Is it true the formula$\newcommand{\Hom}{\operatorname{Hom}}$ $\Hom(\varprojlim G_\alpha ;I) \simeq \varinjlim \Hom(G_\alpha ;I),$ if the group $I$ is injective? We can assume that $G_\alpha$ is finite generated as well.
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2$\begingroup$ Could you clarify what you mean by an injective group? Do you mean an injective abelian group? The only injective objects in the category of (not necessarily abelian) groups are trivial groups, so if that’s what you mean then the answer to your question is trivially “yes”! $\endgroup$– Jeremy RickardCommented Aug 20, 2020 at 11:07
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$\begingroup$ Thank you. Yes, I mean an injective abelian group, but I do not understand why is an injective group trivial. For example, groups of rational numbers $\mathbb{Q}$ is an injective or sometimes is called infinity divisible group. Can you tell me any classical book where I can find it in details. $\endgroup$– AndreyCommented Aug 20, 2020 at 12:18
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$\begingroup$ There are various proofs in, or referred to in, this thread. $\endgroup$– Jeremy RickardCommented Aug 20, 2020 at 12:31
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3$\begingroup$ Re, in particular, as @JeremyRickard says, the key object is in what category you're working. If you embed $\mathbb Q$ in $G = \langle\mathbb Q, x, y \mathrel: [x, y] = 1\rangle$, then the identity map on $\mathbb Q$ does not extend to $G \to \mathbb Q$—we left the category of Abelian groups. $\endgroup$– LSpiceCommented Aug 20, 2020 at 13:12
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No.
$\varprojlim_{n\in\mathbb{N}}\mathbb{Z}/p^n\mathbb{Z}=\mathbb{Z}_p$, the $p$-adic integers.
Take $I=\mathbb{Q}_p$, the $p$-adic rationals.
There is a nonzero map $\mathbb{Z}_p\to\mathbb{Q}_p$, the inclusion, but $\operatorname{Hom}(\mathbb{Z}/p^n\mathbb{Z},\mathbb{Q}_p)=0$ for every $n\in\mathbb{N}$, since $\mathbb{Q}_p$ is torsion-free.
Or alternatively, take $G_n=\mathbb{Z}$, with $G_{n+1}\to G_n$ multiplication by two. Then $\varprojlim G_n=0$ but $\varinjlim\operatorname{Hom}(G_n,\mathbb{Q})\cong\mathbb{Q}$.
answered Aug 20, 2020 at 12:52