# Basic question about the proof of Snake's lemma (WITHOUT elements)? [duplicate]

Unfortunately, I don't know how to do draw commutative diagrams in TeX so I'll hope you're familiar with the statement of the lemma.

We want to show of course that $$0 \rightarrow \ker(f) \rightarrow \ker(g) \rightarrow \ker(h) \rightarrow \newcommand{\coker}{\operatorname{coker}}\coker(f) \rightarrow \coker(g) \rightarrow \coker(h) \rightarrow 0$$ is exact (where $$f, g, h$$ are as usual as in the statement of the lemma). The challenge of the snake's lemmma is to find the map $$\ker(h) \rightarrow \coker(f)$$ and show exactness at this step. But I'm afraid to say I don't see how even $$0 \rightarrow \ker(f) \rightarrow \ker(g) \rightarrow \ker(h)$$ and $$\coker(f) \rightarrow \coker(g) \rightarrow \coker(h) \rightarrow 0$$ is exact. As I stated, I would like to see the proof WITHOUT using elements, so without using Mitchell's embedding theorem, working in a general abelian category.

I know it arises 'naturally' from the fact that the sequences in the hypothesis of the lemma are exact, but I'm not sure when I actually try.

Lemma : Let $$\require{AMScd}\begin{CD}A @>{k}>> B @>{f}>> C \\ @V{\alpha}VV (1) @VV{\beta}V @VV{\gamma}V \\ Ker(f') @>>{ker(f')}> B' @>>{f'}> C' \end{CD}$$ be a commutative diagram in a pointed category, with $$\gamma$$ a monomorphism. Then the square $$(1)$$ is a pullback if and only if $$k=ker(f)$$.
The short proof consists in noticing that because $$\gamma$$ is a monomorphism, $$ker(f)=ker(\gamma f)=ker(f'\beta)$$, and then apply the pullback lemma to the diagram $$\require{AMScd}\begin{CD}A @>{\alpha}>> Ker(f) @>>> 0 \\ @V{k}VV (1) @VV{ker(f')}V @VVV \\ B @>>{\beta}> B' @>>{f'}> C', \end{CD}$$ or alternatively prove directly that the universal properties for the pullback and the kernel coincide.
Once you have this lemma, you can look at the diagram $$\begin{CD} 0 @>>> Ker(f)@>{\tilde{k}}>> Ker(g) @>{\tilde{p}}>> Ker(h)\\ & & @V{ker(f)}VV (2) @VV{ker(g)}V @VV{ker(h)}V \\ 0 @>>> X @>{k}>> Y @>{p}>> Z \\ & @V{f}VV @VV{g}V @VV{h}V \\ 0 @>>> X' @>>{k'}> Y' @>>{p'}> Z'.\end{CD}$$ Since the lower row is exact, $$k'$$ is a monomorphism, and then the lemma implies that $$(2)$$ is a pullback. Since $$ker(h)$$ is also a monomorphism, the other impliciation now shows that $$\tilde{k}$$ is the kernel of $$\tilde{p}$$ and thus the top row is exact. The sequence with cokernels is also exact by duality.