The Stacks project

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

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. $\square$