Firstly, a bit of notation. Let $C$ be a simplicial set. We define, for $x,y \in C$ vertices in $C$

$$Map(x,y) = \{x\}\times_{\Delta^{\{0\}}}Map_{sSet}(\Delta^1,C) \times_{\Delta^{\{1\}}} \{y\} $$

the sSet of arrows from x to y. You can imagine a $n$ simplex here as a cylinder with basis $\Delta^n$, with all $1_x$ on the left base and all $1_y$ on the right base. From this description it is evident that $Map(x,y) \simeq (C_{/y})_x$, the cones over y with all $1_x$ at the basis. Symmetrically, it is equivalent to $(C_{x/})_y$.

As the two sSet are fibers of a right and a left fibration, they are respectively left inner fibrant and right inner fibrant. Thus, the map sets are Kan complexes.

I am searching to generalize this to an arbitrary inner fibration $p:C \to D$, and show that $Map(x,y) \to Map(px,py)$ is a Kan fibration.


  • 1
    $\begingroup$ See this MSE question for some discussion of the result when $D$ is a point. $\endgroup$ – Tim Campion Jun 14 '19 at 17:49
  • $\begingroup$ Just to be clear, the comparison of the different models of mapping spaces is not "evident". It requires some proof. It's not excessively hard, but it's not a one-line argument either. $\endgroup$ – Denis Nardin Jun 14 '19 at 19:25
  • $\begingroup$ I mean, the two inverse functions can be explicitly described. One function takes a cylinder and output the cone taking just one vertex of the right base. The other takes a cone and extend it to a cylinder along identities, using the fact that identities are (co)cartesian. We have some equivalences in the middle because extension along (co)cartesian morphism just yield an equivalence. Isn't something like this? I know it is not a precise proof but close.. $\endgroup$ – Andrea Marino Jun 15 '19 at 13:32

This should be a comment, but I have not enough reputation: if $ C,D $ are $ (\infty,1) $-categories then your statement is proved in Lemma of Higher Topos Theory using the equivalence of mapping spaces given in Corollary (see Remark for the notation).

  • 1
    $\begingroup$ Note that Lurie uses a different model for the mapping spaces, though. $\endgroup$ – Tim Campion Jun 14 '19 at 17:37
  • $\begingroup$ Yeah sorry, I didn't write they are equivalent. I'll edit the post. $\endgroup$ – Alessandro Jun 14 '19 at 19:42
  • 2
    $\begingroup$ But just because the two mapping spaces are equivalent doesn't mean that if one is fibrant, so is the other. This is impicitly claimed in the statement of, but as pointed out by Zhen Lin in the MSE discussion I linked to, it doesn't appear that Lurie actually proves this model is fibrant. $\endgroup$ – Tim Campion Jun 15 '19 at 16:04
  • $\begingroup$ Ah now I see, you're right. Thanks for pointing that $\endgroup$ – Alessandro Jun 16 '19 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.