I have the following problem:

Show or disprove the following: Let $X,Y$ be $\mathbb{Z}$-moduls.

$1. _\mathbb{Z}Hom(X,\mathbb{Z})\tilde{=}_\mathbb{Z}Hom(Y,\mathbb{Z})\Rightarrow X\tilde{=}_\mathbb{Z}Y$

$2. _\mathbb{Z}Hom(\mathbb{Z},X)\tilde{=} _\mathbb{Z}Hom(\mathbb{Z},Y)\Rightarrow X\tilde{=}_\mathbb{Z}Y$

At this point, I'm not sure about either statement. If I wanted to disprove it, I would need to at least be able to show $_\mathbb{Z}Hom(X,\mathbb{Z})\tilde{=} _\mathbb{Z}Hom(Y,\mathbb{Z})$ for some $X,Y$, which I don't know how to do. So I thought about proving it. I know the following theorem:

Let $C$ be a category and $X,Y\in Ob(C)$. Then the following statements are equivalent:

$1. X\tilde{=}_C Y$

$2.$ The functors $Mor_C(X,.),Mor_C(Y,.):C\rightarrow Set$ are naturally isomorph.

$3.$ The functors $Mor_C(.,X),Mor_C(.,Y):C\rightarrow Set$ are naturally isomorph.

So if I can show $2$ or $3$ then the problem is solved, but this basically brings me back to the problem of showing $_\mathbb{Z}Hom(X,\mathbb{Z})\tilde{=}_\mathbb{Z}Hom(Y,\mathbb{Z})...$ Can someone help me?


The issue with applying Yoneda's lemma (your proposed tool) here is that you don't have an isomorphism of functors, but only an isomorphism of some particular abelian groups arising as Hom-sets. The usual intuition is that this should not be enough to guarantee that the objects $X$ and $Y$ are isomorphic. The usual intuition is right for question 1., but wrong for question 2 (as it turns out, the maps from the integers into an abelian group know which group it is).

For 1: Take $X=\mathbf{Z}/2$ and $Y=\mathbf{Z}/3$. Neither of these has any non-zero homomorphism to $\mathbf{Z}$.

For 2: In fact, the map $\mathrm{Hom}(\mathbf{Z},X) \to X$ sending $\phi$ to $\phi(1)$ is an isomorphism, so $X$ and $Y$ are indeed isomorphic.

  • $\begingroup$ For 2, why is the map $Hom(\mathbb{Z},X)\rightarrow X$ an isomorphism? $\endgroup$ – Tobi92sr Apr 27 '18 at 0:34
  • $\begingroup$ @Tobi92sr Given any $x \in X$, the map $m \mapsto mx$ is a homomorphism of abelian groups from $\mathbf{Z}$ to $X$. Thus the map is surjective. If $\phi(1)=\psi(1)$, then $\phi(m)=m \phi(1)=m \psi(1)=\psi(m)$ for all integers $m$, so the map is injective. $\endgroup$ – Stephen Apr 27 '18 at 12:00

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