Dual universal coefficient theorem

For coefficients in an abelian group
Suppose $$X$$ is a topological space and $$M$$ is an abelian group. The dual universal coefficients theorem relates the homology groups of $$X$$ with coefficients in $$\mathbb{Z}$$ and the cohomology groups of $$X$$ with coefficients in $$M$$ as follows:

First, for any $$n \ge 0$$, there is a natural short exact sequence of abelian groups:

$$0 \to \operatorname{Ext}(H_{n-1}(X;\mathbb{Z}),M) \to H^n(X;M) \to \operatorname{Hom}(H_n(X;\mathbb{Z}),M) \to 0$$

Second, the sequence splits (not necessarily naturally), and we get:

$$H^n(X;M) \cong \operatorname{Hom}(H_n(X;\mathbb{Z}),M) \oplus \operatorname{Ext}(H_{n-1}(X;\mathbb{Z}),M)$$

For coefficients in the integers
This is the special case where $$M = \mathbb{Z}$$. In this case, we case:

$$H^n(X;\mathbb{Z}) \cong \operatorname{Hom}(H_n(X;\mathbb{Z}),\mathbb{Z}) \oplus \operatorname{Ext}(H_{n-1}(X;\mathbb{Z}),\mathbb{Z})$$

Related facts

 * Universal coefficient theorem for homology
 * Universal coefficient theorem for cohomology
 * Kunneth formula for homology
 * Kunneth formula for cohomology

Case of free abelian groups
In the case that $$H_{n-1}(X;\mathbb{Z})$$ is a free abelian group, we get:

$$H^n(X;\mathbb{Z}) \cong \operatorname{Hom}(H_n(X;\mathbb{Z}),\mathbb{Z})$$

Further, if $$H_n(X;\mathbb{Z})$$ is finitely generated, then, under these circumstances, $$H^n(X;\mathbb{Z})$$ is simply the torsion-free part of $$H_n(X;\mathbb{Z})$$.

Note that this always applies to the case $$n = 1$$, because $$H_0$$ is a free abelian group of rank equal to the number of connected components. Thus, we get:

$$H^1(X;\mathbb{Z}) \cong \operatorname{Hom}(H_1(X;\mathbb{Z}),\mathbb{Z})$$

In particular, if $$H_1(X;\mathbb{Z})$$ is finitely generated, then $$H^1(X;\mathbb{Z})$$ is free abelian and equals the torsion-free part of $$H_1(X;\mathbb{Z})$$.

In the case that both $$H_{n-1}(X;\mathbb{Z})$$ and $$H_n(X;\mathbb{Z})$$ are free abelian groups, and the latter has finite rank, we get:

$$H^n(X;\mathbb{Z}) \cong H_n(X;\mathbb{Z})$$