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Let $M$ be a simplicial model category, $M^o$ its full subcategory of bifibrant objects. The axioms of a simplicial model category guarantee that $M^o$ is enriched in Kan complexes. Thus the homotopy coherent nerve $X = \mathfrak{N}(M^o)$ is a quasicategory, which we think of as the underlying $\infty$-category of $M$.

Let $x,y \in X_0$, then $\text{Map}_X(x,y)$ is the pullback $\text{Fun}(\Delta^1,X) \times_{X \times X} \partial \Delta^1$ of simplicial sets. This is an $\infty$-groupoid i.e. Kan complex. It is well known that if $\underline{M}(x,y)$ denotes the simplicial mapping space in $M$, then there is a weak equivalence $$\underline{M}(x,y) \to \text{Map}_X(x,y)$$ Just by unravelling the definitions it is clear that $\underline{M}(x,y)_0 = \text{Map}_X(x,y)_0$, but how do I prove that the map is a weak equivalence? I have not seen a direct proof of this fact in the literature.

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    $\begingroup$ Okay found a path to this through Lurie: $\text{Map}_X(x,y) \simeq \text{Hom}^R_X(x,y)$ by Corollary 4.2.1.8 in Higher Topos Theory. By the discussion after Theorem 2.2.0.1 in HTT, $\text{Hom}^R_X(x,y) \simeq \mathfrak{C}(X)(x,y)$. Since $X = \mathfrak{N}(M^o)$, and by Theorem 2.2.0.1 $\mathfrak{C}\mathfrak{N}(M^o) \simeq M^o$. This is a weak equivalence of simplicial categories, which means that $\mathfrak{C} \mathfrak{N}(X)(x,y) \simeq \underline{M}(x,y)$. Thus $\text{Map}_X(x,y) \simeq \underline{M}(x,y)$. I am still gonna leave this question open however in the hopes of a simpler proof. $\endgroup$ Commented Apr 30, 2021 at 15:31
  • $\begingroup$ What you're asking about is substantially easier than 2.2.0.1 (it's easy to see that Hom spaces are compatible with the homotopy coherent nerve; the difficulty is to prove the analogous statement for the left adjoint). See kerodon.net/tag/01LA. $\endgroup$ Commented May 4, 2021 at 18:05
  • $\begingroup$ Ah, perfect, thank you! $\endgroup$ Commented May 4, 2021 at 20:01

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