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Stream: learning: questions

Topic: is this an opfibration?

John Baez (Mar 20 2022 at 20:03):

I've got a question. Let $\mathsf{Graph}$ be the usual category theorist's category of graphs, or [[quivers]]: the category of functors from the category $s,t: E \to V$ to $\mathsf{Set}$.

John Baez (Mar 20 2022 at 20:03):

Let $\mathrm{edge} : \mathsf{Graph} \to \mathsf{Set}$ be the forgetful functor that sends any graph to its set of edges.

John Baez (Mar 20 2022 at 20:04):

I believe $\mathrm{edge}$ is not an opfibration. Can you prove this or disprove it?

Mike Shulman (Mar 20 2022 at 20:55):

Let's see, a graph $E\rightrightarrows V$ is equivalently a single function $E+E \to V$. If we have a function $f:E\to E'$, then we can take the pushout of the span $V \leftarrow E+E \to E'+E'$ to get a set $V'$, and hence a graph $E' \rightrightarrows V'$ with a map from our original one. It looks to me like this would be an opcartesian morphism; does something go wrong?

John Baez (Mar 20 2022 at 21:02):

Thanks! I'm trying to get something to go wrong, but it's not going wrong.

John Baez (Mar 20 2022 at 21:03):

I should switch gears and try to prove it works.

John Baez (Mar 20 2022 at 21:06):

I'm actually trying to settle more general question, maybe I should ask that:

Given a functor $\phi: C \to D$ between small categories, is the precomposition-with-$\phi$ between presheaf categories $\phi^\ast: \hat{D} \to \hat{C}$ always an opfibration?

John Baez (Mar 20 2022 at 21:07):

For some stupid reason I think the answer is no, so I was looking for a counterexample.

Mike Shulman (Mar 20 2022 at 21:47):

There's a general fact (which doesn't seem to be very widely known) that any pushout-preserving isofibration with a fully faithful left adjoint is an opfibration. The proof is kind of follow-your-nose once you suspect that such a thing might be true, and the case of $\rm edge$ is an instance of that. So the answer to your more general question will be yes if $\rm Lan_\phi$ is fully faithful, which is true in particular whenever $\phi$ is fully faithful.

Mike Shulman (Mar 20 2022 at 21:48):

So I would look for a counterexample where $\phi$ is not fully faithful. Perhaps $\phi : 2\to 1$, so that $\phi^*$ is the diagonal functor $\rm Set \to Set\times Set$?

John Baez (Mar 20 2022 at 21:53):

Okay, thanks a lot! For our applications (to structured cospans) we are probably most interested in the case where $\phi$ is fully faithful!

E.g. $\hat{D}$ might be the category of whole-grain Petri nets, and $\hat{C}$ might be the category of sets, and $\phi^*$ might send any whole-grain Petri net to its set of places.

John Baez (Mar 20 2022 at 21:54):

But we also want to know a bit about when we don't give an opfibration, and your proposed counterexample smells good.

Mike Shulman (Mar 20 2022 at 21:58):

I thought the category of whole-grain Petri nets wasn't a presheaf category.

John Baez (Mar 20 2022 at 22:04):

Not with "etale morphisms" it ain't, but there's a larger presheaf category lurking around.

John Baez (Mar 20 2022 at 22:06):

It's very interesting looking at how the people writing software think about this stuff. So far they mainly use morphisms (e.g. of Petri nets) only for composing structured cospans, so a lot of issues are (temporarily) irrelevant.

Morgan Rogers (he/him) (Mar 20 2022 at 22:35):

Mike Shulman said:

So I would look for a counterexample where $\phi$ is not fully faithful. Perhaps $\phi : 2\to 1$, so that $\phi^*$ is the diagonal functor $\rm Set \to Set\times Set$?

Yep, this isn't an opfibration, since we can take the inclusion of the initial object into either of the representables to get a morphism with no lift.

John Baez (Mar 20 2022 at 22:39):

Great, thanks!

John Baez (Mar 20 2022 at 22:40):

Part of my problem is that I'm not at all eager to think about this stuff, I'm doing it just to satisfy a referee... :anguish:

John Baez (Mar 22 2022 at 00:38):

We're going to use your example in the the conclusions of the revised version of Structured versus decorated cospans, @Morgan Rogers (he/him), and acknowledge you. Mike is already prominently acknowledged!

Morgan Rogers (he/him) (Mar 22 2022 at 07:58):

That's very generous for something like this but I'm glad I could save you some work. Thank you!

John Baez (Mar 23 2022 at 23:57):

Sure! You got us out of a jam... I wanted to finish this paper fast. I find it's better to acknowledge people freely rather than treat acknowledgments as a scarce resource.