morphisms of simplicial sets as simplicial sets

Question

I'm preparing a talk about $\infty$-categories for a research seminar, using Luries

http://www.math.harvard.edu/~lurie/papers/croppedtopoi.pdf

as the main source. Here Lurie talks about the sets of morphisms between simplicial sets, like they are simplicial sets themselves (e.g. saying that $\hom_{Set_\Delta}(K,\mathcal{C})$ is an $\infty$-category, if $\mathcal{C}$ is an $\infty$-category.) But as far as I'm aware, he never gives an explicit description of how those sets are made into simplicial sets, and I can't find any source on this. So I want to ask, if someone knows, how this is made possible.

My approach is the following:

Define $\hom_{Set_\Delta}(S,T)_n:=\{f_n:S_n\rightarrow T_n~|~f\in\hom_{Set_\Delta}(S,T)\}$, so that the $n$-simplices are the $n$-th components of natural transformations between $S$ and $T$.

Now my problem is, how I should define the face- and degeneracy maps, since it seems impossible to me to compose in a canonical way with those maps, that come equipped with $S$ and $T$.

Answer 1

You can find this in Goerss-Jardine (§I.5, ``function complexes''), and here's an explanation of how to come up with the right definition.

First of all, we have finite products in $\mathbf{sSet}$: two simplicial sets $X, Y$ are presheaves $\Delta^\circ \to \mathbf{Set}$, so $X\times Y$ is defined as the product in the category of presheaves (``pointwise'': it is given by the product in $\mathbf{Set}$).

Now we would really like $\mathbf{sSet}$ to be cartesian closed: we want to define simplicial sets $\underline{\operatorname{Hom}} (Y,Z)$, in a way that $-\times Y$ is left adjoint to $\underline{\operatorname{Hom}} (Y, -)$. Note that if the adjunction holds, then the $n$-simplices of $\underline{\operatorname{Hom}} (Y,Z)$ must be given by $$\underline{\operatorname{Hom}} (Y,Z)_n \cong \operatorname{Hom}_\mathbf{sSet} (\Delta^n, \underline{\operatorname{Hom}} (Y,Z)) \cong \operatorname{Hom}_\mathbf{sSet} (\Delta^n\times Y, Z)$$ (the first isomorphism is just Yoneda).

So for every $n = 0,1,2,\ldots$ we are forced to define $$\underline{\operatorname{Hom}} (Y,Z)_n = \operatorname{Hom}_\mathbf{sSet} (\Delta^n\times Y, Z).$$ We need to make this into a simplicial set, i.e. $\underline{n} \rightsquigarrow \underline{\operatorname{Hom}} (Y,Z)_n$ should be a functor $\Delta^\circ \to \mathbf{Set}$. But this is obvious, since it is a composition of the following three functors:

• A covariant functor $\Delta \to \mathbf{sSet}$ which sends $\underline{n}$ to $\Delta^n$, and a morphism $\underline{n} \to \underline{m}$ induces a morphism $\colon \Delta^n \to \Delta^m$.

• A covariant functor $-\times Y\colon \mathbf{sSet}\to\mathbf{sSet}$.

• A contravariant functor $\operatorname{Hom}_\mathbf{sSet} (-, Z)\colon \mathbf{sSet}^\circ \to \mathbf{sSet}$.

Answer 2

The definition is rather $$\hom_{Set_\Delta}(S,T)_n:=\operatorname{Hom}_{Set_\Delta}(\Delta^n\times S,T)$$ (here to avoid confusion I write Hom for the ordinary Hom-set and hom for the Hom-simplicial set we're defining). This is a functor $\Delta^{op}\to Set$, since $[n]\mapsto\Delta^n$ is a functor $\Delta\to Set_\Delta$. The usual morphism set is then just the $0$-simplices of this simplicial set, since $S\times\Delta^0\cong S$.

To see why this definition makes sense, note that you would hope that a map $\Delta^n\to \hom_{Set_\Delta}(S,T)$ is equivalent to a map $\Delta^n\times S\to T$. More generally, you can show there is a natural isomorphism between maps $R\to \hom_{Set_\Delta}(S,T)$ and maps $R\times S\to T$ for any $R$, so these hom-objects make $Set_\Delta$ cartesian closed.