Suppose $C$ is a free chain complex over a PID $R$, and $H(C)$ is finitely generated. From the universal coefficient theorem, there is a short exact sequence: $$ 0 \rightarrow \text{Ext}^1_R(H(C), R) \rightarrow H(C^*) \overset{h}{\rightarrow} \text{Hom}_R(H(C), R) \rightarrow 0. $$ My question is, is the following homomorphism $$ \bar{h}: H(C^*)/\text{(tor)} \rightarrow \text{Hom}_R(H(C), R)$$ an isomorphism?
A torsion $[f] \in H(C^*)$ is annihilated by $h$, since if there is an $a \in R$ such that $a[f] = 0$, then $$ah([f])([z]) = a\left< [f], [z]\right> = \left< a[f], [z]\right> = 0,$$ so $h([f]) = 0$ and the map is well-defined. On the other hand, from the structure theorem, we may write $$H(C) \cong R^r \oplus_i R/(a_i)$$ and we have $$\text{Hom}_R(H(C), R) \cong \text{Hom}_R(R^r, R) \oplus_i \text{Hom}_R(R/(a_i), R) \cong R^r$$ and $$\text{Ext}_R(H(C), R) \cong \text{Ext}_R(R^r, R) \oplus_i \text{Ext}_R(R/(a_i), R) \cong \bigoplus_i R/(a_i)$$ so we have $$H(C^*) \cong R^r \oplus_i R/(a_i).$$
So we can say that $\text{Ext}_R(H(C), R)$ is (abstractly) isomorphic to the torsion submodule of $H(C^*)$, but not canonically as discussed in this question.
Thanks in advance.
It appears to me that the image of $\newcommand\Ext{\operatorname{Ext}}\Ext_R^1(H(C),R)$ is necessarily the torsion subgroup of $H(C^*)$. You've already argued that all torsion elements are in the kernel, and every element of $\Ext_R^1(H(C),R)$ has torsion, so its image in $H(C^*)$ will have torsion. Thus by exactness of the sequence, all torsion elements lie in the kernel, hence lie in the image of $\Ext$, and yet also every element of the image of $\Ext$ lies in the torsion subgroup. Thus the image of $\Ext$ is precisely the torsion subgroup. Thus by exactness, $h$ induces the desired isomorphism.
There appears to be no issue of canonicality here, unless the original short exact sequence wasn't canonical (which I'm pretty sure it is), since the scenario here doesn't really match exactly with what you've got in that linked question.