# Inner fibrations are Kan fibrations on Map sets

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.

Thanks!

• See this MSE question for some discussion of the result when $D$ is a point. Jun 14 '19 at 17:49
• 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. Jun 14 '19 at 19:25
• 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.. 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 2.4.4.1 of Higher Topos Theory using the equivalence of mapping spaces given in Corollary 4.2.1.8 (see Remark 1.2.2.5 for the notation).