. Note that $g \circ f' \circ t' = f \circ t \circ \text{pr}_R$ and $g \circ f' \circ s' = f \circ s \circ \text{pr}_R$, where $\text{pr}_R : R' \to R$ is the projection. Thus it makes sense to ask if \begin{equation} \label{equation-verify} \tau \star \text{id}_{\text{pr}_R} = \text{id}_g \star \tau'. \end{equation} Now we make two claims: (1) if Equation (\ref{equation-verify}) holds, then the diagram $2$-commutes, and (2) Equation (\ref{equation-verify}) holds. We omit the proof of both claims. Hints: part (1) follows from the construction of $f = f_{can}$ and $f' = f'_{can}$ in Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}. Part (2) follows by carefuly working through the definitions. \end{proof} \begin{remark} \label{remark-representable-over} Let $\mathcal{Y}$ be an algebraic stack. Consider the following $2$-category: \begin{enumerate} \item An object is a morphism $f : \mathcal{X} \to \mathcal{Y}$ which is representable by algebraic spaces, \item a $1$-morphism $(g, \beta) : (f_1 : \mathcal{X}_1 \to \mathcal{Y}) \to (f_2 : \mathcal{X}_2 \to \mathcal{Y})$ consists of a morphism $g : \mathcal{X}_1 \to \mathcal{X}_2$ and a $2$-morphism $\beta : f_1 \to f_2 \circ g$, and \item a $2$-morphism between $(g, \beta), (g', \beta') : (f_1 : \mathcal{X}_1 \to \mathcal{Y}) \to (f_2 : \mathcal{X}_2 \to \mathcal{Y})$ is a $2$-morphism $\alpha : g \to g'$ such that $(\text{id}_{f_2} \star \alpha) \circ \beta = \beta'$. \end{enumerate} Let us denote this $2$-category $\textit{Spaces}/\mathcal{Y}$ by analogy with the notation of Topologies on Spaces, Section \ref{spaces-topologies-section-procedure}. Now we claim that in this $2$-category the morphism categories $$\Mor_{\textit{Spaces}/\mathcal{Y}}( (f_1 : \mathcal{X}_1 \to \mathcal{Y}), (f_2 : \mathcal{X}_2 \to \mathcal{Y}))$$ are all setoids. Namely, a $2$-morphism $\alpha$ is a rule which to each object $x_1$ of $\mathcal{X}_1$ assigns an isomorphism $\alpha_{x_1} : g(x_1) \longrightarrow g'(x_1)$ in the relevant fibre category of $\mathcal{X}_2$ such that the diagram $$\xymatrix{ & f_2(x_1) \ar[ld]_{\beta_{x_1}} \ar[rd]^{\beta'_{x_1}} \\ f_2(g(x_1)) \ar[rr]^{f_2(\alpha_{x_1})} & & f_2(g'(x_1)) }$$ commutes. But since $f_2$ is faithful (see Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}) this means that if $\alpha_{x_1}$ exists, then it is unique! In other words the $2$-category $\textit{Spaces}/\mathcal{Y}$ is very close to being a category. Namely, if we replace $1$-morphisms by isomorphism classes of $1$-morphisms we obtain a category. We will often perform this replacement without further mention. \end{remark} \section{Points of algebraic stacks} \label{section-points} \noindent Let $\mathcal{X}$ be an algebraic stack. Let $K, L$ be two fields and let $p : \Spec(K) \to \mathcal{X}$ and $q : \Spec(L) \to \mathcal{X}$ be morphisms. We say that $p$ and $q$ are {\it equivalent} if there exists a field $\Omega$ and a $2$-commutative diagram $$\xymatrix{ \Spec(\Omega) \ar[r] \ar[d] & \Spec(L) \ar[d]^q \\ \Spec(K) \ar[r]^p & \mathcal{X}. }$$ \begin{lemma} \label{lemma-equivalence} The notion above does indeed define an equivalence relation on morphisms from spectra of fields into the algebraic stack $\mathcal{X}$. \end{lemma} \begin{proof} It is clear that the relation is reflexive and symmetric. Hence we have to prove that it is transitive. This comes down to the following: Given a diagram $$\xymatrix{ \Spec(\Omega) \ar[r]_b \ar[d]_a & \Spec(L) \ar[d]^q & \Spec(\Omega') \ar[l]^{b'} \ar[d]^{a'} \\ \Spec(K) \ar[r]^p & \mathcal{X} & \Spec(K') \ar[l]_{p'} }$$ with both squares $2$-commutative we have to show that $p$ is equivalent to $p'$. By the $2$-Yoneda lemma (see Algebraic Stacks, Section \ref{algebraic-section-2-yoneda}) the morphisms $p$, $p'$, and $q$ are given by objects $x$, $x'$, and $y$ in the fibre categories of $\mathcal{X}$ over $\Spec(K)$, $\Spec(K')$, and $\Spec(L)$. The $2$-commutativity of the squares means that there are isomorphisms $\alpha : a^*x \to b^*y$ and $\alpha' : (a')^*x' \to (b')^*y$ in the fibre categories of $\mathcal{X}$ over $\Spec(\Omega)$ and $\Spec(\Omega')$. Choose any field $\Omega''$ and embeddings $\Omega \to \Omega''$ and $\Omega' \to \Omega''$ agreeing on $L$. Then we can extend the diagram above to $$\xymatrix{ & \Spec(\Omega'') \ar[ld]_c \ar[d]^{q'} \ar[rd]^{c'} \\ \Spec(\Omega) \ar[r]_b \ar[d]_a & \Spec(L) \ar[d]^q & \Spec(\Omega') \ar[l]^{b'} \ar[d]^{a'} \\ \Spec(K) \ar[r]^p & \mathcal{X} & \Spec(K') \ar[l]_{p'} }$$ with commutative triangles and $$(q')^*(\alpha')^{-1} \circ (q')^*\alpha : (a \circ c)^*x \longrightarrow (a' \circ c')^*x'$$ is an isomorphism in the fibre category over $\Spec(\Omega'')$. Hence $p$ is equivalent to $p'$ as desired. \end{proof} \begin{definition} \label{definition-points} Let $\mathcal{X}$ be an algebraic stack. A {\it point} of $\mathcal{X}$ is an equivalence class of morphisms from spectra of fields into $\mathcal{X}$. The set of points of $\mathcal{X}$ is denoted $|\mathcal{X}|$. \end{definition} \noindent This agrees with our definition of points of algebraic spaces, see Properties of Spaces, Definition \ref{spaces-properties-definition-points}. Moreover, for a scheme we recover the usual notion of points, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-scheme-points}. If $f : \mathcal{X} \to \mathcal{Y}$ is a morphism of algebraic stacks then there is an induced map $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ which maps a representative $x : \Spec(K) \to \mathcal{X}$ to the representative $f \circ x : \Spec(K) \to \mathcal{Y}$. This is well defined: namely $2$-isomorphic $1$-morphisms remain $2$-isomorphic after pre- or post-composing by a $1$-morphism because you can horizontally pre- or post-compose by the identity of the given $1$-morphism. This holds in any (strict) $(2, 1)$-category. If $$\xymatrix{ \mathcal{X} \ar[d] \ar[r] & \mathcal{Y} \ar[d] \\ \mathcal{W} \ar[r] & \mathcal{Z} }$$ is a $2$-commutative diagram of algebraic stacks, then the diagram of sets $$\xymatrix{ |\mathcal{X}| \ar[d] \ar[r] & |\mathcal{Y}| \ar[d] \\ |\mathcal{W}| \ar[r] & |\mathcal{Z}| }$$ is commutative. In particular, if $\mathcal{X} \to \mathcal{Y}$ is an equivalence then $|\mathcal{X}| \to |\mathcal{Y}|$ is a bijection. \begin{lemma} \label{lemma-points-cartesian} Let $$\xymatrix{ \mathcal{Z} \times_\mathcal{Y} \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \ar[d] \\ \mathcal{Z} \ar[r] & \mathcal{Y} }$$ be a fibre product of algebraic stacks. Then the map of sets of points $$|\mathcal{Z} \times_\mathcal{Y} \mathcal{X}| \longrightarrow |\mathcal{Z}| \times_{|\mathcal{Y}|} |\mathcal{X}|$$ is surjective. \end{lemma} \begin{proof} Namely, suppose given fields $K$, $L$ and morphisms $\Spec(K) \to \mathcal{X}$, $\Spec(L) \to \mathcal{Z}$, then the assumption that they agree as elements of $|\mathcal{Y}|$ means that there is a common extension $K \subset M$ and $L \subset M$ such that $\Spec(M) \to \Spec(K) \to \mathcal{X} \to \mathcal{Y}$ and $\Spec(M) \to \Spec(L) \to \mathcal{Z} \to \mathcal{Y}$ are $2$-isomorphic. And this is exactly the condition which says you get a morphism $\Spec(M) \to \mathcal{Z} \times_\mathcal{Y} \mathcal{X}$. \end{proof} \begin{lemma} \label{lemma-characterize-surjective} 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 $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ is surjective, and \item $f$ is surjective (in the sense of Section \ref{section-properties-morphisms}). \end{enumerate} \end{lemma} \begin{proof} Assume (1). Let $T \to \mathcal{Y}$ be a morphism whose source is a scheme. To prove (2) we have to show that the morphism of algebraic spaces $T \times_\mathcal{Y} \mathcal{X} \to T$ is surjective. By Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-surjective} this means we have to show that $|T \times_\mathcal{Y} \mathcal{X}| \to |T|$ is surjective. Applying Lemma \ref{lemma-points-cartesian} we see that this follows from (1). \medskip\noindent Conversely, assume (2). Let $y : \Spec(K) \to \mathcal{Y}$ be a morphism from the spectrum of a field into $\mathcal{Y}$. By assumption the morphism $\Spec(K) \times_{y, \mathcal{Y}} \mathcal{X} \to \Spec(K)$ of algebraic spaces is surjective. By Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-surjective} this means there exists a field extension $K \subset K'$ and a morphism $\Spec(K') \to \Spec(K) \times_{y, \mathcal{Y}} \mathcal{X}$ such that the left square of the diagram $$\xymatrix{ \Spec(K') \ar[r] \ar[d] & \Spec(K) \times_{y, \mathcal{Y}} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ \Spec(K) \ar@{=}[r] & \Spec(K) \ar[r]^-y & \mathcal{Y} }$$ is commutative. This shows that $|X| \to |\mathcal{Y}|$ is surjective. \end{proof} \noindent Here is a lemma explaining how to compute the set of points in terms of a presentation. \begin{lemma} \label{lemma-points-presentation} Let $\mathcal{X}$ be an algebraic stack. Let $\mathcal{X} = [U/R]$ be a presentation of $\mathcal{X}$, see Algebraic Stacks, Definition \ref{algebraic-definition-presentation}. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|\mathcal{X}|$ is the quotient of $|U|$ by this equivalence relation. \end{lemma} \begin{proof} The assumption means that we have a smooth groupoid $(U, R, s, t, c)$ in algebraic spaces, and an equivalence $f : [U/R] \to \mathcal{X}$. We may assume $\mathcal{X} = [U/R]$. The induced morphism $p : U \to \mathcal{X}$ is smooth and surjective, see Algebraic Stacks, Lemma \ref{algebraic-lemma-smooth-quotient-smooth-presentation}. Hence $|U| \to |\mathcal{X}|$ is surjective by Lemma \ref{lemma-characterize-surjective}. Note that $R = U \times_\mathcal{X} U$, see Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-quotient-stack-2-cartesian}. Hence Lemma \ref{lemma-points-cartesian} implies the map $$|R| \longrightarrow |U| \times_{|\mathcal{X}|} |U|$$ is surjective. Hence the image of $|R| \to |U| \times |U|$ is exactly the set of pairs $(u_1, u_2) \in |U| \times |U|$ such that $u_1$ and $u_2$ have the same image in $|\mathcal{X}|$. Combining these two statements we get the result of the lemma. \end{proof} \begin{remark} \label{remark-more-general-presentation} The result of Lemma \ref{lemma-points-presentation} can be generalized as follows. Let $\mathcal{X}$ be an algebraic stack. Let $U$ be an algebraic space and let $f : U \to \mathcal{X}$ be a surjective morphism (which makes sense by Section \ref{section-properties-morphisms}). Let $R = U \times_\mathcal{X} U$, let $(U, R, s, t, c)$ be the groupoid in algebraic spaces, and let $f_{can} : [U/R] \to \mathcal{X}$ be the canonical morphism as constructed in Algebraic Stacks, Lemma \ref{algebraic-lemma-map-space-into-stack}. Then the image of $|R| \to |U| \times |U|$ is an equivalence relation and $|\mathcal{X}| = |U|/|R|$. The proof of Lemma \ref{lemma-points-presentation} works without change. (Of course in general $[U/R]$ is not an algebraic stack, and in general $f_{can}$ is not an isomorphism.) \end{remark} \begin{lemma} \label{lemma-topology-points} There exists a unique topology on the sets of points of algebraic stacks with the following properties: \begin{enumerate} \item for every morphism of algebraic stacks $\mathcal{X} \to \mathcal{Y}$ the map $|\mathcal{X}| \to |\mathcal{Y}|$ is continuous, and \item for every morphism $U \to \mathcal{X}$ which is flat and locally of finite presentation with $U$ an algebraic space the map of topological spaces $|U| \to |\mathcal{X}|$ is continuous and open. \end{enumerate} \end{lemma} \begin{proof} Choose a morphism $p : U \to \mathcal{X}$ which is surjective, flat, and locally of finite presentation with $U$ an algebraic space. Such exist by the definition of an algebraic stack, as a smooth morphism is flat and locally of finite presentation (see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation} and \ref{spaces-morphisms-lemma-smooth-flat}). We define a topology on $|\mathcal{X}|$ by the rule: $W \subset |\mathcal{X}|$ is open if and only if $|p|^{-1}(W)$ is open in $|U|$. To show that this is independent of the choice of $p$, let $p' : U' \to \mathcal{X}$ be another morphism which is surjective, flat, locally of finite presentation from an algebraic space to $\mathcal{X}$. Set $U'' = U \times_\mathcal{X} U'$ so that we have a $2$-commutative diagram $$\xymatrix{ U'' \ar[r] \ar[d] & U' \ar[d] \\ U \ar[r] & \mathcal{X} }$$ As $U \to \mathcal{X}$ and $U' \to \mathcal{X}$ are surjective, flat, locally of finite presentation we see that $U'' \to U'$ and $U'' \to U$ are surjective, flat and locally of finite presentation, see Lemma \ref{lemma-property-spaces-too}. Hence the maps $|U''| \to |U'|$ and $|U''| \to |U|$ are continuous, open and surjective, see Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-surjective} and Lemma \ref{spaces-morphisms-lemma-fppf-open}. This clearly implies that our definition is independent of the choice of $p : U \to \mathcal{X}$. \medskip\noindent Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. By Algebraic Stacks, Lemma \ref{algebraic-lemma-lift-morphism-presentations} we can find a $2$-commutative diagram $$\xymatrix{ U \ar[d]_x \ar[r]_a & V \ar[d]^y \\ \mathcal{X} \ar[r]^f & \mathcal{Y} }$$ with surjective smooth vertical arrows. Consider the associated commutative diagram $$\xymatrix{ |U| \ar[d]_{|x|} \ar[r]_{|a|} & |V| \ar[d]^{|y|} \\ |\mathcal{X}| \ar[r]^{|f|} & |\mathcal{Y}| }$$ of sets. If $W \subset |\mathcal{Y}|$ is open, then by the definition above this means exactly that $|y|^{-1}(W)$ is open in $|V|$. Since $|a|$ is continuous we conclude that $|a|^{-1}|y|^{-1}(W) = |x|^{-1}|f|^{-1}(W)$ is open in $|W|$ which means by definition that $|f|^{-1}(W)$ is open in $|\mathcal{X}|$. Thus $|f|$ is continuous. \medskip\noindent Finally, we have to show that if $U$ is an algebraic space, and $U \to \mathcal{X}$ is flat and locally of finite presentation, then $|U| \to |\mathcal{X}|$ is open. Let $V \to \mathcal{X}$ be surjective, flat, and locally of finite presentation with $V$ an algebraic space. Consider the commutative diagram $$\xymatrix{ |U \times_\mathcal{X} V| \ar[r]_e \ar[rd]_f & |U| \times_{|\mathcal{X}|} |V| \ar[d]_c \ar[r]_d & |V| \ar[d]^b \\ & |U| \ar[r]^a & |\mathcal{X}| }$$ Now the morphism $U \times_\mathcal{X} V \to U$ is surjective, i.e, $f : |U \times_\mathcal{X} V| \to |U|$ is surjective. The left top horizontal arrow is surjective, see Lemma \ref{lemma-points-cartesian}. The morphism $U \times_\mathcal{X} V \to V$ is flat and locally of finite presentation, hence $d \circ e : |U \times_\mathcal{X} V| \to |V|$ is open, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-fppf-open}. Pick $W \subset |U|$ open. The properties above imply that $b^{-1}(a(W)) = (d \circ e)(f^{-1}(W))$ is open, which by construction means that $a(W)$ is open as desired. \end{proof} \begin{definition} \label{definition-topological-space} Let $\mathcal{X}$ be an algebraic stack. The underlying {\it topological space} of $\mathcal{X}$ is the set of points $|\mathcal{X}|$ endowed with the topology constructed in Lemma \ref{lemma-topology-points}. \end{definition} \noindent This definition does not conflict with the already existing topology on $|\mathcal{X}|$ if $\mathcal{X}$ is an algebraic space. \begin{lemma} \label{lemma-space-locally-quasi-compact} Let $\mathcal{X}$ be an algebraic stack. Every point of $|\mathcal{X}|$ has a fundamental system of quasi-compact open neighbourhoods. In particular $|\mathcal{X}|$ is locally quasi-compact in the sense of Topology, Definition \ref{topology-definition-locally-quasi-compact}. \end{lemma} \begin{proof} This follows formally from the fact that there exists a scheme $U$ and a surjective, open, continuous map $U \to |\mathcal{X}|$ of topological spaces. Namely, if $U \to \mathcal{X}$ is surjective and smooth, then Lemma \ref{lemma-topology-points} guarantees that $|U| \to |\mathcal{X}|$ is continuous, surjective, and open. \end{proof} \section{Surjective morphisms} \label{section-surjective} \noindent Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks which is representable by algebraic spaces. In Section \ref{section-properties-morphisms} we have already defined what it means for $f$ to be surjective. In Lemma \ref{lemma-characterize-surjective} we have seen that this is equivalent to requiring $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ to be surjective. This clears the way for the following definition. \begin{definition} \label{definition-surjective} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is {\it surjective} if the map $|f| : |\mathcal{X}| \to |\mathcal{Y}|$ of associated topological spaces is surjective. \end{definition} \noindent Here are some lemmas. \begin{lemma} \label{lemma-composition-surjective} The composition of surjective morphisms is surjective. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-base-change-surjective} The base change of a surjective morphism is surjective. \end{lemma} \begin{proof} Omitted. Hint: Use Lemma \ref{lemma-points-cartesian}. \end{proof} \begin{lemma} \label{lemma-descent-surjective} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Y}' \to \mathcal{Y}$ be a surjective morphism of algebraic stacks. If the base change $f' : \mathcal{Y}' \times_\mathcal{Y} \mathcal{X} \to \mathcal{Y}'$ of $f$ is surjective, then $f$ is surjective. \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-points-cartesian}. \end{proof} \begin{lemma} \label{lemma-surjective-permanence} Let $\mathcal{X} \to \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{Z}$ is surjective so is $\mathcal{Y} \to \mathcal{Z}$. \end{lemma} \begin{proof} Immediate. \end{proof} \section{Quasi-compact algebraic stacks} \label{section-quasi-compact} \noindent The following definition is equivalent with the definition for algebraic spaces by Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-space}. \begin{definition} \label{definition-quasi-compact} Let $\mathcal{X}$ be an algebraic stack. We say $\mathcal{X}$ is {\it quasi-compact} if and only if $|\mathcal{X}|$ is quasi-compact. \end{definition} \begin{lemma} \label{lemma-quasi-compact-stack} Let $\mathcal{X}$ be an algebraic stack. The following are equivalent: \begin{enumerate} \item $\mathcal{X}$ is quasi-compact, \item there exists a surjective smooth morphism $U \to \mathcal{X}$ with $U$ a quasi-compact scheme, \item there exists a surjective smooth morphism $U \to \mathcal{X}$ with $U$ a quasi-compact algebraic space, and \item there exists a surjective morphism $\mathcal{U} \to \mathcal{X}$ of algebraic stacks such that $\mathcal{U}$ is quasi-compact. \end{enumerate} \end{lemma} \begin{proof} We will use Lemma \ref{lemma-characterize-surjective}. Suppose $\mathcal{U}$ and $\mathcal{U} \to \mathcal{X}$ are as in (4). Then since $|\mathcal{U}| \to |\mathcal{X}|$ is surjective and continuous we conclude that $|\mathcal{X}|$ is quasi-compact. Thus (4) implies (1). The implications (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are immediate. Assume (1), i.e., $\mathcal{X}$ is quasi-compact, i.e., that $|\mathcal{X}|$ is quasi-compact. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then since $|U| \to |\mathcal{X}|$ is open we see that there exists a quasi-compact open $U' \subset U$ such that $|U'| \to |X|$ is surjective (and still smooth). Hence (2) holds. \end{proof} \begin{lemma} \label{lemma-finite-disjoint-quasi-compact} A finite disjoint union of quasi-compact algebraic stacks is a quasi-compact algebraic stack. \end{lemma} \begin{proof} This is clear from the corresponding topological fact. \end{proof} \section{Properties of algebraic stacks defined by properties of schemes} \label{section-types-properties} \noindent Any smooth local property of schemes gives rise to a corresponding property of algebraic stacks via the following lemma. Note that a property of schemes which is smooth local is also \'etale local as any \'etale covering is also a smooth covering. Hence for a smooth local property $P$ of schemes we know what it means to say that an algebraic space has $P$, see Properties of Spaces, Section \ref{section-types-properties}. \begin{lemma} \label{lemma-type-property} Let $\mathcal{P}$ be a property of schemes which is local in the smooth topology, see Descent, Definition \ref{descent-definition-property-local}. Let $\mathcal{X}$ be an algebraic stack. The following are equivalent \begin{enumerate} \item for some scheme $U$ and some surjective smooth morphism $U \to \mathcal{X}$ the scheme $U$ has property $\mathcal{P}$, \item for every scheme $U$ and every smooth morphism $U \to \mathcal{X}$ the scheme $U$ has property $\mathcal{P}$, \item for some algebraic space $U$ and some surjective smooth morphism $U \to \mathcal{X}$ the algebraic space $U$ has property $\mathcal{P}$, and \item for every algebraic space $U$ and every smooth morphism $U \to \mathcal{X}$ the algebraic space $U$ has property $\mathcal{P}$. \end{enumerate} If $\mathcal{X}$ is a scheme this is equivalent to $\mathcal{P}(U)$. If $\mathcal{X}$ is an algebraic space this is equivalent to $X$ having property $\mathcal{P}$. \end{lemma} \begin{proof} Let $U \to \mathcal{X}$ surjective and smooth with $U$ an algebraic space. Let $V \to \mathcal{X}$ be a smooth morphism with $V$ an algebraic space. Choose schemes $U'$ and $V'$ and surjective \'etale morphisms $U' \to U$ and $V' \to V$. Finally, choose a scheme $W$ and a surjective \'etale morphism $W \to V' \times_\mathcal{X} U'$. Then $W \to V'$ and $W \to U'$ are smooth morphisms of schemes as compositions of \'etale and smooth morphisms of algebraic spaces, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-smooth} and \ref{spaces-morphisms-lemma-composition-smooth}. Moreover, $W \to V'$ is surjective as $U' \to \mathcal{X}$ is surjective. Hence, we have $$\mathcal{P}(U) \Leftrightarrow \mathcal{P}(U') \Rightarrow \mathcal{P}(W) \Rightarrow \mathcal{P}(V') \Leftrightarrow \mathcal{P}(V)$$ where the equivalences are by definition of property $\mathcal{P}$ for algebraic spaces, and the two implications come from Descent, Definition \ref{descent-definition-property-local}. This proves (3) $\Rightarrow$ (4). \medskip\noindent The implications (2) $\Rightarrow$ (1), (1) $\Rightarrow$ (3), and (4) $\Rightarrow$ (2) are immediate. \end{proof} \begin{definition} \label{definition-type-property} Let $\mathcal{X}$ be an algebraic stack. Let $\mathcal{P}$ be a property of schemes which is local in the smooth topology. We say $\mathcal{X}$ {\it has property $\mathcal{P}$} if any of the equivalent conditions of Lemma \ref{lemma-type-property} hold. \end{definition} \begin{remark} \label{remark-list-properties-local-smooth-topology} Here is a list of properties which are local for the smooth topology (keep in mind that the fpqc, fppf, and syntomic topologies are stronger than the smooth topology): \begin{enumerate} \item locally Noetherian, see Descent, Lemma \ref{descent-lemma-Noetherian-local-fppf}, \item Jacobson, see Descent, Lemma \ref{descent-lemma-Jacobson-local-fppf}, \item locally Noetherian and $(S_k)$, see Descent, Lemma \ref{descent-lemma-Sk-local-syntomic}, \item Cohen-Macaulay, see Descent, Lemma \ref{descent-lemma-CM-local-syntomic}, \item reduced, see Descent, Lemma \ref{descent-lemma-reduced-local-smooth}, \item normal, see Descent, Lemma \ref{descent-lemma-normal-local-smooth}, \item locally Noetherian and $(R_k)$, see Descent, Lemma \ref{descent-lemma-Rk-local-smooth}, \item regular, see Descent, Lemma \ref{descent-lemma-regular-local-smooth}, \item Nagata, see Descent, Lemma \ref{descent-lemma-Nagata-local-smooth}. \end{enumerate} \end{remark} \noindent Any smooth local property of germs of schemes gives rise to a corresponding property of algebraic stacks. Note that a property of germs which is smooth local is also \'etale local. Hence for a smooth local property of germs of schemes $P$ we know what it means to say that an algebraic space $X$ has property $P$ at $x \in |X|$, see Properties of Spaces, Section \ref{section-types-properties}. \begin{lemma} \label{lemma-local-source-target-at-point} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ be a point of $\mathcal{X}$. Let $\mathcal{P}$ be a property of germs of schemes which is smooth local, see Descent, Definition \ref{descent-definition-local-at-point}. The following are equivalent \begin{enumerate} \item for any smooth morphism $U \to \mathcal{X}$ with $U$ a scheme and $u \in U$ with $a(u) = x$ we have $\mathcal{P}(U, u)$, \item for some smooth morphism $U \to \mathcal{X}$ with $U$ a scheme and some $u \in U$ with $a(u) = x$ we have $\mathcal{P}(U, u)$, \item for any smooth morphism $U \to \mathcal{X}$ with $U$ an algebraic space and $u \in |U|$ with $a(u) = x$ the algebraic space $U$ has property $\mathcal{P}$ at $u$, and \item for some smooth morphism $U \to \mathcal{X}$ with $U$ a an algebraic space and some $u \in |U|$ with $a(u) = x$ the algebraic space $U$ has property $\mathcal{P}$ at $u$. \end{enumerate} If $\mathcal{X}$ is representable, then this is equivalent to $\mathcal{P}(\mathcal{X}, x)$. If $\mathcal{X}$ is an algebraic space then this is equivalent to $\mathcal{X}$ having property $\mathcal{P}$ at $x$. \end{lemma} \begin{proof} Let $a : U \to \mathcal{X}$ and $u \in |U|$ as in (3). Let $b : V \to \mathcal{X}$ be another smooth morphism with $V$ an algebraic space and $v \in |V|$ with $b(v) = x$ also. Choose a scheme $U'$, an \'etale morphism $U' \to U$ and $u' \in U'$ mapping to $u$. Choose a scheme $V'$, an \'etale morphism $V' \to V$ and $v' \in V'$ mapping to $v$. By Lemma \ref{lemma-points-cartesian} there exists a point $\overline{w} \in |V' \times_\mathcal{X} U'|$ mapping to $u'$ and $v'$. Choose a scheme $W$ and a surjective \'etale morphism $W \to V' \times_\mathcal{X} U'$. We may choose a $w \in |W|$ mapping to $\overline{w}$ (see Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-surjective}). Then $W \to V'$ and $W \to U'$ are smooth morphisms of schemes as compositions of \'etale and smooth morphisms of algebraic spaces, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-etale-smooth} and \ref{spaces-morphisms-lemma-composition-smooth}. Hence $$\mathcal{P}(U, u) \Leftrightarrow \mathcal{P}(U', u') \Leftrightarrow \mathcal{P}(W, w) \Leftrightarrow \mathcal{P}(V', v') \Leftrightarrow \mathcal{P}(V, v)$$ The outer two equivalences by Properties of Spaces, Definition \ref{spaces-properties-definition-property-at-point} and the other two by what it means to be a smooth local property of germs of schemes. This proves (4) $\Rightarrow$ (3). \medskip\noindent The implications (1) $\Rightarrow$ (2), (2) $\Rightarrow$ (4), and (3) $\Rightarrow$ (1) are immediate. \end{proof} \begin{definition} \label{definition-property-at-point} Let $\mathcal{P}$ be a property of germs of schemes which is smooth local. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. We say $\mathcal{X}$ {\it has property $\mathcal{P}$ at $x$} if any of the equivalent conditions of Lemma \ref{lemma-local-source-target-at-point} holds. \end{definition} \section{Monomorphisms of algebraic stacks} \label{section-monomorphisms} \noindent We define a monomorphism of algebraic stacks in the following way. We will see in Lemma \ref{lemma-monomorphism} that this is compatible with the corresponding $2$-category theoretic notion. \begin{definition} \label{definition-monomorphism} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. We say $f$ is a {\it monomorphism} if it is representable by algebraic spaces and a monomorphism in the sense of Section \ref{section-properties-morphisms}. \end{definition} \noindent First some basic lemmas. \begin{lemma} \label{lemma-base-change-monomorphism} Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a monomorphism. Then $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}$ is a monomorphism. \end{lemma} \begin{proof} This follows from the general discussion in Section \ref{section-properties-morphisms}. \end{proof} \begin{lemma} \label{lemma-composition-monomorphism} Compositions of monomorphisms of algebraic stacks are monomorphisms. \end{lemma} \begin{proof} This follows from the general discussion in Section \ref{section-properties-morphisms} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-monomorphism}. \end{proof} \begin{lemma} \label{lemma-monomorphism} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent: \begin{enumerate} \item $f$ is a monomorphism, \item $f$ is fully faithful, \item the diagonal $\Delta_f : \mathcal{X} \to \mathcal{X} \times_\mathcal{Y} \mathcal{X}$ is an equivalence, and \item there exists an algebraic space $W$ and a surjective, flat morphism $W \to \mathcal{Y}$ which is locally of finite presentation such that $V = \mathcal{X} \times_\mathcal{Y} W$ is an algebraic space, and the morphism $V \to W$ is a monomorphism of algebraic spaces. \end{enumerate} \end{lemma} \begin{proof} The equivalence of (1) and (4) follows from the general discussion in Section \ref{section-properties-morphisms} and in particular Lemmas \ref{lemma-check-representable-covering} and \ref{lemma-check-property-covering}. \medskip\noindent The equivalence of (2) and (3) is Categories, Lemma \ref{categories-lemma-fully-faithful-diagonal-equivalence}. \medskip\noindent Assume the equivalent conditions (2) and (3). Then $f$ is representable by algebraic spaces according to Algebraic Stacks, Lemma \ref{algebraic-lemma-characterize-representable-by-algebraic-spaces}. Moreover, the $2$-Yoneda lemma combined with the fully faithfulness implies that for every scheme $T$ the functor $$\Mor(T, \mathcal{X}) \longrightarrow \Mor(T, \mathcal{Y})$$ is fully faithful. Hence given a morphism $y : T \to \mathcal{Y}$ there exists up to unique $2$-isomorphism at most one morphism $x : T \to \mathcal{X}$ such that $y \cong f \circ x$. In particular, given a morphism of schemes $h : T' \to T$ there exists at most one lift $\tilde h : T' \to T \times_\mathcal{Y} \mathcal{X}$ of $h$. Thus $T \times_\mathcal{Y} \mathcal{X} \to T$ is a monomorphism of algebraic spaces, which proves that (1) holds. \medskip\noindent Finally, assume that (1) holds. Then for any scheme $T$ and morphism $y : T \to \mathcal{Y}$ the fibre product $T \times_\mathcal{Y} \mathcal{X}$ is an algebraic space, and $T \times_\mathcal{Y} \mathcal{X} \to T$ is a monomorphism. Hence there exists up to unique isomorphism exactly one pair $(x, \alpha)$ where $x : T \to \mathcal{X}$ is a morphism and $\alpha : f \circ x \to y$ is a $2$-morphism. Applying the $2$-Yoneda lemma this says exactly that $f$ is fully faithful, i.e., that (2) holds. \end{proof} \begin{lemma} \label{lemma-monomorphism-injective-points} \begin{slogan} Monomorphisms of stacks are injective on points. \end{slogan} A monomorphism of algebraic stacks induces an injective map of sets of points. \end{lemma} \begin{proof} Let $f : \mathcal{X} \to \mathcal{Y}$ be a monomorphism of algebraic stacks. Suppose that $x_i : \Spec(K_i) \to \mathcal{X}$ be morphisms such that $f \circ x_1$ and $f \circ x_2$ define the same element of $|\mathcal{Y}|$. Applying the definition we find a common extension $\Omega$ with corresponding morphisms $c_i : \Spec(\Omega) \to \Spec(K_i)$ and a $2$-isomorphism $\beta : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$. As $f$ is fully faithful, see Lemma \ref{lemma-monomorphism}, we can lift $\beta$ to an isomorphism $\alpha : f \circ x_1 \circ c_1 \to f \circ x_1 \circ c_2$. Hence $x_1$ and $x_2$ define the same point of $|\mathcal{X}|$ as desired. \end{proof} \begin{lemma} \label{lemma-monomorphism-diagonal} Let $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}$ be morphisms of algebraic stacks. If $\mathcal{X} \to \mathcal{X}'$ is a monomorphism then the canonical diagram $$\xymatrix{ \mathcal{X} \ar[r] \ar[d] & \mathcal{X} \times_\mathcal{Y} \mathcal{X} \ar[d] \\ \mathcal{X}' \ar[r] & \mathcal{X}' \times_\mathcal{Y} \mathcal{X}' }$$ is a fibre product square. \end{lemma} \begin{proof} We have $\mathcal{X} = \mathcal{X} \times_{\mathcal{X}'} \mathcal{X}$ by Lemma \ref{lemma-monomorphism}. Thus the result by applying Categories, Lemma \ref{categories-lemma-fibre-product-after-map}. \end{proof} \section{Immersions of algebraic stacks} \label{section-immersions} \noindent Immersions of algebraic stacks are defined as follows. \begin{definition} \label{definition-immersion} Immersions. \begin{enumerate} \item A morphism of algebraic stacks is called an {\it open immersion} if it is representable, and an open immersion in the sense of Section \ref{section-properties-morphisms}. \item A morphism of algebraic stacks is called a {\it closed immersion} if it is representable, and a closed immersion in the sense of Section \ref{section-properties-morphisms}. \item A morphism of algebraic stacks is called an {\it immersion} if it is representable, and an immersion in the sense of Section \ref{section-properties-morphisms}. \end{enumerate} \end{definition} \noindent This is not the most convenient way to think about immersions for us. For us it is a little bit more convenient to think of an immersion as a morphism of algebraic stacks which is representable by algebraic spaces and is an immersion in the sense of Section \ref{section-properties-morphisms}. Similarly for closed and open immersions. 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-immersion} Let $\mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $\mathcal{Z} \to \mathcal{Y}$ be a (closed, resp.\ open) immersion. Then $\mathcal{Z} \times_\mathcal{Y} \mathcal{X} \to \mathcal{X}$ is a (closed, resp.\ open) immersion. \end{lemma} \begin{proof} This follows from the general discussion in Section \ref{section-properties-morphisms}. \end{proof} \begin{lemma} \label{lemma-composition-immersion} Compositions of immersions of algebraic stacks are immersions. Similarly for closed immersions and open immersions. \end{lemma} \begin{proof} This follows from the general discussion in Section \ref{section-properties-morphisms} and Spaces, Lemma \ref{spaces-lemma-composition-immersions}. \end{proof} \begin{lemma} \label{lemma-check-immersion-covering} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. let $W$ be an algebraic space and let $W \to \mathcal{Y}$ be a surjective, flat morphism which is locally of finite presentation. The following are equivalent: \begin{enumerate} \item $f$ is an (open, resp.\ closed) immersion, and \item $V = W \times_\mathcal{Y} \mathcal{X}$ is an algebraic space, and $V \to W$ is an (open, resp.\ closed) immersion. \end{enumerate} \end{lemma} \begin{proof} This follows from the general discussion in Section \ref{section-properties-morphisms} and in particular Lemmas \ref{lemma-check-representable-covering} and \ref{lemma-check-property-covering}. \end{proof} \begin{lemma} \label{lemma-immersion-monomorphism} An immersion is a monomorphism. \end{lemma} \begin{proof} See Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-immersions-monomorphisms}. \end{proof} \noindent The following two lemmas explain how to think about immersions in terms of presentations. \begin{lemma} \label{lemma-immersion-into-presentation} Let $(U, R, s, t, c)$ be a smooth groupoid in algebraic spaces. Let $i : \mathcal{Z} \to [U/R]$ be an immersion. Then there exists an $R$-invariant locally closed subspace $Z \subset U$ and a presentation $[Z/R_Z] \to \mathcal{Z}$ where $R_Z$ is the restriction of $R$ to $Z$ such that $$\xymatrix{ [Z/R_Z] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld]^i \\ & [U/R] }$$ is $2$-commutative. If $i$ is a closed (resp.\ open) immersion then $Z$ is a closed (resp.\ open) subspace of $U$. \end{lemma} \begin{proof} By Lemma \ref{lemma-representable-in-terms-presentations} we get a commutative diagram $$\xymatrix{ [U'/R'] \ar[dr] \ar[rr] & & \mathcal{Z} \ar[ld] \\ & [U/R] }$$ where $U' = \mathcal{Z} \times_{[U/R]} U$ and $R' = \mathcal{Z} \times_{[U/R]} R$. Since $\mathcal{Z} \to [U/R]$ is an immersion we see that $U' \to U$ is an immersion of algebraic spaces. Let $Z \subset U$ be the locally closed subspace such that $U' \to U$ factors through $Z$ and induces an isomorphism $U' \to Z$. It is clear from the construction of $R'$ that $R' = U' \times_{U, t} R = R \times_{s, U} U'$. This implies that $Z \cong U'$ is $R$-invariant and that the image of $R' \to R$ identifies $R'$ with the restriction $R_Z = s^{-1}(Z) = t^{-1}(Z)$ of $R$ to $Z$. Hence the lemma holds. \end{proof} \begin{lemma} \label{lemma-immersion-presentation} Let $(U, R, s, t, c)$ be a smooth groupoid in algebraic spaces. Let $\mathcal{X} = [U/R]$ be the associated algebraic stack, see Algebraic Stacks, Theorem \ref{algebraic-theorem-smooth-groupoid-gives-algebraic-stack}. Let $Z \subset U$ be an $R$-invariant locally closed subspace. Then $$[Z/R_Z] \longrightarrow [U/R]$$ is an immersion of algebraic stacks, where $R_Z$ is the restriction of $R$ to $Z$. If $Z \subset U$ is open (resp.\ closed) then the morphism is an open (resp.\ closed) immersion of algebraic stacks. \end{lemma} \begin{proof} Recall that by Groupoids in Spaces, Definition \ref{spaces-groupoids-definition-invariant-open} (see also discussion following the definition) we have $R_Z = s^{-1}(Z) = t^{-1}(Z)$ as locally closed subspaces of $R$. Hence the two morphisms $R_Z \to Z$ are smooth as base changes of $s$ and $t$. Hence $(Z, R_Z, s|_{R_Z}, t|_{R_Z}, c|_{R_Z \times_{s, Z, t} R_Z})$ is a smooth groupoid in algebraic spaces, and we see that $[Z/R_Z]$ is an algebraic stack, see Algebraic Stacks, Theorem \ref{algebraic-theorem-smooth-groupoid-gives-algebraic-stack}. The assumptions of Groupoids in Spaces, Lemma \ref{spaces-groupoids-lemma-criterion-fibre-product} are all satisfied and it follows that we have a $2$-fibre square $$\xymatrix{ Z \ar[d] \ar[r] & [Z/R_Z] \ar[d] \\ U \ar[r] & [U/R] }$$ It follows from this and Lemma \ref{lemma-check-representable-covering} that $[Z/R_Z] \to [U/R]$ is representable by algebraic spaces, whereupon it follows from Lemma \ref{lemma-check-property-covering} that the right vertical arrow is an immersion (resp.\ closed immersion, resp.\ open immersion) if and only if the left vertical arrow is. \end{proof} \noindent We can define open, closed, and locally closed substacks as follows. \begin{definition} \label{definition-substacks} Let $\mathcal{X}$ be an algebraic stack. \begin{enumerate} \item An {\it open substack} of $\mathcal{X}$ is a strictly full subcategory $\mathcal{X}' \subset \mathcal{X}$ such that $\mathcal{X}'$ is an algebraic stack and $\mathcal{X}' \to \mathcal{X}$ is an open immersion. \item A {\it closed substack} of $\mathcal{X}$ is a strictly full subcategory $\mathcal{X}' \subset \mathcal{X}$ such that $\mathcal{X}'$ is an algebraic stack and $\mathcal{X}' \to \mathcal{X}$ is a closed immersion. \item A {\it locally closed substack} of $\mathcal{X}$ is a strictly full subcategory $\mathcal{X}' \subset \mathcal{X}$ such that $\mathcal{X}'$ is an algebraic stack and $\mathcal{X}' \to \mathcal{X}$ is an immersion. \end{enumerate} \end{definition} \noindent This definition should be used with caution. Namely, if $f : \mathcal{X} \to \mathcal{Y}$ is an equivalence of algebraic stacks and $\mathcal{X}' \subset \mathcal{X}$ is an open substack, then it is not necessarily the case that the subcategory $f(\mathcal{X}')$ is an open substack of $\mathcal{Y}$. The problem is that it may not be a {\it strictly} full subcategory; but this is also the only problem. Here is a formal statement. \begin{lemma} \label{lemma-substack-image} For any immersion $i : \mathcal{Z} \to \mathcal{X}$ there exists a unique locally closed substack $\mathcal{X}' \subset \mathcal{X}$ such that $i$ factors as the composition of an equivalence $i' : \mathcal{Z} \to \mathcal{X}'$ followed by the inclusion morphism $\mathcal{X}' \to \mathcal{X}$. If $i$ is a closed (resp.\ open) immersion, then $\mathcal{X}'$ is a closed (resp.\ open) substack of $\mathcal{X}$. \end{lemma} \begin{proof} Omitted. \end{proof} \begin{lemma} \label{lemma-substacks-presentation} Let $[U/R] \to \mathcal{X}$ be a presentation of an algebraic stack. There is a canonical bijection $$\text{locally closed substacks }\mathcal{Z}\text{ of }\mathcal{X} \longrightarrow R\text{-invariant locally closed subspaces }Z\text{ of }U$$ which sends $\mathcal{Z}$ to $U \times_\mathcal{X} \mathcal{Z}$. Moreover, a morphism of algebraic stacks $f : \mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ if and only if $\mathcal{Y} \times_\mathcal{X} U \to U$ factors through $Z$. Similarly for closed substacks and open substacks. \end{lemma} \begin{proof} By Lemmas \ref{lemma-immersion-into-presentation} and \ref{lemma-immersion-presentation} we find that the map is a bijection. If $\mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ then of course the base change $\mathcal{Y} \times_\mathcal{X} U \to U$ factors through $Z$. Converse, suppose that $\mathcal{Y} \to \mathcal{X}$ is a morphism such that $\mathcal{Y} \times_\mathcal{X} U \to U$ factors through $Z$. We will show that for every scheme $T$ and morphism $T \to \mathcal{Y}$, given by an object $y$ of the fibre category of $\mathcal{Y}$ over $T$, the object $y$ is in fact in the fibre category of $\mathcal{Z}$ over $T$. Namely, the fibre product $T \times_\mathcal{X} U$ is an algebraic space and $T \times_\mathcal{X} U \to T$ is a surjective smooth morphism. Hence there is an fppf covering $\{T_i \to T\}$ such that $T_i \to T$ factors through $T \times_\mathcal{X} U \to T$ for all $i$. Then $T_i \to \mathcal{X}$ factors through $\mathcal{Y} \times_\mathcal{X} V$ and hence through $Z \subset U$. Thus $y|_{T_i}$ is an object of $\mathcal{Z}$ (as $Z$ is the fibre product of $U$ with $\mathcal{Z}$ over $\mathcal{X}$). Since $\mathcal{Z}$ is a strictly full substack, we conclude that $y$ is an object of $\mathcal{Z}$ as desired. \end{proof} \begin{lemma} \label{lemma-open-substacks} Let $\mathcal{X}$ be an algebraic stack. The rule $\mathcal{U} \mapsto |\mathcal{U}|$ defines an inclusion preserving bijection between open substacks of $\mathcal{X}$ and open subsets of $|\mathcal{X}|$. \end{lemma} \begin{proof} Choose a presentation $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation}. By Lemma \ref{lemma-substacks-presentation} we see that open substacks correspond to $R$-invariant open subschemes of $U$. On the other hand Lemmas \ref{lemma-points-presentation} and \ref{lemma-topology-points} guarantee these correspond bijectively to open subsets of $|\mathcal{X}|$. \end{proof} \begin{lemma} \label{lemma-open-image-substack} Let $\mathcal X$ be an algebraic stack. Let $U$ be an algebraic space and $U \to \mathcal X$ a surjective smooth morphism. For an open immersion $V \hookrightarrow U$, there exists an algebraic stack $\mathcal Y$, an open immersion $\mathcal Y \to \mathcal X$, and a surjective smooth morphism $V \to \mathcal Y$. \end{lemma} \begin{proof} We define a category fibred in groupoids $\mathcal Y$ by letting the fiber category $\mathcal{Y}_T$ over an object $T$ of $(\Sch/S)_{fppf}$ be the full subcategory of $\mathcal{X}_T$ consisting of all $y \in \Ob(\mathcal{X}_T)$ such that the projection morphism $V \times_{\mathcal X, y} T \to T$ surjective. Now for any morphism $x : T \to \mathcal X$, the $2$-fibred product $T \times_{x, \mathcal X} \mathcal Y$ has fiber category over $T'$ consisting of triples $(f : T' \to T, y \in \mathcal{X}_{T'}, f^*x \simeq y)$ such that $V \times_{\mathcal X, y} T' \to T'$ is surjective. Note that $T \times_{x, \mathcal X} \mathcal Y$ is fibered in setoids since $\mathcal Y \to \mathcal X$ is faithful (see Stacks, Lemma \ref{stacks-lemma-2-fibre-product-gives-stack-in-setoids}). Now the isomorphism $f^*x \simeq y$ gives the diagram $$\xymatrix{ V \times_{\mathcal X, y} T' \ar[d] \ar[r] & V \times_{\mathcal X, x} T \ar[r] \ar[d] & V \ar[d] \\ T' \ar[r]^f & T \ar[r]^x & \mathcal X }$$ where both squares are cartesian. The morphism $V \times_{\mathcal X, x} T \to T$ is smooth by base change, and hence open. Let $T_0 \subset T$ be its image. From the cartesian squares we deduce that $V \times_{\mathcal X, y} T' \to T'$ is surjective if and only if $f$ lands in $T_0$. Therefore $T \times_{x, \mathcal X} \mathcal Y$ is representable by $T_0$, so the inclusion $\mathcal Y \to \mathcal X$ is an open immersion. By Algebraic Stacks, Lemma \ref{algebraic-lemma-open-fibred-category-is-algebraic} we conclude that $\mathcal{Y}$ is an algebraic stack. Lastly if we denote the morphism $V \to \mathcal X$ by $g$, we have $V \times_{\mathcal X} V \to V$ is surjective (the diagonal gives a section). Hence $g$ is in the image of $\mathcal{Y}_V \to \mathcal{X}_V$, i.e., we obtain a morphism $g' : V \to \mathcal{Y}$ fitting into the commutative diagram $$\xymatrix{ V \ar[r] \ar[d]^{g'} & U \ar[d] \\ \mathcal{Y} \ar[r] & \mathcal{X} }$$ Since $V \times_{g, \mathcal X} \mathcal Y \to V$ is a monomorphism, it is in fact an isomorphism since $(1, g')$ defines a section. Therefore $g' : V \to \mathcal Y$ is a smooth morphism, as it is the base change of the smooth morphism $g : V \to \mathcal{X}$. It is surjective by our construction of $\mathcal{Y}$ which finishes the proof of the lemma. \end{proof} \begin{lemma} \label{lemma-union-open-substacks} Let $\mathcal X$ be an algebraic stack and $\mathcal{X}_i \subset \mathcal X$ a collection of open substacks indexed by $i \in I$. Then there exists an open substack, which we denote $\bigcup_{i\in I} \mathcal{X}_i \subset \mathcal X$, such that the $\mathcal{X}_i$ are open substacks covering it. \end{lemma} \begin{proof} We define a fibred subcategory $\mathcal{X}' = \bigcup_{i \in I} \mathcal{X}_i$ by letting the fiber category over an object $T$ of $(\Sch/S)_{fppf}$ be the full subcategory of $\mathcal{X}_T$ consisting of all $x \in \Ob(\mathcal{X}_T)$ such that the morphism $\coprod_{i \in I} (\mathcal{X}_i \times_{\mathcal X} T) \to T$ is surjective. Let $x_i \in \Ob((\mathcal{X}_i)_T)$. Then $(x_i, 1)$ gives a section of $\mathcal{X}_i \times_{\mathcal X} T \to T$, so we have an isomorphism. Thus $\mathcal{X}_i \subset \mathcal{X}'$ is a full subcategory. Now let $x \in \Ob(\mathcal{X}_T)$. Then $\mathcal{X}_i \times_{\mathcal X} T$ is representable by an open subscheme $T_i \subset T$. The $2$-fibred product $\mathcal{X}' \times_{\mathcal X} T$ has fiber over $T'$ consisting of $(y \in \mathcal{X}_{T'}, f : T' \to T, f^*x \simeq y)$ such that $\coprod (\mathcal{X}_i \times_{\mathcal X, y} T') \to T'$ is surjective. The isomorphism $f^*x \simeq y$ induces an isomorphism $\mathcal{X}_i \times_{\mathcal X, y} T' \simeq T_i \times_T T'$. Then the $T_i \times_T T'$ cover $T'$ if and only if $f$ lands in $\bigcup T_i$. Therefore we have a diagram $$\xymatrix{ T_i \ar[r] \ar[d] & \bigcup T_i \ar[r] \ar[d] & T \ar[d] \\ \mathcal{X}_i \ar[r] & \mathcal{X}' \ar[r] & \mathcal{X} }$$ with both squares cartesian. By Algebraic Stacks, Lemma \ref{algebraic-lemma-open-fibred-category-is-algebraic} we conclude that $\mathcal{X'} \subset \mathcal{X}$ is algebraic and an open substack. It is also clear from the cartesian squares above that the morphism $\coprod_{i \in I} \mathcal{X}_i \to \mathcal{X}'$ which finishes the proof of the lemma. \end{proof} \begin{lemma} \label{lemma-quasi-compact-finite-subcover} Let $\mathcal X$ be an algebraic stack and $\mathcal X' \subset \mathcal X$ a quasi-compact open substack. Suppose that we have a collection of open substacks $\mathcal{X}_i \subset \mathcal X$ indexed by $i \in I$ such that $\mathcal{X}' \subset \bigcup_{i \in I} \mathcal{X}_i$, where we define the union as in Lemma \ref{lemma-union-open-substacks}. Then there exists a finite subset $I' \subset I$ such that $\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i$. \end{lemma} \begin{proof} Since $\mathcal X$ is algebraic, there exists a scheme $U$ with a surjective smooth morphism $U \to \mathcal X$. Let $U_i \subset U$ be the open subscheme representing $\mathcal{X}_i \times_{\mathcal X} U$ and $U' \subset U$ the open subscheme representing $\mathcal{X}' \times_{\mathcal X} U$. By hypothesis, $U'\subset \bigcup_{i\in I} U_i$. From the proof of Lemma \ref{lemma-quasi-compact-stack}, there is a quasi-compact open $V \subset U'$ such that $V \to \mathcal{X}'$ is a surjective smooth morphism. Therefore there exists a finite subset $I' \subset I$ such that $V \subset \bigcup_{i \in I'} U_i$. We claim that $\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i$. Take $x \in \Ob(\mathcal{X}'_T)$ for $T \in \Ob((\Sch/S)_{fppf})$. Since $\mathcal{X}' \to \mathcal{X}$ is a monomorphism, we have cartesian squares $$\xymatrix{ V \times_\mathcal{X} T \ar[r] \ar[d] & T \ar[d]^x \ar@{=}[r] & T \ar[d]^x \\ V \ar[r] & \mathcal{X}' \ar[r] & \mathcal X }$$ By base change, $V \times_{\mathcal X} T \to T$ is surjective. Therefore $\bigcup_{i \in I'} U_i \times_{\mathcal X} T \to T$ is also surjective. Let $T_i \subset T$ be the open subscheme representing $\mathcal{X}_i \times_{\mathcal X} T$. By a formal argument, we have a Cartesian square $$\xymatrix{ U_i \times_{\mathcal{X}_i} T_i \ar[r] \ar[d] & U \times_{\mathcal X} T \ar[d] \\ T_i \ar[r] & T }$$ where the vertical arrows are surjective by base change. Since $U_i \times_{\mathcal{X}_i} T_i \simeq U_i \times_{\mathcal X} T$, we find that $\bigcup_{i \in I'} T_i = T$. Hence $x$ is an object of $(\bigcup_{i\in I'} \mathcal{X}_i)_T$ by definition of the union. Observe that the inclusion $\mathcal{X}' \subset \bigcup_{i \in I'} \mathcal{X}_i$ is automatically an open substack. \end{proof} \begin{lemma} \label{lemma-zariski-open-cover-stack-is-space} Let $\mathcal X$ be an algebraic stack. Let $\mathcal{X}_i$, $i \in I$ be a set of open substacks of $\mathcal{X}$. Assume \begin{enumerate} \item $\mathcal{X} = \bigcup_{i \in I} \mathcal{X}_i$, and \item each $\mathcal{X}_i$ is an algebraic space. \end{enumerate} Then $\mathcal{X}$ is an algebraic space. \end{lemma} \begin{proof} Apply Stacks, Lemma \ref{stacks-lemma-stack-in-setoids-descent} to the morphism $\coprod_{i \in I} \mathcal{X}_i \to \mathcal{X}$ and the morphism $\text{id} : \mathcal{X} \to \mathcal{X}$ to see that $\mathcal{X}$ is a stack in setoids. Hence $\mathcal{X}$ is an algebraic space, see Algebraic Stacks, Proposition \ref{algebraic-proposition-algebraic-stack-no-automorphisms}. \end{proof} \begin{lemma} \label{lemma-zariski-open-cover-stack-is-scheme} Let $\mathcal X$ be an algebraic stack. Let $\mathcal{X}_i$, $i \in I$ be a set of open substacks of $\mathcal{X}$. Assume \begin{enumerate} \item $\mathcal{X} = \bigcup_{i \in I} \mathcal{X}_i$, and \item each $\mathcal{X}_i$ is a scheme \end{enumerate} Then $\mathcal{X}$ is a scheme. \end{lemma} \begin{proof} By Lemma \ref{lemma-zariski-open-cover-stack-is-space} we see that $\mathcal{X}$ is an algebraic space. Since any algebraic space has a largest open subspace which is a scheme, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-subscheme} we see that $\mathcal{X}$ is a scheme. \end{proof} \noindent The following lemma is the analogue of More on Groupoids, Lemma \ref{more-groupoids-lemma-local-source}. \begin{lemma} \label{lemma-local-source} Let $\mathcal{P}, \mathcal{Q}, \mathcal{R}$ be properties of morphisms of algebraic spaces. Assume \begin{enumerate} \item $\mathcal{P}, \mathcal{Q}, \mathcal{R}$ are fppf local on the target and stable under arbitrary base change, \item $\text{smooth} \Rightarrow \mathcal{R}$, \item for any morphism $f : X \to Y$ which has $\mathcal{Q}$ there exists a largest open subspace $W(\mathcal{P}, f) \subset X$ such that $f|_{W(\mathcal{P}, f)}$ has $\mathcal{P}$, and \item for any morphism $f : X \to Y$ which has $\mathcal{Q}$, and any morphism $Y' \to Y$ which has $\mathcal{R}$ we have $Y' \times_Y W(\mathcal{P}, f) = W(\mathcal{P}, f')$, where $f' : X_{Y'} \to Y'$ is the base change of $f$. \end{enumerate} Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks representable by algebraic spaces. Assume $f$ has $\mathcal{Q}$. Then \begin{enumerate} \item[(A)] there exists a largest open substack $\mathcal{X}' \subset \mathcal{X}$ such that $f|_{\mathcal{X}'}$ has $\mathcal{P}$, and \item[(B)] if $\mathcal{Z} \to \mathcal{Y}$ is a morphism of algebraic stacks representable by algebraic spaces which has $\mathcal{R}$ then $\mathcal{Z} \times_\mathcal{Y} \mathcal{X}'$ is the largest open substack of $\mathcal{Z} \times_\mathcal{Y} \mathcal{X}$ over which the base change $\text{id}_\mathcal{Z} \times f$ has property $\mathcal{P}$. \end{enumerate} \end{lemma} \begin{proof} Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Set $U = V \times_\mathcal{Y} \mathcal{X}$ and let $f' : U \to V$ be the base change of $f$. The morphism of algebraic spaces $f' : U \to V$ has property $\mathcal{Q}$. Thus we obtain the open $W(\mathcal{P}, f') \subset U$ by assumption (3). Note that $U \times_\mathcal{X} U = (V \times_\mathcal{Y} V) \times_\mathcal{Y} \mathcal{X}$ hence the morphism $f'' : U \times_\mathcal{X} U \to V \times_\mathcal{Y} V$ is the base change of $f$ via either projection $V \times_\mathcal{Y} V \to V$. By our choice of $V$ these projections are smooth, hence have property $\mathcal{R}$ by (2). Thus by (4) we see that the inverse images of $W(\mathcal{P}, f')$ under the two projections $\text{pr}_i : U \times_\mathcal{X} U \to U$ agree. In other words, $W(\mathcal{P}, f')$ is an $R$-invariant subspace of $U$ (where $R = U \times_\mathcal{X} U$). Let $\mathcal{X}'$ be the open substack of $\mathcal{X}$ corresponding to $W(\mathcal{P}, f)$ via Lemma \ref{lemma-immersion-into-presentation}. By construction $W(\mathcal{P}, f') = \mathcal{X}' \times_\mathcal{Y} V$ hence $f|_{\mathcal{X}'}$ has property $\mathcal{P}$ by Lemma \ref{lemma-check-property-covering}. Also, $\mathcal{X}'$ is the largest open substack such that $f|_{\mathcal{X}'}$ has $\mathcal{P}$ as the same maximality holds for $W(\mathcal{P}, f)$. This proves (A). \medskip\noindent Finally, if $\mathcal{Z} \to \mathcal{Y}$ is a morphism of algebraic stacks representable by algebraic spaces which has $\mathcal{R}$ then we set $T = V \times_\mathcal{Y} \mathcal{Z}$ and we see that $T \to V$ is a morphism of algebraic spaces having property $\mathcal{R}$. Set $f'_T : T \times_V U \to T$ the base change of $f'$. By (4) again we see that $W(\mathcal{P}, f'_T)$ is the inverse image of $W(\mathcal{P}, f)$ in $T \times_V U$. This implies (B); some details omitted. \end{proof} \begin{remark} \label{remark-local-source-warning} Warning: Lemma \ref{lemma-local-source} should be used with care. For example, it applies to $\mathcal{P}=$flat'', $\mathcal{Q}=$empty'', and $\mathcal{R}=$flat and locally of finite presentation''. But given a morphism of algebraic spaces $f : X \to Y$ the largest open subspace $W \subset X$ such that $f|_W$ is flat is {\it not} the set of points where $f$ is flat! \end{remark} \begin{remark} \label{remark-local-source-apply} Notwithstanding the warning in Remark \ref{remark-local-source-warning} there are some cases where Lemma \ref{lemma-local-source} can be used without causing ambiguity. We give a list. In each case we omit the verification of assumptions (1) and (2) and we give references which imply (3) and (4). Here is the list: \begin{enumerate} \item \label{item-rel-dim-leq-d} $\mathcal{Q} =$locally of finite type'', $\mathcal{R} = \emptyset$, and $\mathcal{P} =$relative dimension $\leq d$''. See Morphisms of Spaces, Definition \ref{spaces-morphisms-definition-relative-dimension} and Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-openness-bounded-dimension-fibres} and \ref{spaces-morphisms-lemma-dimension-fibre-after-base-change}. \item \label{item-loc-quasi-finite} $\mathcal{Q} =$locally of finite type'', $\mathcal{R} = \emptyset$, and $\mathcal{P} =$locally quasi-finite''. This is the case $d = 0$ of the previous item, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-quasi-finite-rel-dimension-0}. On the other hand, properties (3) and (4) are spelled out in Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-quasi-finite-part}. \item \label{item-unramified} $\mathcal{Q} =$locally of finite type'', $\mathcal{R} = \emptyset$, and $\mathcal{P} =$unramified''. This is Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-where-unramified}. \item \label{item-flat} $\mathcal{Q} =$locally of finite presentation'', $\mathcal{R} =$flat and locally of finite presentation'', and $\mathcal{P} =$flat''. See More on Morphisms of Spaces, Theorem \ref{spaces-more-morphisms-theorem-openness-flatness} and Lemma \ref{spaces-more-morphisms-lemma-flat-locus-base-change}. Note that here $W(\mathcal{P}, f)$ is always exactly the set of points where the morphism $f$ is flat because we only consider this open when $f$ has $\mathcal{Q}$ (see loc.cit.). \item \label{item-etale} $\mathcal{Q} =$locally of finite presentation'', $\mathcal{R} =$flat and locally of finite presentation'', and $\mathcal{P}=$\'etale''. This follows on combining (\ref{item-unramified}) and (\ref{item-flat}) because an unramified morphism which is flat and locally of finite presentation is \'etale, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-flat-lfp-etale}. \item Add more here as needed (compare with the longer list at More on Groupoids, Remark \ref{more-groupoids-remark-local-source-apply}). \end{enumerate} \end{remark} \section{Reduced algebraic stacks} \label{section-reduced} \noindent We have already defined reduced algebraic stacks in Section \ref{section-types-properties}. \begin{lemma} \label{lemma-reduced-closed-substack} Let $\mathcal{X}$ be an algebraic stack. Let $T \subset |\mathcal{X}|$ be a closed subset. There exists a unique closed substack $\mathcal{Z} \subset \mathcal{X}$ with the following properties: (a) we have $|\mathcal{Z}| = T$, and (b) $\mathcal{Z}$ is reduced. \end{lemma} \begin{proof} Let $U \to \mathcal{X}$ be a surjective smooth morphism, where $U$ is an algebraic space. Set $R = U \times_\mathcal{X} U$, so that there is a presentation $[U/R] \to \mathcal{X}$, see Algebraic Stacks, Lemma \ref{algebraic-lemma-stack-presentation}. As usual we denote $s, t : R \to U$ the two smooth projection morphisms. By Lemma \ref{lemma-points-presentation} we see that $T$ corresponds to a closed subset $T' \subset |U|$ such that $|s|^{-1}(T') = |t|^{-1}(T')$. Let $Z \subset U$ be the reduced induced algebraic space structure on $T'$, see Properties of Spaces, Definition \ref{spaces-properties-definition-reduced-induced-space}. The fibre products $Z \times_{U, t} R$ and $R \times_{s, U} Z$ are closed subspaces of $R$ (Spaces, Lemma \ref{spaces-lemma-base-change-immersions}). The projections $Z \times_{U, t} R \to Z$ and $R \times_{s, U} Z \to Z$ are smooth by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-smooth}. Thus as $Z$ is reduced, it follows that $Z \times_{U, t} R$ and $R \times_{s, U} Z$ are reduced, see Remark \ref{remark-list-properties-local-smooth-topology}. Since $$|Z \times_{U, t} R| = |t|^{-1}(T') = |s|^{-1}(T') = R \times_{s, U} Z$$ we conclude from the uniqueness in Properties of Spaces, Lemma \ref{spaces-properties-lemma-reduced-closed-subspace} that $Z \times_{U, t} R = R \times_{s, U} Z$. Hence $Z$ is an $R$-invariant closed subspace of $U$. By the correspondence of Lemma \ref{lemma-substacks-presentation} we obtain a closed substack $\mathcal{Z} \subset \mathcal{X}$ with $Z = \mathcal{Z} \times_\mathcal{X} U$. Then $[Z/R_Z] \to \mathcal{Z}$ is a presentation (Lemma \ref{lemma-immersion-into-presentation}). Then $|\mathcal{Z}| = |Z|/|R_Z| = |T'|/\sim$ is the given closed subset $T$. We omit the proof of unicity. \end{proof} \begin{lemma} \label{lemma-reduced-stack-determined-by-points} Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{X}' \subset \mathcal{X}$ is a closed substack, $\mathcal{X}$ is reduced and $|\mathcal{X}'| = |\mathcal{X}|$, then $\mathcal{X}' = \mathcal{X}$. \end{lemma} \begin{proof} Choose a presentation $[U/R] \to \mathcal{X}$ with $U$ a scheme. As $\mathcal{X}$ is reduced, we see that $U$ is reduced (by definition of reduced algebraic stacks). By Lemma \ref{lemma-substacks-presentation} $\mathcal{X}'$ corresponds to an $R$-invariant closed subscheme $Z \subset U$. But now $|Z| \subset |U|$ is the inverse image of $|\mathcal{X}'|$, and hence $|Z| = |U|$. Hence $Z$ is a closed subscheme of $U$ whose underlying sets of points agree. By Schemes, Lemma \ref{schemes-lemma-map-into-reduction} the map $\text{id}_U : U \to U$ factors through $Z \to U$, and hence $Z = U$, i.e., $\mathcal{X}' = \mathcal{X}$. \end{proof} \begin{lemma} \label{lemma-map-into-reduction} Let $\mathcal{X}$, $\mathcal{Y}$ be algebraic stacks. Let $\mathcal{Z} \subset \mathcal{X}$ be a closed substack Assume $\mathcal{Y}$ is reduced. A morphism $f : \mathcal{Y} \to \mathcal{X}$ factors through $\mathcal{Z}$ if and only if $f(|\mathcal{Y}|) \subset |\mathcal{Z}|$. \end{lemma} \begin{proof} Assume $f(|\mathcal{Y}|) \subset |\mathcal{Z}|$. Consider $\mathcal{Y} \times_\mathcal{X} \mathcal{Z} \to \mathcal{Y}$. There is an equivalence $\mathcal{Y} \times_\mathcal{X} \mathcal{Z} \to \mathcal{Y}'$ where $\mathcal{Y}'$ is a closed substack of $\mathcal{Y}$, see Lemmas \ref{lemma-base-change-immersion} and \ref{lemma-substack-image}. Using Lemmas \ref{lemma-points-cartesian}, \ref{lemma-monomorphism-injective-points}, and \ref{lemma-immersion-monomorphism} we see that $|\mathcal{Y}'| = |\mathcal{Y}|$. Hence we have reduced the lemma to Lemma \ref{lemma-reduced-stack-determined-by-points}. \end{proof} \begin{definition} \label{definition-reduced-induced-stack} Let $\mathcal{X}$ be an algebraic stack. Let $Z \subset |\mathcal{X}|$ be a closed subset. An {\it algebraic stack structure on $Z$} is given by a closed substack $\mathcal{Z}$ of $\mathcal{X}$ with $|\mathcal{Z}|$ equal to $Z$. The {\it reduced induced algebraic stack structure} on $Z$ is the one constructed in Lemma \ref{lemma-reduced-closed-substack}. The {\it reduction $\mathcal{X}_{red}$ of $\mathcal{X}$} is the reduced induced algebraic stack structure on $|\mathcal{X}|$. \end{definition} \noindent In fact we can use this to define the reduced induced algebraic stack structure on a locally closed subset. \begin{remark} \label{remark-stack-structure-locally-closed-subset} Let $X$ be an algebraic stack. Let $T \subset |\mathcal{X}|$ be a locally closed subset. Let $\partial T$ be the boundary of $T$ in the topological space $|\mathcal{X}|$. In a formula $$\partial T = \overline{T} \setminus T.$$ Let $\mathcal{U} \subset \mathcal{X}$ be the open substack of $X$ with $|\mathcal{U}| = |\mathcal{X}| \setminus \partial T$, see Lemma \ref{lemma-open-substacks}. Let $\mathcal{Z}$ be the reduced closed substack of $\mathcal{U}$ with $|\mathcal{Z}| = T$ obtained by taking the reduced induced closed subspace structure, see Definition \ref{definition-reduced-induced-stack}. By construction $\mathcal{Z} \to \mathcal{U}$ is a closed immersion of algebraic stacks and $\mathcal{U} \to \mathcal{X}$ is an open immersion, hence $\mathcal{Z} \to \mathcal{X}$ is an immersion of algebraic stacks by Lemma \ref{lemma-composition-immersion}. Note that $\mathcal{Z}$ is a reduced algebraic stack and that $|\mathcal{Z}| = T$ as subsets of $|X|$. We sometimes say $\mathcal{Z}$ is the {\it reduced induced substack structure} on $T$. \end{remark} \section{Residual gerbes} \label{section-residual-gerbe} \noindent In the Stacks project we would like to define the {\it residual gerbe} of an algebraic stack $\mathcal{X}$ at a point $x \in |\mathcal{X}|$ to be a monomorphism of algebraic stacks $m_x : \mathcal{Z}_x \to \mathcal{X}$ where $\mathcal{Z}_x$ is a reduced algebraic stack having a unique point which is mapped by $m_x$ to $x$. It turns out that there are many issues with this notion; existence is not clear in general and neither is uniqueness. We resolve the uniqueness issue by imposing a slightly stronger condition on the algebraic stacks $\mathcal{Z}_x$. We discuss this in more detail by working through a few simple lemmas regarding reduced algebraic stacks having a unique point. \begin{lemma} \label{lemma-flat-cover-by-field} Let $\mathcal{Z}$ be an algebraic stack. Let $k$ be a field and let $\Spec(k) \to \mathcal{Z}$ be surjective and flat. Then any morphism $\Spec(k') \to \mathcal{Z}$ where $k'$ is a field is surjective and flat. \end{lemma} \begin{proof} Consider the fibre square $$\xymatrix{ T \ar[d] \ar[r] & \Spec(k) \ar[d] \\ \Spec(k') \ar[r] & \mathcal{Z} }$$ Note that $T \to \Spec(k')$ is flat and surjective hence $T$ is not empty. On the other hand $T \to \Spec(k)$ is flat as $k$ is a field. Hence $T \to \mathcal{Z}$ is flat and surjective. It follows from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-permanence} (via the discussion in Section \ref{section-properties-morphisms}) that $\Spec(k') \to \mathcal{Z}$ is flat. It is clear that it is surjective as by assumption $|\mathcal{Z}|$ is a singleton. \end{proof} \begin{lemma} \label{lemma-unique-point} Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent \begin{enumerate} \item $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton, \item there exists a surjective flat morphism $\Spec(k) \to \mathcal{Z}$ where $k$ is a field, and \item there exists a locally of finite type, surjective, flat morphism $\Spec(k) \to \mathcal{Z}$ where $k$ is a field. \end{enumerate} \end{lemma} \begin{proof} Assume (1). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Then $W$ is a reduced scheme. Let $\eta \in W$ be a generic point of an irreducible component of $W$. Since $W$ is reduced we have $\mathcal{O}_{W, \eta} = \kappa(\eta)$. It follows that the canonical morphism $\eta = \Spec(\kappa(\eta)) \to W$ is flat. We see that the composition $\eta \to \mathcal{Z}$ is flat (see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-composition-flat}). It is also surjective as $|\mathcal{Z}|$ is a singleton. In other words (2) holds. \medskip\noindent Assume (2). Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a surjective flat morphism $\Spec(k) \to \mathcal{Z}$. Then $W \times_\mathcal{Z} \Spec(k)$ is an algebraic space smooth over $k$, hence regular (see Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular}) and in particular reduced. Since $W \times_\mathcal{Z} \Spec(k) \to W$ is surjective and flat we conclude that $W$ is reduced (Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-reduced}). In other words (1) holds. \medskip\noindent It is clear that (3) implies (2). Finally, assume (2). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. The composition $$\Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z}$$ is locally of finite type by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-type} and \ref{spaces-morphisms-lemma-smooth-locally-finite-type}. It is also flat and surjective by Lemma \ref{lemma-flat-cover-by-field}. Hence (3) holds. \end{proof} \noindent The following lemma singles out a slightly better class of singleton algebraic stacks than the preceding lemma. \begin{lemma} \label{lemma-unique-point-better} Let $\mathcal{Z}$ be an algebraic stack. The following are equivalent \begin{enumerate} \item $\mathcal{Z}$ is reduced, locally Noetherian, and $|\mathcal{Z}|$ is a singleton, and \item there exists a locally finitely presented, surjective, flat morphism $\Spec(k) \to \mathcal{Z}$ where $k$ is a field. \end{enumerate} \end{lemma} \begin{proof} Assume (2) holds. By Lemma \ref{lemma-unique-point} we see that $\mathcal{Z}$ is reduced and $|\mathcal{Z}|$ is a singleton. Let $W$ be a scheme and let $W \to \mathcal{Z}$ be a surjective smooth morphism. Choose a field $k$ and a locally finitely presented, surjective, flat morphism $\Spec(k) \to \mathcal{Z}$. Then $W \times_\mathcal{Z} \Spec(k)$ is an algebraic space smooth over $k$, hence locally Noetherian (see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-locally-noetherian}). Since $W \times_\mathcal{Z} \Spec(k) \to W$ is flat, surjective, and locally of finite presentation, we see that $\{W \times_\mathcal{Z} \Spec(k) \to W\}$ is an fppf covering and we conclude that $W$ is locally Noetherian (Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-locally-Noetherian}). In other words (1) holds. \medskip\noindent Assume (1). Pick a nonempty affine scheme $W$ and a smooth morphism $W \to \mathcal{Z}$. Pick a closed point $w \in W$ and set $k = \kappa(w)$. Because $W$ is locally Noetherian the morphism $w : \Spec(k) \to W$ is of finite presentation, see Morphisms, Lemma \ref{morphisms-lemma-closed-immersion-finite-presentation}. Hence the composition $$\Spec(k) \xrightarrow{w} W \longrightarrow \mathcal{Z}$$ is locally of finite presentation by Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-composition-finite-presentation} and \ref{spaces-morphisms-lemma-smooth-locally-finite-presentation}. It is also flat and surjective by Lemma \ref{lemma-flat-cover-by-field}. Hence (2) holds. \end{proof} \begin{lemma} \label{lemma-monomorphism-into-point} Let $\mathcal{Z}' \to \mathcal{Z}$ be a monomorphism of algebraic stacks. Assume there exists a field $k$ and a locally finitely presented, surjective, flat morphism $\Spec(k) \to \mathcal{Z}$. Then either $\mathcal{Z}'$ is empty or $\mathcal{Z}' \to \mathcal{Z}$ is an equivalence. \end{lemma} \begin{proof} We may assume that $\mathcal{Z}'$ is nonempty. In this case the fibre product $T = \mathcal{Z}' \times_\mathcal{Z} \Spec(k)$ is nonempty, see Lemma \ref{lemma-points-cartesian}. Now $T$ is an algebraic space and the projection $T \to \Spec(k)$ is a monomorphism. Hence $T = \Spec(k)$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-toward-field}. We conclude that $\Spec(k) \to \mathcal{Z}$ factors through $\mathcal{Z}'$. Suppose the morphism $z : \Spec(k) \to \mathcal{Z}$ is given by the object $\xi$ over $\Spec(k)$. We have just seen that $\xi$ is isomorphic to an object $\xi'$ of $\mathcal{Z}'$ over $\Spec(k)$. Since $z$ is is surjective, flat, and locally of finite presentation we see that every object of $\mathcal{Z}$ over any scheme is fppf locally isomorphic to a pullback of $\xi$, hence also to a pullback of $\xi'$. By descent of objects for stacks in groupoids this implies that $\mathcal{Z}' \to \mathcal{Z}$ is essentially surjective (as well as fully faithful, see Lemma \ref{lemma-monomorphism}). Hence we win. \end{proof} \begin{lemma} \label{lemma-improve-unique-point} Let $\mathcal{Z}$ be an algebraic stack. Assume $\mathcal{Z}$ satisfies the equivalent conditions of Lemma \ref{lemma-unique-point}. Then there exists a unique strictly full subcategory $\mathcal{Z}' \subset \mathcal{Z}$ such that $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma \ref{lemma-unique-point-better}. The inclusion morphism $\mathcal{Z}' \to \mathcal{Z}$ is a monomorphism of algebraic stacks. \end{lemma} \begin{proof} The last part is immediate from the first part and Lemma \ref{lemma-monomorphism}. Pick a field $k$ and a morphism $\Spec(k) \to \mathcal{Z}$ which is surjective, flat, and locally of finite type. Set $U = \Spec(k)$ and $R = U \times_\mathcal{Z} U$. The projections $s, t : R \to U$ are locally of 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 $$f : \mathcal{Z}' \longrightarrow \mathcal{Z}$$ 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 Lemma \ref{lemma-monomorphism}. By Criteria for Representability, Lemma \ref{criteria-lemma-flat-quotient-flat-presentation} the morphism $U \to \mathcal{Z}'$ is surjective, flat, and locally of finite presentation. Hence $\mathcal{Z}'$ is an algebraic stack which satisfies the equivalent conditions of Lemma \ref{lemma-unique-point-better}. By Algebraic Stacks, Lemma \ref{algebraic-lemma-equivalent} we may replace $\mathcal{Z}'$ by its essential image in $\mathcal{Z}$. Hence we have proved all the assertions of the lemma except for the uniqueness of $\mathcal{Z}' \subset \mathcal{Z}$. Suppose that $\mathcal{Z}'' \subset \mathcal{Z}$ is a second such algebraic stack. Then the projections $$\mathcal{Z}' \longleftarrow \mathcal{Z}' \times_\mathcal{Z} \mathcal{Z}'' \longrightarrow \mathcal{Z}''$$ are monomorphisms. The algebraic stack in the middle is nonempty by Lemma \ref{lemma-points-cartesian}. Hence the two projections are isomorphisms by Lemma \ref{lemma-monomorphism-into-point} and we win. \end{proof} \begin{example} \label{example-distinct} Here is an example where the morphism constructed in Lemma \ref{lemma-improve-unique-point} isn't an isomorphism. This example shows that imposing that residual gerbes are locally Noetherian is necessary in Definition \ref{definition-residual-gerbe}. In fact, the example is even an algebraic space! Let $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group of $\mathbf{Q}$ with the pro-finite topology. Let $$U = \Spec(\overline{\mathbf{Q}}) \times_{\Spec(\mathbf{Q})} \Spec(\overline{\mathbf{Q}}) = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \times \Spec(\overline{\mathbf{Q}})$$ (we omit a precise explanation of the meaning of the last equal sign). Let $G$ denote the absolute Galois group $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with the discrete topology viewed as a constant group scheme over $\Spec(\overline{\mathbf{Q}})$, see Groupoids, Example \ref{groupoids-example-constant-group}. Then $G$ acts freely and transitively on $U$. Let $X = U/G$, see Spaces, Definition \ref{spaces-definition-quotient}. Then $X$ is a non-noetherian reduced algebraic space with exactly one point. Furthermore, $X$ has a (locally) finite type point: $$x : \Spec(\overline{\mathbf{Q}}) \longrightarrow U \longrightarrow X$$ Indeed, every point of $U$ is actually closed! As $X$ is an algebraic space over $\overline{\mathbf{Q}}$ it follows that $x$ is a monomorphism. So $x$ is the morphism constructed in Lemma \ref{lemma-improve-unique-point} but $x$ is not an isomorphism. In fact $\Spec(\overline{\mathbf{Q}}) \to X$ is the residual gerbe of $X$ at $x$. \end{example} \noindent It will turn out later that under mild assumptions on the algebraic stack $\mathcal{X}$ the equivalent conditions of the following lemma are satisfied for every point $x \in |\mathcal{X}|$ (see Morphisms of Stacks, Section \ref{stacks-morphisms-section-existence-residual-gerbes}). \begin{lemma} \label{lemma-residual-gerbe} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$ be a point. The following are equivalent \begin{enumerate} \item there exists an algebraic stack $\mathcal{Z}$ and a monomorphism $\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$, \item there exists a reduced algebraic stack $\mathcal{Z}$ and a monomorphism $\mathcal{Z} \to \mathcal{X}$ such that $|\mathcal{Z}|$ is a singleton and such that the image of $|\mathcal{Z}|$ in $|\mathcal{X}|$ is $x$, \item there exists an algebraic stack $\mathcal{Z}$, a monomorphism $f : \mathcal{Z} \to \mathcal{X}$, and a surjective flat morphism $z : \Spec(k) \to \mathcal{Z}$ where $k$ is a field such that $x = f(z)$. \end{enumerate} Moreover, if these conditions hold, then there exists a unique strictly full subcategory $\mathcal{Z}_x \subset \mathcal{X}$ such that $\mathcal{Z}_x$ is a reduced, locally Noetherian algebraic stack and $|\mathcal{Z}_x|$ is a singleton which maps to $x$ via the map $|\mathcal{Z}_x| \to |\mathcal{X}|$. \end{lemma} \begin{proof} If $\mathcal{Z} \to \mathcal{X}$ is as in (1), then $\mathcal{Z}_{red} \to \mathcal{X}$ is as in (2). (See Section \ref{section-reduced} for the notion of the reduction of an algebraic stack.) Hence (1) implies (2). It is immediate that (2) implies (1). The equivalence of (2) and (3) is immediate from Lemma \ref{lemma-unique-point}. \medskip\noindent At this point we've seen the equivalence of (1) -- (3). Pick a monomorphism $f : \mathcal{Z} \to \mathcal{X}$ as in (2). Note that this implies that $f$ is fully faithful, see Lemma \ref{lemma-monomorphism}. Denote $\mathcal{Z}' \subset \mathcal{X}$ the essential image of the functor $f$. Then $f : \mathcal{Z} \to \mathcal{Z}'$ is an equivalence and hence $\mathcal{Z}'$ is an algebraic stack, see Algebraic Stacks, Lemma \ref{algebraic-lemma-equivalent}. Apply Lemma \ref{lemma-improve-unique-point} to get a strictly full subcategory $\mathcal{Z}_x \subset \mathcal{Z}'$ as in the statement of the lemma. This proves all the statements of the lemma except for uniqueness. \medskip\noindent In order to prove the uniqueness suppose that $\mathcal{Z}_x \subset \mathcal{X}$ and $\mathcal{Z}'_x \subset \mathcal{X}$ are two strictly full subcategories as in the statement of the lemma. Then the projections $$\mathcal{Z}'_x \longleftarrow \mathcal{Z}'_x \times_\mathcal{X} \mathcal{Z}_x \longrightarrow \mathcal{Z}_x$$ are monomorphisms. The algebraic stack in the middle is nonempty by Lemma \ref{lemma-points-cartesian}. Hence the two projections are isomorphisms by Lemma \ref{lemma-monomorphism-into-point} and we win. \end{proof} \noindent Having explained the above we can now make the following definition. \begin{definition} \label{definition-residual-gerbe} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. \begin{enumerate} \item We say the {\it residual gerbe of $\mathcal{X}$ at $x$ exists} if the equivalent conditions (1), (2), and (3) of Lemma \ref{lemma-residual-gerbe} hold. \item If the residual gerbe of $\mathcal{X}$ at $x$ exists, then the {\it residual gerbe of $\mathcal{X}$ at $x$}\footnote{This clashes with \cite{LM-B} in spirit, but not in fact. Namely, in Chapter 11 they associate to any point on any quasi-separated algebraic stack a gerbe (not necessarily algebraic) which they call the residual gerbe. We will see in Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-every-point-residual-gerbe} that on a quasi-separated algebraic stack every point has a residual gerbe in our sense which is then equivalent to theirs. For more information on this topic see \cite[Appendix B]{rydh_etale_devissage}.} is the strictly full subcategory $\mathcal{Z}_x \subset \mathcal{X}$ constructed in Lemma \ref{lemma-residual-gerbe}. \end{enumerate} \end{definition} \noindent In particular we know that $\mathcal{Z}_x$ (if it exists) is a locally Noetherian, reduced algebraic stack and that there exists a field and a surjective, flat, locally finitely presented morphism $$\Spec(k) \longrightarrow \mathcal{Z}_x.$$ We will see in Morphisms of Stacks, Lemma \ref{stacks-morphisms-lemma-noetherian-singleton-stack-gerbe} that $\mathcal{Z}_x$ is a gerbe. Existence of residual gerbes is discussed in Morphisms of Stacks, Section \ref{stacks-morphisms-section-existence-residual-gerbes}. It turns out that $\mathcal{Z}_x$ is a regular algebraic stack as follows from the following lemma. \begin{lemma} \label{lemma-residual-gerbe-regular} A reduced, locally Noetherian algebraic stack $\mathcal{Z}$ such that $|\mathcal{Z}|$ is a singleton is regular. \end{lemma} \begin{proof} Let $W \to \mathcal{Z}$ be a surjective smooth morphism where $W$ is a scheme. Let $k$ be a field and let $\Spec(k) \to \mathcal{Z}$ be surjective, flat, and locally of finite presentation (see Lemma \ref{lemma-unique-point-better}). The algebraic space $T = W \times_\mathcal{Z} \Spec(k)$ is smooth over $k$ in particular regular, see Spaces over Fields, Lemma \ref{spaces-over-fields-lemma-smooth-regular}. Since $T \to W$ is locally of finite presentation, flat, and surjective it follows that $W$ is regular, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-descend-regular}. By definition this means that $\mathcal{Z}$ is regular. \end{proof} \begin{lemma} \label{lemma-residual-gerbe-points} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Assume that the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ exists. Let $f : \Spec(K) \to \mathcal{X}$ be a morphism where $K$ is a field in the equivalence class of $x$. Then $f$ factors through the inclusion morphism $\mathcal{Z}_x \to \mathcal{X}$. \end{lemma} \begin{proof} Choose a field $k$ and a surjective flat locally finite presentation morphism $\Spec(k) \to \mathcal{Z}_x$. Set $T = \Spec(K) \times_\mathcal{X} \mathcal{Z}_x$. By Lemma \ref{lemma-points-cartesian} we see that $T$ is nonempty. As $\mathcal{Z}_x \to \mathcal{X}$ is a monomorphism we see that $T \to \Spec(K)$ is a monomorphism. Hence by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-monomorphism-toward-field} we see that $T = \Spec(K)$ which proves the lemma. \end{proof} \begin{lemma} \label{lemma-residual-gerbe-unique} Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. Let $\mathcal{Z}$ be an algebraic stack satisfying the equivalent conditions of Lemma \ref{lemma-unique-point-better} and let $\mathcal{Z} \to \mathcal{X}$ be a monomorphism such that the image of $|\mathcal{Z}| \to |\mathcal{X}|$ is $x$. Then the residual gerbe $\mathcal{Z}_x$ of $\mathcal{X}$ at $x$ exists and $\mathcal{Z} \to \mathcal{X}$ factors as $\mathcal{Z} \to \mathcal{Z}_x \to \mathcal{X}$ where the first arrow is an equivalence. \end{lemma} \begin{proof} Let $\mathcal{Z}_x \subset \mathcal{X}$ be the full subcategory corresponding to the essential image of the functor $\mathcal{Z} \to \mathcal{X}$. Then $\mathcal{Z} \to \mathcal{Z}_x$ is an equivalence, hence $\mathcal{Z}_x$ is an algebraic stack, see Algebraic Stacks, Lemma \ref{algebraic-lemma-equivalent}. Since $\mathcal{Z}_x$ inherits all the properties of $\mathcal{Z}$ from this equivalence it is clear from the uniqueness in Lemma \ref{lemma-residual-gerbe} that $\mathcal{Z}_x$ is the residual gerbe of $\mathcal{X}$ at $x$. \end{proof} \section{Dimension of a stack} \label{section-dimension} \noindent We can define the dimension of an algebraic stack $\mathcal{X}$ at a point $x$, using the notion of dimension of an algebraic space at a point (Properties of Spaces, Definition \ref{spaces-properties-definition-dimension-at-point}). In the following lemma the output may be $\infty$ either because $\mathcal{X}$ is not quasi-compact or because we run into the phenomenon described in Examples, Section \ref{examples-section-Noetherian-infinite-dimension}. \begin{lemma} \label{lemma-dimension-at-point-well-defined} Let $\mathcal{X}$ be a locally Noetherian algebraic stack over a scheme $S$. Let $x \in |\mathcal{X}|$ be a point of $\mathcal{X}$. Let $[U/R] \to \mathcal{X}$ be a presentation (Algebraic Stacks, Definition \ref{algebraic-definition-presentation}) where $U$ is a scheme. Let $u \in U$ be a point that maps to $x$. Let $e : U \to R$ be the identity'' map and let $s : R \to U$ be the source'' map, which is a smooth morphism of algebraic spaces. Let $R_u$ be the fiber of $s : R \to U$ over $u$. The element $$\dim_x(\mathcal{X}) = \dim_u(U) - \dim_{e(u)}(R_u) \in \mathbf{Z} \cup \infty$$ is independent of the choice of presentation and the point $u$ over $x$. \end{lemma} \begin{proof} Since $R \to U$ is smooth, the scheme $R_u$ is smooth over $\kappa(u)$ and hence has finite dimension. On the other hand, the scheme $U$ is locally Noetherian, but this does not guarantee that $\dim_u(U)$ is finite. Thus the difference is an element of $\mathbf{Z} \cup \{\infty\}$. \medskip\noindent Let $[U'/R'] \to \mathcal{X}$ and $u' \in U'$ be a second presentation where $U'$ is a scheme and $u'$ maps to $x$. Consider the algebraic space $P = U \times_\mathcal{X} U'$. By Lemma \ref{lemma-points-cartesian} there exists a $p \in |P|$ mapping to $u$ and $u'$. Since $P \to U$ and $P' \to U'$ are smooth we see that $\dim_p(P) = \dim_u(U) + \dim_p(P_u)$ and $\dim_p(P) = \dim_{u'}(U') + \dim_p(P_{u'})$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-smoothness-dimension-spaces}. Note that $$R'_{u'} = \Spec(\kappa(u')) \times_\mathcal{X} U' \quad\text{and}\quad P_u = \Spec(\kappa(u)) \times_\mathcal{X} U'$$ Let us represent $p \in |P|$ by a morphism $\Spec(\Omega) \to P$. Since $p$ maps to both $u$ and $u'$ it induces a $2$-morphism between the compositions $\Spec(\Omega) \to \Spec(\kappa(u')) \to \mathcal{X}$ and $\Spec(\Omega) \to \Spec(\kappa(u)) \to \mathcal{X}$ which in turn defines an isomorphism $$\Spec(\Omega) \times_{\Spec(\kappa(u'))} R'_{u'} \cong \Spec(\Omega) \times_{\Spec(\kappa(u))} P_u$$ as algebraic spaces over $\Spec(\Omega)$ mapping the $\Omega$-rational point $(1, e'(u'))$ to $(1, p)$ (some details omitted). We conclude that $$\dim_{e'(u')}(R'_{u'}) = \dim_p(P_u)$$ by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-dimension-fibre-after-base-change}. By symmetry we have $\dim_{e(u)}R_u) = \dim_p(P_{u'})$. Putting everything together we obtain the independence of choices. \end{proof} \noindent We can use the lemma above to make the following definition. \begin{definition} \label{definition-dimension-at-point} Let $\mathcal{X}$ be a locally Noetherian algebraic stack over a scheme $S$. Let $x \in |\mathcal{X}|$ be a point of $\mathcal{X}$. Let $[U/R] \to \mathcal{X}$ be a presentation (Algebraic Stacks, Definition \ref{algebraic-definition-presentation}) where $U$ is a scheme and let $u \in U$ be a point that maps to $x$. We define the {\it dimension of $\mathcal{X}$ at $x$} to be the element $\dim_x(\mathcal{X}) \in \mathbf{Z} \cup \infty$ such that $$\dim_x(\mathcal{X}) = \dim_u(U)-\dim_{e(u)}(R_u).$$ with notation as in Lemma \ref{lemma-dimension-at-point-well-defined}. \end{definition} \noindent The dimension of a stack at a point agrees with the usual notion when $\mathcal{X}$ is a scheme (Topology, Definition \ref{topology-definition-Krull}), or more generally when $\mathcal{X}$ is a locally Noetherian algebraic space (Properties of Spaces, Definition \ref{spaces-properties-definition-dimension-at-point}). \begin{definition} \label{definition-dimension} Let $S$ be a scheme. Let $\mathcal{X}$ be a locally Noetherian algebraic stack over $S$. The {\it dimension} $\dim(\mathcal{X})$ of $\mathcal{X}$ is defined to be $$\dim(\mathcal{X}) = \sup\nolimits_{x \in |\mathcal{X}|} \dim_x(\mathcal{X})$$ \end{definition} \noindent This definition of dimension agrees with the usual notion if $\mathcal{X}$ is a scheme (Properties, Lemma \ref{properties-lemma-dimension}) or an algebraic space (Properties of Spaces, Definition \ref{spaces-properties-definition-dimension}). \begin{remark} \label{remark-dimension-empty-stack} If $\mathcal{X}$ is a nonempty stack of finite type over a field, then $\dim(\mathcal{X})$ is an integer. For an arbitrary locally Noetherian algebraic stack $\mathcal{X}$, $\dim(\mathcal{X})$ is in $Z\cup \{\pm \infty\}$, and $\dim(\mathcal{X}) = -\infty$ if and only if $\mathcal{X}$ is empty. \end{remark} \begin{example} \label{example-dimension-quotient-stack} Let $X$ be a scheme of finite type over a field $k$, and let $G$ be a group scheme of finite type over $k$ which acts on $X$. Then the dimension of the quotient stack $[X/G]$ is equal to $\dim(X)-\dim(G)$. In particular, the dimension of the classifying stack $BG=[\Spec(k)/G]$ is $-\dim(G)$. Thus the dimension of an algebraic stack can be a negative integer, in contrast to what happens for schemes or algebraic spaces. \end{example} \input{chapters} \bibliography{my} \bibliographystyle{amsalpha} \end{document}