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 $\delta : \mathop{\mathrm{Ker}}(\gamma ) \to \mathop{\mathrm{Coker}}(\alpha )$ 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

$\mathop{\mathrm{Ker}}(\alpha ) \xrightarrow {f'} \mathop{\mathrm{Ker}}(\beta ) \xrightarrow {g'} \mathop{\mathrm{Ker}}(\gamma ) \xrightarrow {\delta } \mathop{\mathrm{Coker}}(\alpha ) \xrightarrow {k'} \mathop{\mathrm{Coker}}(\beta ) \xrightarrow {l'} \mathop{\mathrm{Coker}}(\gamma )$

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 $q = \mathop{\mathrm{Coker}}(\alpha ) \amalg _ u v$. 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. $\square$

Comment #162 by on

I get a "parse error" in the second and the third diagram (and also in the second diagram in 08N7).

Comment #163 by on

This should be fixed now by this change. Let me know if it isn't.

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