For any groups $G,H$, we can define the category, in fact a groupoid, $$\underline{\text{Hom}}(G,H)$$ whose objects are group morphisms $G\to H$ and morphisms $(f:G\to H)\to (g:G\to H)$ are elements of $H$ conjugating $f$ into $g$. Let $\{H_i\}_{i\in I}$ be an inverse system of groups. Then of course we have $\text{Hom}(G,\lim_{i\in I} H_i)\cong \lim_{i\in I}\text{Hom}(G, H_i)$. My question is do we also have $$\underline{\text{Hom}}(G,\lim_{i\in I} H_i)\cong \lim_{i\in I}\underline{\text{Hom}}(G, H_i)?$$


1 Answer 1


Sure, this follows from the Yoneda lemma. Let $B : \text{Grp} \to \text{Grpd}$ be the functor sending any group $G$ to the groupoid with one element $\star$ such that $BG(\star, \star) = G$. For any groupoid $X$ we have \begin{align} \text{Grpd}(X, \underline{\text{Hom}}(G,\lim_{i\in I} H_i)) &\cong \text{Grpd}(X, \underline{\text{Grpd}}(BG, B\lim_{i\in I} H_i)) \\&\cong \text{Grpd}(X \times BG, B\lim_{i\in I} H_i) \\&\cong \lim_{i \in I}\text{Grpd}(X \times BG, BH_i) \\&\cong \lim_{i \in I}\text{Grpd}(X, \underline{\text{Grpd}}(BG, BH_i)) \\&\cong \lim_{i \in I}\text{Grpd}(X, \underline{\text{Hom}}(G, H_i)) \\&\cong \text{Grpd}(X, \lim_{i \in I}\underline{\text{Hom}}(G, H_i)) \end{align} using the universal property of the internal hom and the continuity of $B$ and the Yoneda embedding. So by the fully-faithfulness of the Yoneda embedding we have the isomorphism in your question.

To show that $B$ preserves limits it's not too hard to show directly that it preserves products and equalizers. But it also preserves limits because it's a right adjoint. Its left adjoint can be constructed as follows. Let $\text{Grpd}_{1}$ be the full subcategory of $\text{Grpd}$ on the one-element groupoids. Then $B : \text{Grp} \to \text{Grpd}_{1}$ is an equivalence, so it suffices to give a left adjoint to the inclusion $\text{Grpd}_{1} \to \text{Grpd}$, which I'll also abusively call $B$.

For any groupoid $X$, let $X^{\delta}$ be the discrete groupoid on the underlying set of $X$, and let $i_{X} : X^{\delta} \to X$ be the canonical inclusion. The pushout of $i_{X}$ along the unique functor $X^{\delta} \to *$ gives a one-element groupoid $LX$ (it's not hard to show that this has one element, using the fact that the underlying set functor $\text{Grpd} \to \text{Set}$ is a left adjoint). For any one-element groupoid $G$ we therefore have \begin{align} \text{Grpd}_{1}(LX, G) &= \text{Grpd}(LX, G) \\&\cong \text{Grpd}(*, BG) \times_{\text{Grpd}(X^{\delta}, BG)} \text{Grpd}(X, BG) \\&\cong \text{Grpd}(X, BG) \end{align} by the definition of $\text{Grpd}_{1}$, the universal property of $LX$, and the isomorphism $\text{Grpd}(X^{\delta}, BG) \cong * \cong \text{Grpd}(*, BG)$.

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    $\begingroup$ Oh, I didn't know that this construction was the internal Hom in the category of groupoids, thanks! Do you have a reference for the fact that functor (Groups) $\to$ (Groupoids) is continuous? $\endgroup$ Commented Dec 7, 2020 at 12:47
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    $\begingroup$ @curiousmathguy unfortunately I don't, but I will add an argument for it. good call, though, especially since it's not cocontinuous, so I shouldn't have elided the inclusion. $\endgroup$ Commented Dec 7, 2020 at 14:23
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    $\begingroup$ In your last proof, it should be a $\times$, not a $\sqcup$ $\endgroup$ Commented Dec 7, 2020 at 18:41
  • $\begingroup$ @MaximeRamzi oops, you're quite right, thanks $\endgroup$ Commented Dec 7, 2020 at 18:48
  • $\begingroup$ No problem, nice answer ! $\endgroup$ Commented Dec 7, 2020 at 19:33

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