$\text{Ext}(H, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H$ if $H$ is finitely generated From Hatcher, page 196, before corollary 3.3.
We are first given these three properties of the Ext functor:
$\text{Ext}(H \oplus H', G) \cong \text{Ext}(H, G)\oplus\text{Ext}(H′, G)$.
$\text{Ext}(H, G) = 0$ if $H$ is free.
$\text{Ext}(Z_n, G) \cong G/nG.$
Then, Hatcher writes "these three properties imply that $\text{Ext}(H, \mathbb{Z})$ is isomorphic to the torsion subgroup of $H$ if $H$ is finitely generated".
I'm trying to understand why that is.
I know a torsion subgroup is the subgroup composed of all elements of finite order. Also, a finitely generated abelian group $H$ is the direct sum of a free abelian group (which I'll call $H'$) and its torsion subgroup (which I'll call $T$).
So $$\text{Ext}(H)=\text{Ext}(H'\oplus T)$$
From the first property, this equals
$$\text{Ext}(H')\oplus\text{Ext}(T).$$
As $H'$ is free, this equals (from the second property)
$$0\oplus \text{Ext(T)}=\text{Ext(T)}.$$
But why does $\text{Ext}(T)=T$ hold?
 A: The third property (with $G=\mathbb{Z}$) tells you that if $C$ is a finite cyclic group, then $\operatorname{Ext}(C,\mathbb{Z})\cong C$.  Furthermore, any finitely generated torsion abelian group is a direct sum of cyclic groups.  So $T$ is a direct sum of cyclic groups, so by the first and third properties, $\operatorname{Ext}(T,\mathbb{Z})\cong T$.  Note, however, that the isomorphism in the third property is not canonical, so while $\operatorname{Ext}(H,\mathbb{Z})$ is isomorphic to $T$, it is not canonically isomorphic to $T$.
A: To elaborate on my comment, applying $\text{Ext}^{\bullet}(A, -)$ to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$ produces the long exact sequence
$$0 \to \text{Hom}(A, \mathbb{Z}) \to \text{Hom}(A, \mathbb{R}) \to \text{Hom}(A, S^1) \to \text{Ext}^1(A, \mathbb{Z}) \to 0$$
where we can ignore the rest of the sequence because $\text{Ext}^1(A, \mathbb{R})$ vanishes, thanks to the fact that $\mathbb{R}$ is divisible and hence injective. (So is $S^1$; in fact what we've written down above is an injective resolution of $\mathbb{Z}$.)
If $A$ is torsion, then $\text{Hom}(A, \mathbb{R}) = 0$, and the long exact sequence above produces a natural isomorphism
$$\text{Hom}(A, S^1) \cong \text{Ext}^1(A, \mathbb{Z}).$$
The group $\text{Hom}(A, S^1)$ is known as the Pontryagin dual $\widehat{A}$ of $A$. If $A$ is finite it is noncanonically isomorphic to $A$. In general it naturally has a topology making it a profinite abelian group, and every profinite abelian group arises in this way. 
