Homotopy cofinality of $\Delta^{op}$ in $\Delta^{op}\times \Delta^{op}$ There is the usual diagonal inclusion $i:\Delta^{op}\to\Delta^{op}\times \Delta^{op}$ which is easily seen to be cofinal in the $1$-categorical sense, and so one can compute colimits on $\Delta^{op}\times \Delta^{op}$ by simply restricting to the diagonal. 
I've been told that this statement also held in a homotopical sense, and I'm trying to understand that, firstly in the context of model categories. 
So let $\mathscr C$ be a model category (it can be as nice as you want, if combinatorial helps for instance then I'm willing to assume that), and $X: \Delta^{op}\times \Delta^{op}\to \mathscr C$ a functor, I'm trying to understand why $\mathrm{hocolim}(X) = \mathrm{hocolim}(X\circ i)$
My idea was the following : use $\mathrm{colim} = \mathrm{colim}\circ i^*$ (where $i^*$ is precomposition by $i$) and derive that equality; because I was hoping that $i^*$ would be nice enough to not change much. 
My first guess was to use the Reedy model structure on $[\Delta^{op},\mathscr C]$ and $[\Delta^{op}\times \Delta^{op},\mathscr C]$ and show that $i^*$ preserves weak equivalences (this is obvious) and cofibrant objects; but this second point is not clear : when you compute matching objects you'd like $\Delta^{op}_+/[n]\to \Delta^{op}_+/[n]\times \Delta^{op}_+/[n]$ to be cofinal so that the colimit that defines the matching objects in both cases is the same, but that's not the case 
(where $\Delta^{op}_+ = (\Delta_-)^{op}$, $\Delta_-$ being the subcategory of surjective maps; and in this particular case, we take the convention that $\Delta^{op}_+/[n]$ does not contain $id_{[n]}$)
But that doesn't prove that $i^*$ doesn't preserve cofibrant objects (or indeed cofibrations). So that's my first question :

Does $i^*$ preserve cofibrant objects or cofibrations for the Reedy model structure ? For the projective model structure, provided that it exists ? (it does if e.g. $\mathscr C$ is cofibrantly generated and finite coproducts preserve cofibrations)

My second question is, if it's not, then how can one prove that $\mathrm{hocolim}$'s are preserved by $i^*$ ? 
 A: I was actually wrong in my computation of latching objects. If you look at $([r],[s])\overset{+}\to ([n],[n])$ in $\Delta^{op}\times \Delta^{op}$, this amounts to a pair of surjections $[n]\to [r], [n]\to [s]$, so now if you look at the image of $[n]\to [r]\times [s]$ it's isomorphic to some $[l] $ with surjections $[l]\to [r], [l]\to [s]$ that factor $[n]\to [l]\to [r]\times [s]$, so that $(\Delta^{op})_+/[n]$ is in fact cofinal in $(\Delta^{op})_+/[n]\times (\Delta^{op})_+/[n]$
So if $[n]\in\Delta$ and $X\in [\Delta^{op}\times \Delta^{op}, \mathscr C]$, then $L_{[n]}i^*X \cong L_{([n],[n])}X$ (where $L_rX$ denotes the $r$th latching object of $X$), so that $i^*$ preserves cofibrations and weak equivalences in the Reedy model structure. 
It follows that $i^*$ is left-Quillen (the right adjoint being the right Kan extension along $i$, which always exists as $\mathscr C$ is complete), and in fact $\mathbb Li^* = i^*$ because it preserves weak equivalences, so $\mathbb L(\mathrm{colim}\circ i^*) = \mathbb L\mathrm{colim} \circ i^*$, which is the desired claim (the equality follows because both functors are left Quillen (indeed it's clear that the "constant functor" functor $\mathscr C\to [R,\mathscr C]$ is right Quillen if $R$ is a Reedy category))
This is in the context of model categories, I'll just add that one can prove that this holds for $\infty$-categories as well : $\Delta^{op}\to \Delta^{op}\times\Delta^{op}$ is cofinal, and one can prove this using Quillen's theorem A, as is done in Lurie. This relies on a similar observation as above (but somehow "in the other direction") : given $([n],[m])\in\Delta^{op}\times \Delta^{op}$, one looks at $([n],[m])/\Delta^{op}$ and one needs to prove it's contractible. 
This is the same as $\Delta/([n],[m])$, and no one notices that there is an adjunction between this category and the category of subobjects of $[n]\times [m]$, but this category's nerve is precisely the barycentric division of $\Delta^n\times \Delta^m$, which is thus contractible. 
