The singular homology and cohomology of a topological space with coefficients in a zero characteristic field.

Question

I have a field with zero characteristic, like $K=\mathbb{C},\mathbb{R}$ and I want to show that the homology groups and cohomology groups with coefficients in these fields satisfy:

$$H_n(X,K) \approx H_n(X)\otimes K, \ H^n(X,K) \approx {\rm Hom}\ (H_n(X),K)$$

for $n\geq 0$.

Do you have reference for this proof?

Thanks in advance.

Answer 1

In the case of coefficients in the field of charactertistic 0, this is more simple than the universal coefficient theorem. (Although anyone seriously interested in algebraic topology should learn $UCT$ in the general case by heart.)

Let $K$ be a field of charactersitic $0$ and notice that as an abelian group it is a rational vector space and is in particulat flat over $\mathbb{Z}$, ie. the functor $- \otimes _{\mathbb{Z}} K$ is exact.

The singular homology of a space $X$ with coefficients in $\mathbb{R}$ is defined to be the homology of the singular chain complex $C_{n}(X, R) = \bigoplus _{\sigma \in Sing_{n}(X)} R$, where $Sing_{n}(X)$ is the set of $n$-singular simplices of $X$. In particular, we have $C_{n}(X, R) \simeq C_{n}(X, \mathbb{Z}) \otimes R$.

Now if $R$ is flat (for example, $R = K$, $char(K) = 0$), then tensoring with it preserves homology and so we have

$H_{n}(X, R) = H_{n}(C_{\bullet}(X, R)) \simeq H_{n}(C_{\bullet}(X, \mathbb{Z}) \otimes R) \simeq H_{n}(C_{\bullet}(X, \mathbb{Z})) \otimes R = H_{n}(X,\mathbb{Z}) \otimes R$,

where we used flatness in the second-to-last isomorphism.

The version for homology is analogous, since by inspection of the definitions we obtain the cochain complex of $X$ over $K$ by applying $Hom _{Ab}(-, K)$ to $C_{\bullet}(X, \mathbb{Z})$. This functor is again exact, since

$Hom _{Ab}(-, K) \simeq Hom _{K}(- \otimes K, K)$

and so it is a composition of two exact functors (tensoring with $K$ and taking a dual vector space).

Answer 2

For your first question look at Universal coefficient theorem (http://en.wikipedia.org/wiki/Universal_coefficient_theorem )

The second part, if I am not mistaken is just a question in linear algebra. The chain complexes are dual (including the duality of morphisms) and then isomorpism follows from isomorphism of linear spaces $V \subset W, (W/V)^*=W^*/V^*$