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Question: Let $p : E \to B$ be an exponentiable functor of $\infty$-categories. Suppose that for every $b \in B$, the geometric realization of the fiber $|p^{-1}(b)|$ is contractible. Then does $p$ induce an equivalence of geometric realizations $|p| : |E| \to |B|$?

Notes:

  • If $p$ is a locally cartesian fibration or a locally cocartesian fibration, then the answer is yes by Quillen's Theorem A plus the fact that the fibers are reflective in the lax slices (or dually).

  • Since exponentiable fibrations are a common generalization of cartesian and cocartesian fibrations, one might hope for an affirmative answer here.

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The answer is no in general. A counterexample is the codiagonal functor $\Delta^1 \cup_{\partial\Delta^1} \Delta^1 \to \Delta^1$. This is an exponentiable fibration (as is every functor with codomain $\Delta^1$) with contractible fibres, but its domain has geometric realization $S^1$ and its codomain is contractible.

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    $\begingroup$ Sorry when this is obvious, but why is every morphism with codomain $\Delta^1$ exponentiable? $\endgroup$ Commented Oct 22, 2023 at 12:51
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    $\begingroup$ @MartinBrandenburg This follows from the intrinsic characterisation of exponentiable functors (e.g. as "flat inner fibrations" of Appendix B.3 in Lurie's Higher Algebra) which quantifies over factorizations of morphisms in the codomain. Alternatively it follows from Joyal's theorem (Theorem 7.9 in these lecture notes) that the slice category $\mathbf{sSet}/\Delta^1$ equipped with the model structure induced from the model structure for quasi-categories is cartesian closed as a model category. $\endgroup$ Commented Oct 22, 2023 at 18:25

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