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Stream: learning: questions

Topic: Generators vs Covers in Categories

John Onstead (May 08 2025 at 11:22):

Inspired by the previous discussion, I eventually stumbled into this article about "generators" by Mike Shulman. A few different notions of generator are presented, all dealing with some set of objects $F: J \to C$. The first kind is the "basic generator", where the restricted Yoneda embedding is faithful, or where there is some map from the coproduct of the objects into any object of the category that is epic. There are then dense generators which I am familiar with, but also "colimit-dense" generators which I just assumed were the same as dense generators, but apparently not (something I'm still a little confused about- "all objects are colimits of objects in the generator"- that's density, no? How not?) Lastly there are "colimit-closed" generators where the colimit closure is the category itself (ironically, if anything should be called "dense", it's this kind of generator since it seems the most analogous to the notion of density from topology)

John Onstead (May 08 2025 at 11:23):

In any case, these notion of generator seem to almost be ways in which some collection of objects can "cover" all objects in the category (IE, in the first case there seems to even be a sort of "joint epimorphism" condition). This immediately brings to mind the notion of a coverage on a category, making me wonder if there is a connection. To set one up, given a set of objects $F$ in $C$, let an "$F$-coverage" be a cover of $C$ such that for every object $X$ in $C$, one of its covering families is the set of all morphisms from the objects in $F$ to $X$.

John Onstead (May 08 2025 at 11:24):

My question is then: if $F$ is a basic generator, dense generator, colimit-dense generator, or colimit-closed generator, what are the corresponding properties of an $F$-coverage and of its category of sheaves? (if there are any such corresponding properties; "corresponding" meaning that they are properties that hold for $F$-coverages specifically when $F$ is a (whatever type) generator and that never hold when $F$ is not). Also, at which point in this hierarchy does an $F$-coverage become subcanonical?

Mike Shulman (May 08 2025 at 15:57):

John Onstead said:

"all objects are colimits of objects in the generator"- that's density, no?

Density means all objects are colimits of generating objects in a canonical way, not just an arbitrary way.

Mike Shulman (May 08 2025 at 15:58):

I don't understand your notion of F-coverage; are you defining a property of coverages so that for any F there could be many F-coverages?

John Onstead (May 08 2025 at 19:13):

Mike Shulman said:

Density means all objects are colimits of generating objects in a canonical way, not just an arbitrary way.

Ahhh!

John Onstead (May 08 2025 at 19:14):

Mike Shulman said:

I don't understand your notion of F-coverage; are you defining a property of coverages so that for any F there could be many F-coverages?

That's what I was going for, but perhaps to avoid confusion we can just define "the" $F$-coverage to be the coverage where every object $X$ has one covering family, that being the set of all morphisms into it from the objects in $F$. It's meant to be analogous to/an extension of the coverage where the covering family for an object $X$ is the set of all morphisms into it (which I think gives something equivalent, in some sense, to the trivial topology). I guess that coverage would be an $F$-coverage where $F = \mathrm{id}$.

Mike Shulman (May 08 2025 at 20:29):

That won't in general be a coverage.

John Onstead (May 08 2025 at 21:25):

Mike Shulman said:

That won't in general be a coverage.

Hmm... so if not this notion of "$F$-coverage", what coverage would best encapsulate the idea of viewing a "generator" (in any of the above senses) as a set of objects that "covers" all the other objects?

Mike Shulman (May 08 2025 at 23:48):

What's wrong with all the existing definitions of "generator"?

John Onstead (May 09 2025 at 00:12):

Mike Shulman said:

What's wrong with all the existing definitions of "generator"?

Nothing's wrong with them- I guess I just wanted to know if there was a perspective on generators from the coverage point of view.
I think I need to think this through a little bit more!

Morgan Rogers (he/him) (May 09 2025 at 09:00):

John Onstead said:

Lastly there are "colimit-closed" generators where the colimit closure is the category itself (ironically, if anything should be called "dense", it's this kind of generator since it seems the most analogous to the notion of density from topology)

The "closed" is indicating that taking colimits doesn't add anything, as opposed to "dense" meaning that taking colimits produces everything! For example, 1 is colimit-closed (but limit-dense) in FinSet^op.

John Onstead (May 09 2025 at 21:24):

Morgan Rogers (he/him) said:

The "closed" is indicating that taking colimits doesn't add anything, as opposed to "dense" meaning that taking colimits produces everything! For example, 1 is colimit-closed (but limit-dense) in FinSet^op.

That's interesting! So I hope I'm understanding this correctly... being colimit closed means you have to consider the whole category if you want to take colimits of the objects in the generator- and colimits of those and so on- and not have to worry about the colimits not existing. So that if you take some subcategory of the category, taking enough repeated colimits might eventually bring you to an object outside that subcategory. But it doesn't imply that all objects of the whole category can be achieved in this way, expressed as colimits of the objects in the generator.

John Onstead (May 09 2025 at 21:26):

I think I brought up density in topology because of the analogy with closure operators- the topology closure operator applied to a dense subset is the whole topology, while the colimit closure operator applied to a colimit-closed generator is the whole category.

Mike Shulman (May 09 2025 at 21:33):

The analogy with topology can point either way, because classical topological spaces have the property that closure is idempotent: the closure of a set is closed. So saying "X is the smallest closed set containing G" and "X is the closure of G" are the same. But there are weaker notions of "topology" for which this isn't true. For instance, in a [[pretopological space]] there is a notion of "preclosure" that is not idempotent, so saying that a set is "dense" in the sense that its preclosure is the whole space is stronger than saying that the space is the smallest closed set containing those points.

John Onstead (May 13 2025 at 11:30):

Here's a more precise question that I think might be a better starting point. The Yoneda lemma states that an object can be characterized up to isomorphism by all morphisms into it, but this is a "worst case scenario". Often, you can find a more efficient set of morphisms that can do the same job, such as considering only morphisms from the terminal object in $\mathrm{Set}$. More generally you can consider morphisms from a dense subcategory of a category (the singleton is a dense subcategory of $\mathrm{Set}$, making that a special case).

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

My question is: are dense generators the only generators that satisfy this property of their morphisms characterizing all objects up to isomorphism, or can other kinds of generators do that? And fundamentally, what does it even mean for a class of morphisms to determine an object up to isomorphism- if I gave you a random set of morphisms into an object in some random category, how can you tell if it does or does not characterize the object up to isomorphism?

John Onstead (May 13 2025 at 11:32):

I do have a conjecture: given $U_X$ some set of morphisms into $X$, "$U_X$ determines $X$ up to isomorphism" $\iff$ "The sieve generated by $U_X$ is naturally isomorphic to $\mathrm{Hom}(-, X)$". It's a good first guess, since by Yoneda this would directly guarantee that the object is determined uniquely up to isomorphism. It seems to work for the case of the identity- the identity determines an object fully, and indeed the covering sieve generated by the identity is isomorphic to the representable. But I can't seem to figure out how it works for the singleton set, or for any other example...

John Baez (May 13 2025 at 15:30):

John Onstead said:

And fundamentally, what does it even mean for a class of morphisms to determine an object up to isomorphism- if I gave you a random set of morphisms into an object in some random category, how can you tell if it does or does not characterize the object up to isomorphism?

They always characterize the object, not just up to isomorphism but up to equality. Just look at their common target! :crazy:

So, your question needs to be formulated a bit differently....

Kevin Carlson (May 13 2025 at 17:19):

I think in particular you need to let $X$ vary for the question to make any sense. It seems best to focus on the question: when does a class $\mathcal G$ of objects in a category $C$ characterize all the objects of $C$ up to isomorphism? One natural way to flesh this out is as “if the objects of $\mathcal G$ think two $X,Y$ are isomorphic, then they really are.” There are at least two reasonable ways I can think to flesh that out; maybe you’d like to have a crack at it.

John Onstead (May 13 2025 at 19:32):

Kevin Carlson said:

There are at least two reasonable ways I can think to flesh that out; maybe you’d like to have a crack at it.

Sure, I'll give it a try...

John Onstead (May 13 2025 at 19:33):

When I hear about an object that "thinks" two objects are isomorphic, I immediately think of a "local object". This is an object such that its hom functor sends a particular morphism between two objects into a bijection of hom sets. So perhaps this condition that "if the objects of $G$ think two $X, Y$ are isomorphic, then they really are" can be encoded as some sort of local object condition. But the issue is that locality only really makes sense when you start with some class of morphisms, with the class of local objects being determined from that, and also it only makes sense when the morphisms aren't isomorphisms already to begin with.

John Onstead (May 13 2025 at 19:33):

So that's not the answer... I did some more research and uncovered the concept of "reflecting isomorphisms". A presheaf that reflects isomorphisms means that a bijection of homs in Set (thinks that the objects are isomorphic) implies an isomorphism between the objects in the category (they actually are isomorphic). So maybe $G$ is the set of objects such that their homs are all isomorphism reflecting?

Kevin Carlson (May 13 2025 at 20:45):

That's at least one of the directions I was thinking of, but it's not quite a fully precise mathematical definition yet--you mean that each one of their hom functors reflects isomorphisms? If so, then there wasn't much point in having a whole set $\mathcal G,$ was there?

John Onstead (May 13 2025 at 21:27):

Kevin Carlson said:

That's at least one of the directions I was thinking of, but it's not quite a fully precise mathematical definition yet--you mean that each one of their hom functors reflects isomorphisms? If so, then there wasn't much point in having a whole set $\mathcal G,$ was there?

So maybe they somehow are "jointly" isomorphism reflecting? It could be then that the hom of the coproduct of all the objects in $G$ is isomorphism reflecting. But this can be translated into saying that the functor that sends an object $X$ to the product of all sets $\mathrm{Hom}(g, X)$ for $g$ in $G$ is isomorphism reflecting. I hope that's more precise!

Kevin Carlson (May 13 2025 at 21:29):

Yes, that's more like it! Perhaps a bit more neatly, you could say that the restricted Yoneda embedding $C\to [\mathcal G^{\mathrm{op}},\mathbf{Set}]$ reflects isomorphisms. This is half of the usual definition of a "strong generator" or a "strong separator", although there are important examples in which only this half is true, even though it doesn't have a very simple name.

John Onstead (May 13 2025 at 22:20):

Kevin Carlson said:

This is half of the usual definition of a "strong generator" or a "strong separator", although there are important examples in which only this half is true, even though it doesn't have a very simple name.

That's quite interesting... If I am remembering correctly, the full definition of a strong generator is that the restricted Yoneda embedding is faithful and conservative. Interestingly, a basic generator is a set with the other half alone- that the restricted embedding is faithful. So a set of objects with this "restricted embedding is conservative" property may not even be "generators" at all unless the restricted embedding is also faithful. That's certainly something interesting to think about!

John Onstead (May 13 2025 at 22:20):

Also, I'm surprised that the condition doesn't turn out to be entirely equivalent to density, which is what I was initially expecting (though thankfully it does include all examples of density). I guess that means morphisms from a set of objects in a category can fully determine other objects in the category without that set being dense!

Kevin Carlson (May 13 2025 at 23:14):

Yes, very much so! Even as simple a case as $\mathbb{Z}$ in abelian groups is not dense. The fullness of the restricted Yoneda embedding is an exceptionally strong condition.

John Onstead (May 13 2025 at 23:33):

Kevin Carlson said:

Yes, very much so! Even as simple a case as $\mathbb{Z}$ in abelian groups is not dense. The fullness of the restricted Yoneda embedding is an exceptionally strong condition.

$\mathbb{Z}$ is a case where the restricted embedding is conservative? That's interesting... I can see how morphisms from $\mathbb{Z}$ can be used to reconstruct the underlying set of an abelian group, but I'm not sure how you'd also get the abelian group structure on that set as well. That might be an interesting problem for me to think about for a bit!

John Onstead (May 13 2025 at 23:34):

Also in the meantime, I have a follow-up to the connection between conservative functors and localizations. Given a functor $F: C \to D$, define a class of morphisms $W$ in $C$ to be the morphisms $F$ sends to isomorphisms in $D$. Now, define a "conservatization" to be the localization of $C$ (and corresponding functor) at this class of morphisms $W$. My question is: is it true that the statements "the conservatization functor of $F$ is the identity (or maybe something weaker like an equivalence of categories) on $C$" and "$F$ is conservative" are equivalent?

Kevin Carlson (May 13 2025 at 23:44):

John Onstead said:

I can see how morphisms from $\mathbb{Z}$ can be used to reconstruct the underlying set of an abelian group, but I'm not sure how you'd also get the abelian group structure on that set as well.

Well, "conservative" does not in fact say that you can "get the abelian group structure on that set..."!

Kevin Carlson (May 13 2025 at 23:48):

John Onstead said:

Given a functor $F: C \to D$, define a class of morphisms $W$ in $C$ to be the morphisms $F$ sends to isomorphisms in $D$. Now, define a "conservatization" to be the localization of $C$ (and corresponding functor) at this class of morphisms $W$. My question is: is it true that the statements "the conservatization functor of $F$ is the identity (or maybe something weaker like an equivalence of categories) on $C$" and "$F$ is conservative" are equivalent?

Nice question! We can see that you're effectively asking: are the predicates "$F$ doesn't send any non-invertible arrows to isomorphisms" and "Localizing $F$'s domain at every arrow it inverts doesn't change that domain" equivalent? It should be more or less clear that, if $F$ "doesn't invert any arrows", then "inverting everything $F$ inverts" doesn't change anything. The converse is not quite as obvious: perhaps we should phrase it as "If $F$ does invert the non-isomorphism $f$, must the localization of $F$ at everything it inverts be a non-equivalence?" This helps see that an important subquestion, not quite equivalent to the whole question, is: is the canonical functor $C\to C[f^{-1}]$ into the localization at one arrow an equivalence if and only if $f$ is an isomorphism?

Kevin Carlson (May 13 2025 at 23:50):

You might find this a bit easier to get your hands on.

Mike Shulman (May 14 2025 at 00:02):

John Onstead said:

the full definition of a strong generator is that the restricted Yoneda embedding is faithful and conservative. Interestingly, a basic generator is a set with the other half alone- that the restricted embedding is faithful. So a set of objects with this "restricted embedding is conservative" property may not even be "generators" at all unless the restricted embedding is also faithful.

Right. However, here's another fun exercise: prove that if the category has equalizers, then if the restricted Yoneda embedding for some set of objects $G$ is conservative, it is automatically also faithful. So while it's technically true that conservativity of the restricted Yoneda embedding doesn't imply $G$ is a strong generator, in most practical situations there's not much difference.

John Onstead (May 14 2025 at 11:17):

Mike Shulman said:

However, here's another fun exercise: prove that if the category has equalizers, then if the restricted Yoneda embedding for some set of objects $G$ is conservative, it is automatically also faithful.

I think it's a general result that if $F: C \to D$ is conservative, and $C$ has equalizers and $F$ preserves them, then $F$ is also faithful. So this result should follow assuming the restricted Yoneda embedding preserves limits (the normal one does, so hopefully the restricted one does as well!)
I found a proof of the general result on this page. I think the gist is that if $C$ has equalizers, then two parallel morphisms $f,g: A \to B$ are equal only when $\mathrm{id}_B$ is their equalizer. If $F$ preserves equalizers then $F(f) = F(g)$ when $\mathrm{id}_{F(B)}$ is their equalizer in $D$. Then I guess the conservativity of $F$ ensures you can't get something like $F(f) = F(g)$ unless $f = g$ to begin with in $C$, thus ensuring the functor is faithful. My understanding is more than a little spotty but I hope I at least got the gist right!

John Onstead (May 14 2025 at 11:22):

Kevin Carlson said:

This helps see that an important subquestion, not quite equivalent to the whole question, is: is the canonical functor $C\to C[f^{-1}]$ into the localization at one arrow an equivalence if and only if $f$ is an isomorphism?

Ironically, reflecting isomorphisms seems to be helpful here! If the localization functor is an equivalence, then it reflects isomorphisms (as an equivalence is fully faithful). Therefore, any isomorphism that the localization "makes" must have actually already been an isomorphism to begin with! This means that, if the localization is an equivalence, the class of arrows must have consisted only of isomorphisms (or have been empty). Conversely, if we localize a non-isomorphism, then the localization functor must send a non-isomorphism to an isomorphism (since it is the universal such functor that sends the non-isomorphism to an isomorphism). Therefore, it cannot be conservative and thus cannot be an equivalence. On the other hand if we localize an isomorphism, the functor sends isomorphisms to isomorphisms, and so is conservative. I don't think that fully proves it must be an equivalence, but I'm not sure how I'd do that.

Kevin Carlson (May 14 2025 at 16:19):

You've proved that if the localization is an equivalence, the inverted arrows are isomorphisms, and if the inverted arrows are not all isomorphisms, then the localization is not an equivalence--but those two statements are just contrapositives of each other, not converses!

John Onstead (May 14 2025 at 19:03):

Kevin Carlson said:

You've proved that if the localization is an equivalence, the inverted arrows are isomorphisms, and if the inverted arrows are not all isomorphisms, then the localization is not an equivalence--but those two statements are just contrapositives of each other, not converses!

Ahh... well I don't know how to prove the converse. It's "intuitively true" in that, if the whole purpose of a localization is to invert morphisms, and the morphisms you want to invert are already invertible, then the process should do nothing. But gut feelings commonly lead astray in the world of math, and I'm not sure what a good proof would look like.

James Deikun (May 14 2025 at 19:06):

You should probably be making use of the universal property of a localization here.

John Onstead (May 14 2025 at 19:16):

James Deikun said:

You should probably be making use of the universal property of a localization here.

Thanks for the tip!

John Onstead (May 14 2025 at 19:16):

So the universal property says that for any functor $F: C \to D$ that sends the morphisms in the class to isomorphisms, there is a unique functor $G: C[W^{-1}] \to D$ such that the composition with $L: C \to C[W^{-1}]$ is $F$. Now, let's say that the class $W$ consists only of isomorphisms. The identity would be another such functor that sends all morphisms in this class to isomorphisms (trivially). Thus, there is a (unique) functor $G: C[W^{-1}] \to C$ such that $G \circ L = \mathrm{id}_C$.

John Onstead (May 14 2025 at 19:36):

I think I also have to show $L \circ G \cong \mathrm{id}_{C[W^{-1}]}$ to show this is a true equivalence. I'm not sure how exactly to do this, but perhaps this is where the uniqueness comes into play (since we only used the "existence" feature of the universal property above). Maybe something like... both $L \circ G$ and $\mathrm{id}_{C[W^{-1}]}$ precompose with $L$ to give $L$. So by some uniqueness there must be an isomorphism of the form $L \circ G \cong \mathrm{id}_{C[W^{-1}]}$. So I think this shows that there is an equivalence. Hopefully the proof is now complete!

James Deikun (May 14 2025 at 19:48):

Yeah, that works. Perhaps, though, it would have been even easier to show that the identity fulfills the universal property of the localization, though!

John Onstead (May 14 2025 at 20:43):

So to summarize, localizing at a class of isomorphisms gives an equivalence, and any equivalence that is a localization comes from localizing at a class of isomorphisms (or at an empty class). Going all the way back, I guess this proves my original idea that a functor is conservative if and only if the localization at the class of morphisms it sends to isomorphisms is an equivalence. So if it's not an equivalence, I guess such a thing would be a handy way to, in some sense, "measure" the failure of a functor to be conservative (similarly to how the codensity monad measures the failure of a functor to be codense)!

Kevin Carlson (May 14 2025 at 20:57):

Yes! And there's a natural follow-on question: if a functor $F:C\to D$ is not conservative, then $F$ has a nontrivial factorization $F':C[I^{-1}]\to D$ through the localization $C\to C[I^{-1}]$ at the class $I$ of arrows $F$ inverts. Is $F'$ then always conservative?

John Onstead (May 14 2025 at 21:38):

It's interesting you bring this up since it was something I was thinking about. $F'$ seems to follow from the universal property of localization, since both, by definition, send the same class of morphisms in $C$ to isomorphisms. Intuitively, it seems $F'$ should be conservative since the localization turns all the "counterexample" morphisms in $C$ (the non-isomorphisms that get sent to isomorphisms) into isomorphisms. But again this is a case of it being hard to formally prove.

John Onstead (May 14 2025 at 21:38):

One approach might be this. Let's assume there's a non-isomorphism $L(f)$ in the localization that gets mapped by $F'$ to an isomorphism in $D$. Since $F'(L(f)) = F(f)$, this means $F'L(f))$ is an isomorphism and thus $F(f)$ is an isomorphism. But we defined the localization so that any $f$ such that $F(f)$ is an isomorphism is an element of the class we invert with respect to. Thus, $f$ must be in this class, and thus $L(f)$ must be an isomorphism. But this contradicts the assumption that $f$ was a non-isomorphism. I don't know if this is enough since it assumes that the localization is always full (IE, all morphisms in the localization are of form $L(f)$)

James Deikun (May 14 2025 at 21:49):

Say you have the category that looks like $\bullet \to \bullet \leftarrow \bullet \to \bullet$. Then you take the functor $F$ into the walking split idempotent that sends the first arrow to the section, the second (leftward) arrow to the identity, and the third arrow to the retraction. What does $F'$ look like?

John Onstead (May 14 2025 at 23:49):

Hmm... well, only the middle morphism would get localized since it's the only one that $F$ sends to an identity. This gives a new category with 4 objects, and the same morphisms as before, but with four additional morphisms. First, the inverse between the second and third objects. Second, the composition between the first arrow and the inverse of the second. Third, the composition of the inverse and the third morphism. And finally, the composition of the first, inverse, and third morphism. So no, the localization is not always full, since we have 4 different morphisms that do not get mapped into by any from the original category.

John Onstead (May 14 2025 at 23:51):

I might be wrong about this, but it still seems that $F'$ is conservative. The only morphisms that get mapped to the identity in the walking split idempotent from the localization is still the isomorphism, and not any other morphisms.

James Deikun (May 15 2025 at 00:36):

What about the composition of the first, inverse, and third morphism? What morphism does it get mapped to?

John Onstead (May 15 2025 at 01:32):

It gets mapped to the composition of the section and the retraction (I think? This is the only morphism from the first to last object I can think of). It's not an isomorphism, as an isomorphism is both a section and a retraction, not the composition of a section with a retraction.

James Deikun (May 15 2025 at 03:31):

What makes a section a section, or a retraction a retraction?

John Onstead (May 15 2025 at 03:37):

James Deikun said:

What makes a section a section, or a retraction a retraction?

A section is a right inverse to some morphism, while a retraction is a left inverse. For the case of a section, given a morphism $f: A \to B$, $s: B \to A$ is a section if $f \circ s = \mathrm{id}_B$

James Deikun (May 15 2025 at 03:40):

Ok, so in the walking split idempotent, which has only 5 morphisms, which morphism is the section the section of?

John Onstead (May 15 2025 at 11:10):

James Deikun said:

Ok, so in the walking split idempotent, which has only 5 morphisms, which morphism is the section the section of?

It's not the retraction, is it? I'm a little confused, so just so we are on the same page... which category are you thinking of by "walking split idempotent"? The one I'm thinking of has 3 objects and a few more than 5 morphisms, maybe I'm thinking of the wrong category entirely.

James Deikun (May 15 2025 at 11:12):

You are. The walking split idempotent has 2 objects, the large and small object, and 5 morphisms: the idempotent, the section, the retraction, and two identities. I wonder which category you are thinking of.

John Onstead (May 15 2025 at 11:15):

Kevin Carlson said:

Well, "conservative" does not in fact say that you can "get the abelian group structure on that set..."!

I wanted to come back to this because I want some more clarification...

John Onstead (May 15 2025 at 11:16):

Given a strong generator $F$ and an object $X$, $X$ will receive a wholly unique pattern of arrows from the objects in $F$ (since, as we covered, morphisms from $F$ determine objects in $C$ uniquely up to isomorphism). What this should mean is that, starting with some pattern of morphisms from the strong generator, you should be able to "go backwards" and reconstruct the object (if such an object exists of course). For instance, in the case of $\mathrm{Set}$, a set with $n$ elements corresponds to there being $n$ morphisms from the singleton. Going backwards, if we started with a pattern of $n$ morphisms from the singleton, we can construct an object in $\mathrm{Set}$ that is a set with $n$ elements. In other words, the fact that morphisms from a strong generator can determine an object up to isomorphism should automatically imply the object can be reconstructed from said morphisms, as said morphisms, in some sense, contain all the necessary information about the object.

John Onstead (May 15 2025 at 11:17):

So my question is: is this in fact true? And if it's not, then how is it at all possible for you to be able to characterize something up to isomorphism without actually being able to construct said thing? It would be antithetical for instance to the whole idea of universal constructions, which are constructed entirely by some property which characterizes them up to isomorphism!

James Deikun (May 15 2025 at 11:39):

Up to unique isomorphism. This is important!

Mike Shulman (May 15 2025 at 13:54):

John Onstead said:

the fact that morphisms from a strong generator can determine an object up to isomorphism should automatically imply the object can be reconstructed from said morphisms

No, I don't think so. The phrase "determine an object up to isomorphism" does not have a precise meaning. The notion of "strong generator" does have a precise meaning, but it doesn't imply that one can reconstruct an object in this way. For instance, it could be that $\cal G$ is a strong generator and two objects $X$ and $Y$ have isomorphic homsets $\hom(G,X) \cong \hom(G,Y)$ for all $G\in \cal G$, even naturally in $G$, and yet $X\not\cong Y$. What you need for this sort of "reconstruction" is really density.

Kevin Carlson (May 15 2025 at 18:08):

You also haven't concretely defined the critical phrase "pattern of morphisms"; if you did, I think it might be clearly why Mike says that your notion of reconstruction needs density, or something similar.

John Onstead (May 15 2025 at 18:42):

Mike Shulman said:

For instance, it could be that $\cal G$ is a strong generator and two objects $X$ and $Y$ have isomorphic homsets $\hom(G,X) \cong \hom(G,Y)$ for all $G\in \cal G$, even naturally in $G$, and yet $X\not\cong Y$.

But I thought this was implied by the conservativity of the reduced Yoneda embedding?

Kevin Carlson said:

You also haven't concretely defined the critical phrase "pattern of morphisms"; if you did, I think it might be clearly why Mike says that your notion of reconstruction needs density, or something similar.

I guess the closest formal term would be something like "cocone", but that already requires you to know the codomain object. So I guess I was thinking of an assignment of a set to every object in the strong generator, that's meant to represent the set of morphisms from that object into some object in the category (if it exists) we want to find.

John Onstead (May 15 2025 at 18:45):

James Deikun said:

You are. The walking split idempotent has 2 objects, the large and small object, and 5 morphisms: the idempotent, the section, the retraction, and two identities. I wonder which category you are thinking of.

Ah, I see. I was thinking of a category with three objects where there was a section from the first to second and a retraction from the second to the third. My bad. I think I see the point now: composing the section and retraction one way gives the identity, while composing it the other way gives the idempotent.

Kevin Carlson (May 15 2025 at 19:00):

John Onstead said:

I guess the closest formal term would be something like "cocone", but that already requires you to know the codomain object. So I guess I was thinking of an assignment of a set to every object in the strong generator, that's meant to represent the set of morphisms from that object into some object in the category (if it exists) we want to find.

And don't you want that set to vary functorially for maps between the generators? So, you're just giving a presheaf on the generators!

Kevin Carlson (May 15 2025 at 19:02):

John Onstead said:

But I thought this was implied by the conservativity of the reduced Yoneda embedding?

This gets back to an "at least two" I mentioned earlier. Conservativity of $F:\mathcal C\to\mathcal D$ does not say that, if $F(x)\cong F(y)$, then $x\cong y.$ This is sometimes a useful property ("injectivity on isomorphism classes," or "reflecting the existence of an isomorphism") but is considerably less important than conservativity, which says that if $F(f)$ is an isomorphism, then so is $f.$ It's much easier to get conservativity, where you assume $f$ was already handed to you.

John Onstead (May 15 2025 at 19:24):

Kevin Carlson said:

And don't you want that set to vary functorially for maps between the generators? So, you're just giving a presheaf on the generators!

Viewing objects of $C$ as presheaves on the generator means $C$ embeds into the generator's presheaf category. But this is the same as density, right? Since that's the condition where the restricted Yoneda embedding is fully faithful.

Kevin Carlson (May 15 2025 at 19:28):

That's the idea, yeah.