Let $B$ be a weakly contractible category (that is, it has a weakly contractible nerve), and let $F:E\to B$ be a Grothendieck fibration. Suppose further that the ordinary fibers of the Grothendieck fibration, $F^{-1}(b)$ for each $b\in \mathrm{Ob}(B)$ are themselves weakly contractible. Does it follow that $E$ is weakly contractible?
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asked Jul 24, 2011 at 5:12
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6$\begingroup$ Does Quillen's theorem A (ncatlab.org/nlab/show/geometric+realization+of+categories) imply this? It does if the fibers $F^{-1}(b)$ are the same as the geometric realizations of the comma categories $F / b$. $\endgroup$– Evan JenkinsCommented Jul 24, 2011 at 5:27
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3$\begingroup$ I think that because the map is supposed to be a fibration $F^{-1}b$ should be weakly equivalent to $F/b$ so indeed. $\endgroup$– Torsten EkedahlCommented Jul 24, 2011 at 6:57
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$\begingroup$ To elaborate on Torsten's point (also see the reference in my answer): Any functor admitting an adjoint is a weak (Thomason) equivalence, and it is not hard to show that the property of being a prefibration (every map $a\to F(x)$ lifts to a weakly cartesian map $y\to x$ (weak cartesianness is a universal factorization only over identities) is equivalent to showing that the canonical functor from the fiber $F^{-1}(b)\to b\F$, the coslice category, admits a right adjoint. $\endgroup$– Harry GindiCommented Jul 24, 2011 at 8:09
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1 Answer
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This follows easily from proposition 2.1.10 of La théorie de l'homotopie de Grothendieck (Astérisque, 301) by G. Maltsiniotis.
The statement there is that given any functor $u:A\to B$ in $\mathrm{Cat}/C$ between (Grothendieck) prefibrations $A\to C$ and $B\to C$ whose fibers $u\times_C c:A\times_C c \to B\times_C c$ for all $c\in \mathrm{Ob}(C)$ are $W$-equivalences with respect to a fixed fundamental localizer $W$ on $Cat$, the functor $u$ is a colocal $W$-equivalence over $C$ (and therefore a $W$-equivalence).
The case in my question then follows by letting $W$ be the minimal fundamental localizer, by letting $C=B$, and letting the prefibration $B\to C$ be the identity map.
answered Jul 24, 2011 at 7:00