Going through old notes I see following problem.
$X$ and $Y$ are modules over $\mathbb{Z}$
i) $ \operatorname{Hom}_{\mathbb{Z}}(X,\mathbb{Z})\cong _{\mathbb{Z}} \operatorname{Hom}_{\mathbb{Z}}(Y,\mathbb{Z})$ follows $X \cong _{\mathbb{Z}}Y$
ii) $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},X)\cong _{\mathbb{Z}} \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z},Y)$ follows $X \cong _{\mathbb{Z}}Y$
It says i) is always true while ii) isn't necessarily true. However I don't see this conclusion at all? Anyone has a idea why its like that?
I would say the contrary: i) is false and ii) is true.
A counter-example to i): $$\operatorname{Hom}_{\mathbf{Z}}(\mathbf{Z}/m\mathbf{Z},\mathbf{Z})=\{0\}=\operatorname{Hom}_{\mathbf{Z}}(\mathbf{Z}/n\mathbf{Z},\mathbf{Z}),$$ however $\;\mathbf{Z}/m\mathbf{Z}\simeq \mathbf{Z}/n\mathbf{Z}$ only if $m=n$.
ii) is trivial, as for any ring $R$ and any $R$-module $M$, one has a canonical isomorphism: $$\operatorname{Hom}_{R}(R,M)\simeq M,$$ which maps a linear map $u$ (from $R$ to $M$) to $u(1)\in M$.