Let
$$0\to V'\to V\to V''\to 0$$
be a short exact sequence of abelian groups.
Let $X$ be a topological space. How to construct long exact sequences in singular homology and cohomology
$$\cdots \to H_i(X;V')\to H_i(X;V)\to H_i(X;V'')\to H_{i-1}(X;V')\to\cdots,$$
$$\cdots \to H^i(X;V')\to H^i(X;V)\to H^i(X;V'')\to H^{i+1}(X;V')\to\cdots?$$
My first idea was to prove, that we have short exact sequences of singular chain (co-)complexes
$$0\to C_i(X;V')\to C_i(X;V)\to C_i(X;V'')\to 0,$$
$$0\to C^i(X;V')\to C^i(X;V)\to C^i(X;V'')\to 0$$
and then use the snake lemma as usual. But here we are in a different situation, because we have the one topological space and different coefficients. Therefore I think my idea above isn't useful, or am I wrong? Could you give me a hint?
Best Answer
Tensor product is right exact: if $0 \to V' \to V \to V'' \to 0$ is an exact sequence of abelian groups, then $M \otimes V' \to M \otimes V \to M \otimes V'' \to 0$ is an exact sequence too (I'm looking at tensor product over $\mathbb{Z}$; if $f : X \to Y$ is a morphism, then the induced morphism $M \otimes X \to M \otimes Y$ is given on generators by $m \otimes x \mapsto m \otimes f(x)$).
But by definition $C_i(X)$ is a free abelian group, and a free abelian group if flat, meaning that $M \otimes -$ is actually exact (and not just right exact). So you do get an exact sequence $$0 \to C_i(X) \otimes V' \to C_i(X) \otimes V \to C_i(X) \otimes V'' \to 0,$$ and by definition $C_i(X;A) := C_i(X) \otimes A$. One can also check directly that the exact sequence above commutes with differentials, so you get a short exact sequence of chain complexes and can apply the snake lemma as usual.
For cohomology the idea is the same: a free abelian group is projective (one has a chain of implications "free $\implies$ projective $\implies$ flat"), and by definition this means that $\hom(P,-)$ is an exact functor (in general it's only left exact), and you get a short exact sequence: $$0 \to \hom(C_i(X), V') \to \hom(C_i(X),V) \to \hom(C_i(X),V'') \to 0$$ where by definition $C^i(X;A) := \hom(C_i(X), A)$, and this short exact sequence commutes with differentials.
PS: This long exact sequence is how you construct the Bockstein homomorphism, for example. It's the connecting homomorphism $H_i(X;V'') \to H_{i-1}(X;V')$.