The value of the left adjoint of the diagonal functor $\delta^*:\mathsf{bisSet}\to\mathsf{sSet}$ at $\Delta^m\boxtimes \Delta^n$

Question

I'm currently reading the proof of Theorem 5.5.7 in Cisinski's book Higher Categories and Homotopical Algebra. There's one detail I don't understand, and I need someone's help.

Let us introduce some notations. We write $\mathsf{sSet}$ and $\mathsf{bisSet}$ for the categories of simplicial sets and bisimplicial sets, respectively. If $X$ and $Y$ are simplicial sets, then the external product $X\boxtimes Y$ is defined by $X\boxtimes Y_{m,n}=X_m\times Y_n$. There is the diagonal functor $\delta^*:\mathsf{bisSet}\to\mathsf{sSet}$ which takes a bisimplicial set $X$ to its diagonal $\delta^*(X)_n=X_{n,n}$, and this functor has a left adjoint $\delta_!$ and a right adjoint $\delta_\ast$.

The theorem concerns the diagonal model structure on $\mathsf{bisSet}$, whose weak equivalences are created by $\delta^*$ and whose cofibrations are the monomorphisms, as well as the fact that the pairs $(\delta_!, \delta^*)$ and $(\delta^*,\delta_\ast)$ are both Quillen equivalences. In the proof, the author claims that the counit $\delta_!\delta^*(X)\to X$ is a weak equivalence for $X$ representable, by using the equality

$$\delta_!\delta^*(\Delta^m\boxtimes\Delta^n)=\Delta^m\times \Delta^n\boxtimes \Delta^m\times \Delta^n.$$

It is this equality that I don't understand. Sure, we do have $\delta_!(\Delta^m)=\Delta^m\boxtimes \Delta^m$ (by the Yoneda lemma), so the above formula holds when $m$ or $n$ is $0$. But other than this very special case, I see no reason why the above formula holds. Can someone explain why the above formula is valid?

---

Here are some thoughts:

• By adjunction, a map $\Delta^m\times \Delta^n\to Y$ gives rise to a map $$\delta_!\delta^*(\Delta^m\boxtimes \Delta^n)\to Y.$$ But with the above formula, I don't see an obvious choice for such a map.

• Since $\delta_!(S)=\operatorname{colim}_{\Delta^k\to S}\Delta^k\boxtimes \Delta^k$, we also have the canonical map $\delta_!\delta^*(X)\to X\boxtimes X$. Maybe this map is an isomorphism in some special case, so let's consider this possibility. The map is epic iff for every pair of simplices $(x,y)\in X_k\times Y_l$, we can find some simplex $z\in X_p$ and maps $f:[p]\to[k]$ and $g:[q]\to[l]$ such that $f^*z=x$ and $g^*z=y$. Alas, I don't see why this is the case $X=\Delta^m\times \Delta^n$. Showing that the canonical map is monic seems even more daunting.

Answer 1

I think the claimed isomorphism is false. Specifically, I think we don't even have $\delta_! \delta^* (\Delta^1 \boxtimes \Delta^1) \cong (\Delta^1 \times \Delta^1) \boxtimes (\Delta^1 \times \Delta^1)$. (But maybe it doesn't matter for the actual claim at hand, which is that something is a weak equivalence.)

By analogy with ordinary simplicial sets, let us say that an $(m, n)$-cell of a bisimplicial set is degenerate if it is the image of $(m', n')$-cell under some bisimplicial operator, where $m' \le m$, $n' \le n$, but $(m', n') \ne (m, n)$. So, for example, a degenerate $(1, 1)$-cell could be the image of a $(0, 0)$-cell, a $(0, 1)$-cell, or a $(1, 0)$-cell. It is straightforward to see an $(m, n)$-cell of $X \boxtimes Y$ is non-degenerate if and only if the corresponding $m$-cell of $X$ and $n$-cell of $Y$ are both non-degenerate.

Now, consider $\Delta^1 \times \Delta^1$. We have a pushout diagram of the form below in the category of simplicial sets: $$\require{AMScd} \begin{CD} \Delta^1 @>>> \Delta^2 \\ @VVV @VVV \\ \Delta^2 @>>> \Delta^1 \times \Delta^1 \end{CD}$$ Since $\delta_!$ preserves colimits, and $\delta_! \Delta^k \cong \Delta^k \boxtimes \Delta^k$, we obtain a pushout diagram of the form below in the category of bisimplicial sets: $$\begin{CD} \Delta^1 \boxtimes \Delta^1 @>>> \Delta^2 \boxtimes \Delta^2 \\ @VVV @VVV \\ \Delta^2 \boxtimes \Delta^2 @>>> \delta_! (\Delta^1 \times \Delta^1) \end{CD}$$ Since all the arrows in the diagram are monomorphisms, we can straightforwardly compute the number of non-degenerate cells in $\delta_! (\Delta^1 \times \Delta^1)$: $$ 2 \times \begin{pmatrix} 9 & 9 & 3 \\ 9 & 9 & 3 \\ 3 & 3 & 1 \end{pmatrix} - \begin{pmatrix} 4 & 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 14 & 16 & 6 \\ 16 & 17 & 6 \\ 6 & 6 & 2 \end{pmatrix} $$ On the other hand, the number of non-degenerate cells of $(\Delta^1 \times \Delta^1) \boxtimes (\Delta^1 \times \Delta^1)$ is: $$\begin{pmatrix} 16 & 20 & 8 \\ 20 & 25 & 10 \\ 8 & 10 & 4 \end{pmatrix}$$ So there is no chance that $\delta_! (\Delta^1 \times \Delta^1) \to (\Delta^1 \times \Delta^1) \boxtimes (\Delta^1 \times \Delta^1)$ is an isomorphism.