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Given a category $\mathcal{C}$, by a category over $\mathcal{C}$, I mean a category $\mathcal{D}$ along with a functor $\pi_{\mathcal{D}}:\mathcal{D}\rightarrow \mathcal{C}$.

Consider the category $Cat/C$, of categories over $\mathcal{C}$. I am assuming the familiarity with the notion of a fibered category over $\mathcal{C}$ (if not, please see Definition 3.5 of Angelo Vistoli's Notes on Grothendieck topologies,fibered categoriesand descent theory).

Question is the following:

  • Is there a (interesting/non-trivial) model structure on $Cat/\mathcal{C}$, in which fibered categories are the fibrant objects?

Or, are the terms "fibered category" and "fibrations" just two unrelated terms that sound same?

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    $\begingroup$ I think the terminology represents "two unrelated terms that sound the same" since both are kinda inspired by fibrations in topological spaces (like, for every point in the base space there is a fiber above it). That said, the answer to your question could still be "yes" and I'm curious to see if anyone knows. $\endgroup$
    – David White
    Commented Nov 10, 2020 at 14:43
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    $\begingroup$ Please check arxiv.org/pdf/math/0110247.pdf (A homotopy theory for stacks by Sharon Hollander) $\endgroup$
    – Adittya Chaudhuri
    Commented Nov 10, 2020 at 14:48
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    $\begingroup$ Is any retract in $\mathbf{Cat}/C$ of a fibered category over $C$ a fibered category over $C$? $\endgroup$
    – Alexander Campbell
    Commented Nov 10, 2020 at 23:39
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    $\begingroup$ One concrete issue, related to @TimCampion's first comment, is that the "inclusion" of fibered categories into $\mathit{Cat}/C$ is not full: the "natural" choice of morphism between fibered categories is a functor that preserves cartesian arrows. This is not generally what happens for the fibrant objects in a model category. $\endgroup$
    – Mike Shulman
    Commented Nov 11, 2020 at 1:57
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    $\begingroup$ @AlexanderCampbell β€œIs any retract in π‚πšπ­/𝐢 of a fibered category over 𝐢 a fibered category over 𝐢” looks more than a comment and less than an answer :) Thanks, did not thought about it. I will think and respond. $\endgroup$
    – Praphulla Koushik
    Commented Nov 11, 2020 at 2:28

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