# Limit of Hom-groupoid

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 $$\underline{\text{Hom}}(G,\lim_{i\in I} H_i)\cong \lim_{i\in I}\underline{\text{Hom}}(G, H_i)?$$

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 $$\text{Grpd}(X^{\delta}, BG) \cong * \cong \text{Grpd}(*, BG)$$.
• 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? Dec 7 '20 at 12:47
• In your last proof, it should be a $\times$, not a $\sqcup$ Dec 7 '20 at 18:41