snake lemma

Last updated January 27, 2022

Statement

In an abelian category, consider a commutative diagram

where the rows are exact sequences and 0 is the Zero object.

Then there is an exact sequence $$\text{ker }a\to\text{ker }b\to\text{ker }c\overset{d}{\longrightarrow}\text{coker }a\to\text{coker }b\to\text{coker }c.$$ The map $d$ is a homomorphism called the connecting homomorphism.

Corollaries

1. If the two sequences are short exact (not just right/left exact) then the induced sequence is too, i.e. the sequence $$0\to\text{ker }a\to\text{ker }b\to\text{ker }c\overset{d}{\longrightarrow}\text{coker }a\to\text{coker }b\to\text{coker }c\to 0$$ is exact.

Proofs 1 It suffices to show that $\text{ker }a\to\text{ker }b$ is injective and $\text{coker }b\to\text{coker }c$ is surjective. First we will show that $\text{ker }a\to\text{ker b}$ is injective. By assumption, $f$ is injective, so its kernel is trivial, hence $\text{ker }a\to\text{ker }b$ is injective.

Now we will show that $\text{coker }b\to\text{coker }c$ is surjective. By assumption $g'$ is surjective. To show that the induced map $B'/\text{im }b\to C'/\text{im }c$ is surjective, it suffices to show that $g'(\text{im }b)\subset\text{im }c$. Let $y\in\text{im }b$. Then there is $x\in B$ such that $b(x)=y$. Then also $c(g(x))=g'(b(x))=g'(y)$, hence the preimage of $g'(y)$ by $c$ is $g(x)\in C$. So $g'(y)\in\text{im }c$, and since our choice of $y\in\text{im }b$ was arbitrary this shows that $g'(\text{im }b)\subset\text{im }c$.