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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeSep 8th 2015

added to conservative functor the proposition saying that pullback along strong epis is a conservative functor (if strong epis pull back).

How about the $\infty$-version?

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeSep 8th 2015

It’s easy to show in HoTT that pullback along surjections (effective epis) is conservative. I don’t know how general of ∞-categories that can be made sense of in.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeSep 9th 2015

Thanks. I’d just need it in $\infty$-toposes.

Maybe you have the leisure to add a remark along these lines to conservative (∞,1)-functor?

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeSep 11th 2015

Done.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeSep 11th 2015

Thanks!!

1. At conservative functor we had the statement that a conservative functor reflects all those (co-)limits which it preserves. I qualified this statement to refer only to those (co-)limits which at all exist in the source category.

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeMar 25th 2016

According to preserved limit (and all uses of the term I’m familiar with), only limits that exist can be preserved, so I don’t think the qualification is necessary.

• CommentRowNumber8.
• CommentAuthorIngoBlechschmidt
• CommentTimeMar 26th 2016
• (edited Mar 26th 2016)

I thought so – that it only would be a language problem, I’m sure we agree on the formal mathematical statement – but can’t quite reconcile the wording with my understanding. Here is an explicit example:

Let $\mathcal{C}$ be any groupoid in which there is an object $X$ with at least two endomorphisms. Such a category doesn’t possess a terminal object. Therefore the unique functor $F : \mathcal{C} \to \mathbb{1}$ into the terminal category vacuously preserves terminal objects. Since any morphism in $\mathcal{C}$ is invertible, this functor is also conservative. But it doesn’t reflect terminal objects:

Consider the cone $X$ on the empty diagram in $\mathcal{C}$. Its image under $F$ is a limiting cone, since $F(X)$ is a terminal object in $\mathbb{1}$. But the considered cone is not.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeMar 27th 2016

I would say that in that case $F$ does not preserve terminal objects, or rather that it is ill-typed to even ask whether it does, because its domain does not have terminal objects.

• CommentRowNumber10.
• CommentAuthorTodd_Trimble
• CommentTimeMar 27th 2016

Maybe the reconciliation is that the functor $F: \mathcal{C} \to \mathbb{1}$ of #8, or its prolongation $y \circ F: \mathcal{C} \to Set$ along the Yoneda embedding $y: \mathbb{1} \to Set$, is not a flat functor. See the section in preserved limit which discusses the case where limits in the domain don’t exist.

So I presume that a suitable resolution for #6 is to say that a conservative functor $F: \mathcal{C} \to \mathcal{D}$ reflects limits that are (hypothetically) preserved, in the sense of that section. Is it true that if $F$ is conservative, then so is the corresponding left Kan $L F: Set^{\mathcal{C}^{op}} \to Set^{\mathcal{D}^{op}}$?

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeMar 27th 2016

I’m not sure that that’s true. Isn’t it possible to have a flat and conservative functor whose codomain has finite limits but whose domain does not?

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeMar 27th 2016

Oh well, I guess you’re right. If $B G$ is a group, then flat functors into $Set$ correspond to torsors, and any functor $B G \to Set$ is conservative.

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeMar 27th 2016

To me, asking whether a functor preserves finite limits when its domain doesn’t have finite limits is like asking whether a function is a group homomorphism when its domain is not a group: it’s an ill-typed question. The fact that flat functors are “functors that would preserve finite limits if they existed” is a nice way to think about them, but being flat is still a different thing from preserving finite limits.

• CommentRowNumber14.
• CommentAuthorIngoBlechschmidt
• CommentTimeMar 27th 2016
• (edited Mar 27th 2016)

Could the perspective of “structure vs. property” help here? I totally agree that the question “is this function a group homomorphism?” is ill-typed unless source and target are groups; being a group homomorphism means being compatible with the group structure, which has to be given for the question to make sense.

However, at the moment, my personal reading of “$F:\mathcal{C}\to\mathcal{D}$ preserves $\mathcal{I}$-shaped limits” is still the following, which I believe is well-typed irregardless of whether $\mathcal{I}$-shaped limits actually exist in the source category: “For any diagram $G:\mathcal{I}\to\mathcal{C}$, the image under $F$ of any $G$-cone which happens to be a limiting cone is again a limiting cone.”

Maybe the difference is in saying “$F$ preserves limiting cones” (which should make sense even if there are no limiting cones) vs. “$F$ preserves limits” (which shouldn’t make sense if no limits are specified).

• CommentRowNumber15.
• CommentAuthorMike Shulman
• CommentTimeMar 28th 2016

I agree that structure vs property is relevant; but “having finite limits” is actually (like “being a group”) a structure, not a property, precisely because not all functors between categories with finite limits preserve them. It is a special sort of structure called a “property-like structure”, meaning that it is “unique when it exists” (more precisely, it is preserved by all equivalences), but it is still a structure.

I’m not just making a philosophical point either; this is an empirical observation about mathematical language. I’ve never ever heard a mathematician talk about a functor “preserving finite limits” when it wasn’t known or implicitly assumed that its domain and codomain had them.

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeNov 24th 2017

Added statement that every fully faithful functor is conservative: here.

• CommentRowNumber17.
• CommentAuthorjesse
• CommentTimeNov 25th 2017

I also added the statement that for pretoposes, conservative pretopos morphisms are precisely those which induce injective maps on subobject lattices. (This is the notion of conservativity which is used in Makkai-Reyes.)

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeNov 25th 2017

Thanks! Is it necessary to have pretoposes there? Is one direction or the other, at least, true more generally? (I should know that, but I don’t have time to think about it right now…)

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeNov 25th 2017

Also: how did you manage to edit the nLab? It’s not letting me edit any pages right now, trying to save a page just takes me back to the edit page.

• CommentRowNumber20.
• CommentAuthorjesse
• CommentTimeNov 25th 2017

@Mike #19: that’s strange—I edited the page the usual way. I tried editing the Sandbox just now and while it took a while to accept the edit, it ended up working.

Re: #18, I believe the proof only requires that you can talk about images of maps as subobjects and that the functor preserves images and meets in the subobject lattices.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeNov 25th 2017

how did you manage to edit the nLab? It’s not letting me edit any pages right now, trying to save a page just takes me back to the edit page.

If that’s so since that incident from Friday, then deleting cookies should help.

That’s anyway what happened for me: After Adeel had fixed the editing behaviour the problem did remain on my end, but clearing cookies made it go away.

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeNov 26th 2017

Yay, thanks Urs!

Jesse #20: Makes sense. So it’s enough to talk about a regular functor between regular categories?

• CommentRowNumber23.
• CommentAuthorDavidRoberts
• CommentTimeOct 20th 2020

Gave more precise reference to a cited result from the Elephant.

• CommentRowNumber24.
• CommentAuthorvarkor
• CommentTimeMay 26th 2021

Mention that monadic functors are conservative.