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