I have been using the following statement of the UCT:
$$H_{i}(C;G) \cong H_{i}(C) \otimes G \oplus \operatorname{Tor}(H_{i-1}(C),G).$$
But now I want to calculate the cohomology ring of a certain map and I know its homology groups, is there a statement of UCT that help me in calculating the cohomology groups knowing the homology groups? and how this form is related to the form of UCT I mentioned above?
Best Answer
Given an abelian group $G$ and a chain complex $(C_\bullet, \partial_\bullet)$, the universal coefficient theorem for cohomology is given by the short exact sequence
$$0\to \operatorname{Ext}(H_{n-1}(C),G) \to H^n(C_n,G) \to \operatorname{Hom}(H_n(C),G)\to 0$$ where $H_*(C)$ is homology with integral coefficients, i.e. $$H_*(C) = H_*(C,\mathbb{Z})$$
The upshot of this is that it allows you to compute cohomology with coefficients in $G$ purely from homology with integral coefficients.
Note that obviously your homology and cohomology are dual exactly if $\operatorname{Ext} (H_{n-1}(C),G)$ vanishes, which is precisely the case whenever $H_{n-1}(C)$ is free abelian.
Therefore, if your homology group $H_{n-1}(C)$ happens to be free abelian, you have
$$H^n(C_n,G) \cong \operatorname{Hom}(H_n(C),G).$$