The kernel-cokernel sequence of a composite is a snake

Consider the morphism between the canonical split short exact sequences tikzcd 0 \arrow[r] & A \arrow[r, "j_A"] \arrow[d, "f"'] & A\oplus B \arrow[r, "\pi_B"] \arrow[d, "m"'] & B \arrow[r] \arrow[d, "g"'] & 0 \\ 0 \arrow[r] & B \arrow[r, "j_B"'] & B\oplus C \arrow[r, "\pi_C"'] & C \arrow[r] & 0 where, in biproduct matrix notation, $m=\left(\begin{array}{cc}f& -1_B\\ 0& g\end{array}\right),m(a,b)=(f(a)-b,g(b))\mathrm{.}$ The two squares commute, so [L1] applies.

The map $a\mapsto (a,f(a))$ identifies $\ker (gf)$ with $\ker (m)$: the equations $m(a,b)=0$ are exactly $b=f(a)$ and $gf(a)=0$. Dually, the map $B\oplus C⟶\mathrm{coker}(gf),(b,c)⟼[c+g(b)]$ identifies $\mathrm{coker}(m)$ with $\mathrm{coker}(gf)$. Under these identifications, the snake sequence of step 1.1 is exactly $0\to \ker (f)\to \ker (gf)\to \ker (g)\to \mathrm{coker}(f)\to \mathrm{coker}(gf)\to \mathrm{coker}(g)\to 0.$

The maps in step 2.1 are the canonical comparison maps of [L2], so the kernel-cokernel sequence of a composite is a special case of the snake lemma.