Suppose we have two short exact sequences in an abelian category $$0 \to A \mathrel{\overset{f}{\to}} B \mathrel{\overset{g}{\to}} C \to 0 $$ $$0 \to A' \mathrel{\overset{f'}{\to}} B' \mathrel{\overset{g'}{\to}} C' \to 0 $$ and morphisms $a : A \to A', b : B \to B', c : C \to C'$ making the obvious diagram commute. The snake lemma states that there is then an exact sequence $$0 \to \ker a \to \ker b \to \ker c \to \operatorname{coker} a \to \operatorname{coker} b \to \operatorname{coker} c \to 0$$ where the morphisms between the kernels are induced by $f$ and $g$ while the maps between the cokernels are induced by $f'$ and $g'$.

It is not hard to show that the morphisms induced by $f, g, f', g'$ exist, are unique, and that the sequence is exact at $\ker a, \ker b, \operatorname{coker} b, \operatorname{coker} c$. With the use of a somewhat large diagram shown here, we can even construct the connecting morphism $d : \ker c \to \operatorname{coker} a$. However, I'm stuck showing exactness at $\ker c$ and $\operatorname{coker} a$. I thought Freyd might have had an element-free proof in his book, but it turns out he proves it by diagram chasing and invoking the Mitchell embedding theorem [pp. 98–99]. Is there a direct proof?