12
$\begingroup$

The category of elements of a functor $F:\mathcal C\to\mathsf{Set}$ can be obtained as the strict pullback in with the forgetful functor of pointed sets $\mathsf{Set_*}\to\mathsf{Set}$:

$$ \begin{array}{ccc} \int F & \xrightarrow{} & \mathsf{Set}_*\\ \downarrow & \ulcorner & \downarrow \\ \mathcal C & \xrightarrow{} & \mathsf{Set} \end{array} $$

We could also have taken a lax pullback (aka a comma square) with the functor $\ast\to\mathsf{Set}$: $$ \begin{array}{ccc} F\downarrow \ast & \xrightarrow{} & \ast\\ \downarrow & & \downarrow \\ \mathcal C & \xrightarrow{} & \mathsf{Set} \end{array} $$

Then $F\downarrow\ast = \int F$.

This feels very much like a homotopy pullback, where we replace the map by a fibration and then take the strict pullback. In fact, the forgetful functor $\mathsf{Set}_*\to\mathsf{Set}$ is a discrete opfibration. Also, while $\mathsf{Set}_*$ isn't equivalent to a point, it is ``contractible'' in the sense that there is a natural transformation $\ast\Rightarrow\text{id}_{\mathsf{Set}_*}$.

  • Can this remark be formalized? i.e./e.g. is there a model structure on $\mathsf{Cat}$ or $\mathsf{2Cat}$ where homotopy pullbacks are lax 2-pullbacks? What about pseudolimits?

  • The Grothendieck construction of a pseudofunctor $F:\mathcal{C}\to\mathsf{Cat}$ can be defined as an analogous strict 2-pullback in the 2-category of categories:

$$ \begin{array}{ccc} \int F & \xrightarrow{} & \mathsf{Cat}_{*,\mathcal l}\\ \downarrow & \ulcorner & \downarrow \\ \mathcal C & \xrightarrow{} & \mathsf{Set} \end{array}, $$

where $\mathsf{Cat}_{*,\mathcal l}$ is the (lax) category of pointed categories. Can we draw a diagram analogous to the 2nd diagram above? If so, what is the answer to the previous question in this case?

edit: It seems that Gray analyzed this question in his 1980 monograph "Closed categories, lax limits and homotopy limits". His proposition 4.6.5 asserts that the lax limit of a functor $F:\mathcal C\to \mathsf{Cat}$ can indeed be calculated through the corresponding homotopy limit of $NF$. He even analyzes the particular case of comma squares in Example 4.6.7.

The diagram in the proof of that proposition is really similar to the picture painted by Fernando Muro in this answer, so I wonder if there is a Quillen equivalence there. Maybe it's even Thomason's model structure, perhaps even trivially, but I'm not kowledgeable enough to answer that. I'll leave the question open in case people have deeper insights related or not to this edit).

$\endgroup$
0

1 Answer 1

5
$\begingroup$

The analogy is of course a correct/useful analogy, but I think any model structure for which the literal statement is correct must have $Set_* \simeq *$, so it would be a bit too coarse to do anything reasonable with categories as categories. As Alexander Campbell points out in the comments below, in the Thomason model stucture (which uses categories to model homotopy types) we do have $Set_* \simeq *$, but it destroys (what I guessed is) the point of you question.

Indeed, your assumption guarantees the existence of a homotopy pullback square:

$$\require{AMScd} \begin{CD}Set_* @>>> * \\ @VVV @VVV \\ Set @>>> Set \end{CD}$$

But $Set \to Set$ is also a weak equivalence (it is an isomorphism !), so $$\require{AMScd} \begin{CD}* @>>> * \\ @VVV @VVV \\ Set @>>> Set \end{CD}$$ is also a homotopy pullback, which implies $Set_*\simeq *$.

Pseudo-limits (the kind where the diagrams commute up to isomorphism rather than just $2$-cells) can be expressed as homotopy limits, though. But for this kind of (op)lax limit, it seems like one need some notion of $2$-model category to make it work. I don't know more about this though.

$\endgroup$
2
  • 1
    $\begingroup$ The questioner may be satisfied with an answer involving something like the Thomason model structure, in which $Set_*$ and $*$ are of course weakly equivalent. $\endgroup$ Sep 24, 2023 at 10:38
  • $\begingroup$ @AlexanderCampbell : that's a good point. I guess I should add something to the effect that it is too coarse to remember "category theory" $\endgroup$ Sep 24, 2023 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.