An important part of the snake lemma is the naturality of the ker-coker sequence produced by it. However, no source seems to state or prove this part for arbitrary abelian categories. However, it is needed, for example, to product long exact homology sequence.

Wikipedia claims the following:

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I would like to understand a proof of this fact without using Freyd-Mitchell embedding.

Edit: this is different from existing questions, since my question concerns naturality of the ker-coker sequences, which the other questions do not adress.


1 Answer 1


We note that if you accept that $\ker$ and $\newcommand\coker{\operatorname{coker}}\coker$ are functorial, the only square we need to prove commutes is the one involving the boundary map.

Let's recall how we define the boundary map $d$ in an element-free fashion.

Let $f$ and $g$ be the nonzero maps in the top row, $f'$, $g'$ the nonzero maps in the bottom row, so we have $$ \require{AMScd} \begin{CD} @. A @>f>> B @>g>> C @>>> 0 \\ @. @VaVV @VbVV @VcVV @. \\ 0 @>>> A' @>f'>> B' @>g'>> C' @. \\ \end{CD} $$

Then $g$ induces an epimorphism from $\ker cg$ to $\ker c$. Then $b$ gives a map $\ker cg$ to $B'$, and $g'b=cg$, so $b|_{\ker cg}$ lifts to a unique map $d_0 : \ker cg\to A'$.

Then if we let $q_a : A'\to \coker a$, and consider $q_ad_0 : \ker cg \to \coker a$, observe that $f$ lifts to $f_0 : A\to \ker cg$ by the universal property of the kernel, and you can check that $\ker c = \coker f_0$. Now $d_0$ was the lift of $b|_{\ker cg}$ along $f'$ to $A'$, so $d_0\circ f_0$ is the lift of $bf=f'a$ along $f'$. Therefore $d_0\circ f_0 =a$. Hence $q_ad_0f_0=q_aa=0$.

As a result, $q_ad_0$ extends along $g|_{\ker cg}$ to give a unique map $d:\ker c\to \coker a$.

A summary and some diagrams:

First we enlarge the diagram like so: $$ \require{AMScd} \begin{CD} @. A @>f_0 >> \ker cg @>g|_{\ker{cg}}>> \ker c @>>> 0\\ @. @| @VVV @VVV @.\\ @. A @>f>> B @>g>> C @>>> 0 \\ @. @VaVV @VbVV @VcVV @. \\ 0 @>>> A' @>f'>> B' @>g'>> C' @. \\ @. @Vq_A VV @. @. \\ @. \coker a @. @. @.\\ \end{CD} $$ Then $d_0$ and $d$ are diagonal maps and can't be drawn, but $d_0$ is the unique map $\ker cg\to A'$ such that $f'd_0 = b|_{\ker cg}$, and $d$ is the unique map $\ker c\to \coker a$ such that $d(g|_{\ker cg}) = q_ad_0$.


Naturality is a result of the construction. For every new map we produce in this construction, you can check that if we had two copies of the snake diagram with a translation between them, then the appropriate square between the constructed maps in each diagram must also commute, since at every stage we construct the new map by either applying a functor, like $\ker$ or $\coker$ to a map already in our diagram, taking a composite of maps in our diagram, or extending/lifting along a kernel/cokernel, and all of these operations produce commutative squares. (Or thought of another way, all of these constructions are functorial in the original diagram in that they take a commutative diagram and enlarge it, and also send translations of the original diagram to translations of the larger diagram).

Note: I apologize for the limited diagrams, but MSE has very limited diagram drawing capabilities, and diagonal morphisms can't be drawn. You may want to draw the diagrams on paper yourself.


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