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Stream: learning: questions

Topic: When do joint epimorphisms sense structure?

John Onstead (Jun 03 2025 at 11:12):

In $\mathrm{Top}$, an example of a joint epimorphism is an open cover, from which one can recover not just the underlying set of a space, but also its structure. Another example of a joint epimorphism is the set of morphisms from the point into a topological space (since it is a generator). However, this only recovers the underlying set of a space, and is not sensitive to the structure on it. And so that's my question: what is it about the first joint epimorphism that the second joint epimorphism did not have, that allows for the recovery of the full object? (I'm hypothesizing something like "it must be jointly regularly (or effectively) epimorphic".)

John Baez (Jun 03 2025 at 12:43):

Since Top has coproducts we can always replace a joint epimorphism by a single epimorphism out the coproduct of the domains. Then your question becomes: what's the difference, categorically speaking, between an open cover of a space $X$

$\bigsqcup_{\alpha} U_\alpha \to X$

and the map from $X$ with the discrete topology to $X$

$\bigsqcup_{x \in X} \{x\} \to X$ ?

One big difference, as you guessed, is that the former is a regular epimorphism and the latter is not.

In terms ordinary mortals can understand, the former map lets us transfer the topology on $\bigsqcup_{\alpha} U_\alpha$ to a topology on $X$, namely the strongest topology that makes this map continuous, and we get the usual topology on $X$ this way. In the latter case we can do the same thing but we don't get the usual topology on $X$: we get the discrete topology.

Mike Shulman (Jun 03 2025 at 16:12):

In fact the former is not just a regular epimorphism but an effective epimorphism, which says that the parallel pair it's a coequalizer of is its [[kernel pair]].

Mike Shulman (Jun 03 2025 at 16:14):

John Baez said:

the former map lets us transfer the topology on $\bigsqcup_{\alpha} U_\alpha$ to a topology on $X$, namely the strongest topology that makes this map continuous, and we get the usual topology on $X$ this way.

Maybe it's worth pointing out that while in Top this property is the same as being a regular epimorphism, it could be different in other categories. The official name for this property is being an [[opcartesian arrow]] or [[final lift]] with respect to the forgetful functor to Set. (So, in particular, it is defined only in reference to that functor, whereas being a regular epimorphism is fully internal to Top.)

John Onstead (Jun 03 2025 at 21:20):

John Baez said:

One big difference, as you guessed, is that the former is a regular epimorphism and the latter is not.

Mike Shulman said:

In fact the former is not just a regular epimorphism but an effective epimorphism, which says that the parallel pair it's a coequalizer of is its [[kernel pair]].

Ah, I'm glad my hypothesis was spot on!

John Onstead (Jun 03 2025 at 21:20):

Mike Shulman said:

Maybe it's worth pointing out that while in Top this property is the same as being a regular epimorphism, it could be different in other categories.

That raises an interesting question. What if we wanted to go beyond $\mathrm{Top}$ into a completely arbitrary (perhaps still concrete) category, and we wanted to know if some particular sink (set of morphisms into an object) were sufficient to reconstruct the object? What is the most general condition in which that can be the case? For instance, it seems obvious that being jointly effective epimorphic is sufficient, since we can directly reconstruct the object from this sink by taking the coequalizer of the "joint kernel" (the canonical pair of arrows from the coproduct of pullbacks of pairs of morphisms in the sinks- the overlaps- into the coproduct itself). But "sufficient" (if) does not imply "sufficient + necessary" (if and only if), and I want to know the most general necessary condition. Is it being a joint regular epimorphism, a joint strict epimorphism, a final lift, or some other property?

Mike Shulman (Jun 03 2025 at 21:24):

What exactly do you mean by "reconstruct"?

John Baez (Jun 03 2025 at 21:25):

Don't you have to specify a method by which you plan to reconstruct the object from a set of morphisms into that object, before you can answer whether it's possible? After all, you can always just take the target of all those morphisms.

Mike Shulman (Jun 03 2025 at 21:25):

(Unless the set of morphisms is empty, I suppose.)

John Baez (Jun 03 2025 at 21:26):

Okay, maybe we can agree that you can't reconstruct an object from the empty set of morphisms... except in the case of a one-object category.

Mike Shulman (Jun 03 2025 at 22:45):

Well, even then it depends on what you're going to do with the morphisms. If you're going to take the quotient of their joint kernel, then in the case of the empty set of morphisms you'll get the initial object of any category.

Mike Shulman (Jun 03 2025 at 22:47):

Also it depends a bit on how you formalize a sink. If you say that a sink is a set of morphisms with the property that the targets of any two of them are equal, then from the empty set of morphisms you can't do a whole lot. But if you say that a sink is an object together with a set of morphisms having that object as their target (which is kind of what you have to do if you want to avoid talking about equality of objects), then even if the set of morphisms is empty you still have a target object.

Mike Shulman (Jun 03 2025 at 22:47):

But this is getting a bit silly, sorry for derailing the conversation.

John Onstead (Jun 03 2025 at 23:11):

Mike Shulman said:

What exactly do you mean by "reconstruct"?

John Baez said:

Don't you have to specify a method by which you plan to reconstruct the object from a set of morphisms into that object, before you can answer whether it's possible? After all, you can always just take the target of all those morphisms.

With sets, once you specify all the elements of the set, you have fully specified the set itself. In an arbitrary category, you can think of the morphisms into an object as the "generalized elements" of the object. The question becomes when specifying a particular set of generalized elements is sufficient to do the same thing to the object as specifying all the elements of a set.

John Onstead (Jun 03 2025 at 23:15):

The notion of the coequalizer of the joint kernel does this effectively: one can think of it taking certain generalized elements of an object and gluing them together along their overlaps to get some object in the category (IE, a "joint image" so to speak). If this happens to be isomorphic to the original object, it implies that the set of generalized elements you started with were sufficient such that gluing them together along overlaps would recover the original object.

John Baez (Jun 03 2025 at 23:21):

John Onstead said:

With sets, once you specify all the elements of the set, you have fully specified the set itself. In an arbitrary category, you can think of the morphisms into an object as the "generalized elements" of the object. The question becomes when specifying a particular set of generalized elements is sufficient to do the same thing to the object as specifying all the elements of a set.

Yes, but you're thinking about material set theory, where I can know something is an element of some set without knowing that set. In category theory you can't know a morphism without knowing its target. So what do you mean by "specifying some set of generalized elements $f_\alpha : x_\alpha \to x$" without already having $x$? Do you have some idea of what $f_\alpha : x_\alpha \to \qquad$ means?

David Egolf (Jun 03 2025 at 23:54):

Perhaps a related question: In a category $C$ with an object $x$, we get a category $C/x$ of morphisms to $x$. Then, we can imagine taking various subcategories of $C/x$. If $A$ is a subcategory of $C/x$ for some unknown $x$, we could ask: When is $A$ "big enough" so that we can identity $x$ up to isomorphism from $A$?

In other words: When given a category $A$ that is known to be (up to equivalence?) a subcategory of $C/x$ for some unknown $x$, for which $A$ can we figure out $x$ up to isomorphism?

David Egolf (Jun 03 2025 at 23:55):

(I don't know how closely this relates to what @John Onstead has in mind!)

John Onstead (Jun 03 2025 at 23:59):

David Egolf said:

(I don't know how closely this relates to what John Onstead has in mind!)

I think it relates! I think the trick would be to ensure the assignment of subcategories of $C/x$ is consistent across every object- that is, you can organize the data into a pseudofunctor $C \to \mathrm{Cat}$ (if compatible with composition) or $C^{op} \to \mathrm{Cat}$ (if compatible with pullback) with a natural transformation to the slice category functor $C/(-)$ such that every component is fully faithful. That way the theory of Grothendieck fibrations becomes relevant!

John Onstead (Jun 04 2025 at 00:07):

John Baez said:

So what do you mean by "specifying some set of generalized elements $f_\alpha : x_\alpha \to x$" without already having $x$? Do you have some idea of what $f_\alpha : x_\alpha \to \qquad$ means?

I still think you need to have $x$! I'm not trying to construct $x$ without knowledge of $x$, I'm trying to determine the conditions on which a set of morphisms into $x$ is a "good probe" of $x$. Let me try to put this another way. One can think of a generalized element $f: y \to x$ as "picking out" a "part" of $x$. Obviously, if I have enough parts of something, I can put those parts together and get the full thing. So I want to know, for which set of parts is this possible?

James Deikun (Jun 04 2025 at 00:28):

I really think the notion of jointly opcartesian sink with regard to a functor is exactly the thing you're looking for. Jointly epimorphic is precisely a "good probe" if you remember all the information about how the incoming morphisms fit together inside the target object, but if you forget some of that information via a functor it might take more to make a probe "good".

Mike Shulman (Jun 04 2025 at 01:50):

John Onstead said:

One can think of a generalized element $f: y \to x$ as "picking out" a "part" of $x$. Obviously, if I have enough parts of something, I can put those parts together and get the full thing. So I want to know, for which set of parts is this possible?

I think the most natural reading of this for me leads to the notion of jointly effective-epimorphic sink. A colimit is the most obvious way of "putting together parts" to make a larger object in a category. And in general since the parts will "overlap" you have to specify some amount of "gluings" or "identifications", and the joint kernel pair is arguably the most natural way to extract that information from the sink.

John Onstead (Jun 04 2025 at 02:09):

James Deikun said:

I really think the notion of jointly opcartesian sink with regard to a functor is exactly the thing you're looking for. Jointly epimorphic is precisely a "good probe" if you remember all the information about how the incoming morphisms fit together inside the target object, but if you forget some of that information via a functor it might take more to make a probe "good".

That's interesting, and I've been reading more on opcartesian morphisms/final sinks. It makes me wonder more about the relation between these and joint (regular) epimorphisms. As mentioned above, one is internal while the other is relative, but they seem to converge at least for $\mathrm{Top}$ and perhaps other categories. In addition, the nlab page states a semi final sink relative to the canonical functor $C \to *$ is a colimit. It seems it'd be more accurate to say it's a coproduct since a sink is a discrete cocone and a colimiting discrete cocone is a coproduct. But it does raise an interesting question: when is it that a colimit can be given as a final sink and vice versa?

John Onstead (Jun 04 2025 at 02:16):

Mike Shulman said:

I think the most natural reading of this for me leads to the notion of jointly effective-epimorphic sink. A colimit is the most obvious way of "putting together parts" to make a larger object in a category. And in general since the parts will "overlap" you have to specify some amount of "gluings" or "identifications", and the joint kernel pair is arguably the most natural way to extract that information from the sink.

That sounds reasonable! My only worry then is that this intuition really only seems to work, at least best, for regular categories, since those are the categories for which every sink is guaranteed to have a "joint kernel" you can coequalize and get meaningful results out of (IE, you can do this construction in any bicomplete category, but I'm not sure if it gives exactly what you'd "hope for" if that category isn't also regular). In some regard this makes sense, as regular categories are categories with a good notion of "image", and the coequalizer of the joint kernel is basically sort of like a "joint image". But if the category isn't regular, then we can't always "glue together" morphisms in this way and compare the result with the object itself. It doesn't help it's provably impossible to embed a non-regular category into a regular category in such a way it preserves regular epimorphisms (due to the requirement a regular epimorphism in a regular category be stable under pullback). Would you know of any remedies that can aid in understanding joint epimorphisms better in a non-regular category?

Mike Shulman (Jun 04 2025 at 02:26):

It is actually possible to define an effective-epimorphic sink in any category, without assuming any limits or colimits. First note that the colimits are no problem: you can assert that a particular cocone is a colimit without having assumed a priori that the category has any colimits. Second, you can replace the pairwise pullbacks by the family of all possible squares (of which the pullback, if it existed, would be the terminal one). When you compile this out you get:

A sink $\{ f_i : x_i \to y \}$ is effective-epimorphic if for any object $z$, and any family of morphisms $\{ g_i : x_i\to z\}$ having the property that for any $i,j$, object $w$, and morphisms $h:w\to x_i$ and $k:w\to x_j$ such that $g_i h = g_j k$, there exists a unique morphism $\ell : y\to z$ such that $\ell f_i = g_i$ for all $i$.

Mike Shulman (Jun 04 2025 at 02:28):

Now I can't resist mentioning the usual next thing to do: define a sink $\{ f_i : x_i \to y \}$ to be universally effective-epimorphic if for any morphism $c : u \to y$, there exists an effective-epimorphic sink $\{ d_j : v_j \to u\}$ such that for any $j$ there exists an $i$ and a morphism $e : v_j \to x_i$ such that $f_i e = c d_j$. Then the collection of all universally effective-epimorphic sinks in a category forms the [[canonical topology]] on that category.

John Onstead (Jun 04 2025 at 11:22):

Mike Shulman said:

A sink $\{ f_i : x_i \to y \}$ is effective-epimorphic if for any object $z$, and any family of morphisms $\{ g_i : x_i\to z\}$ having the property that for any $i,j$, object $w$, and morphisms $h:w\to x_i$ and $k:w\to x_j$ such that $g_i h = g_j k$, there exists a unique morphism $\ell : y\to z$ such that $\ell f_i = g_i$ for all $i$.

I think I'll have to sit and draw this one out. But looks interesting!

Mike Shulman said:

Then the collection of all universally effective-epimorphic sinks in a category forms the [[canonical topology]] on that category.

That's interesting- I always wondered why the canonical topology used universal rather than merely effective epimorphisms. But now it makes more sense, since it seems the universal ones are defined to be stable under pullback.

John Onstead (Jun 10 2025 at 11:11):

Mike Shulman said:

A sink $\{ f_i : x_i \to y \}$ is effective-epimorphic if for any object $z$, and any family of morphisms $\{ g_i : x_i\to z\}$ having the property that for any $i,j$, object $w$, and morphisms $h:w\to x_i$ and $k:w\to x_j$ such that $g_i h = g_j k$, there exists a unique morphism $\ell : y\to z$ such that $\ell f_i = g_i$ for all $i$.

Ok, I've drawn it out and it makes sense. It seems the first part of the definition is specifying the sink is joint epimorphic, and then the second part is adding on a universal property to make it effective. But I'm curious: given an "effective epimorphic" sink defined by this universal property, will that universal property be preserved by a colimit preserving functor? For instance, let's say I had a bicomplete category $D$, a (faithful) colimit preserving functor $C \to D$, and a sink in $C$ satisfying the above universal property. I'd like to be able to transfer it over into $D$ where I can now actually have the colimits and limits to write down the usual "coequalizer of kernel pair" representation for this joint effective epi sink.

Mike Shulman (Jun 10 2025 at 15:19):

No, preserving effective-epimorphy in this sense is different from preserving colimits. Although if the category does have pullbacks, then you can use the joint-kernel-pair definition of effective-epimorphy, and a functor that preserves pullbacks and colimits will then preserve it.

James Deikun (Jun 10 2025 at 15:59):

Would a flat functor that preserves colimits preserve effective-epimorphy?

Mike Shulman (Jun 10 2025 at 17:19):

Yes!

John Onstead (Jun 10 2025 at 19:43):

What would even be an example of a flat functor that preserves colimits, let alone a faithful (or better yet fully faithful) functor into a bicomplete category that is flat and preserves colimits? None of the Yoneda embeddings work- the usual one doesn't preserve colimits, and the dual one doesn't preserve limits so it can't be flat. Furthermore there's no idea of a sheaf such that the sheafified Yoneda embedding would be both colimit preserving and fully faithful. It almost seems like a no-win scenario!

John Baez (Jun 10 2025 at 19:51):

John Onstead said:

What would even be an example of a flat functor that preserves colimits,...

If $R\mathsf{Mod}$ is the category of modules of some commutative ring and $M$ is some module, the functor $M \otimes - : R\mathsf{Mod} \to R\mathsf{Mod}$ always preserves colimits because it's a left adjoint. This functor is flat if $M$ is a flat module. The link provides lots of examples of flat modules, hence lots of examples of what you seek - and in a context which, I believe, is near the origin of the term 'flat'.

John Onstead (Jun 10 2025 at 21:26):

John Baez said:

The link provides lots of examples of flat modules, hence lots of examples of what you seek - and in a context which, I believe, is near the origin of the term 'flat'.

That's quite interesting! It always seems a lot of terminology used in category theory comes from other places, usually abstract algebra or topology, so it's interesting to see where the term of "flat" comes from.

John Onstead (Jun 10 2025 at 21:33):

After some thought, I realized that maybe trying to find an embedding that preserves colimits and pullbacks might be too strong and unnecessary of a condition to require. TL;DR: there's a sense in which the effectiveness of a sink can be preserved in spite of a functor not preserving pullbacks!
For instance, let's say you had a joint effective epimorphism in a category $C$ without all pullbacks or all colimits (but still where the effectivity is given by the coequalizer of the joint kernel). Then the embedding $C \to [C, \mathrm{Set}]^{op}$ will not preserve it (since pullbacks aren't preserved), but it does preserve all joint regular epimorphisms since it preserves colimits. But since this category has all pullbacks, automatically the preserved regular epimorphism becomes effective, since the two concepts converge in any category with pullbacks. It doesn't matter that the embedding does not preserve pullbacks, the only thing that matters is that the domain category has pullbacks in the first place.

John Onstead (Jun 10 2025 at 21:36):

So my revised question is: given a sink $[U_i \to X]$ in a category $C$ without limits or colimits satisfying the universal property of being effective epimorphic as described above, will the coequalizer of the joint kernel of the sink $y([U_i \to X])$ as computed in $[C, \mathrm{Set}]^{op}$ always be isomorphic to $y(X)$? (with $y$ the dual Yoneda embedding)

Mike Shulman (Jun 11 2025 at 04:29):

John Onstead said:

Furthermore there's no idea of a sheaf such that the sheafified Yoneda embedding would be both colimit preserving and fully faithful.

Of course, that's part of why people usually pass to universally effective-epimorphic sinks, obtaining a Grothendieck topology for which we can consider sheaves.

Mike Shulman (Jun 11 2025 at 04:31):

John Onstead said:

So my revised question is: given a sink $[U_i \to X]$ in a category $C$ without limits or colimits satisfying the universal property of being effective epimorphic as described above, will the coequalizer of the joint kernel of the sink $y([U_i \to X])$ as computed in $[C, \mathrm{Set}]^{op}$ always be isomorphic to $y(X)$? (with $y$ the dual Yoneda embedding)

I think basically the same argument that you gave in your previous comment should also show this.

John Onstead (Jun 11 2025 at 09:07):

Mike Shulman said:

I think basically the same argument that you gave in your previous comment should also show this.

Great, seems reasonable. Thanks!

John Onstead (Jun 11 2025 at 11:22):

As I was working on this topic, I realized there might be an interesting connection to draw between joint effective sinks and descent, and I want to know if I'm on to something. Of course, there's an obvious connection: a sieve is literally the coequalizer of the joint kernel for some sink in the presheaf category, so localizing at them is in some sense "making" a sink effective epimorphic. And as mentioned above, the universal joint effective epimorphic sinks are those in the canonical topology. But there's a more subtle connection I wanted to go over (that might be related to this).

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

Let $C$ be a category with a topology such that the sink $[U_i \to X]$ is a part of it. Now, define a stack with respect to this topology $C^{op} \to \mathrm{Cat}$, and then apply the Grothendieck construction to get a fibration $F: E \to C$. My question is: given a sink $S$ in $E$, if $F(S) = [U_i \to X]$ (one of the sinks in the topology), does this automatically imply that $S$ is joint regular or effective epimorphic in $E$? (If $C$ or $E$ doesn't have the required pullbacks or coproducts, then this extends to using the definition of joint effective epi by the universal property given above). Furthermore, what is the relation between sinks $S$ with this property and descent data? It seems there should be a connection- descent data involves transition functions giving agreement on overlaps, and taking the coequalizer of a joint kernel pair also involves gluing on overlaps. These seem too close for coincidence!

Mike Shulman (Jun 11 2025 at 15:48):

Well, a sink is effective-epi exactly when all representable presheaves (or their corresponding discrete fibrations) satisfy (effective) descent with respect to it. That's where I would say the two notions of "gluing" meet.

John Onstead (Jun 11 2025 at 19:30):

I understand, but I am a little confused about how this helps with the above problem. Specifically, I'm asking about the stack itself (which we are already assuming to satisfy effective descent with respect to some sinks in the base category) and about the conditions in which the stack's corresponding Grothendieck fibration contains effective epi sinks "over" the sinks in the base category's topology.

John Onstead (Jun 11 2025 at 19:33):

For instance, a motivating problem for this could be vector bundles. We know a locally trivial vector bundle has a morphism into it from a set of trivial bundles, and this corresponding sink sits "over" some open cover in the category of topological spaces (as vector bundles are a stack on topological spaces). My question in this context would be if this particular sink of trivial bundles actually genuinely "covers" the locally trivial bundle via being effective epi.

Mike Shulman (Jun 11 2025 at 20:10):

I was answering your second question:
John Onstead said:

Furthermore, what is the relation between sinks $S$ with this property and descent data?

not the first one about epimorphy in the total category of a stack.

Mike Shulman (Jun 11 2025 at 20:15):

For the first one, I think the condition $F(S) = [U_i \to X]$ certainly won't be enough, e.g. the domains of all the morphisms in $S$ could be the initial objects of their fibers. But I suspect it would work if you assume also that all the morphisms in $S$ are cartesian.

John Onstead (Jun 11 2025 at 20:42):

Mike Shulman said:

I was answering your second question:

Ah, my apologies.

John Onstead (Jun 11 2025 at 20:44):

Mike Shulman said:

But I suspect it would work if you assume also that all the morphisms in $S$ are cartesian.

I can't help but wonder if this ties into the definition of a stack in terms of effective descent data- that is, the descent data can be seen as pullbacks of some global data along the morphisms of a sink in the topology of the site the stack is defined over.

Mike Shulman (Jun 11 2025 at 20:50):

Certainly!

John Onstead (Jun 13 2025 at 11:31):

As I was reviewing the previous discussion on generators, I wanted to know if there was a way to approach that subject from the joint epi perspective. Given a functor (that might be a generator) $F: D \to C$, let an "$F$-sink" of an object $X$ of $C$ be the sink of all morphisms from objects in the image of $F$ into $X$. It's already well established that $F$ is the weakest notion of generator (where the restricted Yoneda is faithful) exactly when every $F$-sink is a joint epimorphism. It's also established that $F$ is an extremal ("strong") generator precisely when every $F$-sink is a joint extremal epimorphism. (these follow from this article)

John Onstead (Jun 13 2025 at 11:33):

My question is about the other kinds of epimorphism. What is $F$ called when every $F$-sink is joint regular or joint effective epimorphic? Since any colimit is computed by a coproduct followed by a coequalizer, every colimiting cocone is a joint regular epi. Since any object in $C$ can be computed from colimits of objects in a colimit-dense generator, it seems that if $F$ is a colimit-dense generator (including dense generators), then automatically every $F$-sink is joint regular epi. I'm wondering then if these two conditions- $F$ is a colimit-dense generator and every $F$-sink is joint regular epi- are wholly equivalent in the same way plain generators are related to plain joint epis. I'm also wondering then if "$F$ is a dense generator" is equivalent to "every $F$-sink is joint effective epi" in that same way (after all, the coequalizer of the joint kernel pair acts as a "canonical" way to do colimits, perhaps this relates to the fact density implies every object is given via a canonical colimit)

Mike Shulman (Jun 13 2025 at 16:59):

It certainly won't work for density because density also involves the morphisms in $D$ whereas your notion of $F$-sink depends only on the objects of $D$.

Mike Shulman (Jun 13 2025 at 17:01):

I don't think it'll work for colimit-dense either, since colimit-density allows iterated colimits to reach any object, whereas joint regular epimorphy of $F$-sinks is a "one-step" statement. Moreover, just being a regular epi as a property doesn't say anything about what the domain of the morphisms it coequalizes is, whereas colimit-dense would require that to also come from $D$.

John Onstead (Jun 13 2025 at 21:28):

Mike Shulman said:

I don't think it'll work for colimit-dense either, since colimit-density allows iterated colimits to reach any object, whereas joint regular epimorphy of $F$-sinks is a "one-step" statement.

Ah I think there might be a mixup with the language, which I should have clarified. When I said "colimit-dense", I meant that any object can be written as a colimit of objects in the generator in some way (maybe not in a canonical way). I'm pretty sure iterated colimits come in only when it comes to "colimit generators", which are the closure of some generator under colimits (IE, the smallest subcategory closed under colimits containing the image of the generator).

John Onstead (Jun 13 2025 at 21:28):

Interestingly, I found this article which I think uses perhaps cleaner terminology. Using the terminology of this article, it seems a "regular generator" includes (or might be equivalent to) the article's "presenting generators", and "colimit-dense" as I described above is renamed "naive colimit generator". Colimit generators are thus renamed "iterated colimit generators". The article seems to confirm that every naive colimit generator is a regular generator, but the converse does not always seem to be true interestingly enough!

Mike Shulman (Jun 14 2025 at 02:55):

Oh, sorry, I misremembered the terminology. But I'm still doubtful, because joint regular epimorphy says that the map from the coproduct is a coequalizer of some pair of maps into it, whereas colimit-density would require something like that those maps being coequalized are also in the image of $F$.

John Onstead (Jun 14 2025 at 04:30):

That makes sense; indeed the linked article shows a "naive" generator/colimit-dense generator is a special case of a presenting generator, but says the converse is not always true. Interestingly enough, even though presenting generators require the domain of the coequalizer to be a coproduct of objects in the generator, and a regular generator does not, the article seems to suggest the two are still equivalent. For instance, if $d$ is the domain of the coequalizer, we can use the "plain" generator condition to generate an epimorphism from a coproduct of generator objects into $d$, then compose with the pair of morphisms from $d$ into the codomain coproduct. Since coequalizers are unchanged by precomposing with an epimorphism, this properly allows us to realize any regular generator as a presenting generator. Quite cool!

Mike Shulman (Jun 14 2025 at 04:55):

I guess that works!

John Onstead (Jun 17 2025 at 11:16):

Since it was brought up earlier, I actually did want to get more into iterated colimit generators, since I find them interesting for a number of reasons. First, in one of the articles it states that in a category with certain properties (it's well copowered, and also bicomplete) a strong/extremal generator converges with an iterated colimit generator. My first question about this is: does this result generalize from generators to arbitrary sinks? That is, in a well copowered bicomplete category, does the existence of some joint strong (or extremal, but those are the same in a complete category) epimorphic sink $U_i \to X$ imply that $X$ can be constructed as an iterated colimit of the $U_i$?

Mike Shulman (Jun 17 2025 at 16:02):

I kind of doubt it, but you'd have to look at the proof and see whether it generalizes.

John Onstead (Jun 17 2025 at 19:39):

Mike Shulman said:

you'd have to look at the proof and see whether it generalizes.

It doesn't seem like there'd be any major differences in the proof if we move from generators to sieves. Here's how it might be done. Given some joint strong epimorphic sink $U_i \to X$, we can convert this situation into a generator situation. Of course, in most cases, the $U_i$ will not be strong generators of the whole category, but that's not important- they will be strong generators of a subcategory of $C$ that will include our object $X$! So all we have to do is take that subcategory, recognize $U_i$ as strong generators of that subcategory, and then apply the proof there. If all the conditions are met, since $X$ is in the subcategory, it follows that it can be constructed as an iterated colimit.

John Onstead (Jun 17 2025 at 19:39):

The only issue is that the subcategory might not be cocomplete and finitely complete. If this proof works out then it obviously would be cocomplete (since by definition of colimit closure it would be the smallest subcategory closed under colimits containing $U_i$). But maybe it just suffices to take the needed colimits and limits in $C$.

Mike Shulman (Jun 17 2025 at 21:09):

John Onstead said:

The only issue is that the subcategory might not be cocomplete and finitely complete.

Yes, that's where I would expect it to go wrong.

John Onstead (Jun 17 2025 at 21:20):

I think I see the issue, so this seems to work best when the subcategory inclusion is not only fully faithful (so it reflects colimits/limits), but also preserves them (so that it creates limits). But this brings up another interesting question (and for some reason I can't seem to find any concrete answers to it online, which is honestly a bit confusing to me!)

John Onstead (Jun 18 2025 at 03:14):

What are the necessary requirements of a functor to preserve the strongness of an epimorphism? By this, I mean that if $f$ is strong, then $F(f)$ is also strong in the codomain category, not necessarily that the strongness is preserved "on the nose" (like how I showed above that effectiveness can be preserved by a functor that does not directly preserve effective epis). Of course, as a bare minimum, the functor has to preserve epis, but what else is needed- does it also need to preserve monos (since strong epis are defined in terms of them)? Or is there some other condition that is needed?

Mike Shulman (Jun 18 2025 at 03:21):

Actually I don't think such a functor has to preserve epis either; it might take all strong epis to strong epis, but take some non-strong epis to maps that aren't even epi.

Mike Shulman (Jun 18 2025 at 03:21):

I don't really know of a condition other than "preserves strong epis" (which is, of course, a condition).

Mike Shulman (Jun 18 2025 at 03:22):

Unless you add extra hypotheses, such as assuming that all strong epis are regular and the functor preserves coequalizers.

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

Mike Shulman said:

I don't really know of a condition other than "preserves strong epis" (which is, of course, a condition).

That's certainly quite odd this property hasn't been studied too well! There's no idea of a "lifting property/orthogonality-preserving" functor? That seems weird especially with the extensive use of these lifting properties in model category theory, where it might make sense to study such a notion as a potential characteristic of a "model structure-preserving" functor.

John Onstead (Jun 18 2025 at 11:26):

Maybe it'd be easier to consider preservation of extremal epis. Let's say you had a full embedding $D \to C$ that preserves monomorphisms- then within the image of $D$ there cannot be any violations of the extremal property. However, given an extremal $f: A \to B$ in $D$, it might in $C$ factor into a monomorphism $X \to B$ where $X$ is an object outside the image of $D$. There doesn't seem on the surface to be anything requiring this mono to be an iso, which could lead to violations of the extremal condition, so it doesn't seem being monomorphism preserving (or even mono creating) is sufficient here either. Would you know of any equivalent characterizations of an extremal preserving functor?

James Deikun (Jun 18 2025 at 11:27):

Normally a "model-structure preserving" functor preserves either the left class or the right class of each factorization system, not some combination of both. (c.f. [[Quillen adjunction]])

Mike Shulman (Jun 18 2025 at 14:15):

John Onstead said:

That's certainly quite odd this property hasn't been studied too well!

I didn't mean it hasn't been studied, rather that "preserves strong epis" is the condition itself that is studied.

John Onstead (Jun 18 2025 at 20:23):

James Deikun said:

Normally a "model-structure preserving" functor preserves either the left class or the right class of each factorization system, not some combination of both. (c.f. [[Quillen adjunction]])

I see, I suppose that's why adjunctions play a role, so that the left adjoint preserves the left class (cofibration) and the right adjoint preserves the right class (fibrations).

John Onstead (Jun 18 2025 at 20:24):

Mike Shulman said:

I didn't mean it hasn't been studied, rather that "preserves strong epis" is the condition itself that is studied.

Ah, I see. So what would be some interesting examples of functors that preserve strong epis? For instance, does the standard example of a colimit preserving embedding $C \to [C, \mathrm{Set}]^{op}$ preserve strong epis?

Mike Shulman (Jun 18 2025 at 20:48):

Probably not unless they coincide with regular epis. (-:

Mike Shulman (Jun 18 2025 at 20:49):

To be honest I think the closest thing that's been studied significantly are "regular functors" between regular categories, which are those that preserve finite limits and strong epis. But of course in a regular category, strong epis coincide with regular epis. It's tricky to think of how you would prove that some functor preserves strong epis unless they coincide with regulare epis or it has a right adjoint that preserves monos.

John Onstead (Jun 18 2025 at 22:53):

Mike Shulman said:

unless they coincide with regulare epis or it has a right adjoint that preserves monos.

Ah, this is the kind of condition I was looking for. Now it makes more sense why adjoints are used to preserve lifting properties in model categories as mentioned above! Though it's quite sad it won't work to embed categories into nicer ones, since if the embedding has a right adjoint, the preservation of limits means that the category already has to have a lot of limits to begin with, and certainly not all categories will satisfy that.

John Onstead (Jun 19 2025 at 00:01):

I wanted to do an exercise to find more examples of extremal epi preservation, and so I wanted to prove the embedding $F: C \to \mathrm{Cauchy}(C)$ preserves extremal epis. Here's my attempt: The objects of the Cauchy completion of $C$ are pairs $(X, i)$ where $i$ is an idempotent. Let's assume $F$ does not preserve extremal epis, this means for some extremal epi $f: A \to B$ there exists $(X, i)$ such that $F(f)$ factors $m: (X, i) \to F(B) = (B, \mathrm{id}_B)$, with $m$ a non-isomorphism monomorphism. Breaking down $m$ into a commutative square in $C$ we get that $\mathrm{id}_B \circ U(m) = U(m) \circ i$, thus a commutative triangle $U(m) = U(m) \circ i$

John Onstead (Jun 19 2025 at 00:01):

We note that $U(m)$ is a monomorphism in $C$. Furthermore, with the other part of the factorization $g: F(A) \to (X, i)$, we realize $U(m) \circ U(g) = f$, and since $f$ is extremal epi by hypothesis, $U(m)$ must be an isomorphism. Thus, the commutative triangle above gains an inverse, and we conclude that $m$ must be an isomorphism. Therefore, we reject the hypothesis that $m$ is a nontrivial monomorphism and confirm that $F$ preserves extremal epimorphisms.

John Onstead (Jun 19 2025 at 00:01):

Hopefully there's no mistakes, please let me know if there are any!

Mike Shulman (Jun 19 2025 at 00:09):

I think the only "mistake" is an unnecessary use of proof by contradiction. (-:

John Onstead (Jun 19 2025 at 01:58):

Mike Shulman said:

I think the only "mistake" is an unnecessary use of proof by contradiction. (-:

Thanks! I guess I'm more used to setting up problems via a proof by contradiction :)

John Onstead (Jun 19 2025 at 01:58):

I promise this will be relevant to the discussion, but here's another thing I was wondering about: characterizing subfunctors of representables (sieves) in $[C^{op}, \mathrm{Set}]$ that preserve limits (send colimits in $C$ to limits in $\mathrm{Set}$). Of course, because Yoneda preserves monomorphisms and representables preserve limits, every subobject in $C$ will satisfy this condition. In addition, any subobject in the Karoubi envelope of $C$ will too, since it shares the same presheaf category as $C$. But is that it? I have a conjecture that it is: a subfunctor of a representable preserves limits if and only if it is representable in some way, IE, an object in the Karoubi envelope of $C$. I've tried to prove this explicitly but I'm not making much progress. Any thoughts?

Mike Shulman (Jun 19 2025 at 04:16):

On one hand, if $C^{\rm op}$ satisfies the hypotheses of some adjoint functor theorem, then any limit-preserving functor $P:C^{\rm op} \to \rm Set$ has a left adjoint $F$, and hence is representable by $F1$.

On the other hand, it seems unlikely that this will be true in general: if $C^{op}$ doesn't have very many limits at all, then knowing that some functor preserves all of them isn't very useful.

John Onstead (Jun 19 2025 at 11:16):

Mike Shulman said:

On the other hand, it seems unlikely that this will be true in general: if $C^{op}$ doesn't have very many limits at all, then knowing that some functor preserves all of them isn't very useful.

Hmm, I see. I'll think more on it. But I do have something interesting to share that I realized while thinking about extremal epis in presheaf categories: The Yoneda embedding does not preserve extremal epimorphisms, ironically due to the Yoneda lemma itself! Given any morphism $f: A \to B$ in $C$, its inclusion into a sieve $S$ automatically induces a natural transformation $y(A) \to S$ (by the Yoneda lemma) that factors $y(f)$ through the sieve inclusion. Therefore, if $f$ is an extremal epimorphism included in some sieve, $F(f)$ will factor through the sieve with the non-iso sieve monomorphism into $y(X)$- exactly the condition that $y(f)$ is NOT an extremal epimorphism! How ironic!

John Onstead (Jun 19 2025 at 11:18):

Here's another approach to understanding strong/extremal preservation. Let $G: J \to C$ be the full subcategory spanned by objects that form a strong generator of $C$. The restricted Yoneda functor $y_G: C \to [J^{op}, \mathrm{Set}]$ is by definition both conservative and faithful. Say that a functor $F: C \to D$ "preserves" the strong generator if $F \circ G$ is a strong generator- that is, $y_{F \circ G}: D \to [J^{op}, \mathrm{Set}]$ is conservative and faithful. Can we say anything in general about functors $F$ that do this (other than the obvious fact that if $F$ preserves strong/extremal epis, it will also preserve strong generators)? After all I suspect that $F$ can fail to preserve strong/extremal epis yet still preserve strong generators, making it a weaker condition that more functors might satisfy.

Mike Shulman (Jun 19 2025 at 15:10):

John Onstead said:

Therefore, if $f$ is an extremal epimorphism included in some sieve, $F(f)$ will factor through the sieve with the non-iso sieve monomorphism into $y(X)$

Right, as long as $f$ is contained in some proper sieve. Do you know a simpler condition on $f$ that ensures it is contained in some proper sieve?

Mike Shulman (Jun 19 2025 at 15:11):

John Onstead said:

other than the obvious fact that if $F$ preserves strong/extremal epis, it will also preserve strong generators

Only if it's also essentially surjective, no?

John Onstead (Jun 19 2025 at 20:27):

Mike Shulman said:

Right, as long as $f$ is contained in some proper sieve. Do you know a simpler condition on $f$ that ensures it is contained in some proper sieve?

Isn't it true that any general morphism $f$ is contained in the sieve generated by it? That is, given a morphism $f$, the canonical sieve containing it would consist of all morphisms into its codomain that factor through $f$.

John Onstead (Jun 19 2025 at 20:28):

Mike Shulman said:

Only if it's also essentially surjective, no?

Oh shoot! I forgot to include the condition that $F: C \to D$ should also be a strong generator in $D$ (or better yet a dense generator). That way if it preserves strong epis, you can compose the joint strong epis from the objects in the image of $F$ into $D$ with the joint strong epis from the objects in the image of $G$ into those of $C$. Unlike regular epis the strong ones compose, and so with this assumption I think it should be true that if $F$ is both a strong generator and preserves strong epis, then it will preserve strong generators. So to restate my question properly, it's what the conditions are on a strong generator $F$ to preserve another strong generator into its domain- that is, when is the composition of two strong generators also a strong generator?

Mike Shulman (Jun 19 2025 at 20:38):

John Onstead said:

Mike Shulman said:

Right, as long as $f$ is contained in some proper sieve. Do you know a simpler condition on $f$ that ensures it is contained in some proper sieve?

Isn't it true that any general morphism $f$ is contained in the sieve generated by it?

Yes, but that sieve might not be proper.

Mike Shulman (Jun 19 2025 at 20:39):

John Onstead said:

when is the composition of two strong generators also a strong generator?

I don't know, but it sounds like a question that someone might have studied.

John Onstead (Jun 19 2025 at 23:23):

Mike Shulman said:

Yes, but that sieve might not be proper.

What is a proper sieve? Even when I specify "category theory" in my Google search it only turns up links to buy kitchen sieves!

John Onstead (Jun 19 2025 at 23:23):

Mike Shulman said:

I don't know, but it sounds like a question that someone might have studied.

I've looked and I haven't found much, but I'll keep at it. In the meantime, I've made the simplifying assumption that $F$ be dense and have some progress. If $G: J \to C$ is strong, then $y_G: C \to [J^{op}, \mathrm{Set}]$ is conservative and faithful. Also, $y_G = G^* \circ y$ where $G^*: [C^{op}, \mathrm{Set}] \to [J^{op}, \mathrm{Set}]$ is composition by $G$. Of course, $y$ is fully faithful, but it's possible to compose a fully faithful functor with a non-faithful, non-conservative one to get a faithful conservative functor so long as the "bad stuff" happens outside the image of $C$. We then find that if $F: C \to D$ is dense, there is a fully faithful functor $D \to [C^{op}, \mathrm{Set}]$, and $F \circ G$ is strong precisely when the composition of this embedding with $G^*$ is conservative and faithful. Therefore, $F$ preserves the strong generator if and only if $D$ lies within $E$, the maximal full subcategory of $[C^{op}, \mathrm{Set}]$ such that the composite of its inclusion and $G^*$ is faithful and conservative.

Mike Shulman (Jun 19 2025 at 23:24):

John Onstead said:

What is a proper sieve?

Same as a proper subset or a proper subobject: a sieve that doesn't contain all the morphisms with a given codomain.

Mike Shulman (Jun 19 2025 at 23:26):

You claimed that "$F(f)$ will factor through the sieve with the non-iso sieve monomorphism into $y(X)$", but the monomorphism is only non-iso if the sieve is proper.

John Onstead (Jun 19 2025 at 23:34):

Mike Shulman said:

Same as a proper subset or a proper subobject: a sieve that doesn't contain all the morphisms with a given codomain.

Ah ok, so a non-maximal sieve.

Mike Shulman said:

You claimed that "$F(f)$ will factor through the sieve with the non-iso sieve monomorphism into $y(X)$", but the monomorphism is only non-iso if the sieve is proper.

Well, $f$ probably shouldn't be an isomorphism. But maybe even if $f$ is a split epimorphism this might happen. If $f$ is a split epi, then there is a map $s: B \to A$ such that $f \circ s = \mathrm{id}_B$. So for any map into $B$, given by $g: X \to B$, we have $g = \mathrm{id}_B \circ g = (f \circ s) \circ g$. But that means every $g$ will factor through $f$ and be included. So I think as long as $f$ isn't a split epi it will have a proper sieve.

Mike Shulman (Jun 19 2025 at 23:51):

Right!

Mike Shulman (Jun 19 2025 at 23:51):

So the Yoneda embedding doesn't preserve non-split extremal epis. But it does preserve split epis, since any functor does.

John Onstead (Jun 20 2025 at 02:12):

Mike Shulman said:

So the Yoneda embedding doesn't preserve non-split extremal epis. But it does preserve split epis, since any functor does.

Ah, that makes sense! I was actually thinking of a version of this using effective epis. By definition, a morphism where the "sieve" (coequalizer of a joint kernel pair) that includes it is an isomorphism is effective epi. The only effective epis in the image of the Yoneda embedding are the split epis, since any functor preserves them. So those are precisely the morphisms that make the sieve an isomorphism.

John Onstead (Jun 20 2025 at 04:09):

Let $G: J \to C$ be a strong generator and $G^*: [C^{op}, \mathrm{Set}] \to [J^{op}, \mathrm{Set}]$ the induced functor. It has a right adjoint $R: [J^{op}, \mathrm{Set}] \to [C^{op}, \mathrm{Set}]$ given by right Kan extension. Let $E$ denote the subcategory of $[C^{op}, \mathrm{Set}]$ on the fixed points of the monad $R \circ G^*$ (IE, the fixed points of the adjunction). The question is: is the resulting functor $E \to [J^{op}, \mathrm{Set}]$ conservative and faithful (again keeping in mind that $G$ is a strong separator)?

John Onstead (Jun 20 2025 at 07:40):

Scratch that, this construction wouldn't work for what I want regardless. Here's a counter-example I found. Let $C = \mathrm{Ab}$ and $\mathbb{Z}: * \to \mathrm{Ab}$ be the functor picking out the integers, which are a strong generator in abelian groups. The restricted yoneda embedding $\mathrm{Ab} \to [*, \mathrm{Set}] = \mathrm{Set}$ is just the usual forgetful functor. The induced functor (yes, ignore size concerns) $[\mathrm{Ab}^{op}, \mathrm{Set}] \to \mathrm{Set}$ has right adjoint $R$, but whatever it does (probably not the free group functor!) the resulting $E$ cannot be the maximal subcategory of $[\mathrm{Ab}^{op}, \mathrm{Set}]$ such that the functor into $\mathrm{Set}$ is conservative and faithful, which is what I wanted. That's because if it lands in the representables, the representables would be a larger example, and if it lands outside, then it will miss some representables, which would all need to be in a hypothetical maximal subcategory. Thus, either way we fail, and so we have a counterexample to $E$ always being maximal in this regard.

John Onstead (Jun 20 2025 at 20:39):

I promise this is my last question on this topic, but I figured I'd give it one last try! My question is simply if the following conjecture is true or not. If the result isn't known, please provide a strategy that I might be able to use to derive the proof, since everything I've already tried on my own led to dead ends.

John Onstead (Jun 20 2025 at 20:39):

Conjecture: For any strong generator $G: J \to C$, there exists a reflective subcategory $i: E \to [C^{op}, \mathrm{Set}]$ such that the Yoneda embedding factors into fully faithful $y: C \to E$ and such that $G^* \circ i$ is faithful and conservative.

(Corollary: For any strong generator, there exists an embedding into a bicomplete category- and quite possibly a locally presentable and hence well copowered category- that preserves the strongness of the generator. Corollary of the corollary: There's always a canonical way to enlarge a category so that the notions of strong and colimit closed generator converge.)

Mike Shulman (Jun 20 2025 at 21:22):

I don't have any idea whether that's true. I don't immediately see any reason to think it might be.

Morgan Rogers (he/him) (Jun 23 2025 at 13:42):

That conjecture has the flavour of a 'denseness of sites' condition, cf for example Caramello's paper, section 4. Since these are topos-theoretic results, this is specifically for the case of lex reflective subcategories, and I expect that one would need universality of the cocones for that case, as discussed previously.

John Onstead (Jun 24 2025 at 01:18):

Morgan Rogers (he/him) said:

That conjecture has the flavour of a 'denseness of sites' condition, cf for example Caramello's paper, section 4. Since these are topos-theoretic results, this is specifically for the case of lex reflective subcategories, and I expect that one would need universality of the cocones for that case, as discussed previously.

Interesting, I'll have to check that out! But yes, in the above "conjecture" the reflective subcategory need not be a sheaf topos. I'll even take a non-reflective subcategory so long as it's cocomplete and finitely complete.

John Onstead (Jun 24 2025 at 01:19):

Also as an update, while I've continued to work on this, in my work I've come across a related problem that I want to investigate more. So I might post about that at some point soon!