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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.

user26857
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R_D
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2 Answers2

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The sequence $$N\stackrel{h}\to N'\stackrel{\pi}\to\operatorname{Coker}h\to0$$ is exact, so $$0\to \operatorname{Hom}(\operatorname{Coker}h,\mathbb Z) \stackrel{\pi^*}{\to}\operatorname{Hom}(N',\mathbb Z)\stackrel{h^*}{\to} \operatorname{Hom}(N,\mathbb Z)$$ is also exact; see here. But $\operatorname{Coker}h$ is a finite abelian group, so $\operatorname{Hom}(\operatorname{Coker}h,\mathbb Z)=0$. Then $\ker h^*=\operatorname{Im} \pi^*=0$.

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user26857
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  • I don't understand the statement Hom(finite abelian group, $\mathbb Z$) =0. I can only say that the Cokernel is the direct sum of finitely many cyclic groups. Then what? – R_D Dec 09 '15 at 18:07
  • @Rise In your question it is written: "If $h$ has finite cokernel...". What is this supposed to mean? (Btw, a finite direct sum of finitely many cyclic groups must be a finite group, unless you assume that the direct sum contains some copies of $\mathbb Z$.) – user26857 Dec 09 '15 at 18:22
  • it means the Cokernel is a finite set. OK let's forget what I said about finite direct sum. Can you give me a hint to prove Hom(cokernel $h, \mathbb Z$) = 0? Thanks in advance. – R_D Dec 09 '15 at 18:35
  • @Rise If $(G,+)$ is a finite group, $n=|G|$, and $f:G\to\mathbb Z$ is a group homomorphism, then $nf(x)=f(nx)=f(0)=0\implies f(x)=0$ for all $x\in G$. ($nx=0$ follows from Lagrange's theorem – user26857 Dec 09 '15 at 18:39
  • This is very nice. Thank you for your help. – R_D Dec 09 '15 at 18:44
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Note that $h^{*}$ acts on hom sets. By the Yoneda Lemma, that each $h:N\rightarrow N'$ induces a natural transformation $h^{*}:[N',-]\rightarrow [N,-]$ defined by $h^{*}(m:N'\rightarrow M)=m\circ h$.

Then, it is not hard to show that $h^{*}$ is monic (injective) iff $h$ is epic (surjective).

Matematleta
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