0
$\begingroup$

Let $$\begin{array}{ccccccccc} A & \rightarrow & X \\ i\downarrow & & \downarrow p \\ B & \xrightarrow{v} & S \end{array} $$ be a commutative diagram of simplicial sets, with $p$ an inner fibration. Two solutions $f,g$ to this lifting problems are called homotopic relative to $A$ over $S$ provided that they are equivalent in the fiber of the map $$X^B\rightarrow X^A\times_{S^A}S^B.$$

Now, it is easy to see that this imply the existence of a map $\Delta^1\times B\rightarrow X$ satisfying certain obvious properties (spelled out, for instance, here: https://ncatlab.org/nlab/show/minimal+inner+fibration), amongst which is that the map $$F|_{\{b\}\times\Delta^1}\rightarrow X_{v(b)}$$ is an equivalence for every $b$.

How can you prove the converse? I found an argument for the case when $A=\emptyset$ and $S=*$, but not for the general case.

NB: Chances are it is proven at some point in HTT, but so far, I have only found this assertion without proof.

Improve this question
$\endgroup$
2
  • $\begingroup$ I followed the ncatlab link in your question. In order to match the exposition there, do you mean $F: B\times \Delta[1] \to X$ rather than $\Delta^1 \times X \to X$? In that ncatlab page, it already says the existence of $F$ is equivalent to the first meaning of homotopic relative to $A$ over $S$. $\endgroup$
    – David White
    Commented Mar 10, 2019 at 18:09
  • $\begingroup$ You're definitely right! I know that these are equivalent, but I can't figure out a proof for the general case. $\endgroup$
    – user09127
    Commented Mar 10, 2019 at 18:30

1 Answer 1

Reset to default
0
$\begingroup$

This fact is indeed standard. Because $\Delta^1 \times B$ is a cylinder object for $B$, the equivalent formulation (the hypothesis of the converse direction you're asking for) is just explicitly spelling out the homotopy that shows the two maps are equivalent in that fiber. A good place to learn about relative homotopy in a general category $C$ (including $C = sSet/S$) is Isaev's paper On Fibrant Objects in Model Categories. Specifically, Proposition 2.4 might help you (and, the picture later on that page might help you unpack what both notions mean). The argument you wrote down for $S = \ast$ will also work fiber by fiber to prove the general case. Isaev's paper will help you shift from thinking about $A = \emptyset$ to thinking about a general object $A$.

To be even clearer, using the order from Lurie's book and the ncatlab page:

  • The first two conditions on $F$ are saying it's a homotopy between $f$ and $g$.
  • The next condition is saying it's a homotopy over $S$.
  • The next condition is saying it's a homotopy relative to $A$.
  • The last condition is giving the equivalence in the fiber.

This information explicitly tells me I can deform $f$ to $g$ in the fiber of $X^B \to X^A \times_{S^A} S^B$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .