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 tikzcd P \arrow[r, "\pi'"] \arrow[d, "\pi"'] & Y \arrow[d, "b"] \\ K \arrow[r, "k_\gamma"'] & Z. 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, it is 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 $\ker (\beta )\to \ker (\gamma )$.

Conversely, let $u$ be a member of $K$ with $\delta u\equiv 0$. Because $\pi$ is epic, [L5] gives a member $n$ of $P$ with $\pi n\equiv u$. Then $q_{\alpha }rn=\delta \pi n\equiv \delta u\equiv 0,$ so the cokernel property of $q_{\alpha }$ gives a member $x$ of $X$ with $\alpha x\equiv rn$. Hence $\beta {\pi }^{\mathrm{\prime }}n=krn\equiv k\alpha x=\beta ax\mathrm{.}$ Applying [L6] to ${\pi }^{\mathrm{\prime }}n$ and $ax$ with respect to $\beta$, obtain a member $y$ of $Y$ with $\beta y\equiv 0$ and $by\equiv b{\pi }^{\mathrm{\prime }}n\mathrm{.}$ Since $b{\pi }^{\mathrm{\prime }}n=k_{\gamma }\pi n\equiv k_{\gamma }u$, the member $y$ factors through $\ker (\beta )$ and maps to $u$ in $\ker (\gamma )$. Thus every member of $\ker (\delta )$ lies in the image of $\ker (\beta )\to \ker (\gamma )$, so the sequence is exact at $\ker (\gamma )$.

The exactness at $\mathrm{coker}(\alpha )$ is the formal dual of step 3.3 in the opposite abelian category. By [L7], that dual exactness transports back to the statement that $\mathrm{im}(\delta )=\ker (\mathrm{coker}(\alpha )\to \mathrm{coker}(\beta ))\mathrm{.}$

Hence the displayed six-term sequence is exact under the weaker Stacks hypotheses, with the additional endpoint monic and epic clauses already proved in step 3.1.