# Proof of $H^k(X,\mathbf k) = H^k(X,\mathbb Z) \otimes \mathbf k$

Let $X$ be a compact manifold and denote $H^k(X,G)$ the $k$-th cohomology group with coefficients in the abelian group $G$.

Using Cech cohomology one can prove that there is a natural isomorphism $H^k(X,\mathbf k) \simeq H^k(X,\mathbb Z) \otimes \mathbf k$ for every field $\mathbf k$ of characteristic $0$ (see for instance the book "Hodge Theory and Complex Algebraic Geometry I", C.Voisin p 157.)

Is there a way to proove this fact without using sheaf-cohomology? I have looked around for a "Universal Coefficient Theorem" for cohomology but the only one I know relates $H^k(X,\mathbf k)$ with $H_k(X,\mathbb Z)$ and $H_{k-1}(X,\mathbb Z)$, but not with $H^k(X,\mathbb Z)$.

-

If $S_{\bullet}$ is the singular chain complex for $X$, then $H^i(X,A)$ is computed by the cochain complex $Hom(S_{\bullet},A)$. Now there is natural map $Hom(S_{\bullet},\mathbb Z) \otimes k \to Hom(S_{\bullet},k),$ and hence, passing to cohomology, a map $$H^{\bullet}(X,\mathbb Z)\otimes k \to H^{\bullet}(X,k).$$ This will not be an isomorphism for arbitrary spaces $X$. (E.g. if $X$ is just a disjoint union of a set $I$ of points, then this is the natural map $(\mathbb Z^I)\otimes k \to k^I$, which is not an isomorphism in general if $I$ is infinite.)
However, if the homology groups of $X$ with $\mathbb Z$-coefficients are finitely generated, then this will be an isomorphism, as follows from the universal coefficient theorem for cohomology that you mentioned in your question. (Use the fact that the universal coefficient short exact sequence is natural in $A$; that $Ext(H_{i-1},A)$ is torsion when $A = \mathbb Z$ --- since $H_{i-1}$ is f.g. by assumption --- and vanishes when $A = k$, since $k$ is of char. zero; and that $Hom(H_i,\mathbb Z)\otimes k \cong Hom(H_i,k)$, since $H_i$ is f.g.)