# How are these homology groups turned into cohomology groups?

The below proof is from Rotman's Algebraic Topology and my question is:

How does the universal coefficient theorem apply to $A^*$ to get the cohomology sequence here? How specifically do you use apply a short exact sequence of HOMOLOGY groups to $A^*$ and get the COHOMOLOGY groups in large text?

If $X$ is a space of finite type and if $G$ is an abelian group, then there is an exact sequence for eery $n \ge 0$ : $$0 \rightarrow H^n(X) \otimes G \rightarrow ^ {\alpha} H^n(X;G) \rightarrow(H^{n+1}(X), G) \rightarrow 0$$ where $\alpha : (\text{cls } z) \oplus g \mapsto \text{cls } zg$.

Since $X$ has finite type, there's a free chain complex $C_*$ of finite type with $H_*(C_*)=H_*(X)$. If $A^*=\text{Hom}(C_*, Z)$, then the universal coefficient theorem for homology applies because $A^*$ is a free chain complex and there is thus a split short exact sequence

## $$0 \rightarrow H^n(A^*) \otimes G \rightarrow ^{\alpha} H^n(A^* \otimes G) \rightarrow \text{Tor }(H^{n+1}(A^*), G) \rightarrow 0$$

and the proof follows from the identities: $$H^n(\text{Hom }(S_*(X), Z)) \approx H^n(X) \text{ and } H^n(A^* \otimes G) \approx H^n(X;G)$$

The "universal coefficient theorem for homology" should be a purely algebraic statement which applies to any chain complex of free abelian groups. If the differentials happen to increase degree instead of decrease it, that won't affect the validity of the theorem, just the grading on the homology group in the Tor term. In this setting, $A^*$ is such a chain complex (with the differentials increasing degree), so the theorem applies.