$\alpha : H^{n}(\mathrm{Hom}(K,G)) \to \mathrm{Hom}(H_{n}(K),G)$ is an isomorphism Let $G$ is divisible group and abelian , $K$ is a chain complex . The map $\alpha$ in the title above takes idea from inner product . To be more specific , $x \in H_{n}(K)$ and $u \in H^{n}(\mathrm{Hom}(K,G))$ the inner product $<u,x>$ is an element of $G$ obtained according to the following simple prescription : Choose a representative cocycle $u' \in \mathrm{Hom}(H_{n}(K),G)$ for $u$ and $x' \in K_{n}$ for $x$ then $<u,x> = u'(x')$ ( one can prove this definition is independent of choices . Then my question is : 
$$\alpha :  H^{n}(\mathrm{Hom}(K,G)) \to \mathrm{Hom}(H_{n}(K),G)$$
$$(\alpha u)(x) = <u,x>$$
is an isormophism when $G$ is a divisible group 

By the universal coefficient theorem, the obstruction to isomorphism
will be an Ext group: $\text{Ext}^1(H_{n-1}(X),G)$. But because
$G$ is divisible, this Ext group vanishes, as divisible groups are
injective in the category of Abelian groups.
