TITLE: Proving the snake lemma without a diagram chase
QUESTION [43 upvotes]: 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?

REPLY [5 votes]: The salamander lemma described by George Bergman and summarised by nlab & the secret blogging seminar help simplify proofs of the basic diagram chases, including the 3x3, four, snake and long exact diagrams.
However the Secret Blogging Seminar says: 

If you don’t like diagram chases, it’s likely that you still won’t like them once you know the Salamander lemma. The salamanders chase the diagrams for you, but you still have to chase the salamanders. I think the salamander proofs are easier to explain (once you know the Salamander lemma), and it’s easier to see where you use the hypotheses. For example, it is totally clear that the argument for the 3x3 lemma can prove the 20x20 lemma as well.