Snake lemma under the weaker Stacks hypotheses

Because the top row is exact and ends in $0$, the map $b$ is epic by [L1]. Because the bottom row is exact and begins at $0$, the map $k$ is monic by [L1]. Choose a kernel $k_{\gamma }:K\to Z$ of $\gamma$ and a cokernel $q_{\alpha }:U\to Q$ of $\alpha$. Form the pullback $P$ displayed above. By [L3], $\pi$ is epic. Since $l\beta {\pi }^{\mathrm{\prime }}=\gamma b{\pi }^{\mathrm{\prime }}=\gamma k_{\gamma }\pi =0,$ the kernel property of $k$ gives a unique map $r:P\to U$ such that $kr=\beta {\pi }^{\mathrm{\prime }}\mathrm{.}$

Let $j:J\to P$ be a kernel of $\pi$. By [L3], the induced map $J\to \ker (b)$ is an isomorphism. Exactness of the top row at $Y$ says that $\ker (b)$ is the image of $a$, so there is an epimorphism $e:X\to J$ with ${\pi }^{\mathrm{\prime }}je=a\mathrm{.}$ Then $krje=\beta {\pi }^{\mathrm{\prime }}je=\beta a=k\alpha ,$ and monicity of $k$ gives $rje=\alpha$. Therefore $q_{\alpha }rje=q_{\alpha }\alpha =0.$ Because $e$ is epic, $q_{\alpha }rj=0$. Since $\pi$ is epic, [L6] makes it the cokernel of its kernel $j$, so there is a unique morphism $\delta :K\to Q$ with $\delta \pi =q_{\alpha }r\mathrm{.}$

Applying [L4] to the given diagram gives exactness of $\ker (\alpha )\to \ker (\beta )\to \ker (\gamma )$ and of $\mathrm{coker}(\alpha )\to \mathrm{coker}(\beta )\to \mathrm{coker}(\gamma )\mathrm{.}$ If $a$ is monic, then the sequence $0\to X\stackrel{a}{\to }Y\stackrel{b}{\to }Z$ is exact, so the same theorem gives that $\ker (\alpha )\to \ker (\beta )$ is monic. Dually, if $l$ is epic, then $\mathrm{coker}(\beta )\to \mathrm{coker}(\gamma )$ is epic. Thus only exactness at $\ker (\gamma )$ and at $\mathrm{coker}(\alpha )$ remains.

Let $i:\ker (\beta )\to Y$ be a kernel of $\beta$, and let $s:\ker (\beta )\to K$ be the induced map with $k_{\gamma }s=bi$. Because $\beta i=0$, the pair $(i,s)$ factors through the pullback, giving $t:\ker (\beta )\to P$ with ${\pi }^{\mathrm{\prime }}t=i,\pi t=s\mathrm{.}$ Then $krt=\beta {\pi }^{\mathrm{\prime }}t=\beta i=0,$ so monicity of $k$ gives $rt=0$. Therefore $\delta s=\delta \pi t=q_{\alpha }rt=0,$ which proves that $\delta$ kills the image of $\ker (\beta )\to \ker (\gamma )$.

Let $u:T\to K$ be a kernel of $\delta$. Pull back $\pi$ along $u$, obtaining an epimorphism $e_0:E_0\to T$ and $v:E_0\to P$ with $\pi v=u{e}_0$. Then $q_{\alpha }rv=\delta u{e}_0=0$. By [L5], $\mathrm{im}(\alpha )=\ker (q_{\alpha })$, so $rv$ factors through the image inclusion of $\alpha$. Pull back the epic image projection $X\to \mathrm{im}(\alpha )$ along that factorization. This gives an epimorphism $e_1:E\to E_0$ and $x:E\to X$ with $\alpha x=rv{e}_1$.

Let $q_{\beta }:V\to Q_{\beta }$ be a cokernel of $\beta$, and let $c:Q\to Q_{\beta }$ be induced by $k$, so $c{q}_{\alpha }=q_{\beta }k$. Since $kr=\beta {\pi }^{\mathrm{\prime }}$, one has $c\delta \pi =q_{\beta }kr=q_{\beta }\beta {\pi }^{\mathrm{\prime }}=0$, and epicity of $\pi$ gives $c\delta =0$. Let $t:R\to Q$ be a kernel of $c$. Pull back the epimorphism $q_{\alpha }:U\to Q$ along $t$, giving an epimorphism $d_0:D_0\to R$ and $v:D_0\to U$ with $q_{\alpha }v=t{d}_0$. Then $q_{\beta }kv=ct{d}_0=0$, so $kv$ factors through $\mathrm{im}(\beta )=\ker (q_{\beta })$. Pull back the epic image projection $Y\to \mathrm{im}(\beta )$ to obtain an epimorphism $d_1:D\to D_0$ and $y:D\to Y$ with $\beta y=kv{d}_1$.

Put $z={\pi }^{\mathrm{\prime }}v{e}_1-ax:E\to Y$. Since $\beta z=krv{e}_1-k\alpha x=0$, it factors through a kernel $i:K_{\beta }\to Y$ of $\beta$ as $z=iw$. Let $s:K_{\beta }\to K$ be the induced arrow. Then $k_{\gamma }sw=biw=bz=b{\pi }^{\mathrm{\prime }}v{e}_1=k_{\gamma }u{e}_0{e}_1$. Monicity of $k_{\gamma }$ gives $sw=u{e}_0{e}_1$. Since $e_0{e}_1$ is epic, $\mathrm{coker}(s)u=0$. By [L6], $u$ factors through $\mathrm{im}(s)$; with $\delta s=0$ from step 3.2, this proves $\mathrm{im}(s)=\ker (\delta )$.

The square and bottom-row relation give $\gamma by=l\beta y=lkv{d}_1=0$. Thus $by$ factors through $k_{\gamma }:K\to Z$ as $by=k_{\gamma }x$ for a unique $x:D\to K$. The pair $(y,x)$ gives $w:D\to P$ with ${\pi }^{\mathrm{\prime }}w=y$ and $\pi w=x$. Since $krw=\beta y=kv{d}_1$ and $k$ is monic, $rw=v{d}_1$. Hence $\delta x=q_{\alpha }rw=q_{\alpha }v{d}_1=t{d}_0{d}_1$. The composite $d_0{d}_1$ is epic, so $\mathrm{coker}(\delta )t=0$. By [L6], $t$ factors through $\mathrm{im}(\delta )$; together with $c\delta =0$, this proves $\mathrm{im}(\delta )=\ker (c)$.

The two central nodes are exact by steps 4.1 and 4.2; the other nodes and the endpoint clauses follow from step 3.1. Thus the stated sequence is exact.