The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

12.5 Abelian categories

An abelian category is a category satisfying just enough axioms so the snake lemma holds. An axiom (that is sometimes forgotten) is that the canonical map $\mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f)$ of Lemma 12.3.12 is always an isomorphism. Example 12.3.13 shows that it is necessary.

Definition 12.5.1. A category $\mathcal{A}$ is abelian if it is additive, if all kernels and cokernels exist, and if the natural map $\mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f)$ is an isomorphism for all morphisms $f$ of $\mathcal{A}$.

Lemma 12.5.2. Let $\mathcal{A}$ be a preadditive category. The additions on sets of morphisms make $\mathcal{A}^{opp}$ into a preadditive category. Furthermore, $\mathcal{A}$ is additive if and only if $\mathcal{A}^{opp}$ is additive, and $\mathcal{A}$ is abelian if and only if $\mathcal{A}^{opp}$ is abelian.

Proof. The first statement is straightforward. To see that $\mathcal{A}$ is additive if and only if $\mathcal{A}^{opp}$ is additive, recall that additivity can be characterized by the existence of a zero object and direct sums, which are both preserved when passing to the opposite category. Finally, to see that $\mathcal{A}$ is abelian if and only if $\mathcal{A}^{opp}$ is abelian, observes that kernels, cokernels, images and coimages in $\mathcal{A}^{opp}$ correspond to cokernels, kernels, coimages and images in $\mathcal{A}$, respectively. $\square$

Definition 12.5.3. Let $f : x \to y$ be a morphism in an abelian category.

  1. We say $f$ is injective if $\mathop{\mathrm{Ker}}(f) = 0$.

  2. We say $f$ is surjective if $\mathop{\mathrm{Coker}}(f) = 0$.

If $x \to y$ is injective, then we say that $x$ is a subobject of $y$ and we use the notation $x \subset y$. If $x \to y$ is surjective, then we say that $y$ is a quotient of $x$.

Lemma 12.5.4. Let $f : x \to y$ be a morphism in an abelian category $\mathcal{A}$. Then

  1. $f$ is injective if and only if $f$ is a monomorphism, and

  2. $f$ is surjective if and only if $f$ is an epimorphism.

Proof. Proof of (1). Recall that $\mathop{\mathrm{Ker}}(f)$ is an object representing the functor sending $z$ to $\mathop{\mathrm{Ker}}(\mathop{Mor}\nolimits _\mathcal {A}(z, x) \to \mathop{Mor}\nolimits _\mathcal {A}(z, y))$. Thus $\mathop{\mathrm{Ker}}(f)$ is $0$ if and only if $\mathop{Mor}\nolimits _\mathcal {A}(z, x) \to \mathop{Mor}\nolimits _\mathcal {A}(z, y)$ is injective for all $z$ if and only if $f$ is a monomorphism. The proof of (2) is similar. $\square$

In an abelian category, if $x \subset y$ is a subobject, then we denote

\[ y/x = \mathop{\mathrm{Coker}}(x \to y). \]

Lemma 12.5.5. Let $\mathcal{A}$ be an abelian category. All finite limits and finite colimits exist in $\mathcal{A}$.

Proof. To show that finite limits exist it suffices to show that finite products and equalizers exist, see Categories, Lemma 4.18.4. Finite products exist by definition and the equalizer of $a, b : x \to y$ is the kernel of $a - b$. The argument for finite colimits is similar but dual to this. $\square$

Example 12.5.6. Let $\mathcal{A}$ be an abelian category. Pushouts and fibre products in $\mathcal{A}$ have the following simple descriptions:

  1. If $a : x \to y$, $b : z \to y$ are morphisms in $\mathcal{A}$, then we have the fibre product: $x \times _ y z = \mathop{\mathrm{Ker}}((a, -b) : x \oplus z \to y)$.

  2. If $a : y \to x$, $b : y \to z$ are morphisms in $\mathcal{A}$, then we have the pushout: $x \amalg _ y z = \mathop{\mathrm{Coker}}((a, -b) : y \to x \oplus z)$.

Definition 12.5.7. Let $\mathcal{A}$ be an additive category. We say a sequence of morphisms

\[ \ldots \to x \to y \to z \to \ldots \]

in $\mathcal{A}$ is a complex if the composition of any two (drawn) arrows is zero. If $\mathcal{A}$ is abelian then we say a sequence as above is exact at $y$ if $\mathop{\mathrm{Im}}(x \to y) = \mathop{\mathrm{Ker}}(y \to z)$. We say it is exact if it is exact at every object. A short exact sequence is an exact complex of the form

\[ 0 \to A \to B \to C \to 0. \]

In the following lemma we assume the reader knows what it means for a sequence of abelian groups to be exact.

Lemma 12.5.8. Let $\mathcal{A}$ be an abelian category. Let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a complex of $\mathcal{A}$.

  1. $M_1 \to M_2 \to M_3 \to 0$ is exact if and only if

    \[ 0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_3, N) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_2, N) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_1, N) \]

    is an exact sequence of abelian groups for all objects $N$ of $\mathcal{A}$, and

  2. $0 \to M_1 \to M_2 \to M_3$ is exact if and only if

    \[ 0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_2) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_1) \]

    is an exact sequence of abelian groups for all objects $N$ of $\mathcal{A}$.

Proof. Omitted. Hint: See Algebra, Lemma 10.10.1. $\square$

Definition 12.5.9. Let $\mathcal{A}$ be an abelian category. Let $i : A \to B$ and $q : B \to C$ be morphisms of $\mathcal{A}$ such that $0 \to A \to B \to C \to 0$ is a short exact sequence. We say the short exact sequence is split if there exist morphisms $j : C \to B$ and $p : B \to A$ such that $(B, i, j, p, q)$ is the direct sum of $A$ and $C$.

Lemma 12.5.10. Let $\mathcal{A}$ be an abelian category. Let $0 \to A \to B \to C \to 0$ be a short exact sequence.

  1. Given a morphism $s : C \to B$ left inverse to $B \to C$, there exists a unique $\pi : B \to A$ such that $(s, \pi )$ splits the short exact sequence as in Definition 12.5.9.

  2. Given a morphism $\pi : B \to A$ right inverse to $A \to B$, there exists a unique $s : C \to B$ such that $(s, \pi )$ splits the short exact sequence as in Definition 12.5.9.

Proof. Omitted. $\square$

Lemma 12.5.11. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]

be a commutative diagram.

  1. The diagram is cartesian if and only if

    \[ 0 \to w \xrightarrow {(g, f)} x \oplus y \xrightarrow {(k, -h)} z \]

    is exact.

  2. The diagram is cocartesian if and only if

    \[ w \xrightarrow {(g, -f)} x \oplus y \xrightarrow {(k, h)} z \to 0 \]

    is exact.

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 $i : \mathop{\mathrm{Ker}}(v) \to x \oplus y$ be the canonical injection. By Example 12.5.6, the diagram is cartesian if and only if there exists an isomorphism $r : \mathop{\mathrm{Ker}}(v) \to w$ 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 $r : \mathop{\mathrm{Ker}}(v) \to w$ with $u \circ r = i$. But given $r : \mathop{\mathrm{Ker}}(v) \to w$, 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. $\square$

Lemma 12.5.12. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]

be a commutative diagram.

  1. If the diagram is cartesian, then the morphism $\mathop{\mathrm{Ker}}(f)\to \mathop{\mathrm{Ker}}(k)$ induced by $g$ is an isomorphism.

  2. If the diagram is cocartesian, then the morphism $\mathop{\mathrm{Coker}}(f)\to \mathop{\mathrm{Coker}}(k)$ induced by $h$ is an isomorphism.

Proof. Suppose the diagram is cartesian. Let $e:\mathop{\mathrm{Ker}}(f)\to \mathop{\mathrm{Ker}}(k)$ be induced by $g$. Let $i:\mathop{\mathrm{Ker}}(f)\to w$ and $j:\mathop{\mathrm{Ker}}(k)\to x$ be the canonical injections. There exists $t:\mathop{\mathrm{Ker}}(k)\to w$ with $f\circ t=0$ and $g\circ t=j$. Hence, there exists $u:\mathop{\mathrm{Ker}}(k)\to \mathop{\mathrm{Ker}}(f)$ with $i\circ u=t$. It follows $g\circ i\circ u\circ e=g\circ t\circ e=j\circ e=g\circ i$ and $f\circ i\circ u\circ e=0=f\circ i$, hence $i\circ u\circ e=i$. Since $i$ is a monomorphism this implies $u\circ e=\text{id}_{\mathop{\mathrm{Ker}}(f)}$. Furthermore, we have $j\circ e\circ u=g\circ i\circ u=g\circ t=j$. Since $j$ is a monomorphism this implies $e\circ u=\text{id}_{\mathop{\mathrm{Ker}}(k)}$. This proves (1). Now, (2) follows by duality. $\square$

Lemma 12.5.13. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]

be a commutative diagram.

  1. If the diagram is cartesian and $k$ is an epimorphism, then the diagram is cocartesian and $f$ is an epimorphism.

  2. If the diagram is cocartesian and $g$ is a monomorphism, then the diagram is cartesian and $h$ is a monomorphism.

Proof. Suppose the diagram is cartesian and $k$ is an epimorphism. Let $u = (g, f) : w \to x \oplus y$ and let $v = (k, -h) : x \oplus y \to z$. As $k$ is an epimorphism, $v$ is an epimorphism, too. Therefore and by Lemma 12.5.11, the sequence $0\to w\overset {u}\to x\oplus y\overset {v}\to z\to 0$ is exact. Thus, the diagram is cocartesian by Lemma 12.5.11. Finally, $f$ is an epimorphism by Lemma 12.5.12 and Lemma 12.5.4. This proves (1), and (2) follows by duality. $\square$

Lemma 12.5.14. Let $\mathcal{A}$ be an abelian category.

  1. If $x \to y$ is surjective, then for every $z \to y$ the projection $x \times _ y z \to z$ is surjective.

  2. If $x \to y$ is injective, then for every $x \to z$ the morphism $z \to z \amalg _ x y$ is injective.

Proof. Immediately from Lemma 12.5.4 and Lemma 12.5.13. $\square$

Lemma 12.5.15. Let $\mathcal{A}$ be an abelian category. Let $f:x\to y$ and $g:y\to z$ be morphisms with $g\circ f=0$. Then, the following statements are equivalent:

  1. The sequence $x\overset {f}\to y\overset {g}\to z$ is exact.

  2. For every $h:w\to y$ with $g\circ h=0$ there exist an object $v$, an epimorphism $k:v\to w$ and a morphism $l:v\to x$ with $h\circ k=f\circ l$.

Proof. Let $i:\mathop{\mathrm{Ker}}(g)\to y$ be the canonical injection. Let $p:x\to \mathop{\mathrm{Coim}}(f)$ be the canonical projection. Let $j:\mathop{\mathrm{Im}}(f)\to \mathop{\mathrm{Ker}}(g)$ be the canonical injection.

Suppose (1) holds. Let $h:w\to y$ with $g\circ h=0$. There exists $c:w\to \mathop{\mathrm{Ker}}(g)$ with $i\circ c=h$. Let $v=x\times _{\mathop{\mathrm{Ker}}(g)}w$ with canonical projections $k:v\to w$ and $l:v\to x$, so that $c\circ k=j\circ p\circ l$. Then, $h\circ k=i\circ c\circ k=i\circ j\circ p\circ l=f\circ l$. As $j\circ p$ is an epimorphism by hypothesis, $k$ is an epimorphism by Lemma 12.5.13. This implies (2).

Suppose (2) holds. Then, $g\circ i=0$. So, there are an object $w$, an epimorphism $k:w\to \mathop{\mathrm{Ker}}(g)$ and a morphism $l:w\to x$ with $f\circ l=i\circ k$. It follows $i\circ j\circ p\circ l=f\circ l=i\circ k$. Since $i$ is a monomorphism we see that $j\circ p\circ l=k$ is an epimorphism. So, $j$ is an epimorphisms and thus an isomorphism. This implies (1). $\square$

Lemma 12.5.16. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ x \ar[r]^ f \ar[d]^\alpha & y \ar[r]^ g \ar[d]^\beta & z \ar[d]^\gamma \\ u \ar[r]^ k & v \ar[r]^ l & w } \]

be a commutative diagram.

  1. If the first row is exact and $k$ is a monomorphism, then the induced sequence $\mathop{\mathrm{Ker}}(\alpha ) \to \mathop{\mathrm{Ker}}(\beta ) \to \mathop{\mathrm{Ker}}(\gamma )$ is exact.

  2. If the second row is exact and $g$ is an epimorphism, then the induced sequence $\mathop{\mathrm{Coker}}(\alpha ) \to \mathop{\mathrm{Coker}}(\beta ) \to \mathop{\mathrm{Coker}}(\gamma )$ is exact.

Proof. Suppose the first row is exact and $k$ is a monomorphism. Let $a:\mathop{\mathrm{Ker}}(\alpha )\to \mathop{\mathrm{Ker}}(\beta )$ and $b:\mathop{\mathrm{Ker}}(\beta )\to \mathop{\mathrm{Ker}}(\gamma )$ be the induced morphisms. Let $h:\mathop{\mathrm{Ker}}(\alpha )\to x$, $i:\mathop{\mathrm{Ker}}(\beta )\to y$ and $j:\mathop{\mathrm{Ker}}(\gamma )\to z$ be the canonical injections. As $j$ is a monomorphism we have $b\circ a=0$. Let $c:s\to \mathop{\mathrm{Ker}}(\beta )$ with $b\circ c=0$. Then, $g\circ i\circ c=j\circ b\circ c=0$. By Lemma 12.5.15 there are an object $t$, an epimorphism $d:t\to s$ and a morphism $e:t\to x$ with $i\circ c\circ d=f\circ e$. Then, $k\circ \alpha \circ e=\beta \circ f\circ e=\beta \circ i\circ c\circ d=0$. As $k$ is a monomorphism we get $\alpha \circ e=0$. So, there exists $m:t\to \mathop{\mathrm{Ker}}(\alpha )$ with $h\circ m=e$. It follows $i\circ a\circ m=f\circ h\circ m=f\circ e=i\circ c\circ d$. As $i$ is a monomorphism we get $a\circ m=c\circ d$. Thus, Lemma 12.5.15 implies (1), and then (2) follows by duality. $\square$

Lemma 12.5.17. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ & x \ar[r]^ f \ar[d]^\alpha & y \ar[r]^ g \ar[d]^\beta & z \ar[r] \ar[d]^\gamma & 0 \\ 0 \ar[r] & u \ar[r]^ k & v \ar[r]^ l & w } \]

be a commutative diagram with exact rows.

  1. There exists a unique morphism $\delta : \mathop{\mathrm{Ker}}(\gamma ) \rightarrow \mathop{\mathrm{Coker}}(\alpha )$ such that the diagram

    \[ \xymatrix{ y \ar[d]_\beta & y \times _ z \mathop{\mathrm{Ker}}(\gamma ) \ar[l]_{\pi '} \ar[r]^{\pi } & \mathop{\mathrm{Ker}}(\gamma ) \ar[d]^\delta \\ v \ar[r]^{\iota '} & \mathop{\mathrm{Coker}}(\alpha ) \amalg _ u v & \mathop{\mathrm{Coker}}(\alpha ) \ar[l]_\iota } \]

    commutes, where $\pi $ and $\pi '$ are the canonical projections and $\iota $ and $\iota '$ are the canonical coprojections.

  2. The induced sequence

    \[ \mathop{\mathrm{Ker}}(\alpha )\overset {f'} \to \mathop{\mathrm{Ker}}(\beta ) \overset {g'}\to \mathop{\mathrm{Ker}}(\gamma )\overset {\delta }\to \mathop{\mathrm{Coker}}(\alpha ) \overset {k'}\to \mathop{\mathrm{Coker}}(\beta ) \overset {l'}\to \mathop{\mathrm{Coker}}(\gamma ) \]

    is exact. If $f$ is injective then so is $f'$, and if $l$ is surjective then so is $l'$.

Proof. As $\pi $ is an epimorphism and $\iota $ is a monomorphism by Lemma 12.5.13, uniqueness of $\delta $ is clear. Let $p=y\times _ z\mathop{\mathrm{Ker}}(\gamma )$ and $q=\mathop{\mathrm{Coker}}(\alpha )\amalg _ uv$. Let $h:\mathop{\mathrm{Ker}}(\beta )\to y$, $i:\mathop{\mathrm{Ker}}(\gamma )\to z$ and $j:\mathop{\mathrm{Ker}}(\pi )\rightarrow p$ be the canonical injections. Let $p:u\to \mathop{\mathrm{Coker}}(\alpha )$ be the canonical projection. Keeping in mind Lemma 12.5.13 we get a commutative diagram with exact rows

\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ker}}(\pi ) \ar[r]^ j & p \ar[r]^{\pi } \ar[d]_{\pi '} & \mathop{\mathrm{Ker}}(\gamma ) \ar[d]_ i \ar[r] & 0 \\ & x \ar[r]^ f \ar[d]_\alpha & y \ar[r]^ g \ar[d]_\beta & z \ar[d]_\gamma \ar[r] & 0 \\ 0 \ar[r] & u \ar[r]^ k \ar[d]_ p & v \ar[r]^ l \ar[d]_{\iota '} & w & \\ 0 \ar[r] & \mathop{\mathrm{Coker}}(\alpha ) \ar[r]^\iota & q & & } \]

As $l \circ \beta \circ \pi ' = \gamma \circ i \circ \pi = 0$ and as the third row of the diagram above is exact, there is an $a:p\to u$ with $k \circ a = \beta \circ \pi '$. As the upper right quadrangle of the diagram above is cartesian, Lemma 12.5.12 yields an epimorphism $b : x \to \mathop{\mathrm{Ker}}(\pi )$ with $\pi ' \circ j \circ b = f$. It follows $k \circ a \circ j \circ b = \beta \circ \pi ' \circ j \circ b = \beta \circ f = k \circ \alpha $. As $k$ is a monomorphism this implies $a \circ j \circ b = \alpha $. It follows $p \circ a \circ j \circ b = p \circ \alpha = 0$. As $b$ is an epimorphism this implies $p\circ a\circ j=0$. Therefore, as the top row of the diagram above is exact, there exists $\delta : \mathop{\mathrm{Ker}}(\gamma ) \to \mathop{\mathrm{Coker}}(\alpha )$ with $\delta \circ \pi = p \circ a$. It follows $\iota \circ \delta \circ \pi = \iota \circ p \circ a = \iota ' \circ k \circ a = \iota ' \circ \beta \circ \pi '$ as desired.

As the upper right quadrangle in the diagram above is cartesian there is a $c : \mathop{\mathrm{Ker}}(\beta ) \to p$ with $\pi ' \circ c = h$ and $\pi \circ c = g'$. It follows $\iota \circ \delta \circ g' = \iota \circ \delta \circ \pi \circ c = \iota ' \circ \beta \circ \pi ' \circ c = \iota ' \circ \beta \circ h = 0$. As $\iota $ is a monomorphism this implies $\delta \circ g' = 0$.

Next, let $d : r \to \mathop{\mathrm{Ker}}(\gamma )$ with $\delta \circ d = 0$. Applying Lemma 12.5.15 to the exact sequence $p \overset {\pi }\to \mathop{\mathrm{Ker}}(\gamma ) \to 0$ and $d$ yields an object $s$, an epimorphism $m : s \to r$ and a morphism $n : s \to p$ with $\pi \circ n = d \circ m$. As $p \circ a \circ n = \delta \circ d \circ m = 0$, applying Lemma 12.5.15 to the exact sequence $x \overset {\alpha }\to u \overset {p}\to \mathop{\mathrm{Coker}}(\alpha )$ and $a \circ n$ yields an object $t$, an epimorphism $\varepsilon : t \to s$ and a morphism $\zeta : t \to x$ with $a \circ n \circ \varepsilon = \alpha \circ \zeta $. It holds $\beta \circ \pi ' \circ n \circ \varepsilon = k \circ \alpha \circ \zeta = \beta \circ f \circ \zeta $. Let $\eta = \pi ' \circ n \circ \varepsilon - f \circ \zeta : t \to y$. Then, $\beta \circ \eta = 0$. It follows that there is a $\vartheta : t \to \mathop{\mathrm{Ker}}(\beta )$ with $\eta = h \circ \vartheta $. It holds $i \circ g' \circ \vartheta = g \circ h \circ \vartheta = g \circ \pi ' \circ n \circ \varepsilon - g \circ f \circ \zeta = i \circ \pi \circ n \circ \varepsilon = i \circ d \circ m \circ \varepsilon $. As $i$ is a monomorphism we get $g' \circ \vartheta = d \circ m \circ \varepsilon $. Thus, as $m \circ \varepsilon $ is an epimorphism, Lemma 12.5.15 implies that $\mathop{\mathrm{Ker}}(\beta ) \overset {g'}\to \mathop{\mathrm{Ker}}(\gamma ) \overset {\delta }\to \mathop{\mathrm{Coker}}(\alpha )$ is exact. Then, the claim follows by Lemma 12.5.16 and duality. $\square$

Lemma 12.5.18. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ & & & x\ar[ld]\ar[rr]\ar[dd]^(.4)\alpha & & y\ar[ld]\ar[rr]\ar[dd]^(.4)\beta & & z\ar[ld]\ar[rr]\ar[dd]^(.4)\gamma & & 0\\ & & x'\ar[rr]\ar[dd]^(.4){\alpha '} & & y'\ar[rr]\ar[dd]^(.4){\beta '} & & z'\ar[rr]\ar[dd]^(.4){\gamma '} & & 0 & \\ & 0\ar[rr] & & u\ar[ld]\ar[rr] & & v\ar[ld]\ar[rr] & & w\ar[ld] & & \\ 0\ar[rr] & & u'\ar[rr] & & v'\ar[rr] & & w' & & & } \]

be a commutative diagram with exact rows. Then, the induced diagram

\[ \xymatrix@C=15pt{ \mathop{\mathrm{Ker}}(\alpha ) \ar[r] \ar[d] & \mathop{\mathrm{Ker}}(\beta ) \ar[r] \ar[d] & \mathop{\mathrm{Ker}}(\gamma ) \ar[r]^(.45){\delta } \ar[d] & \mathop{\mathrm{Coker}}(\alpha ) \ar[r] \ar[d] & \mathop{\mathrm{Coker}}(\beta ) \ar[r] \ar[d] & \mathop{\mathrm{Coker}}(\gamma ) \ar[d] \\ \mathop{\mathrm{Ker}}(\alpha ') \ar[r] & \mathop{\mathrm{Ker}}(\beta ') \ar[r] & \mathop{\mathrm{Ker}}(\gamma ') \ar[r]^(.45){\delta '} & \mathop{\mathrm{Coker}}(\alpha ') \ar[r] & \mathop{\mathrm{Coker}}(\beta ') \ar[r] & \mathop{\mathrm{Coker}}(\gamma ') } \]

commutes.

Proof. Omitted. $\square$

Lemma 12.5.19. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ w \ar[r] \ar[d]^\alpha & x \ar[r] \ar[d]^\beta & y \ar[r] \ar[d]^\gamma & z \ar[d]^\delta \\ w' \ar[r] & x' \ar[r] & y' \ar[r] & z' } \]

be a commutative diagram with exact rows.

  1. If $\alpha , \gamma $ are surjective and $\delta $ is injective, then $\beta $ is surjective.

  2. If $\beta , \delta $ are injective and $\alpha $ is surjective, then $\gamma $ is injective.

Proof. Assume $\alpha , \gamma $ are surjective and $\delta $ is injective. We may replace $w'$ by $\mathop{\mathrm{Im}}(w' \to x')$, i.e., we may assume that $w' \to x'$ is injective. We may replace $z$ by $\mathop{\mathrm{Im}}(y \to z)$, i.e., we may assume that $y \to z$ is surjective. Then we may apply Lemma 12.5.17 to

\[ \xymatrix{ & \mathop{\mathrm{Ker}}(y \to z) \ar[r] \ar[d] & y \ar[r] \ar[d] & z \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathop{\mathrm{Ker}}(y' \to z') \ar[r] & y' \ar[r] & z' } \]

to conclude that $\mathop{\mathrm{Ker}}(y \to z) \to \mathop{\mathrm{Ker}}(y' \to z')$ is surjective. Finally, we apply Lemma 12.5.17 to

\[ \xymatrix{ & w \ar[r] \ar[d] & x \ar[r] \ar[d] & \mathop{\mathrm{Ker}}(y \to z) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & w' \ar[r] & x' \ar[r] & \mathop{\mathrm{Ker}}(y' \to z') } \]

to conclude that $x \to x'$ is surjective. This proves (1). The proof of (2) is dual to this. $\square$

reference

Lemma 12.5.20. Let $\mathcal{A}$ be an abelian category. Let

\[ \xymatrix{ v \ar[r] \ar[d]^\alpha & w \ar[r] \ar[d]^\beta & x \ar[r] \ar[d]^\gamma & y \ar[r] \ar[d]^\delta & z \ar[d]^\epsilon \\ v' \ar[r] & w' \ar[r] & x' \ar[r] & y' \ar[r] & z' } \]

be a commutative diagram with exact rows. If $\beta , \delta $ are isomorphisms, $\epsilon $ is injective, and $\alpha $ is surjective then $\gamma $ is an isomorphism.

Proof. Immediate consequence of Lemma 12.5.19. $\square$


Comments (3)

Comment #367 by Fan on

The line after Lemma 3.16, normally I would identify Coker as instead of .

Comment #1330 by jojo on

I think that the comment above is right and I don't believe it has been fixed (now it's lemma 12.5.4).


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