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Stream: learning: questions

Topic: Confusion with conservative adjoints

John Onstead (Jun 24 2025 at 11:16):

There's a claim that if a functor $R: C \to D$ has a left adjoint and $C$ is finitely complete, $R$ is conservative if and only if every counit component is strong epi. Interestingly, this implies that if $C$ is finitely complete, any conservative functor with a left adjoint must be faithful, since being faithful in an adjunction is the same counit requirement but without the "strongness" in the epis.

John Onstead (Jun 24 2025 at 11:17):

Let's set up the following situation. Let $G: J \to C$ be a plain separator and $G^*: [C^{op}, \mathrm{Set}] \to [J^{op}, \mathrm{Set}]$ the induced functor. It has a left adjoint given by $L$, a left Kan extension, with counit $\epsilon: L \circ G^* \to \mathrm{id}$. Define $D$ as the full subcategory of $[C^{op}, \mathrm{Set}]$ on objects $X$ such that $\epsilon_X$ are epimorphisms. Now, since we've made the components of the counit all epimorphisms, the resulting functor $G^*_D: D \to [J^{op}, \mathrm{Set}]$ must be faithful. So far so good, and we can even determine that $C$ embeds into $D$ if and only if $G$ is a separator of some form, since the restricted Yoneda functor is $y_G = G^* \circ y$ and the composite of two faithful functors is faithful.

John Onstead (Jun 24 2025 at 11:18):

But here comes the spanner in the works: In a presheaf category, all epimorphisms are strong epimorphisms! Thus, by our starting claim above, the functor $G^*_D$ must be conservative and so every plain generator should be strong- which is obviously false. So what is going on here? The only possibility I can see is that $D$ might not in fact be finitely complete! But if this is the answer, I want to understand why this limits condition is needed. I know a strong epi might not be an actual epi unless the category has equalizers, but we already know that all $\epsilon_X$ in $D$ are both strong epis and actual epis just as they were in the presheaf category due to the fact that fully faithful functors reflect epis and strong epis. So then where exactly do things break down?

Mike Shulman (Jun 24 2025 at 15:07):

Did you look at the proof?

John Onstead (Jun 24 2025 at 19:55):

Mike Shulman said:

Did you look at the proof?

Yes and that's what inspired my question. The proof says "Because $C$ has finite limits and $\epsilon_Z$ is a strong epi, it must be an epi, so this implies that $f = f'$." But in my case above, we are restricting the presheaf category to the category $D$ where all $\epsilon_Z$ are necessarily epimorphisms. Since $D \to [C^{op}, \mathrm{Set}]$ reflects epis, it means every $\epsilon_Z$ is also an epi in $D$. So that's my confusion- it seems to me we don't have to assume that $D$ has finite limits to know that $\epsilon_Z$ is an epi, and therefore the overall proof should follow even if $D$ doesn't have finite limits (unless I missed another part of the proof where the finite limits are needed). So why doesn't it?

Mike Shulman (Jun 24 2025 at 21:18):

Here's a first place that I'm suspicious:
John Onstead said:

we can even determine that $C$ embeds into $D$ if and only if $G$ is a separator of some form, since the restricted Yoneda functor is $y_G = G^* \circ y$ and the composite of two faithful functors is faithful.

That argument gives you "only if". Where do you get "if"?

John Onstead (Jun 24 2025 at 21:58):

Mike Shulman said:

That argument gives you "only if". Where do you get "if"?

I'm not sure what you mean. The functor $G^*_D: D \to [J^{op}, \mathrm{Set}]$ is faithful, and since $C \to D$ factors the Yoneda embedding, it implies that the restricted Yoneda embedding must also be faithful (since the composite of these two faithful functors is faithful as mentioned). Therefore, $J$ must be a separator by definition, since its restricted Yoneda functor is faithful.

John Onstead (Jun 24 2025 at 21:58):

Maybe it's the other direction you're asking about? In that case, $D$ is the maximal subcategory of presheaves that $G^*$ acts faithfully on. Since it is the maximal such subcategory, any other subcategory of presheaves that $G^*$ acts faithfully on must factor through the embedding of $D$ into presheaves. If $J$ is a separator, then $G^*$ acts faithfully on the image of the Yoneda embedding by definition, therefore the Yoneda embedding must factor through $D$.

Mike Shulman (Jun 24 2025 at 22:37):

"if" is $\Leftarrow$, "only if" is $\Rightarrow$. So yes, it was the "other" direction I was asking about.

Mike Shulman (Jun 24 2025 at 22:39):

John Onstead said:

$D$ is the maximal subcategory of presheaves that $G^*$ acts faithfully on.

No, there is no such thing. $D$ is the maximal subcategory of presheaves on which the counit is epi. That makes sense because the counit being epi is a property of a single object, so you can consider the category of all such objects. But faithfulness is not a property of a single object, it's a property pertaining to pairs of objects, so you can't consider "the category of all objects on which $G^*$ acts faithfully".

Mike Shulman (Jun 24 2025 at 22:40):

It's true that a right adjoint is faithful globally if and only if all components of the counit are epi, but that equivalence doesn't "localize" to compare properties of individual objects or subcategories.

Mike Shulman (Jun 24 2025 at 22:43):

If you trace through the proof, what you get is that a particular component of the counit $L R X \to X$ is epi if and only if for any object $Y$ and any morphisms $f,g:X\to Y$, if $Rf = Rg$ then $f=g$. In other words, $R$ is "faithful on all morphisms with domain $X$". So the category of objects whose counit is epi is equivalently the category of all objects such that $R$ (in this case, $G^*$) is faithful on all morphisms with them as domain.

But if you have a subcategory, like the image of Yoneda, on which $G^*$ is faithful in the sense that it detects equality between morphisms whose domain and codomain lie in that subcategory, that doesn't imply that $G^*$ is faithful on all morphisms whose domains lie in that subcategory, and hence it doesn't imply that the components of the counit at objects in that subcategory are epi.

John Onstead (Jun 24 2025 at 23:51):

Mike Shulman said:

So the category of objects whose counit is epi is equivalently the category of all objects such that $R$ (in this case, $G^*$) is faithful on all morphisms with them as domain.

But if you have a subcategory, like the image of Yoneda, on which $G^*$ is faithful in the sense that it detects equality between morphisms whose domain and codomain lie in that subcategory, that doesn't imply that $G^*$ is faithful on all morphisms whose domains lie in that subcategory, and hence it doesn't imply that the components of the counit at objects in that subcategory are epi.

Ohhh I think this was where my mistake was, I guess it had nothing to do with limits after all. So the problem is, if there's a subcategory that $G^*$ acts faithfully on, it might not be true that a presheaf in this subcategory also lies in $D$ unless every morphism out of that object into any other presheaf outside of this subcategory is also acted on faithfully. So in some sense, the condition of a presheaf being in $D$ is "too strong" for what I'm trying to do.

John Onstead (Jun 24 2025 at 23:56):

Mike Shulman said:

That makes sense because the counit being epi is a property of a single object, so you can consider the category of all such objects. But faithfulness is not a property of a single object, it's a property pertaining to pairs of objects, so you can't consider "the category of all objects on which $G^*$ acts faithfully".

So instead, to characterize faithfulness of $G^*$, I'll probably need to consider pairs of presheaves. For instance, if I have some subcategory in mind, I'll need to check that every pair of presheaves in that subcategory is acted on faithfully. It sounds like a harder characterization since there's a lot more to check. For instance, if I have some pair of presheaves $X$ and $Y$, I'll have to make sure that no two natural transformations between $X$ and $Y$ have the same components on all objects in the image of $G$. If so, then "restricting" by postcomposing with $G$, the very action of $G^*$, will map these two transformations into each other. But that sounds like it'd be very difficult to translate into a condition on the pair of presheaves $X$ and $Y$ themselves, as opposed to their natural transformations.

Mike Shulman (Jun 24 2025 at 23:57):

Indeed.

John Onstead (Jun 25 2025 at 01:17):

Maybe I can better understand this with a concrete example. For instance, let $P(C) = \mathrm{Pullback}(C^{op}, \mathrm{Set})$, the category of pullback preserving presheaves, and thus presheaves that send pushouts in $C$ to pullbacks in $\mathrm{Set}$. Since epis are a pushout property, $C \to P(C)$ preserves epis, and since $C$ is dense in $P(C)$, any separator in $C$ is also one in $P(C)$. Therefore, $P(C)$ is a good example of a subcategory of presheaves that $G^*$ acts faithfully on.

John Onstead (Jun 25 2025 at 01:17):

My question would then be: specifically, what is it about this property of turning pushouts into pullbacks implies that no two natural transformations between such presheaves act the same on objects in a separator of $C$? Put another way, this is asking why natural transformations between such presheaves are uniquely set/determined by their components at these objects.

Mike Shulman (Jun 25 2025 at 01:23):

Preserving pushouts implies preserving individual epis, but I don't see why that would imply preserving jointly epimorphic families.

John Onstead (Jun 25 2025 at 05:16):

Mike Shulman said:

Preserving pushouts implies preserving individual epis, but I don't see why that would imply preserving jointly epimorphic families.

I guess one could consider instead $C \to L(C)$, where $L(C) = \mathrm{Lim}(C^{op}, \mathrm{Set})$. These presheaves send every colimit in $C$ to limits in $\mathrm{Set}$. So the embedding also preserves coproducts and thus jointly epimorphic families (since such a family is summarized by a single epi out of the coproduct).

John Onstead (Jun 25 2025 at 05:17):

I tried myself for a bit to solve this. First, let's make a simplifying assumption that our separator has one object $X$, and we can take enough coproducts of $X$ to get an object $UX$ that covers all objects in the category with epis. A presheaf $F$ with the above property would turn this epi into a mono into a product in $\mathrm{Set}$. I'm not sure where to go from here however.

Mike Shulman (Jun 25 2025 at 05:19):

John Onstead said:

(since such a family is summarized by a single epi out of the coproduct).

Assuming that $C$ has small coproducts and the family is small.

John Onstead (Jun 25 2025 at 05:20):

Given a natural transformation $\eta: F \to G$ between presheaves, the component $\eta_{UX}$ (between the products of $F(X)$ and $G(X)$) is determined itself by $\eta_X$, and $\eta_{UX}$ will always appear in some naturality square with any component $\eta_c$ since any $c$ has the epimorphism from $UX$. But this would be true for any presheaf $F$ and $G$, so it doesn't point out what is different when they are limit preserving.

Mike Shulman (Jun 25 2025 at 05:21):

John Onstead said:

the component $\eta_{UX}$ (between the products of $F(X)$ and $G(X)$) is determined itself by $\eta_X$

(when $F$ and $G$ preserve products)

Mike Shulman (Jun 25 2025 at 05:23):

When they are additionally epi-preserving, the maps $Fc \to F(UX)$ are monos. Thus, comparing the two naturality squares for two transformations $F\rightrightarrows G$, if their $UX$-components are the same, so must their $c$-components be.

John Onstead (Jun 25 2025 at 05:23):

Mike Shulman said:

(when $F$ and $G$ preserve products)

Oh, right. Sorry! But what is the significance that these products are preserved, how does that help in the proof?

John Onstead (Jun 25 2025 at 05:24):

Mike Shulman said:

Thus, comparing the two naturality squares for two transformations $F\rightrightarrows G$, if their $UX$-components are the same, so must their $c$-components be.

Ah, I see. Thanks!

Mike Shulman (Jun 25 2025 at 06:01):

Does your second message mean that you understand everything now? Or do you still have a question?

John Onstead (Jun 26 2025 at 01:02):

Mike Shulman said:

Does your second message mean that you understand everything now? Or do you still have a question?

It means I understand this now- thanks for the help!

Mike Shulman (Jun 26 2025 at 01:11):

Great! You're welcome.