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I have encountered a problem in understanding Lurie's proof of the following fact:

"Given a left fibration between simplicial sets $q:X \to S$, there exists a functor $$ho(S) \to Ho(sSet)$$ which is defined as follows: a point $s \in S$ is sent to the fiber $X_s$ of $q$ over $s$, and an arrow $f:s \to s'$ in $S'$ is sent to the homotopy class of a lift in

enter image description here

restricted to $\{1\} \times X_s$, which will be denoted by $f_!:X_s \to X_{s'}$".

The author wants to characterize the map $$[f_!]\circ (\cdot): Ho(sSet)(\cdot, X_s) \to Ho(sSet)(\cdot, X_{s'})$$ so as to obtain the desired result (except for the independence from the homotopy class of $f$, which I guess is left to the reader).

The problem is I don't see why it holds that, given a $K\in sSet$ and arrows $\eta:K \to X_s, \ \eta':K \to X_{s'}$, $$\eta'\simeq f_! \circ \eta \iff \exists p:K \times \Delta[1] \to X$$ such that $p_{|K \times \{0\}}\simeq \eta, \ p_{|K \times \{1\}}\simeq \eta'$ and $$q \circ p= K \times \Delta[1] \to \Delta[1] \to S$$ where the last arrow is $f$.

Thanks in advance for any help, which will of course be highly appreciated.

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  • $\begingroup$ I want to comment that the pullback of a right horn along a left fibration is a right anodyne map. One may find this statement in Rezk's lecture notes ' Stuff about quasicategories'. But I think he/she does not provide a proof. $\endgroup$
    – user12580
    Aug 16, 2021 at 15:51

1 Answer 1

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Consider the diagram

enter image description here

where the top left horizontal arrow is given by $\eta$. Observe that homotopy classes of lifts of the map $X_s \times \Delta^1 \rightarrow S$ are the same as homotopy classes of lifts of the map $K \times \Delta^1 \rightarrow S$. This follows from the same argument showing that the homotopy class of $f_!$ only depends on $f$, together with the fact that the left hand square is a pullback. (I'll include the argument below for completeness).

Given that, we're done. A lift of $K \times \Delta^1 \rightarrow S$ is precisely the map $p$ as stated, and the above observation says the lift is homotopic to the composition $K \times \Delta^1 \rightarrow X_s \times \Delta^1 \rightarrow X$, which is the same as saying that $[\eta'] = [f_!] \circ [\eta]$.


Okay, now for the promised argument which says that homotopy classes of lifts in a diagram like

enter image description here

only depend on the map $f$. But such a lift is equivalent to a section of $X \times_S (Y \times \Delta^1)$ over $Y \times \Delta^1$ which is specified at $Y \times \{0\}$. This pullback may be obtained by first pulling back $X$ to $\Delta^1$, where the fibration trivializes up to homotopy. Thus $X \times_S(Y \times \Delta^1)$ is trivial up to homotopy so that sections are determined up to homotopy by maps $Y \times \Delta^1 \rightarrow X_s$. Since we've already specified the map on $Y \times \{0\}$, there is only one extension up to homotopy as $Y \times \{0\} \rightarrow Y \times \Delta^1$ is an equivalence.

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  • $\begingroup$ (Alternatively I can quote Corollary 2.1.2.9 of HTT if you don't like that argument at the end.) $\endgroup$ Mar 7, 2015 at 13:47
  • $\begingroup$ First of all thanks, I just have two questions: I like the argument at the end, but it seems to me that you are using the fact that the pullback of two homotopic maps along a left fibration gives two homotopic maps. Is it true also for left fibrations (I knew it for Kan fibrations)? $\endgroup$ Mar 7, 2015 at 14:31
  • $\begingroup$ Secondly, I agree that having a lift for $X_s \times \Delta[1] \to S$ gives me a lift for $K \times \Delta[1] \to S$, but how to obtain the converse in order to establish the equivalence you mentioned? $\endgroup$ Mar 7, 2015 at 14:33
  • $\begingroup$ For your second question: Applying the last argument for $Y = K$ and $Y = X_s$ shows that in both cases the homotopy class of a lift is determined by $f$, i.e. there is only one homotopy class of lift for $K \times \Delta^1 \rightarrow S$ so it must be the one obtained by composiiton. $\endgroup$ Mar 7, 2015 at 18:56
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    $\begingroup$ For the first question: I believe the exact thing I need is the aforementioned corollary or 2.1.2.10, I can try to rewrite this more clearly later. $\endgroup$ Mar 7, 2015 at 19:01

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