# How does this statement reduces to the Yoneda lemma?

Context: I am reading the following page in the nLab, which is about simplicial presheaves, i.e Functors $$\mathscr{C}^{\mathrm{op}} \to [\mathbf{\Delta}^{\mathrm{op}}, \mathbf{Set}]$$. Equivalently, they are functors $$\mathbf{\Delta}^{op} \to [\mathscr{C}^{op}, \mathbf{Set}]$$. I will denote this functor category as $$\mathrm{sPre}(\mathscr{C})$$. This has a model structure where weak equivalences are object-wise weak equivalences of simplicial sets. One has an embedding $$\mathscr{C} \to \mathrm{sPre}(\mathscr{C})$$, which sends an object $$X$$ to $$Y(X) = \left(U \mapsto ([n] \mapsto \mathrm{Hom}_{\mathscr{C}}(U,X))\right)$$. Denote by $$[A,B]$$ the internal hom of this category, when it is viewed as enriched over the category of simplicial sets.
Given $$X$$ a simplicial presheaf, denote by $$X_n$$ the presheaf $$X_n$$ viewed as a simplicial presheaf. One has $$X \simeq \mathrm{Hocolim}_n X_n$$, and given $$A$$: \begin{align*} [X,A] &\simeq [\mathrm{Hocolim}X_n, A]\\ &\simeq \mathrm{Holim}_n[X_n, A] \end{align*} Each presheaf $$X_n$$ is a colimit of representables, say $$X_n = \mathrm{colim}Y(U_{i,n})$$, so that \begin{align*} [X,A] &\simeq \mathrm{Holim}_n\lim_i[Y(U_{i,n}), A] \end{align*}

Problem; Then, the nlab claims that by the Yoneda lemma, one has $$[X,A] \simeq \mathrm{Holim}_n\lim A(U_{i,n})$$. I guess the underlying claim is that $$[Y(A), B] = B(A)$$ for any simplicial presheaf $$B$$. But given that we are working with mapping spaces, and that $$Y$$ is not the "usual" Yoneda embedding, I do not know how to apply the Yoneda lemma in this situation to get the claim.

I tried to read a bit on the enriched Yoneda lemma, but I did not manage to relate the "enriched" yoneda embedding to the functor $$Y$$ we are given here.

So my question is: how to apply the Yoneda lemma in this setting to get the last claim, namely $$[Y(A), B] = B(A)$$?

• Your representable presheaves have discrete values, and the subcategory of $sSet$ consisting of discrete simplicial sets is essentially $Set$. Now given that a map $X\to Y$ with $X$ discrete is the same as a map $X\to Y_0$, you can probably just apply the usual Yoneda. I did not check the details, but it should work. You should get something like $[Y(A), B]_n = [Y(A), B_n]$ with the obvious abuse of notation – Maxime Ramzi Oct 27 '19 at 17:29

I managed to work it out, it is essentially what Max said in his comment, although with some extra computational steps, namely: \begin{align*} [Y(A), B]_n = \mathrm{Hom}_{\mathrm{sPre}(\mathscr{C})}(Y(A) \otimes \Delta^n, B) \end{align*} An object of the later set is a natural transform $$Y(A) \otimes \Delta^n \to B$$ so for every $$X \in \mathscr{C}$$, we are given a map of simplicial sets $$\mathrm{const}(\mathrm{Hom}(X,A)) \times \Delta^n \to B(X)$$, which corresponds to a map of simplicial sets $$\mathrm{const}(\mathrm{Hom}(X,A)) \to \mathrm{Map}_{\mathrm{sSet}}(\Delta^n, B(X))$$, since $$\mathrm{const}(\mathrm{Hom}(X,A))$$ is discrete, it corresponds to a map $$\mathrm{Hom}(X,A) \to \mathrm{Map}_{\mathrm{sSet}}(\Delta^n, B(X))_0 = \mathrm{Hom}_{\mathrm{Set}}(\Delta^n, B(X))$$ (Kerodon, tag 00FV).
But by Yoneda, $$\mathrm{Hom}_{\mathrm{sSet}}(\Delta^n, B(X)) = B_n(X)$$. So we have $$\mathrm{Hom}(X,A) \to B_n(X)$$, and we have naturality in $$X$$ because we used adjunctions all alongs, that are natural in $$X$$. So by Yoneda again, we can identify $$[Y(A),B]_n$$ with $$B_n(A)$$, as desired.
It should be checked as well that this identification behaves well with respect to $$n$$, i.e that the faces and degeneraties of the simplicial sets are identified as well, but I'll leave that under the rug for now, if there is an elegant way to do it, I'm curious to know it.