Do Homotopy limits commute with right Quillen functors

292 Views Asked by At

In particular I'm interested in this situation. Let $sSet^{2}$ denote the category of bisimplicial sets with diagonal model structure ($ F: X \rightarrow Y$ is a weak equivalence of bisimplicial sets if $ dF: dX \rightarrow dY$ is a weak equivalence of simplicial sets and cofibrations are monomorphisms) and $sSet$ be the category of simplicial sets with usual model structure. Then we know $d$ (the diagonal functor) is a right Quillen functor. Note that diagonal $dX$ of a bisimplical set $X$ is a simplicial set with $(dX)_{n}$ = $X_{n,n}$.

Now following is the proof of the claim $\forall X \in {(sSet^2)}^I $ (diagram category of bisimplicial sets) $$ d(holimX)\simeq (holim (dX)).$$

We denote the induced diagonal functor on diagram category (which is again a right Quillen functor) by $d$ as well.

Proof: Let $X'$ be fibrant replacement of $X$ in the injective model structure on ${sSet^2}^I$. Hence the homotopy limit $holimX$ is weakly equivalent to limit $limX'$ . Since this is a diagonal weak equivalence we have, $$ d(holimX)\simeq d(limX').$$ Furthermore, $d$ is a right adjoint, hence the R.H.S. above is weakly equivalent to $lim(dX')$. Moreover $d$ is right Quillen functor hence it preserves fibrant objects, so $dX'$ is fibrant. Again the limit $lim(dX')$ is weakly equivalent to the homotopy limit $holim(dX)$.

I have two questions

Is the statement of the claim correct?

If yes , then is this proof correct?