Homological Algebra – Proving the Snake Lemma Without Diagram Chase

abelian-categoriescategory-theorycommutative-algebrahomological-algebra

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?

Best Answer

You can always "diagram chase" in any abelian category, without invoking any embedding theorem, using arguments with subobjects, as in MacLane's book.

In any case, you can also construct the boundary map as follows:

We are given a map $b: B \to B'$. Let $B'' \hookrightarrow B$ denote the preimage in $B$ of $\ker c$. (If you want to desribe this in more categorical terms, it is the kernel of the composite $B \to C \to C'$.)

Then the map $B''\hookrightarrow B \rightarrow B'$ factors through the monomorphism $A' \hookrightarrow B'$ (using the fact that $A' =\ker(B' \to C')\, \, $). This then induces a map on quotients $ B''/A \to A'/\operatorname{im}A$, which is precisely the desired map $\ker c \to\operatorname{coker}a.$

Checking the various exactness claims is just a matter of using all the relevant universal properties of kernels, cokernels, quotients, etc.

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