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Let $X$ be a simplicial set. It is well known that a model for the path object can be given by the mapping space $\mathrm{Hom}(\Delta[1], X)$. In particular this offers a fiber replacement for the diagonal map $X \to X \times X$ by taking the composition $$ X \overset{\sim}{\to} \mathrm{Hom}(\Delta[1], X) \twoheadrightarrow X \times X $$ Where the first map sends each simplex in $X$ to the constant homotopy at that simplex and it is a weak equivalence.

Is $X \to \mathrm{Hom}(\Delta[k], X)$ a model for a higher path object? That is,

  1. Is it true that the map $X \to \mathrm{Hom}(\Delta[k], X)$ sending each simplex in $X$ to the constant $k$-homotopy at that simplex is a weak equivalence? If so, does it additionally admit a homotopy inverse?

  2. Does this provide a fiber replacement for the higher diagonal $X \to X^{k+1}$?

I am mainly interested in an answer to 1, as 2 would likely follow immediately.

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    $\begingroup$ 1. Yes, because $\Delta[k]$ is contractible. 2. Yes, at least if $X$ is a Kan complex, because the inclusion of vertices in $\Delta[k]$ is a cofibration. $\endgroup$
    – Fernando Muro
    Commented May 10 at 15:16
  • $\begingroup$ @FernandoMuro I suspected this must be the reason but I don't see how to use the fact that $\Delta[k]$ is contractible to prove this. Do you know of a reference? $\endgroup$
    – SetR
    Commented May 10 at 15:31
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    $\begingroup$ The map $\Delta[k]\rightarrow *$ is a homotopy equivalence. $\endgroup$
    – Fernando Muro
    Commented May 10 at 15:52

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