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It is elementary that for modules over a ring $R$, $\text{Hom}_R(\bigoplus_i A_i,B)=\prod_i \text{Hom}_R (A_i,B)$, and $\text{Hom}_R(A,\prod_i B_i)=\prod_i \text{Hom}_R(A,B_i)$. I am curious that what can we say about $ \text{Hom}_R (\prod_i A_i, B)$? Is there a result concerning this situation? Actually I want to check that $\text{Hom}_{\Bbb Z}(\prod_{n\in \Bbb N} \Bbb Z,\Bbb Z)$ is whether countable or not.

user302934
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    The hom functor is left exact on the second entry. On the first entry, the map $\mathrm{Hom}_R(\prod A_i, B)\to\mathrm{Hom}_R(\oplus A_i,B)$ is by restriction, and you can have two different maps with the same restriction; just take $B=(\prod A_i)/(\oplus A_i)$ and consider the canonical projection and the zero map. – Arturo Magidin May 06 '21 at 23:51
  • @ArturoMagidin Oh I see. I have deleted that part. Thanks – user302934 May 06 '21 at 23:53
  • I think you just misinterpreted what the left exactness means in that case: if you have an exact sequence, the image is left exact. So if you have a surjection $f\colon X\to Y$, then the induced map $\mathrm{Hom}(Y,B)\to\mathrm{Hom}(X,B)$ is injective, since the map is given by sending $h\colon Y\to B$ to $h\circ f$, and $h\circ f= g\circ f$ implies $h=g$ when $f$ is surjective. – Arturo Magidin May 06 '21 at 23:55
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    it is the case that $\operatorname{Hom}_\mathbb{Z}(\mathbb{Z}^\mathbb{N},\mathbb{Z})$ is countable, and it is in fact isomorphic to $\mathbb{Z}^{\oplus\mathbb{N}}$, but this is quite a non-trivial result. see eg here – Atticus Stonestrom May 06 '21 at 23:58
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    @AtticusStonestrom Thanks! – user302934 May 07 '21 at 00:58

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