Universal coefficient theorem

Remarks

Here $H_n(X)$ is singular homology with integer coefficients and $H^n(X; G)$ is singular cohomology; Tor and Ext are the derived functors of $\otimes$ and $\Hom$. For the proof, the cycles $Z_n$ and boundaries $B_n$ of $C$ are free, since submodules of free modules over a principal ideal domain are free. So $0 \to Z_\bullet \to C_\bullet \to B_{\bullet-1} \to 0$ splits in each degree. Apply $- \otimes M$ or $\Hom(-, M)$ and take the long exact sequence; its connecting maps are induced by the inclusions $B_n \to Z_n$. Their kernels and cokernels are the Tor and Ext terms. Example: for the real projective plane $P$, $H_0(P) = \Z$, $H_1(P) = \Z/2\Z$, $H_2(P) = 0$. So $H^1(P; \Z) \cong \Hom(\Z/2\Z, \Z) \oplus \operatorname{Ext}^1(\Z, \Z) = 0$ and $H^2(P; \Z) \cong \Hom(0, \Z) \oplus \operatorname{Ext}^1(\Z/2\Z, \Z) \cong \Z/2\Z$. This shows that cohomology is not simply the dual of homology: the torsion of $H_1$ moves up to $H^2$.

Going deeper

For a student

The homology part of the theorem is a special case of the Künneth theorem, with the second complex a module in degree $0$. It is used constantly, for example to compute cohomology from homology. It also shows that a map inducing isomorphisms on integral homology induces isomorphisms on homology and cohomology with all coefficients. Over a field $k$, it gives $H^n(X; k) \cong \Hom_k(H_n(X; k), k)$.