Let $N$ and $N'$ be finite rank free $\mathbb Z$-modules. Let $M=\operatorname{Hom}\_{\mathbb Z \text{-mod}}(N,\mathbb Z)$ and $M'=\operatorname{Hom}\_{\mathbb Z \text{-mod}}(N',\mathbb Z)$ . 

Suppose I am given a $\mathbb Z$-module homomorphism $h:N\rightarrow N'$. Then I can define a map $h^\*:M'\rightarrow M$ by $m\mapsto m\circ h$ . 

**If $h$ has finite cokernel then is it true that $h^\*$ is injective?**

My attempt - 

$\ker h^\*=\{f:f\circ h = 0\}$ where $0$ here is the map that sends everything to $0\in\mathbb Z$. Then is it obvious that $f$ is the zero map? I was thinking $f\circ h=0\Rightarrow f\circ h(x)=f(x)h(x)=0$ for all $x\in \mathbb Z$. Since $\mathbb Z$ is an integarl domain this means that one among $f(x)$ and $h(x)$ must be $0$. But suppose for some $x$ if $h(x)=0$ then it could happen that $f(x)\neq0$ so that means $f$ need not be the zero map. Also I am not sure how to use the finiteness of cokernel. Any help will be appreciated.

Thank you

**EDIT** I just realised that I wrote the composition $f\circ h$ as the pointwise product which is wrong. So now I am completely stumped on how to prove $f$ is 0.