Composition in quasi-categories

Question

Let $C$ be a quasi-category. Then $C^{\Delta^2}\to C^{\Lambda_1^2}$ is a trivial fibration and we may choose a section $s$. The map $C^{\Lambda_1^2}\xrightarrow s C^{\Delta^2} \to C^{\{0,2\}}$ defines a composition on $C$. Of course, the face map also induces a map $C^{\Delta^2}\to C^{\{0,2\}}$, so by precomposing the first map with $C^{\Delta^2}\to C^{\Lambda_1^2}$, we get two vertices of the quasi-category $\operatorname{Fun}(C^{\Delta^2},C^{\{0,2\}})$. I would like to find a natural equivalence between these two functors; i.e. an equivalence $\Delta^1\to\operatorname{Fun}(C^{\Delta^2},C^{\{0,2\}})$ connecting these two vertices. I know this can be done because it’s implicit in the proof of Lemma 2.3.4 of Land’s book. (I’m aware that the sections of $C^{\Delta^2}\to C^{\Lambda_1^2}$ assemble into a contractible Kan complex, but the face map in general doesn’t factor via $C^{\Delta^2}\to C^{\Lambda_1^2}$.)

Answer 1

I will write $[X, Y]$ for the exponential object $Y^X$.

As you say, for a quasicategory $C$, the inclusion $\Lambda^2_1 \hookrightarrow \Delta^2$ induces a trivial Kan fibration $p : [\Delta^2, C] \to [\Lambda^2_1, C]$. Let $s : [\Lambda^2_1, C] \to [\Delta^2, C]$ be any section of $p : [\Delta^2, C] \to [\Lambda^2_1, C]$. Let $d_1 : [\Delta^2, C] \to [\Delta^1, C]$ be the morphism induced by the coface $\delta^1 : \Delta^1 \to \Delta^2$. Your question is, are $d_1$ and $d_1 \circ s \circ p$ equivalent? The answer is yes.

It suffices to show that $\textrm{id}_{[\Delta^2, C]}$ and $s \circ p$ are equivalent. Consider the following commutative diagram, $$\require{AMScd} \begin{CD} [\Delta^2, C] \amalg [\Delta^2, C] @>{\partial h}>> [\Delta^2, C] \\ @V{\iota}VV @VV{p}V \\ [\Delta^2, C] \times I @>>{p \circ \pi_1}> [\Lambda^2_1, C] \end{CD}$$ where $\partial h : [\Delta^2, C] \amalg [\Delta^2, C] \to [\Delta^2, C]$ is defined by $\textrm{id}_{[\Delta^2, C]}$ on one half and $s \circ p$ on the other half, $I$ is any contractible Kan complex with two points, and $\iota : [\Delta^2, C] \amalg [\Delta^2, C] \to [\Delta^2, C] \times I$ is the obvious inclusion. (The diagram does commute, because $p = p \circ s \circ p$.) Since $p : [\Delta^2, C] \to [\Lambda^2_1, C]$ is a trivial Kan fibration, it has the rlp wrt $\iota : [\Delta^2, C] \amalg [\Delta^2, C] \to [\Delta^2, C] \times I$, so we get a morphism $h : [\Delta^2, C] \times I \to [\Delta^2, C]$ such that $p \circ h = p \circ \pi_1$ and (more crucially) $h \circ \iota = \partial h$. Rearranging things, we get a morphism $I \to [[\Delta^2, C], [\Delta^2, C]]$ exhibiting an equivalence between $\textrm{id}_{[\Delta^2, C]}$ and $s \circ p$.

Actually, there was nothing special about $p : [\Delta^2, C] \to [\Lambda^2_1, C]$ needed other than the fact that it is a trivial fibration, and there was nothing special about $I$ needed other than the fact that it has (at least) two points. This argument can easily be adapted to show that, in a general model category, if $p : E \to B$ is a trivial fibration and $s : B \to E$ is a section of $p : E \to B$, then $s \circ p$ is homotopic to $\textrm{id}_E$ fibrewise over $p : E \to B$. (Slogan: a trivial fibration has contractible fibres.)