Factorization of a map from a contractible Kan complex through a Kan complex

Question

Suppose we are given a contractible Kan complex $S$ and a map of simplicial set $f : S \rightarrow T$. Under what conditions can we say that $f$ factors through the largest Kan complex $Z$ contained in $T$?

I am asking this question because it pops up in Proposition 2.2.5.7 of Higher Topos Theory. In the proof we are given two maps $f,g : K \rightarrow \mathcal{C}$ where $K$ is an arbritary simplicial set and $\mathcal{C}$ an $\infty$-category. We also have a a contractible Kan complex $S$ and a map $h : S \rightarrow \mathcal{C}^K$ such that $h(x) = f, h(y) = g$ where $x,y$ are two distinct vertices of $S$. Lurie claims that the map $h$ then factors through the largest Kan complex $Z$ of $\mathcal{C}^K$ and I do not see why.

Answer 1

First note that $\mathcal C^K$ is again an $\infty$-category. Then this follows from the fact that passing from an $\infty$-category to the largest Kan complex contained in it is a functor right adjoint to the inclusion of Kan complexes into all $\infty$-categories. This is proved in Prop 1.2.5.3 of HTT.

That is, adjointness tells us more generally that if $S$ is a Kan complex and $X$ is an $\infty$-category, then any map $S \to X$ factors through the largest Kan complex contained in $X$. (The ordinary categorical counterpart of this statement is that if $S$ is a groupoid and $X$ is a category, then any functor $S \to X$ factors through the maximal subgroupoid of $X$, i.e. the groupoid of isomorphisms in $X$. In the same way, the 1.2.5.3 says that the largest Kan complex contained in an $\infty$-category $X$ is given by throwing out all 1-cells which are not equivalences.)