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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 $Y^X$)?

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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”.

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  • $\begingroup$ 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 '17 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 '17 at 14:29

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