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?
Best Answer
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 $T$.