Slice and Internal Hom; On the Definition of a Map $\underline{\operatorname{Hom}}(A,X/x)\to \underline{\operatorname{Hom}}(A,X)/x$

Question

Recall that the join of two simplicial sets $X,Y$ are defined by $$(X\ast Y)_n=X_n\amalg X_{n-1}\times Y_0\amalg\cdots\amalg X_{0}\times Y_{n-1}\amalg Y_n,$$ with face and degeneracy maps defined appropriately. The join functor $-\ast T:\mathsf{sSet}\to T/\mathsf{sSet}$ can be shown to preserve colimits, and hence has a right adjoint which we denote by $(t:T\to X) \mapsto X/t$. In Proposition 4.2.12 of the book Higher Categories and Homotopical Algebra by Denis-Charles Cisinski, it is noted that given a simplicial set $A$, an $\infty$-category $X$, and its object $x$, there is a canonical map

$$\underline{\operatorname{Hom}}(A,X/x)\to \underline{\operatorname{Hom}}(A,X)/x$$

which the author claims to be an equivalence of $\infty$-categories. (Here $\underline{\operatorname{Hom}}$ is the internal hom in $\mathsf{sSet}$.) I understand the idea of what this means intuitively by doing everything in $\mathsf{Cat} $ instead of $\mathsf{sSet}$, but sadly, I cannot seem to figure out how the above map is defined in the case of $\mathsf{sSet}.$ Could someone enlighten me?

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Here are some thoughts, which led nowhere:

• Of course, by adjunction, we can also specify the above map by a pointed map $$\underline{\operatorname{Hom}}(A,X/x)\ast \Delta^0\to\underline{\operatorname{Hom}}(A,X),$$ but I don't have an obvious guess for this map either.

• An $n$-simplex of $\underline{\operatorname{Hom}}(A,X/x)$ is a map $\Delta^n\times A\to X/x$, while an $n$-simplex of $\underline{\operatorname{Hom}}(A,X)/x$ is a pointed map $\Delta^n\ast\Delta^0\to \underline{\operatorname{Hom}}(A,X)$. The latter corresponds to a map $\Delta^{n+1}\times A\to X$ which restricts on $\{n+1 \}\times A$ to the constant map at $x$. So what we must come up with is a way to assign to each map $$\Delta^n\times A\to X/x$$ a map $$\Delta^{n+1}\times A\to X$$ whose restriction to $\{n+1 \}\times A$ to the constant map at $x$. But I still don't see how one achieves this.

Answer 1

I think I've figured this out. For simplicial sets $A,B$ and a nonnegative integer $n\geq0$, there is a map $$(A\ast \Delta^0)\times B \to (A\times B)\ast\Delta^0$$ whose degree $k$-part (where $k\geq 0$ arbitrary) $(\coprod_{i=0}^kA_i\amalg \ast)\times B_k\to(\coprod_{i=0}^k A_i\times B_i)\amalg \ast$ maps:

• the summand $A_i\times B_k$ to the summand $A_i\times B_i$ by $(a,b)\mapsto(a,b\vert_{[i]})$; and
• the summand $\ast \times B_k$ to the summand $\ast$.

The definition seems quite arbitrary, but in fact it is quite canonical if one does everything in $\mathsf{Cat}$ instead (so that we consider a certain functor $([n]\ast[0])\times C\to([n]\times C)\ast\Delta^0$ for a small category $C$). The desired map $$\underline{\operatorname{Hom}}(A,X/x)\to \underline{\operatorname{Hom}}(A,X)/x$$ is then obtained as follows: An $n$-simplex of the left hand side is a pointed map $\sigma:(\Delta^n\times A)\ast \Delta^0\to X$. The composition $$\kappa_\sigma:(\Delta^n \ast\Delta^0)\times A\to (\Delta^n\times A)\ast\Delta^0\xrightarrow{\sigma}X$$ represents an $n$-simplex of $\underline{\operatorname{Hom}}(A,X)/x$. The association $\sigma\mapsto\kappa_\sigma $ gives the desired map.