7
$\begingroup$

Let $X,Y$ be two $(\infty,2)$-categories, viewed as two fibrant objects in $\mathrm{Fun}(\Delta^{op},\mathrm{Set}_\Delta)$ with the complete Segal model structure (one uses the Joyal model structure on $\mathrm{Set}_\Delta$ to define this). It has been proved that the $(\infty,1)$-category 2-$\mathrm{Cat}$ of $(\infty,2)$-categories is Cartesian closed. The category of bisimplicial sets $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta)$ is also Cartesian closed.

My question is:

Does $Y^X$ in 2-$\mathrm{Cat}$ coincide with $Y^X$ in $\mathrm{Fun}(\Delta^{op}, \mathrm{Set}_\Delta)$, i.e. is the latter one fibrant?

If not, is there a reference about the relation between the two (more specific than saying the first $Y^X$ is a fibrant replacement of the latter $Y^X$)?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Yes because the model structure on Fun(Δ^op, sSet) is a cartesian model structure. This follows from the fact that the Joyal model structure is cartesian, and §10, §11 in Rezk's paper “A model for the homotopy theory of homotopy theory”.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks Dmitri! I am not very familiar with model categories. Does Cartesian model structure implies something like $Y^X$ is fibrant, when $Y$ is fibrant and $X$ is cofibrant? $\endgroup$
    – Xin Jin
    Jul 23, 2017 at 5:31
  • $\begingroup$ Yes, internal hom is a right Quillen functor in a cartesian model structure, which implies your claim. $\endgroup$
    – Dmitri Pavlov
    Jul 23, 2017 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.