stacks/stacks-project

Fixes

Thanks to Fan. See
http://stacks.math.columbia.edu/tag/08N2#comment-380
 @@ -263,7 +263,7 @@ \section{The definition of a triangulated category} \noindent If $H : \mathcal{D} \to \mathcal{A}$ is a homological functor we often write $H^n(X) = H(X[n])$ so that $H(X) = H^0(X)$. Our discussion of TR2 above implies that says that a distinguished triangle Our discussion of TR2 above implies that a distinguished triangle $(X, Y, Z, f, g, h)$ determines a long exact sequence \begin{equation} \label{equation-long-exact-cohomology-sequence} @@ -527,9 +527,14 @@ \section{Elementary results on triangulated categories} \end{lemma} \begin{proof} Immediate from Lemma \ref{lemma-cone-triangle-unique-isomorphism} and TR1. By TR1 the triangle $(X, X, 0, 1, 0, 0)$ is distinguished. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. By TR3 there is a map of distinguished triangles $(1, f, 0) : (X, X, 0) \to (X, Y, Z)$. If $f$ is an isomorphism, then $(1, f, 0)$ is an isomorphism of triangles by Lemma \ref{lemma-third-isomorphism-triangle} and $Z = 0$. Conversely, if $Z = 0$, then $(1, f, 0)$ is an isomorphism of triangles as well, hence $f$ is an isomorphism. \end{proof} \begin{lemma} @@ -565,14 +570,9 @@ \section{Elementary results on triangulated categories} (X \oplus X', Y \oplus Y', Q, f \oplus f', g'', h''). } $$Let W be any object in \mathcal{D} and apply the functor H_W = \Hom_\mathcal{D}(W, -) to this diagram. By Lemma \ref{lemma-representable-homological} (applied three times) we deduce that H_W(c) : H_W(Z \oplus Z') \to H_W(Q) is an isomorphism. Hence c is an isomorphism and we conclude that (1) holds. By Lemma \ref{lemma-third-isomorphism-triangle} we find that c is an isomorphism and we conclude that (1) holds. \medskip\noindent Assume (1). We will show that (X, Y, Z, f, g, h) is a distinguished @@ -594,18 +594,18 @@ \section{Elementary results on triangulated categories} \longrightarrow (X, Y, Q, f, g'', h'')$$ Applying $H_W$ and using the above we once again see that $H_W(c) : H_W(Z) \to H_W(Q)$ is an isomorphism and we conclude that $c$ is an isomorphism. Hence we win. By Lemma \ref{lemma-third-isomorphism-triangle} we find that $c$ is an isomorphism and we conclude that (2) holds. \end{proof} \begin{lemma} \label{lemma-split} Let $\mathcal{D}$ be a pre-triangulated category. Let $(X, Y, Z, f, g, h)$ be a distinguished triangle. \begin{enumerate} \item If $h = 0$, then there exists a left inverse $s : Z \to Y$ to $g$. \item For any left inverse $s : Z \to Y$ of $g$ the map \item If $h = 0$, then there exists a right inverse $s : Z \to Y$ to $g$. \item For any right inverse $s : Z \to Y$ of $g$ the map $f \oplus s : X \oplus Z \to Y$ is an isomorphism. \item For any objects $X', Z'$ of $\mathcal{D}$ the triangle $(X', X' \oplus Z', Z', (1, 0), (0, 1), 0)$ is distinguished. @@ -648,10 +648,10 @@ \section{Elementary results on triangulated categories} Hence (3) $\Rightarrow$ (1), (2). Next we prove (1) $\Rightarrow$ (3). Suppose first that $f : X \to Y$ is a monomorphism, i.e., its kernel is zero. By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$ and by Lemma \ref{lemma-representable-homological} we see that $h = 0$. Then By TR1 there exists a distinguished triangle $(X, Y, Z, f, g, h)$. By Lemma \ref{lemma-composition-zero} the composition $h[-1] \circ f = 0$. As $f$ is a monomorphism we see that $h[-1] = 0$ and hence $h = 0$. Then Lemma \ref{lemma-split} implies that $Y = X \oplus Z$, i.e., we see that (3) holds. Next, assume $f$ has a kernel $K$. As $K \to X$ is a monomorphism we @@ -836,7 +836,7 @@ \section{Elementary results on triangulated categories} \end{lemma} \begin{proof} The if'' part is clear. Assume $(F(X), F(Y), F(Z))$ is The only if'' part is clear. Assume $(F(X), F(Y), F(Z))$ is distinguished in $\mathcal{D}'$. Pick a distinguished triangle $(X, Y, Z', f, g', h')$ in $\mathcal{D}$. By Lemma \ref{lemma-cone-triangle-unique-isomorphism} @@ -1174,7 +1174,7 @@ \section{Localization of triangulated categories} Lemma \ref{lemma-third-object-zero}. Hence $F(j)$ is an isomorphism by the same lemma, i.e., $j \in S$. Finally, $j \circ a = j \circ i \circ d = 0$ as $j \circ i = 0$. Thus $j \circ f = j \circ d$ and we see that LMS3 holds. Thus $j \circ f = j \circ g$ and we see that LMS3 holds. The proof of RMS3 is dual. \end{proof}
 @@ -478,32 +478,32 @@ \section{Abelian categories} \begin{enumerate} \item The diagram is cartesian if and only if $$0\to w\xrightarrow{(g,-f)}x\oplus y\xrightarrow{(k,-h)}z 0 \to w \xrightarrow{(g, f)} x \oplus y \xrightarrow{(k, -h)} z$$ is exact. \item The diagram is cocartesian if and only if $$w\xrightarrow{(g,-f)}x\oplus y\xrightarrow{(k,-h)}z\to 0 w \xrightarrow{(g, -f)} x \oplus y \xrightarrow{(k, h)} z \to 0$$ is exact. \end{enumerate} \end{lemma} \begin{proof} Let $u=(g,-f):w\to x\oplus y$ and $v=(k,-h):x\oplus y\to z$. Let $p:x\oplus y\to x$ and $q:x\oplus y\to y$ be the canonical projections. Let $i:\text{Ker}(v)\to x\oplus v$ be the canonical Let $u = (g, f) : w \to x \oplus y$ and $v = (k, -h) : x \oplus y \to z$. Let $p : x \oplus y \to x$ and $q : x \oplus y \to y$ be the canonical projections. Let $i : \text{Ker}(v) \to x \oplus y$ be the canonical injection. By Example \ref{example-fibre-product-pushouts}, the diagram is cartesian if and only if there exists an isomorphism $r:\text{Ker}(v)\to w$ with $-f\circ r=q\circ i$ and $g\circ r=p\circ i$. The sequence $0\to w\overset{u}\to x\oplus y\overset{v}\to z$ is exact if and only if there exists an isomorphism $r:\text{Ker}(v)\to w$ with $u\circ r=i$. But given $r:\text{Ker}(v)\to w$, we have $-f\circ r=q\circ i$ and $g\circ r=p\circ i$ if and only if $q\circ u\circ r=-f\circ r=q\circ i$ and $p\circ u\circ r=g\circ r=p\circ i$, hence if and only if $u\circ r=i$. This proves (1), and then (2) follows by duality. $r : \text{Ker}(v) \to w$ with $f \circ r = q \circ i$ and $g \circ r = p \circ i$. The sequence $0 \to w \overset{u} \to x \oplus y \overset{v} \to z$ is exact if and only if there exists an isomorphism $r : \text{Ker}(v) \to w$ with $u \circ r = i$. But given $r : \text{Ker}(v) \to w$, we have $f \circ r = q \circ i$ and $g \circ r = p \circ i$ if and only if $q \circ u \circ r= f \circ r = q \circ i$ and $p \circ u \circ r = g \circ r = p \circ i$, hence if and only if $u \circ r = i$. This proves (1), and then (2) follows by duality. \end{proof} \begin{lemma} @@ -931,7 +931,7 @@ \section{Extensions} \begin{definition} \label{definition-ext-group} Let $\mathcal{A}$ be an abelian category. Let $A, C \in \Ob(\mathcal{A})$. Let $A, B \in \Ob(\mathcal{A})$. The set of isomorphism classes of extensions of $B$ by $A$ is denoted  @@ -2928,7 +2928,7 @@ \section{Homotopy and the shift functor} \label{lemma-ses-termwise-split-homotopy} Notation and assumptions as in Lemma \ref{lemma-ses-termwise-split} above. Suppose $\{s'_n : C_n \to B_n\}$ is a second choice of splittings. Write $s'_n = s_n + \pi_n \circ h_n$ for some unique Write $s'_n = s_n + i_n \circ h_n$ for some unique morphisms $h_n : C_n \to A_n$. The family of maps $\{h_n : C_n \to A[-1]_{n + 1}\}$ is a homotopy between the associated morphisms