$\newcommand{\cosimp}[1]{{#1}^\bullet} \newcommand{\sSet}{\mathsf{sSet}} \newcommand{\Set}{\mathsf{Set}}$ [I realize that the post is imposing.However, all the content of the problem is in the gray area. Everything else is my attemp to solve the problem.]
I'm asked to prove the following, where $\Delta$ is the simplex category (finite ordinals and non decreasing maps), $[k]$ denotes the ordinal $\{0<1<\dots<k\}$, and $\Delta[k] = \hom_\Delta(-,[k])$.
Given a cosimplicial simplicial set $\cosimp X$, show that the equalizers (in the category $\sSet$ of simplicial sets) of $$ \prod_{n=0,1,2} {(X^n)}^{\Delta[n]} \rightrightarrows \prod_{\substack{[m] \to [n] \\ n,m=0,1,2}} {(X^n)}^{\Delta[m]} $$ and of $$ \prod_{n=0,1,2} {(X^n)}^{\Delta[n]} \rightrightarrows \prod_{\substack{[m] \to [n] \\ n=0,1,2 \\ m=0,1}} {(X^n)}^{\Delta[m]} $$ are isomorphic.
[Where the arrows are the obvious ones when writing $((X^k)^{\Delta[\ell]})_r$ as $\hom_{\sSet}(\Delta[\ell] \times \Delta[r],X^k)$ (with the help of Yoneda).]
That is, one can throw away arrows with domain $[2]$ in the product on the right.
This problem arose in the study of totalization and 2-truncated totalization of cosimplicial simplicial sets and cosimplicial simplicial groupoids. More precisely, I came into this problem when trying to write the details of Hollander's article A homotopy theory for stacks (theorem 2.9).
Actually the 2-truncated totalization $\operatorname{Tot}_2(\cosimp X)$ is this common equalizer, but one can explicit it more easily when ditching the index $m=2$.
I really do not know how to proceed. It seems that the ordinal $2$ is not that important and that I think it can be done with a fixed $k \in \mathbb N$ (in this case, we make the same statement but dicthing the arrows with domain $[k]$). But I can be wrong about that...
What I did is the following : as equalizers (actually all limits) of $\sSet$ are computed degreewise, we are left to show the (natural in $r$) isomorphism for every degree $r \in \mathbb N$. Then we know how to compute equalizers in $\Set$ : it is the subset of the domain on which the two arrows agree. One inclusion is then trivial : the equalizer of $$ \prod_{n=0,1,2} \hom_{\sSet}(\Delta[n] \times \Delta[r],X^n) \rightrightarrows \prod_{\substack{[m] \to [n] \\ n,m=0,1,2}} \hom_{\sSet}(\Delta[m] \times \Delta[r],X^n) $$ is clearly a subset of the equalizer of $$ \prod_{n=0,1,2} \hom_{\sSet}(\Delta[n] \times \Delta[r],X^n) \rightrightarrows \prod_{\substack{[m] \to [n] \\ n=0,1,2 \\ m=0,1}} \hom_{\sSet}(\Delta[m] \times \Delta[r],X^n). $$ The converse, I can not prove. As I see it, I must prove that if $(\varphi^0,\varphi^1,\varphi^2)$ are maps of simplicial sets such that $(\varphi^0,\varphi^1)$ forms a map of $1$-truncated cosimplicial simplicial sets, then automatically $(\varphi^0,\varphi^1,\varphi^2)$ is a map of $2$-truncated cosimplicial simplicial sets. But I can't manage to do it. Any hint ?
Edit. Ok, so I tried a little harder and I have found something of a counterexample to the statement... I'm pretty sure the statement is correct though. But helping me spot the error in the following might get me in the right road to the proof.
I consider an element in the second equalizer $$(\varphi^0 \colon \Delta[0]\times\Delta[r] \to X^0, \varphi^1 \colon \Delta[1]\times\Delta[r] \to X^1, \varphi^2 \colon \Delta[2]\times\Delta[r] \to X^2).$$ It verifies for any $\vartheta \colon [m] \to [n],\, m\in\{0,1\},\, n\in\{0,1,2\}$, $$ \varphi^n \circ (\Delta[\vartheta], \mathbf 1_{\Delta[r]}) = \cosimp X(\vartheta) \circ \varphi^m . \tag{$\star$} $$
Now, I want $(\star)$ for $\vartheta \colon [m] \to [n]$ with $m$ posssibly equal to $2$. If I show that, I can conclude that my $(\varphi^i)_{i=0,1,2}$ is actually also in the first equalizer. Proving the statement. Of course, for $\vartheta$, I can restrict to the generating maps of $\Delta$ : I have to check $(\star)$ for $\vartheta = \mathbf 1_{[2]}$ (trivial), $s^0$ and $s^1$. Recall
$$ s^0 \colon [2] \to [1] : 0\mapsto 0, 1 \mapsto 0, 2 \mapsto 1 , \\
s^1 \colon [2] \to [1] : 0\mapsto 0, 1 \mapsto 1, 2 \mapsto 1 .$$
Let's concentrate on $s^1$ ($s^0$ is similar) and note $\cosimp Y$ for $\Delta[\bullet]\times\Delta[r]$. Observe that $s^1$ has a "right inverse"
$$ d^2 \colon [1] \to [2] : 0 \mapsto 0, 1\mapsto 1$$
such that $s^1 \circ d^2 = \mathbf 1_{[1]}$. So
$$ \varphi^1 \circ \cosimp Y(s^1) = \cosimp X(s^1) \circ \cosimp X(d^2) \circ \varphi^1 \circ \cosimp Y(s^1).$$
But now the term $\cosimp X(d^2) \circ \varphi^1$ is know to be $\varphi^2 \circ \cosimp Y(d^2)$ because the domain of $d^2$ is $[1]$ (so not $[2]$). I replace in the previous equation :
$$ \varphi^1 \circ \cosimp Y(s^1) = \cosimp X(s^1) \circ \varphi^2 \circ \cosimp Y(d^2) \circ \cosimp Y(s^1).$$
So the condition for $(\star)$ to hold is that $\cosimp Y(d^2 \circ s^1) = \mathbf 1_{Y^2}$.
[Edit 2 : my mistake is obviously here, $\cosimp X(s^1) \circ \varphi^2$ is a priori not invertible.]
But $\cosimp Y(d^2 \circ s^1)$ is the simplicial map defined in degree $\ell \in \mathbb N$ as $$ \hom_\Delta([\ell],[2]) \times \hom_\Delta([\ell],[r]) \to \hom_\Delta([\ell],[2]) \times \hom_\Delta([\ell],[r]) \colon (\mu, \nu) \mapsto (d^2 \circ s^1 \circ \mu,\nu).$$ For example, in degree $2$, $(\mathbf 1_{[2]}, \nu)$ is send to $(d^2 \circ s^1,\nu)$. So for $(\star)$ to be true, one must have $d^2 \circ s^1 = \mathbf 1_{[2]}$ which is false...
Anyone spotting the problem?