Dual universal coefficient theorem

A more detailed page on the same theorem, but from a purely algebraic perspective, is at Groupprops:Dual universal coefficient theorem for group cohomology

Statement

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 $n\geq 0$, 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

Particular cases

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:

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