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 $\mathcal{A}$.
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.
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.
We say $f$ is injective if $\mathop{\mathrm{Ker}}(f) = 0$.
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
$f$ is injective if and only if $f$ is a monomorphism,
$f$ is surjective if and only if $f$ is an epimorphism, and
$f$ is an isomorphism if and only if $f$ is injective and surjective.
Proof. Proof of (1). Recall that $\mathop{\mathrm{Ker}}(f)$ is an object representing the functor sending $z$ to $\mathop{\mathrm{Ker}}(\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, x) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, y))$, see Definition 12.3.9. Thus $\mathop{\mathrm{Ker}}(f)$ is $0$ if and only if $\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, x) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, y)$ is injective for all $z$ if and only if $f$ is a monomorphism. The proof of (2) is similar. For the proof of (3) note that an isomorphism is both a monomorphism and epimorphism, which by (1), (2) proves $f$ is injective and surjective. If $f$ is both injective and surjective, then $x = \mathop{\mathrm{Coim}}(f)$ and $y = \mathop{\mathrm{Im}}(f)$ whence $f$ is an isomorphism. $\square$
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:
If $a : x \to y$, $b : z \to y$ are morphisms in $\mathcal{A}$, then we have the fibre product: $x \times _ y z = \mathop{\mathrm{Ker}}((a, -b) : x \oplus z \to y)$.
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. Consider a sequence of morphisms in $\mathcal{A}$. We say such a sequence is a complex if the composition of any two consecutive (drawn) arrows is zero. If $\mathcal{A}$ is abelian then we say a complex of the first type above is exact at $y$ if $\mathop{\mathrm{Im}}(x \to y) = \mathop{\mathrm{Ker}}(y \to z)$ and we say a complex of the second kind is exact at $x_ i$ where $1 < i < n$ if $\mathop{\mathrm{Im}}(x_{i - 1} \to x_ i) = \mathop{\mathrm{Ker}}(x_ i \to x_{i + 1})$. We a sequence as above is exact or is an exact sequence or is an exact complex if it is a complex and exact at every object (in the first case) or exact at $x_ i$ for all $1 < i < n$ (in the second case). There are variants of these notions for sequences of the form A short exact sequence is an exact complex of the form
\[ \ldots \to x \to y \to z \to \ldots \quad \text{or}\quad x_1 \to x_2 \to \ldots \to x_ n \]
\[ \ldots \to x_{-3} \to x_{-2} \to x_{-1} \quad \text{and}\quad x_1 \to x_2 \to x_3 \to \ldots \]
\[ 0 \to A \to B \to C \to 0. \]
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 $\mathcal{A}$.
$M_1 \to M_2 \to M_3 \to 0$ is exact if and only if
is an exact sequence of abelian groups for all objects $N$ of $\mathcal{A}$, and
$0 \to M_1 \to M_2 \to M_3$ is exact if and only if
is an exact sequence of abelian groups for all objects $N$ of $\mathcal{A}$.
\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_3, N) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_2, N) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(M_1, N) \]
\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_1) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_2) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(N, M_3) \]
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.
Given a morphism $s : C \to B$ right 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.
Given a morphism $\pi : B \to A$ left 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 be a commutative diagram.
The diagram is cartesian if and only if
is exact.
The diagram is cocartesian if and only if
is exact.
\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]
\[ 0 \to w \xrightarrow {(g, f)} x \oplus y \xrightarrow {(k, -h)} z \]
\[ w \xrightarrow {(g, -f)} x \oplus y \xrightarrow {(k, h)} z \to 0 \]
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 be a commutative diagram.
If the diagram is cartesian, then the morphism $\mathop{\mathrm{Ker}}(f)\to \mathop{\mathrm{Ker}}(k)$ induced by $g$ is an isomorphism.
If the diagram is cocartesian, then the morphism $\mathop{\mathrm{Coker}}(f)\to \mathop{\mathrm{Coker}}(k)$ induced by $h$ is an isomorphism.
\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]
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 be a commutative diagram.
If the diagram is cartesian and $k$ is an epimorphism, then the diagram is cocartesian and $f$ is an epimorphism.
If the diagram is cocartesian and $g$ is a monomorphism, then the diagram is cartesian and $h$ is a monomorphism.
\[ \xymatrix{ w\ar[r]^ f\ar[d]_ g & y\ar[d]^ h\\ x\ar[r]^ k & z } \]
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.
If $x \to y$ is surjective, then for every $z \to y$ the projection $x \times _ y z \to z$ is surjective.
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:
The sequence $x\overset {f}\to y\overset {g}\to z$ is exact.
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 be a commutative diagram.
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.
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.
\[ \xymatrix{ x \ar[r]^ f \ar[d]^\alpha & y \ar[r]^ g \ar[d]^\beta & z \ar[d]^\gamma \\ u \ar[r]^ k & v \ar[r]^ l & w } \]
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 be a commutative diagram with exact rows.
There exists a unique morphism $\delta : \mathop{\mathrm{Ker}}(\gamma ) \to \mathop{\mathrm{Coker}}(\alpha )$ such that the diagram
commutes, where $\pi $ and $\pi '$ are the canonical projections and $\iota $ and $\iota '$ are the canonical coprojections.
The induced sequence
is exact. If $f$ is injective then so is $f'$, and if $l$ is surjective then so is $l'$.
\[ \xymatrix{ & x \ar[r]^ f \ar[d]^\alpha & y \ar[r]^ g \ar[d]^\beta & z \ar[r] \ar[d]^\gamma & 0 \\ 0 \ar[r] & u \ar[r]^ k & v \ar[r]^ l & w } \]
\[ \xymatrix{ y \ar[d]_\beta & y \times _ z \mathop{\mathrm{Ker}}(\gamma ) \ar[l]_{\pi '} \ar[r]^{\pi } & \mathop{\mathrm{Ker}}(\gamma ) \ar[d]^\delta \\ v \ar[r]^{\iota '} & \mathop{\mathrm{Coker}}(\alpha ) \amalg _ u v & \mathop{\mathrm{Coker}}(\alpha ) \ar[l]_\iota } \]
\[ \mathop{\mathrm{Ker}}(\alpha ) \xrightarrow {f'} \mathop{\mathrm{Ker}}(\beta ) \xrightarrow {g'} \mathop{\mathrm{Ker}}(\gamma ) \xrightarrow {\delta } \mathop{\mathrm{Coker}}(\alpha ) \xrightarrow {k'} \mathop{\mathrm{Coker}}(\beta ) \xrightarrow {l'} \mathop{\mathrm{Coker}}(\gamma ) \]
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 $q = \mathop{\mathrm{Coker}}(\alpha ) \amalg _ u v$. Let $h : \mathop{\mathrm{Ker}}(\beta ) \to y$, $i : \mathop{\mathrm{Ker}}(\gamma ) \to z$ and $j : \mathop{\mathrm{Ker}}(\pi ) \to p$ be the canonical injections. Let $\pi '' : u \to \mathop{\mathrm{Coker}}(\alpha )$ be the canonical projection. Keeping in mind Lemma 12.5.13 we get a commutative diagram with exact rows
\[ \xymatrix{ 0 \ar[r] & \mathop{\mathrm{Ker}}(\pi ) \ar[r]^ j & p \ar[r]^{\pi } \ar[d]_{\pi '} & \mathop{\mathrm{Ker}}(\gamma ) \ar[d]_ i \ar[r] & 0 \\ & x \ar[r]^ f \ar[d]_\alpha & y \ar[r]^ g \ar[d]_\beta & z \ar[d]_\gamma \ar[r] & 0 \\ 0 \ar[r] & u \ar[r]^ k \ar[d]_{\pi ''} & v \ar[r]^ l \ar[d]_{\iota '} & w & \\ 0 \ar[r] & \mathop{\mathrm{Coker}}(\alpha ) \ar[r]^\iota & q & & } \]
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 $\pi '' \circ a \circ j \circ b = \pi '' \circ \alpha = 0$. As $b$ is an epimorphism this implies $\pi '' \circ a \circ j = 0$. Therefore, as the top row of the diagram above is exact, there exists $\delta : \mathop{\mathrm{Ker}}(\gamma ) \to \mathop{\mathrm{Coker}}(\alpha )$ with $\delta \circ \pi = \pi '' \circ a$. It follows $\iota \circ \delta \circ \pi = \iota \circ \pi '' \circ a = \iota ' \circ k \circ a = \iota ' \circ \beta \circ \pi '$ 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 $p \xrightarrow {\pi } \mathop{\mathrm{Ker}}(\gamma ) \to 0$ 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 $\pi '' \circ a \circ n = \delta \circ d \circ m = 0$, applying Lemma 12.5.15 to the exact sequence $x \xrightarrow {\alpha } u \xrightarrow {p} \mathop{\mathrm{Coker}}(\alpha )$ 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 $\mathop{\mathrm{Ker}}(\beta ) \xrightarrow {g'} \mathop{\mathrm{Ker}}(\gamma ) \xrightarrow {\delta } \mathop{\mathrm{Coker}}(\alpha )$ 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 be a commutative diagram with exact rows. Then, the induced diagram commutes.
\[ \xymatrix{ & & & x\ar[ld]\ar[rr]\ar[dd]^(.4)\alpha & & y\ar[ld]\ar[rr]\ar[dd]^(.4)\beta & & z\ar[ld]\ar[rr]\ar[dd]^(.4)\gamma & & 0\\ & & x'\ar[rr]\ar[dd]^(.4){\alpha '} & & y'\ar[rr]\ar[dd]^(.4){\beta '} & & z'\ar[rr]\ar[dd]^(.4){\gamma '} & & 0 & \\ & 0\ar[rr] & & u\ar[ld]\ar[rr] & & v\ar[ld]\ar[rr] & & w\ar[ld] & & \\ 0\ar[rr] & & u'\ar[rr] & & v'\ar[rr] & & w' & & & } \]
\[ \xymatrix@C=15pt{ \mathop{\mathrm{Ker}}(\alpha ) \ar[r] \ar[d] & \mathop{\mathrm{Ker}}(\beta ) \ar[r] \ar[d] & \mathop{\mathrm{Ker}}(\gamma ) \ar[r]^(.45){\delta } \ar[d] & \mathop{\mathrm{Coker}}(\alpha ) \ar[r] \ar[d] & \mathop{\mathrm{Coker}}(\beta ) \ar[r] \ar[d] & \mathop{\mathrm{Coker}}(\gamma ) \ar[d] \\ \mathop{\mathrm{Ker}}(\alpha ') \ar[r] & \mathop{\mathrm{Ker}}(\beta ') \ar[r] & \mathop{\mathrm{Ker}}(\gamma ') \ar[r]^(.45){\delta '} & \mathop{\mathrm{Coker}}(\alpha ') \ar[r] & \mathop{\mathrm{Coker}}(\beta ') \ar[r] & \mathop{\mathrm{Coker}}(\gamma ') } \]
Proof. Omitted. $\square$
Lemma 12.5.19. Let $\mathcal{A}$ be an abelian category. Let be a commutative diagram with exact rows.
If $\alpha , \gamma $ are surjective and $\delta $ is injective, then $\beta $ is surjective.
If $\beta , \delta $ are injective and $\alpha $ is surjective, then $\gamma $ is injective.
\[ \xymatrix{ w \ar[r] \ar[d]^\alpha & x \ar[r] \ar[d]^\beta & y \ar[r] \ar[d]^\gamma & z \ar[d]^\delta \\ w' \ar[r] & x' \ar[r] & y' \ar[r] & z' } \]
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. $\square$
Lemma 12.5.20. Let $\mathcal{A}$ be an abelian category. Let be a commutative diagram with exact rows. If $\beta , \delta $ are isomorphisms, $\epsilon $ is injective, and $\alpha $ is surjective then $\gamma $ is an isomorphism.
\[ \xymatrix{ v \ar[r] \ar[d]^\alpha & w \ar[r] \ar[d]^\beta & x \ar[r] \ar[d]^\gamma & y \ar[r] \ar[d]^\delta & z \ar[d]^\epsilon \\ v' \ar[r] & w' \ar[r] & x' \ar[r] & y' \ar[r] & z' } \]
Proof. Immediate consequence of Lemma 12.5.19. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (9)
Comment #367 by Fan on
Comment #1330 by jojo on
Comment #1350 by Johan on
Comment #4308 by dpw on
Comment #4469 by Johan on
Comment #6211 by Hans Schoutens on
Comment #6350 by Johan on
Comment #8377 by Xueping Huang on
Comment #8982 by Stacks project on