# Intuition behind Snake Lemma

I've been struggling with this for some time. I can prove the Snake Lemma, but I don't really “understand” it. By that I mean if no one told me Snake Lemma existed, I would not even attempt to prove it. I believe I am missing some important intuition that tells me that the lemma “could” be true before actually trying to prove it. Please give me some pointers.

-
It is indeed a strange thing... Exterior differentiation of differential forms, and boundary operators on spaces, were perhaps the plausible and "more physical/intuitive" antecedents of the abstracted form. –  paul garrett Aug 14 '12 at 19:32
youtube.com/watch?v=etbcKWEKnvg –  Will Jagy Aug 14 '12 at 19:47
@WillJagy Yes I've seen that. It's entertaining. Doesn't really help though :) –  Tunococ Aug 14 '12 at 19:48
In that case, you were wise to ask here. –  Will Jagy Aug 14 '12 at 19:49

A special case is easy to see: Imagine $A \leq A' \leq B=B'$, and $C=B/A$ and $C'=B'/A'$ with the obvious maps from a module $M$ to $M'$. The kernels are $0$, $0$, and $A'/A$. The cokernels are $A'/A$, $0$, $0$. The last kernel and the first cokernel are linked.

The snake lemma is then just one of the isomorphism theorems. As you deform $B$ and $B'$ more, how do the kernels and cokernels deform? The last kernel and first cokernel no longer need be isomorphic, but the kernel and cokernel of that linking (snakey) map can be described in terms of the kernels and cokernels already there. A specific relation is the snake lemma.

## Example deformation

An example deformation might be helpful: distort the $A' \to B' \to C'$ sequence by quotienting out by some $M \leq B$ (imagine $M=IB$ for some ideal $I$, so we are tensoring the second line with $R/I$). How does this change the kernels and cokernels?

The first line is $$0 \to A \to B \to B/A \to 0,$$ and the second line becomes $$0 \to (A'+M)/M \to B'/M \to B'/(A'+M) \to 0$$ so the kernels are $A \cap M$, $M$, and $(A'+M)/A$ and the cokernels become $(A'+M)/(A+M)$, $0$, and $0$. The last kernel and the first cokernel are related, but not equal. One clearly has the relation $0 \to (A+M)/A \to (A'+M)/A \to (A'+M)/(A+M) \to 0$ where the last two nonzero terms are the last kernel and the first cokernel. The first term is weird though. We apply another isomorphism theorem to write $(A+M)/A \cong M/(A\cap M)$ and then the solution is clear: We already have $0 \to A \cap M \to M \to M/(M\cap A) \to 0$ so we splice these two together to get the snake lemma: $$0 \to A \cap M \to M \to (A'+M)/A \to (A'+M)/(A+M) \to 0 \to 0 \to 0$$

-
Ok, your first special case is pretty helpful. Let me try to explain what I understand. $A \le A' \le B = B'$ corresponds to growing $A$ to $A'$. The elements in $\text{coker}(A \to A')$ correspond to the growth of $A'$ over $A$. Since $B = B'$, $C$ must reduce in size. $\ker(C \to C')$ measures the shrinkage $C \to C'$, and so $\ker(C \to C') \cong \operatorname{coker}(A \to A')$. Does this make sense? –  Tunococ Aug 14 '12 at 23:32
@Tunococ: exactly. The first special case should be intuitive. The next section is “what happens in a less intuitive case.” We had to check if $\ker(C\to C') \cong \ker(A \to A')$ was always true. It wasn't but at least we could “compare” the two modules. In a category “compare” is a codeword for “homomorphism”, and in abelian category that means exact sequences. –  Jack Schmidt Aug 15 '12 at 13:18
@JackSchmidt: after staring at your deformed example for a while trying to see the kernels of the vertical maps, I noticed I can't figure out what you mean by $A'M$. These are modules so there's no element-wise product $\{am, a\in A', m\in M\}$, and neither the direct nor tensor product makes sense to read here-what do you mean? –  Kevin Carlson Sep 20 '12 at 12:26
@Kevin: Try $A'+ M =\{a+m : a \in A', m \in M\}$ to be the join of $A'$ and $M$. The notation is standard for multiplicative groups (where I normally work), but $A'+M$ is usually used for additive groups. –  Jack Schmidt Sep 20 '12 at 12:39
Thanks for the quick reply, and for the fact that'll surely be good to know again. –  Kevin Carlson Sep 20 '12 at 12:46
show 1 more comment

All these diagram chasing lemmas (snake lemma, $3x3$ lemma, four lemma, five lemma, etc.) follow directly from the "salamander lemma" due to George Bergman, see salamander lemma.

And that is pretty transparent. It is so transparent that for instance it is immediate to see (which no textbook ever mentions) that there are just as easily 4x4 lemmas, 5x5 lemmas, 6x6 lemmas. etc. In other words: once you see the simple idea of the salamander lemma, you can come up yourself with more such diagram chasing lemmas at will.

-
Nice answer! +1 –  Matt N. Sep 20 '12 at 12:01
I can't resist. If you talk about derived functors and the snake lemma, you can go a step further. One of my favorite homological algebra exercises is: compute the derived functors of the kernel functor $\ker\colon \mathscr{A}^\to \to \mathscr{A}$ (seen as a left exact functor from the category of morphisms of $\mathscr{A}$ to $\mathscr{A}$)... Interestingly, none of the homological algebra texts I ever read seem to contain this. A pity. –  t.b. Aug 15 '12 at 7:54