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Consider two simplicial sets $K$ and $L$. Their mapping space (or mapping complex) is the internal hom of simplicial sets, i.e. $\underline{\mathrm{Hom}}(K,L)$, where $$ \underline{\mathrm{Hom}}(K,L)_n := \mathrm{Hom}_{\mathsf{SSet}}(K \times \Delta[n], L). $$

There is a (I presume classical) result in simplicial algebraic topology, that states

(1) If $L$ is a Kan complex, and $K$ a simplicial set, then $\underline{\mathrm{Hom}}(K,L)$ is a Kan complex.

This is for example present in J. P. May, Simplicial objects in algebraic topology, Theorem I.6.9. A Kan complex is a simplicial set $L$ where the horn projections $p^m_j: L_m \to \Lambda^m_j(L)$ are surjective for any $m \ge 1$, $0\le j \le m$. If in addition, all these are bijective for all $m > n$, we say that $L$ is an $n$-groupoid. So in this language, a Kan complex is an $\infty$-groupoid.

I am currently trying to specialize the above result to the statement

(2) If $L$ is an $n$-groupoid, and $K$ a simplicial set, then $\underline{\mathrm{Hom}}(K,L)$ is an $n$-groupoid.

I have some ideas on how to adapt the proof of (1) to show this, but since I am not so familiar with the algebraic topology literature, I am wondering two things.

  1. Are there other proofs of (1) in more modern introductions to the subject since May's lecture notes?

  2. Does (2) already appear in the literature? Is it already known?

Improve this question
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  • $\begingroup$ (2) follows from Lemma 2.14 of arxiv.org/abs/0905.4044 , but there may well be earlier references. $\endgroup$
    – Jon Pridham
    Commented Mar 10, 2023 at 19:44
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    $\begingroup$ You can also find (2) as a special case of 2.3.4.7-8 in Higher Topos Theory, which proves the analogous result for $n$-categories. $\endgroup$
    – Kevin Carlson
    Commented Mar 10, 2023 at 21:25
  • $\begingroup$ Jon, why would Lemma 2.14 help? I didn't get it.. $\endgroup$
    – Chenchang Zhu
    Commented Mar 13, 2023 at 16:52
  • $\begingroup$ @JonPridham after thinking for some time, I do not see how it should follow from Lemma 2.14, either. Could you elaborate a bit? $\endgroup$
    – SetR
    Commented Mar 21, 2023 at 18:55
  • $\begingroup$ Take $Y$ to be a point and observe that the condition is satisfied by $A \times \Lambda^{m,k} \to A \times \Delta^m$ for $m>n$ and arbitrary $A$. $\endgroup$
    – Jon Pridham
    Commented Mar 22, 2023 at 19:14

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