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.



A characterization of maps that are homotopic relative to $A$ over $S$
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](https://arxiv.org/pdf/1312.4327.pdf). 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 $X^B \to X^A \times\_{S^A} S^B$.


