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

$0\to \mathrm{E}\mathrm{x}\mathrm{t}(H_{n-1}(X;\mathbb{Z}),M)\to H^n(X;M)\to \mathrm{H}\mathrm{o}\mathrm{m}(H_n(X;\mathbb{Z}),M)\to 0$

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

$H^n(X;M)\cong \mathrm{H}\mathrm{o}\mathrm{m}(H_n(X;\mathbb{Z}),M)\oplus \mathrm{E}\mathrm{x}\mathrm{t}(H_{n-1}(X;\mathbb{Z}),M)$

For coefficients in the integers

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

$H^n(X;\mathbb{Z})\cong \mathrm{H}\mathrm{o}\mathrm{m}(H_n(X;\mathbb{Z}),\mathbb{Z})\oplus \mathrm{E}\mathrm{x}\mathrm{t}(H_{n-1}(X;\mathbb{Z}),\mathbb{Z})$

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:

$H^n(X;\mathbb{Z})\cong \mathrm{H}\mathrm{o}\mathrm{m}(H_n(X;\mathbb{Z}),\mathbb{Z})$

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:

$H^1(X;\mathbb{Z})\cong \mathrm{H}\mathrm{o}\mathrm{m}(H_1(X;\mathbb{Z}),\mathbb{Z})$

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:

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