# A short exact sequence of abelian groups induces a long exact sequence in (co)homology with coefficients

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?

• Well, it would work if you did actually have short exact sequences of (co)chain complexes. Why do you think you don't? Commented Nov 23, 2015 at 8:06
• ok. I don't know how to obtain the short exact sequences of chaincomplexes/ cochaincomplexes. Another thing is that the homfunctor $Hom(X,-)$ don't need to be exact for a topological space $X$, if I'm right Commented Nov 23, 2015 at 8:12
• If $V \overset{f}{\to }V'$, then there is a map $C_i(X, V) \to C_i(X, V')$ given by $\sum a_i \sigma_i \mapsto \sum f(a_i) \sigma_i$. One can check that the sequence on $C_i(X, \cdot)$ is short exact.
– user99914
Commented Nov 23, 2015 at 8:18
• ok, thanks! I will try it Commented Nov 24, 2015 at 6:16

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 $H_i(X;V'') \to H_{i-1}(X;V')$.
• The standard definitions immediately imply that $C \otimes (-)$ is exact when $C$ is flat and $\mathrm{Hom}(C, -)$ is exact when $C$ is projective. There is no need to bring in Tor or Ext. Commented Nov 23, 2015 at 9:34