2
$\begingroup$

Basically my question is:

Does the Grothendieck construction make sense for (some?) functors valued in $\mathrm{Ho}\,(\mathbf{Cat})$?

By $\mathrm{Ho}\,(\mathbf{Cat})$, I mean the category whose objects are small categories and morphisms are natural isomorphism classes of functors. Otherwise said, this is what Bénabou would call the classifying category of the bicategory $\mathbf{Cat}$ in Introduction to bicategories (definiton 7.2).

To frame my question in a slightly more formal setting, remark that for any category $\mathcal A$, a pseudofunctor $F : \mathcal A \to \mathbf{Cat}$ induces a functor $\bar F : \mathcal A \to \operatorname{Ho}\,(\mathbf{Cat})$. Conversely, given a functor $G : \mathcal A \to \operatorname{Ho}\,(\mathbf{Cat})$, is it always possible to construct a pseudofunctor $\mathbb G : \mathcal A \to \mathbf{Cat}$ such that $\bar{\mathbb G} = G$ ?

I suspect this is false, as it would mean we can magically find coherence cells from an "incoherent" pseudofunctor, but I'm not sure how to craft a counter example. If it is indeed false, is there nice conditions on $G$ to make it true?

(If it makes any difference, I am actually more interested in the case where $\mathbf{Cat}$ is replaced by $\mathbf{Adj}$, whose objects are categories and morphisms are adjunctions going in the direction of the left adjoint.)

Edit. Let me add a little more context. Consider a category $\mathcal A$ with a distinguished class of morsphism $W$ such that the Gabriel-Zisman localization $\pi:\mathcal A \to W^{-1}\mathcal A$ exists (in the same universe). This settings enjoys the following property: any functor $F:\mathcal A \to \mathcal B$ such that $F(W)$ lands in the isosmorphisms of $\mathcal B$ can be factored (uniquely) through the localization $\pi$. Now replacing $\mathcal B$ by a bicategory $\mathcal K$ and accordingly replacing "isomorphisms" by "equivalences" in the previous statement, one can wonder for which pseudofunctors $F:\mathcal A \to \mathcal K$ such a factorization still (uniquely) exists. A naive approach is to consider the induced functor $\bar F: \mathcal A \to \underline{\operatorname C}\mathcal K$ valued in the classifying category of $\mathcal K$, defined as the category with same objects as $\mathcal K$ and with $\underline{\operatorname C}\mathcal K \,(A,B)$ the set of 1-cells $A\to B$ in $\mathcal K$ up to invertible 2-cells. This functor sends $W$ to isos, so there is a factorization $\bar F = G \pi$ for some (unique) $G : W^{-1}\mathcal A \to \underline{\operatorname C}\mathcal K$. What conditions are necessary on $F$ so that $G = \bar {\mathbb G}$ for some pseudofunctor $\mathbb G : W^{-1}\mathcal A \to \mathcal K$. In the above, I was concentrating on the case $\mathcal K = \mathbf{Cat}$.

$\endgroup$
  • 1
    $\begingroup$ Well, what if $\mathcal A=\mathrm{Ho}({\bf Cat})$ and $G=id$? So, we're basically looking for a right inverse for ${\bf Cat}\to \mathrm{Ho}({\bf Cat})$. $\endgroup$ – Berci Feb 26 '18 at 23:21
  • $\begingroup$ @Berci Well a pseudofunctor that acts like a right inverse. Any easy way to see this is not possible? (I was actually in a framework with $\mathcal A$ small, but an answer to this question would be a good start.) $\endgroup$ – Pece Feb 27 '18 at 12:24
  • $\begingroup$ Are you sure you want $\overline{\mathbb{G}} = G$, or is it enough to have $\overline{\mathbb{G}} \cong G$? $\endgroup$ – Hurkyl Feb 27 '18 at 18:04
  • $\begingroup$ @Hurkyl I guess $\cong$ is a good start. $\endgroup$ – Pece Feb 28 '18 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.