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Stream: learning: questions

Topic: Hints for Leinster's "Basic Category Theory" exercises

Ryan Schwiebert (Jul 13 2024 at 00:13):

This is exercise 2.1.16

Let G be a group considered as a category and [G,Set] be the category of functors from G to Set (a.k.a left G-sets). What interesting functors are there between Set and [G,Set]? which if any are adjoint to each other?

I get how open ended questions like this can be beneficial but I also hate how easy it would be to completely miss intended insights.

The only candidates I spotted are the functor (call it U) which "forgets" its G action. and just returns the underlying set, and in the other direction, the functor that imbues any set with the trivial G action, call it T.

I'm still trying to puzzle out whether or not they're adjoint in one order or the other, but firstly I just wanted to ask what, if any, other functors I am missing.

By the way, is there some trick to spot which order they might be adjoint in? Since free -| forgetful, one might wonder if U is the right adjoint. But free functors for algebras seem like they are "elaborating" and the trivial action is kind of a dumbing-down, so maybe it goes on the right.

Or I could be completely off and they're not adjoint in any way :/ But since any set function can be regarded as a G-equivariant map between two sets with trivial G actions, it seems like it is an adjoint relationship.

Your input is appreciated!

David Egolf (Jul 13 2024 at 01:08):

When trying to think of more functors between $\mathsf{Set}$ and $[G, \mathsf{Set}]$, the first question that comes to mind for me is this: What are the representable functors from $[G, \mathsf{Set}]$ to $\mathsf{Set}$?

If we pick some $\rho: G \to \mathsf{Set}$, then what is the functor $[G, \mathsf{Set}](\rho, -): [G, \mathsf{Set}] \to \mathsf{Set}$? Does this give anything interesting in some special cases, maybe? (In particular, I'm curious what happens in the case where $\rho$ sends each element of $G$ to the same identity function).

I don't know if this will be helpful... But after a little bit of searching online (for example, I found this), it seems like there is some connection between a functor being representable and having a left adjoint. So maybe this is relevant!

Kevin Carlson (Jul 13 2024 at 01:29):

I’m not sure how much you know about coproducts, but one good way to find left adjoints out of Sets uses the fact that every set is a coproduct of copies of a one-element set (so the whole left adjoint, since it must preserve coproducts, is determined by where it sends a singleton.)

Kevin Carlson (Jul 13 2024 at 01:31):

As for finding adjoint relationships with functors you’ve already named, just think about the adjunction relationship. What’s a map from a set $S$ into some $U(X),$ and what about in the other direction? Could you replace it with one involving some $G$-action related to $S$?

Ryan Schwiebert (Jul 13 2024 at 03:08):

IIRC both of these are salient to material later in chapter 2, but I was trying to do it with just what was given in and before this section. I do plan to move on regardless of my progress with this exercise, though. This one and 2.1.17 are interesting, but they both put me in the same fraught mindset of "what on earth am I looking for?"

Ryan Schwiebert (Jul 13 2024 at 03:13):

David Egolf said:

In particular, I'm curious what happens in the case where ρ sends each element of G to the same identity function

I think what you're describing is "giving X the trivial G action," the functor T I proposed.

Ryan Schwiebert (Jul 13 2024 at 03:18):

Kevin Carlson said:

What’s a map from a set S into some U(X), and what about in the other direction? Could you replace it with one involving some G-action related to S?

AFAICT there isn't any obvious G action you can give to just any set beyond the trivial action. For example, if X is a 2 element set and G is order 3, there isn't anything else, right? I've been interpreting this exercise as if G cannot be varied, if that makes a difference.

David Egolf (Jul 13 2024 at 03:40):

Ryan Schwiebert said:

David Egolf said:

In particular, I'm curious what happens in the case where ρ sends each element of G to the same identity function

I think what you're describing is "giving X the trivial G action," the functor T I proposed.

I'm interested in the case where $\rho$ specifies a trivial action. I think more specifically I'm curious what $[G, \mathsf{Set}](\rho, -):[G,\mathsf{Set}] \to \mathsf{Set}$ is like when $\rho:G \to \mathsf{Set}$ is acting on a set with a single element.

Let $*$ be the single object of $G$, when $G$ is viewed as a category. When $\rho$ acts on a singleton set, $\rho(*)$ is a singleton set. Further, in this case a natural transformation from $\rho$ to some $\rho': G \to \mathsf{Set}$ amounts to a function $\alpha_*: \rho(*) \to \rho'(*)$ such that this square commutes for all $g \in G$:
naturality square

A function from $\rho(*)$ to $\rho'(*)$ amounts to choosing an element of $\rho'(*)$, because $\rho(*)$ has only one element. So, a natural transformation from $\rho$ to $\rho'$ I think amounts to the choice of an element of $\rho'(*)$ that doesn't change when acted on by any element in $G$, using the action $\rho'$.

Building on this, the set of natural transformations from $\rho$ to $\rho'$, namely $[G,\mathsf{Set}](\rho, \rho')$, I think is in bijection with the set $\{x \in \rho'(*) | \rho'(g)(x)=x \textrm{ for all } g \in G\}$.

If I didn't make a mistake, it seems like there is a functor $[G, \mathsf{Set}](\rho, -): [G, \mathsf{Set}] \to \mathsf{Set}$ that sends each group action $\rho'$ (acting on some set $\rho'(*)$) to the set of elements (of $\rho'(*)$) that are invariant under that action.

John Baez (Jul 13 2024 at 06:21):

Ryan Schwiebert said:

IIRC both of these are salient to material later in chapter 2, but I was trying to do it with just what was given in and before this section. I do plan to move on regardless of my progress with this exercise, though. This one and 2.1.17 are interesting, but they both put me in the same fraught mindset of "what on earth am I looking for?"

I read that question and thought: "oh, Leinster is trying to get people to think about the forgetful functor sending each G-set to its underlying set and the functor sending each set to the free G-set on that set, and show they're adjoints to each other". I believe this is what should flash through the mind of any category theorist in the first second after reading this question. Then there are other things to do, like check if these functors really are adjoint, and look for others.

I wonder if at this point in his book Leinster has mentioned that the words "underlying" or "forgetful" should scream out RIGHT ADJOINT!!! and the word "free" should scream out LEFT ADJOINT!!!

John Baez (Jul 13 2024 at 06:24):

If not, maybe he's trying to get you to learn this through experience.

John Baez (Jul 13 2024 at 06:25):

But when I teach category theory, I just come out and say it - and illustrate it with a bunch of examples. As you basically mention, one of the first steps in getting good at adjoints is being able to quickly smell the difference between a left adjoint and a right adjoint.

Ryan Schwiebert (Jul 13 2024 at 13:41):

David Egolf said:

it seems like there is a functor G,Set:[G,Set]→Set that sends each group action ρ′ (acting on some set ρ′(∗)) to the set of elements (of ρ′(∗)) that are invariant under that action.

Oh! Yes, that's a very natural candidate isn't it! And related to trivial G actions... thanks, I will investigate it.

Oscar Cunningham (Jul 13 2024 at 15:09):

So so far we've mentioned four functors:

1. The underlying set of a G-set.
2. The trivial action
3. The fixed points
4. John mentioned the free G-set on a set, but didn't actually construct what it was.

I can think of a fifth functor as well

John Baez (Jul 13 2024 at 16:00):

Oscar Cunningham said:

So so far we've mentioned four functors:

1. The underlying set of a G-set.
2. The trivial action
3. The fixed points
4. John mentioned the free G-set on a set, but didn't actually construct what it was.

It's a great exercise to do that, if one doesn't already know what it is.

Another good puzzle: show that functor 3 is adjoint to another functor on this list!

Of course, because it's easy to find functors from $\mathsf{Set}$ to $\mathsf{Set}$, we can compose the functors on your list with such functors and get tons more functors from $\mathsf{Set}$ to $[G,\mathsf{Set}]$ or from $[G,\mathsf{Set}]$ to $\mathsf{Set}$.

John Baez (Jul 13 2024 at 16:08):

So it's good that Leinster didn't ask us to find all functors going back and forth between $\mathsf{Set}$ and $[G,\mathsf{Set}]$ because we'd be here all night!

Kevin Carlson (Jul 14 2024 at 01:59):

John Baez said:

But when I teach category theory, I just come out and say it - and illustrate it with a bunch of examples. As you basically mention, one of the first steps in getting good at adjoints is being able to quickly smell the difference between a left adjoint and a right adjoint.

This is maybe not the best possible motivating example of something with the right adjoint smell though, since [secrets about this forgetful functor and its adjoint(s)!]

Kevin Carlson (Jul 14 2024 at 02:01):

Ryan Schwiebert said:

Kevin Carlson said:

What’s a map from a set S into some U(X), and what about in the other direction? Could you replace it with one involving some G-action related to S?

AFAICT there isn't any obvious G action you can give to just any set beyond the trivial action. For example, if X is a 2 element set and G is order 3, there isn't anything else, right? I've been interpreting this exercise as if G cannot be varied, if that makes a difference.

Who said it had to be an action on the same set as $X,$ though? Lots of functors between concrete categories change the underlying set.

John Baez (Jul 14 2024 at 07:47):

Kevin Carlson said:

John Baez said:

But when I teach category theory, I just come out and say it - and illustrate it with a bunch of examples. As you basically mention, one of the first steps in getting good at adjoints is being able to quickly smell the difference between a left adjoint and a right adjoint.

This is maybe not the best possible motivating example of something with the right adjoint smell though, since [secrets about this forgetful functor and its adjoint(s)!]

Clearly neither of us want to give away all the details of we're really talking about here, so as not to spoil Ryan's fun. I still think it's a useful to have a sense of smell, so that you walk into an adjunction and instantly have a sense of who is the right adjoint and who is the left adjoint. But then you learn that a right adjoint can have its own right adjoint, which Lawvere called a 'far-right' or 'fascist' functor.

Morgan Rogers (he/him) (Jul 15 2024 at 08:53):

I actually think that Leinster was trying to appeal to the intuition of a reader who has taken a course in group theory here! From that perspective, there is a functor "more obvious" than fixed points that goes from left $G$-sets to sets!
I don't know if this was @Oscar Cunningham 's "fifth functor", but I feel entitled to hint that I expect the list of functors that Leinster was expecting you to come up with should have six functors in it :wink:
The final one is the trickiest because it involves sets of functions, and requires some more advanced algebra intuition to find. If you get stuck at 5, I'm sure someone will further hint at what I'm talking about.

Ryan Schwiebert (Jul 16 2024 at 11:28):

Kevin Carlson said:

Who said it had to be an action on the same set as X, though? Lots of functors between concrete categories change the underlying set.

Certainly I didn't suggest that... I only meant to clarify it sounded like G was to be left alone.

Ryan Schwiebert (Jul 16 2024 at 11:47):

Morgan Rogers (he/him) said:

I expect the list of functors that Leinster was expecting you to come up with should have six functors in it :wink:
The final one is the trickiest because it involves sets of functions, and requires some more advanced algebra intuition to find. If you get stuck at 5, I'm sure someone will further hint at what I'm talking about.

I definitely had a graduate course in group theory including actions on sets, and feel like I've gotten a lot of use out of it over the years, but I haven't the foggiest idea of what I'm missing. I'll probably kick myself when I hear it, but that's how hindsight works...

I think whatever benefits the "guess what functors the author is thinking of" game are over for me at this point. It's coming at the expense of the rest of the exercise and the rest of the book, and if I dwell on this part the more it will only turn into an assignment remembered as a cruel joke. I'd like to give myself time for the next interesting part of the task (investigating their adjointness.) So feel free to "spoil" 5+6 for me.

Oscar Cunningham (Jul 16 2024 at 12:40):

The one I was thinking of sends a group action to its set of orbits.

Morgan Rogers (he/him) (Jul 16 2024 at 16:12):

The sixth one is the "cofree action" sending X to X^G

David Egolf (Jul 16 2024 at 16:51):

I was curious about how one could think up the functor that sends a group action to its set of orbits. I found this interesting quote from "Category Theory in Context" (p. 108):

The limit of a functor $X : \mathsf{B}G → \mathsf{Set}$ is the set of $G$-fixed points... and the colimit is the set of $G$-orbits... .

We obtain functors by taking these limits or colimits, as indicated by these statements (see here, for example):

• Given a category of $D$-shaped diagrams $[D,C]$, there is a functor $:[D, C] \to C$ that takes the limit of each diagram, provided those limits exist in $C$.
• Similarly, given a category of $D$-shaped diagrams $[D,C]$, there is a functor $:[D, C] \to C$ that takes the colimit of each diagram, provided those colimits exist in $C$.

If I had thought to explore these general statements in the context of this exercise, presumably I would have discovered the functor that sends a group action to its set of orbits!

Oscar Cunningham (Jul 16 2024 at 20:25):

The other way you could think it up is by finding it as an adjoint to one of the other functors.

John Baez (Jul 17 2024 at 12:46):

Yes, it's all the functors on your list of functors going between $\mathsf{Set}$ and $[G,\mathsf{Set}]$ and for each one either find both a left and a right adjoint, or prove some of these adjoints don't exist. (The latter is usually easy by finding a colimit or limit that the functor doesn't preserve.)

Ryan Schwiebert (Jul 18 2024 at 03:00):

@John Baez Pretty sure I'm convinced Free -| forgetful and Trivial -| Fixed. I'm still thinking about the gaps that I haven't mentioned, but I thought it would be good to confirm the following points in case I am wrong:

1. the free group action on X has underlying set G\times X with action h(g,x):=(hg,x)
2. The initial object of [G, Set] is the unique action on the emptyset and the terminal objects are the unique actions on singletons.

John Baez (Jul 18 2024 at 07:41):

All that is correct, @Ryan Schwiebert! Good work.

Are you familiar with the set of orbits of a G-set X? This set is also called the quotient X/G. Does the functor from [G,Set] to Set sending a G-set to its set of orbits have a left adjoint? Does it have a right adjoint?

Ryan Schwiebert (Jul 20 2024 at 02:15):

@John Baez Huh! It took me a while but my findings are that $\mathcal O: [G,Set]\to Set$ (my notation for the "orbit functor" ) is left adjoint to the trivial action functor T!

I was not expecting that until I realized the trivial action sort of makes the number of orbits maximal, and if one is to have a G equivariant map into such an action, the fibers are the orbits. From there things seemed to flow naturally.

I think I don't have the energy right now to investigate if $\mathcal O$ or $F$ have left adjoints, or if $\Phi$ ( my notation for the fixed point functor) or U have right adjoints. The only suggested tool at this point is preservation of initials and terminals (the full fact about limits/colimits has not been reached yet, and that doesn't seem to put a dent in any of those problems.)

The other half of the exercise asks the same question but with the categories [G, Vect_k] and Vect_k. I assume we have at least all the same functors and adjoint relationships discovered. I thought proving so might be possible with composition of adjunctions, say free/forgetful functors between Vect and Set but actually I wasn't getting anywhere with that.

Are there any surprising ones that are new?

John Baez (Jul 20 2024 at 08:02):

By the way @Ryan Schwiebert, due to inflation you need to use double dollars, not mere dollar signs, to get LaTeX to work here. You can edit your post that way and make it beautiful.

John Baez (Jul 20 2024 at 08:17):

@Ryan Schwiebert- I think it would be nice to organize all the functors you've found in one or more lists, where within each list each functor is left adjoint to the previous one. I'm having trouble extracting this information from your comments. But these are the 6 functors I've seen people point out here, and the adjunctions that I've seen mentioned:

"set of orbits of a G-set" is left adjoint to "trivial G-set on a set".

"free G-set on a set" is left adjoint to "underlying set of a G-set".

"set of fixed points of a G-set"

"G-set of functions from a set to G"

Can you organize these all into a single list of 6 functors, each left adjoint to the previous one? Or is this impossible? If it's impossible, how close can you get?

I hope experts don't dive in and solve this puzzle unless you ask them to!

John Baez (Jul 20 2024 at 08:19):

But Morgan Rogers (he/him)was already nudging you and winking at you:

I feel entitled to hint that I expect the list of functors that Leinster was expecting you to come up with should have six functors in it :wink:

Ryan Schwiebert (Jul 20 2024 at 18:56):

@John Baez So far I've got $F\dashv U$ and $\mathcal O\dashv T\dashv \Phi$. Mixing fonts/symbology for these things is suboptimal I know but I'm trying to help remind myself which is which.

I haven't considered "G-set of functions from a set to G," but I guess I should give it a whirl too. (I dub thee $C$ for "cofree" for the purposes of posting...)

John Baez (Jul 20 2024 at 19:33):

Thanks! You should try to keep making those two chains of adjoints longer and see if you can merge them into one. And I'm pretty sure "G-set of functions from a set to G" is the adjoint of something. (The name "cofree" positively screams that it's a right adjoint.)

Morgan Rogers (he/him) (Jul 20 2024 at 21:55):

Ryan Schwiebert said:

I haven't considered "G-set of functions from a set to G," but I guess I should give it a whirl too. (I dub thee $C$ for "cofree" for the purposes of posting...)

That's quite an appropriate name..!

Reid Barton (Jul 22 2024 at 20:51):

John Baez said:

"G-set of functions from a set to G"

I think you mean "G-set of functions from G to a set", right?

John Baez (Jul 22 2024 at 21:06):

That's what I should have said. There's also a G-set of functions from a set to G, but that's not so interesting in the conversations here.

Ryan Schwiebert (Jul 24 2024 at 03:33):

Well, I have to give up, since I think I'm again approaching the border between "let's struggle with it" and "ragequit." I just can't intuit the correct combination of action and transpose.

I had thought that the correct action for $X^G$ would be, given $f:G\to X$ and $h\in G$, the map $h\cdot f$ given by $(h\cdot f)(g)=f(hg)$ and

I had thought that the transpose of a map $\alpha: U(A)\to X$ would be given by $\bar\alpha(x)$ given by $\bar\alpha(x)(g):=\alpha(gx)$ but AFAICT with those definitions $\bar\alpha$ is not G equivariant.

If you use $(h\cdot f)(g)=f(h^{_1}g)$ then I think it doesn't work as a G action. I thought also of inverting the $g$ in the transpose map but that too seemed to misorder things.

John Baez (Jul 24 2024 at 09:04):

Ryan Schwiebert said:

Well, I have to give up, since I think I'm again approaching the border between "let's struggle with it" and "ragequit." I just can't intuit the correct combination of action and transpose.

I had thought that the correct action for $X^G$ would be, given $f:G\to X$ and $h\in G$, the map $h\cdot f$ given by $(h\cdot f)(g)=f(hg)$

This formula doesn't make $X^G$ into a left $G$-set, since it obeys

$h \cdot ( h' \cdot f) = (h' h) \cdot f$

rather than

$h \cdot ( h' \cdot f) = (h h' ) \cdot f$

You're instead getting a right $G$-set. If you're shooting for a left $G$-set - and I seem to recall we're working with left $G$-sets in this conversation - then I think there are two ways to do it.

John Baez (Jul 24 2024 at 09:08):

There's a total of four formulas you could try:

1) $(h \cdot f)(g) = f(h g)$

2) $(h \cdot f)(g) = f(h^{-1} g)$

3) $(h \cdot f)(g) = f(g h)$

4) $(h \cdot f)(g) = f(g h^{-1})$

John Baez (Jul 24 2024 at 09:09):

It looks like you tried 1) and 2), though there's a typo in your second try.

Ryan Schwiebert (Jul 24 2024 at 11:40):

Right... I overlooked half the actions because of tunnel vision on the left and either way was checking things wrongly:

$(gh)\cdot f(x)=f(gh\cdot x)=f(g\cdot (h\cdot x)=g\cdot f(h\cdot x)$

and from there I should not have thought it was $g\cdot (h\cdot f(x))$ but rather $h\cdot(g\cdot f)(x)$ :confounded:

Investigation resumed... thank you

John Baez (Jul 24 2024 at 12:01):

This stuff about left and right actions always offers plenty of scope for getting confused. In fact any group has 3 left actions on itself and 3 right actions, not counting trivial actions. (The kind we're not concerned with here involve conjugation.)

Ryan Schwiebert (Jul 25 2024 at 02:14):

@John Baez With what I think is the "right choice" ($(g\cdot f)(x)=f(xg)$) I think everything works to show $U\dashv C$.

If these two lists were to link up then it'd be one of $\Phi\dashv F$ or $C\dashv \mathcal O$. I've stared at them for quite a while today thinking about candidates for transpose maps.

The latter one seems less probable to me. It seems like, given a left action with no fixed points (like $G$ acting on itself) $Set(\Phi(A), B)$ would offer no maps while $[G,Set](A, F(B))$ could have lots.

For the more plausible one, I had at least one stab for producing a map from $C(A)$ to $B$ given a map from A to $\mathcal O(B)$, but I have trouble seeing how the other direction would go.

I guess I'm asking for a hint: is this latter adjunction supposed to work? Or am I totally off base with the ideas above?

Ryan Schwiebert (Jul 25 2024 at 02:25):

As a side note, I find it completely bonkers these functors fit together even as nicely as we've seen so far. Although, I guess Set, G and [G, Set] are all fairly "pure" categories so maybe one should expect exceptionally good behavior...

John Baez (Jul 25 2024 at 06:58):

Yes, $C$ is right adjoint to the underlying set functor from [G,Set] to Set, which is why it deserves to be called "cofree". Good work!

John Baez (Jul 25 2024 at 07:07):

I guess I'm asking for a hint: is this latter adjunction supposed to work? Or am I totally off base with the ideas above?

I don't really know; I'd have to think about it. I just find it hard to believe that the two chains of adjunctions you've found so far don't link together. You say that Set and [G,Set] are "fairly pure" categories but that's a ridiculous understatement: they are exceptional jewels in the firmament of mathematics, among the best things in the universe, so we should expect that everything works out wonderfully well.

I could throw around some jargon and say we're dealing with an essential geometric morphism between Boolean presheaf categories. Basically this means we're in a situation that's really famous for having interesting long-ish chains of adjoint functors. Someone like @Morgan Rogers (he/him) or @Reid Barton could jump in and say a lot more, but they've wisely refrained from doing that, merely giving us a few hints here and there, because they both know it's better if we fiddle around and work out this stuff from scratch, while they watch us and smile knowingly.

John Baez (Jul 25 2024 at 07:09):

Since I'm thinking about fifteen things at a time these days, I tend to forget what you've done unless I see it summarized in a nice little chart. So let me make that chart.

You've got two chains of adjunctions:

$F\dashv U \dashv C$ and $\mathcal O\dashv T\dashv \Phi$

so there are two possible ways to glue them together into one really long chain.

John Baez (Jul 25 2024 at 07:11):

Okay, I agree that the "cofree G-set on a set $X$",

$C(X) = X^G$

looks like it might possibly be adjoint to the "set of orbits of a G-set $Y$",

$\mathcal{O}(Y) = Y/G$

John Baez (Jul 25 2024 at 07:12):

And given your two chains of adjunctions, the only possibility is that $C$ is left adjoint is $\mathcal{O}$.

John Baez (Jul 25 2024 at 07:14):

So, as you've already said, our job is to think about whether

$\mathrm{hom}_{\mathrm{[G,Set]}} (X^G, Y) \cong \mathrm{hom}_{\mathrm{Set}} (X, Y/G)$

John Baez (Jul 25 2024 at 07:15):

If you can find a map going one way, that's already a huge sign that it's going to work. We are standing in a gold mine here. And when you're standing in a gold mine and you see something shiny in the dirt, there's a good chance you've found gold. What's your idea for a map going one way?

Morgan Rogers (he/him) (Jul 25 2024 at 09:33):

I should also note that if you want to show that adjoints to some of these functors don't exist, you need to:

1. assume that the group is non-trivial, since when the group is trivial $[G,\mathrm{Set}]$ is equivalent to $\mathrm{Set}$ and all of the functors we've discussed are naturally isomorphic to the identity
2. consider some limits and colimits which might fail to be preserved by the respective functors (recalling that left adjoints preserve colimits and right adjoints preserve limits, so failure to preserve these shows that an adjoint cannot exist). To avoid spoilers I'll just say that binary products or coproducts will do for at least one of the examples (but maybe just one :stuck_out_tongue_wink: ), but for another you'll need to consider a coequalizer.

Ryan Schwiebert (Jul 25 2024 at 19:11):

John Baez said:

that's a ridiculous understatement: they are exceptional jewels in the firmament of mathematics,

You'd have the experience to say so, so I believe! When making my comment, I was thinking that other things like categories of groups and modules may be "more crystalline."

John Baez (Jul 25 2024 at 20:42):

If you like [[toposes]], a category of G-sets is a very simple and beautiful example. @David Egolf is writing articles here about about another classic example: the category of sheaves on a topological space.

If you like [[abelian categories]], a category of R-modules for a ring is a very typical example.

John Baez (Jul 25 2024 at 20:44):

And if you don't know what toposes and abelian categories are, these examples are very good ways to start getting a feel for them.

John Baez (Jul 25 2024 at 20:50):

The category of groups is wholly different from either of these, but it's a good example of an [[algebraic category]].

Ryan Schwiebert (Jul 27 2024 at 13:15):

@John Baez Well, given an arrow $\alpha : X^G\to A$ it seemed plausible that $\bar\alpha :X\to A/G$ might be given by $\bar\alpha(x)=\pi(\alpha(c_x))$ where $c_x$ is the function on $G$ that is constantly $x$ and $\pi$ projects elements of $A$ to their orbits.

But it seems less and less likely that this is invertible. The candidate in the other direction seemed to require picking $\theta : A/G\to A$ (needing to 'choose' already made me doubtful). Given $\beta:X\to A/G$ The candidate $\bar\beta: X^G\to A$ I tried was $\bar\beta(f)= \theta(\beta(f(1)))$. This didn't work out, more or less I think, because we need not have $\theta\pi\neq Id_A$. I had been hoping something about $G$ equivariant maps coming out of $X^G$ would make the choice of $\theta$ irrelevant, but I haven't seen why.

I think I'll my leads on this have dried up.

John Baez (Jul 27 2024 at 13:21):

I would find this easier to understand if you told me the domain and codomain of $\overline{\alpha}$, and what $x$ is an element of. I can try to figure out what are the only possibilities that make sense, but it's considered good style to introduce maps by telling us their source and target, and elements by saying what they're elements of. Then the reader can focus on more interesting issues, instead of the detective work needed to figure out those things.

Ryan Schwiebert (Jul 27 2024 at 13:22):

Morgan Rogers (he/him) said:

1. consider some limits and colimits which might fail to be preserved by the respective functors

I'm trying to stick with the tools I've been given thus far in the book (I don't officially have the fact about adjoints preserving limits/colimits, or equalizers yet.) But I may circle back later to try this hint out. I think the last thing I'd like to try before I move on is to determine if $C\dashv \mathcal O$ is true or not.

Ryan Schwiebert (Jul 27 2024 at 13:25):

@John Baez ok, i edited to specify those.

John Baez (Jul 27 2024 at 13:44):

Thanks. So you're looking for a "candidate in the other direction": a recipe which when given an arbitrary function $f: X \to A/G$ produces a $G$-equivariant map $\underline{f}: X^G \to A$.

In other words, given an arbitrary function $f: X \to A/G$ and an arbitrary function $h: G \to X$ we need to produce an element of $A$, which will be called $\underline{f}(h)$. And then we need to check that

$\underline{f}(g h ) = g \underline{f}(h)$

to show $\underline{f}$ is $G$-equivariant.

John Baez (Jul 27 2024 at 13:46):

As you note, it seems like a severe uphill climb to get an element of $A$ out of $f$ and $h$ - it looks bad.

John Baez (Jul 27 2024 at 13:47):

$f$ and $h$ are begging to be composed, and that'll give $f \circ h: G \to A/G$. But what use is that?

John Baez (Jul 27 2024 at 13:51):

So this looks bad. But as Morgan was hinting, reasoning of this general sort - fiddling around with the things we can do - can't really prove that a functor has no left adjoint. This sort of reasoning lets us either find the left adjoint if it exists, or convince us that it doesn't exist if we're unable to find it. But to put the nail in the coffin, and show that some functor has no left adjoint, we need to assume our functor has a left adjoint and get a contradiction somehow. (This is legitimate even intuitionistically!) And the simplest way is often to show our functor fails to preserve some colimit.

Ryan Schwiebert (Jul 27 2024 at 15:33):

I'm definitely glad you all convinced me to press harder on this problem, but I think the time has come for me to move on. There's just one more problem in this section I'd like to tangle with, and I'll type it up tonight, hopefully.

John Baez (Jul 27 2024 at 17:26):

Good! By the way, I think you've seen a great example of this phenomenon:

Whenever we have a functor between categories $f: C \to D$ we get a functor

$f^\ast: [D,\mathsf{Set}] \to [C, \mathsf{Set}]$

given in the obvious way: given $h : D \to \mathsf{Set}$ we say

$f^\ast(h) = h \circ f$

Such a functor $f^\ast: [D,\mathsf{Set}] \to [C, \mathsf{Set}]$ always has both a left adjoint and a right adjoint!

In particular we can think of a group as a one-object category and apply this idea to groups.

Any group $G$ has homomorphisms to and from the trivial group $1$, say

$i: 1 \to G$

and

$t: G \to 1$

But $[1,\mathsf{Set}] \cong \mathsf{Set}$. So, we get functors

$i^\ast : [G, \mathsf{Set}] \to \mathsf{Set}$

and

$t^\ast: \mathsf{Set} \to [G, \mathsf{Set}]$

each of which has both a left and a right adjoint.

This accounts for the 6 functors you found, and why they form two separate adjoint strings, each 3 long.

Morgan Rogers (he/him) (Jul 27 2024 at 17:53):

Since Ryan is moving on, it's worth pointing out that the converse is true too: a functor between categories of presheaves on idempotent complete categories is necessarily induced by a functor... which gives another way of deducing whether or not more of the functors are adjoint!

Nathanael Arkor (Jul 27 2024 at 18:39):

Presumably you mean a cocontinuous functor between categories of presheaves?

David Egolf (Jul 27 2024 at 19:17):

John Baez said:

Such a functor $f^\ast: [D,\mathsf{Set}] \to [C, \mathsf{Set}]$ always has both a left adjoint and a right adjoint!

I am curious as to how one could prove this! This seems like a very handy result!

I'm also curious if this result generalizes to the case where we switch out $\mathsf{Set}$ for some other suitably nice category.

Peva Blanchard (Jul 27 2024 at 20:01):

@David Egolf They relate to left and right Kan extensions (I think). And, if I remember correctly, they are also named $\exists_f$ and $\forall_f$. The reason behind those names is that it also relates to logic. Another interpretation exists in terms of data migration. I have to dig a bit to find the papers I read that from, but @Ryan Wisnesky gave a very good talk at the Zulip CT Seminar.

David Egolf (Jul 27 2024 at 20:13):

Ah, thanks @Peva Blanchard! With the help of your comment, I was able to find the following in "Category Theory in Context" (Proposition 6.1.5):

If, for fixed $K:C \to D$ and $E$, the left and right Kan extensions of any functor $F:C \to E$ along $K$ exist, then these define left and right adjoints to the pre-composition functor $K^*:E^D \to E^C$.

John Baez (Jul 27 2024 at 20:42):

Yes, and these particular right and left Kan extensions can be described very explicitly, for example using ends and coends. So you can actually 'compute' them.

Btw, @David Egolf and @Peva Blanchard, I'll talk about one of these adjoints in Topos Theory (Part 4) when I discuss functors between categories of sheaves (which generalize presheaf categories in a sense). The nice kind of map between categories of sheaves is called a 'geometric morphism', and it consists of a functor called the 'inverse image' and its right adjoint, called the 'direct image'.

John Baez (Jul 27 2024 at 20:55):

When the inverse image functor also has a left adjoint, we say we've got an 'essential' geometric morphism.

John Baez (Jul 27 2024 at 20:56):

So, the stuff we've been doing in this thread can be seen as a warmup for topos theory.

Morgan Rogers (he/him) (Jul 28 2024 at 15:47):

Nathanael Arkor said:

Presumably you mean a cocontinuous functor between categories of presheaves?

Nope, those are more general and induced by profunctors.

Nathanael Arkor (Jul 28 2024 at 16:25):

I think I'm misunderstanding: how is a cocontinuous functor more general than an arbitrary functor?

John Baez (Jul 28 2024 at 16:33):

A cocontinuous functor from presheaves on A to presheaves on B is more general than an arbitrary functor from A to B.

Morgan Rogers (he/him) (Jul 28 2024 at 22:24):

Aha I now understand the correction you were making! I did indeed mean "a functor between categories of presheaves with a left and a right adjoint" or (in the context of the preceding messages) "such a functor"

Ryan Schwiebert (Jul 30 2024 at 22:20):

Just as an update, the exercise I went on to says:
Exhibit a chain of functors $\Lambda\dashv\Pi\dashv\Delta\dashv\Gamma\dashv\nabla$
Between $Set$ and $[\mathcal O(X)^{op}, Set]$ where $\mathcal O(X)$ is the poset of open subsets of a topological space $X$, and $\Delta: Set\to[\mathcal O(X)^{op}, Set]$ sends a set $A$ to a functor which sends each open subset of $X$ to $A$ and (I think) each $O(X)^{op}$ morphism to the identity morphism on $A$.

I got candidates for $\Pi, \Gamma$ already that seem like they will work out. Will drop back in if I get stuck.

John Baez (Jul 31 2024 at 10:38):

Cool. You are already an expert on chains of adjunctions between presheaf categories and the category Set, though you were looking at presheaves on the group $G \cong G^{\rm op}$ instead of on the poset $\mathcal{O}(X)$.

Ryan Schwiebert (Aug 04 2024 at 01:06):

Using $\Gamma$ defined by $\Gamma(B)=B(\emptyset)$ and $\Pi$ defined by $\Pi(B)=B(X)$, I was able to show $\Pi\dashv \Delta\dashv\Gamma$. (Yes, I remembered to define these functors on arrows too, I just am abbreviating.)

I haven't gotten very far with candidates for $\nabla, \Lambda$, although given a set $C$, $Hom(-, C)$ and $Hom(C,-)$ crossed my mind as possible presheaves on $\mathcal O(X)^{op}$. I did not catch a connection yet with those. Maybe the fact the elements of the poset are sets is a red herring?

I'm circling back to understand this comment fully for inspiration in case it helps me out here.

John Baez (Aug 04 2024 at 07:16):

That comment implies that you'll get a functor

$[\mathcal{O}(X)^{\textrm{op}}, \mathsf{Set}] \to \mathsf{Set}$

with both a left and right adjoint from any functor

$1 \to \mathcal{O}(X)^{\textrm{op}}$

It also implies that you'll get a functor

$\mathsf{Set} \to [\mathcal{O}(X)^{\textrm{op}}, \mathsf{Set}]$

with both a left and right adjoint from any functor

$\mathcal{O}(X)^{\textrm{op}} \to 1$

John Baez (Aug 04 2024 at 07:23):

Do you see what all the functors

$1 \to \mathcal{O}(X)^{\textrm{op}}$

are, and what all the functors

$\mathcal{O}(X)^{\textrm{op}} \to 1$

are? They're both rather easy to describe.

Jencel Panic (Aug 04 2024 at 08:47):

Sorry for the not-very-smart question, but I am having trouble visualizing what would a functor between Set and [G,Set] constitute?

Kevin Carlson (Aug 04 2024 at 21:05):

Whether or not it's very smart, it's not a very clear question. Such a functor is constituted by, of course, a choice of $G$-set for every set and a choice of $G$-set morphism for every corresponding set morphism respecting identities and composition. Are you asking for an example of such a functor, maybe?

John Baez (Aug 04 2024 at 21:46):

Six examples were discussed in this thread; they were listed here but described individually earlier.

Jencel Panic (Aug 05 2024 at 12:12):

Kevin Carlson said:

Whether or not it's very smart, it's not a very clear question. Such a functor is constituted by, of course, a choice of $G$-set for every set and a choice of $G$-set morphism for every corresponding set morphism...

i.e. a natural transformation, I suppose?

Ryan Schwiebert (Aug 05 2024 at 16:37):

Jencel Panic said:

i.e. a natural transformation, I suppose

Right. A natural transformation because that's the type of arrow in the target category of the functor. But specifically in this case, that transformation amounts to a $G$ equivariant map between $G$-sets.

Ryan Schwiebert (Aug 06 2024 at 11:37):

@John Baez From the unique functor $\mathcal C\to 1$, the induced functor $[1, Set]\to [\mathcal C, Set]$ in my case is $\Delta$.

For the (multiple) functors $E:1\to\mathcal C$, they become "evaluate at $X$ functors" meaning $E^\ast(B):=B(X)$ where $X=E(1)$. The two other functors I found were of that type.

Unfortunately I still am still at a loss for candidates for $\nabla$ and $\Lambda$...

John Baez (Aug 06 2024 at 11:58):

I'm losing track of notation... let me go back to earlier comments and try to figure out what you're talking about.

Okay, one thing you want is candidate for some functor called $\nabla$, and all you know about it is that it should be right adjoint to some functor called $\Gamma$. So if I can figure out what $\Gamma$ is, I'll know what you're trying to do!

John Baez (Aug 06 2024 at 12:00):

Okay, piecing together the puzzle, it seems you said

$\Gamma : [\mathcal{O}(X)^{\rm{op}}, \mathsf{Set}] \to \mathsf{Set}$

is the functor that sends any $B: \mathcal{O}(X)^{\rm{op}}\to \mathsf{Set}$ to $B(\emptyset)$.

John Baez (Aug 06 2024 at 12:03):

So you are looking for a functor

$\nabla : \mathsf{Set} \to [\mathcal{O}(X)^{\rm{op}}, \mathsf{Set}]$

with a natural isomorphism

$\mathrm{hom}(\Gamma(B), C) \cong \mathrm{hom}(B, \nabla(C))$

Is that right?

John Baez (Aug 06 2024 at 12:06):

If I haven't screwed up already, this means you are looking for a functor $\nabla : \mathsf{Set} \to [\mathcal{O}(X)^{\rm{op}}, \mathsf{Set}]$ with a natural isomorphism

$\mathrm{hom}(B(\emptyset), C) \cong \mathrm{hom}(B, \nabla(C))$

for every $B: \mathcal{O}(X)^{\rm{op}}\to \mathsf{Set}$ , $C \in \mathsf{Set}$.

John Baez (Aug 06 2024 at 12:07):

(It would be really helpful if you posed self-contained questions. So far I've just been figuring out what one of your two questions is.)

John Baez (Aug 06 2024 at 12:18):

Hmm, maybe there's something wrong. If I'm reading you right, you've got a functor $\Delta$ that sends any set $S$ to the constant presheaf on $X$ with value equal to $S$. You said the functor $\Gamma$ that's right adjoint to $\Delta$ has

$\Gamma(B) = B(\emptyset)$

for every presheaf $B$ on $X$. But I seem to be getting

$\Gamma(B) = B(X)$.

John Baez (Aug 06 2024 at 12:19):

I could be mixed up. But if we get $\Gamma$ wrong, we won't be able to calculate its right adjoint correctly. So this is worth straightening out.

John Baez (Aug 06 2024 at 12:36):

In other words, you seem to be claiming that for every $S \in \mathsf{Set}$ and every $B: \mathcal{O}(X)^{\rm{op}}\to \mathsf{Set}$ we have a natural isomorphism

$\mathrm{hom}(\Delta S, B) \cong \mathrm{hom}(S, B(\emptyset))$

where $\Delta S \in \mathcal{O}(X)^{\rm{op}}\to \mathsf{Set}$ is defined by $\Delta(S)(U) = S$ for all open sets $U \subseteq X$, while I seem to be getting

$\mathrm{hom}(\Delta S, B) \cong \mathrm{hom}(S, B(X))$

Kevin Carlson (Aug 06 2024 at 21:02):

I agree that $\Gamma(B)=B(\emptyset)$ seems wrong.

Ryan Schwiebert (Aug 07 2024 at 01:55):

I thought I was being very meticulous when I wrote this and this but maybe it's worth linking them together.

Ryan Schwiebert (Aug 07 2024 at 02:37):

Given a morphism of sets $f:A\to \Gamma(B)$, I set out to produce the transpose map (a natural transformation) $\bar f:\Delta A \to B$.

For each open set $S$ of $X$, we need to specify a set morphism $A\to B(S)$. The reason I chose $\emptyset$ is because there's a unique arrow $a_S:\emptyset \to S$ in $\mathcal O(X)^{op}$, and applying $B$ to this arrow gives you $B(a_S): B(\emptyset)\to B(S)$ which you can follow $f$ with to get a morphism $A\to B(S)$, i.e. the $S$ component of $\bar f$ is $\bar f_S:=B(c_S)\circ f$

If instead we use $X$ you can only have an arrow $a'_s:S\to X$ and application of $B$ gets you $B(a'_s):B(S)\to B(X)$, and that doesn't give you anything to combine with $f:A\to \Gamma(B)$ if $\Gamma(B)=B(X)$.

The inverse of the transpose map was given by the $\emptyset$ component of each natural transformation, that is for natural transformation $g: B\to \Delta(A)$ defining $\bar g:=g_\emptyset$, and this seemed to work out for me.)

Ryan Schwiebert (Aug 07 2024 at 02:39):

Dually the choice of $\Pi(B)=B(X)$ works out as a left adjoint to $\Delta$ (for me at least.) Have I made some error in the preceding post? @John Baez @Kevin Carlson Progressing to the other two will be impossible if I already chose poorly here :)

Ryan Schwiebert (Aug 07 2024 at 03:00):

(As an aside, if $F\dashv G$, where $F:\mathcal C\to \mathcal D$, then can we also say the presheaf functors John defined above $F^\ast$ and $G^\ast$ are adjoint in some order too? The reason I ask is that I proved earlier that a left adjoint to $\mathcal C\to 1$ amounts to an initial object of $\mathcal C$, and a right adjoint amounts to a terminal object of $\mathcal C$, and that seems to jive with the top and bottom of $\mathcal O(X)^{op}$ winding up in my functors.)

(Also mind you both I'm just using the interesting presheaf theorem as a dowsing rod: I don't want to use it officially until I've gotten to the point of proving it, which I'm not actively doing.)

Ryan Schwiebert (Aug 07 2024 at 03:28):

Ugh, after thinking about it more I think I know what happened: I was interpreting $b\leq a$ as $a\to b$. That's not officially how it's done though, is it. $\emptyset$, being initial in $\mathcal O(X)$ would have the arrows departing wouldn't it. Which means in the opposite poset I wanted arrows departing from $X$.

John Baez (Aug 07 2024 at 07:56):

Ryan Schwiebert said:

For each open set $S$ of $X$, we need to specify a set morphism $A\to B(S)$. The reason I chose $\emptyset$ is because there's a unique arrow $a_S:\emptyset \to S$ in $\mathcal O(X)^{op}$...

I believe you slipped here: there's a unique arrow from $\emptyset$ to $S$ in $\mathcal O(X)$.

Ryan Schwiebert (Aug 08 2024 at 02:36):

@John Baez Yep... I agree, as mentioned in my last paragraph. So I think we can proceed with my old $\Gamma$ and $\Pi$ flipped: $\Gamma(B):=B(X)$ and $\Pi(B):=B(\emptyset)$.

So now I have a set morphism $f:B(X)\to C$ and aim to create a natural transformation $B\to \nabla(C)$. For an object $S\in \mathcal O(X)^{op}$, I need to bridge this: $B(S)\to \cdots \to (\nabla C)(S)$. Since $X$ precedes all the $S$'s I'm kind of at a loss of what to try.

Apparently I used up all my instincts on finding $\Gamma$ and $\Pi$. Those seemed straightforward. I've got this rudder of definitions but it's not much use in these doldrums...

John Baez (Aug 08 2024 at 09:04):

An interesting fact is that if a functor $L$ has a right adjoint $R$, this fact determines $R$, and you can actually crank out what $R$ must be using the natural isomorphism

$\mathrm{hom}(L x, y) \cong \mathrm{hom}(x, R y)$

John Baez (Aug 08 2024 at 09:05):

Simply put, this isomorphism lets you figure out what the morphisms from any $x$ to $R y$ are, and this in turn lets you figure out what $R y$ must be.

John Baez (Aug 08 2024 at 09:09):

So in some sense there's no "creativity" required for determining a right adjoint if one has been told it exists: it's a calculation, not a matter of guesswork. I find this reassuring... though there is creativity required in finding a nice description of the right adjoint.

John Baez (Aug 08 2024 at 09:13):

Let's try it a bit. We've got this functor

$\Gamma : [\mathcal{O}(X)^{\rm{op}}, \mathsf{Set}] \to \mathsf{Set}$

defined on objects by

$\Gamma(B) = B(X)$

for every $B: \mathcal{O}(X)^{\rm{op}}\to \mathsf{Set}$, and similarly for morphisms.

John Baez (Aug 08 2024 at 09:16):

Now we want a functor

$\nabla : \mathsf{Set} \to [\mathcal{O}(X)^{\rm{op}}, \mathsf{Set}]$

with a natural isomorphism

$\mathrm{hom}(\Gamma(B), S) \cong \mathrm{hom}(B, \nabla(S))$

i.e.

$\mathrm{hom}(B(X), S) \cong \mathrm{hom}(B, \nabla(S))$

for every presheaf $B$ and set $S$.

John Baez (Aug 08 2024 at 09:25):

What are some things we can figure out about $\nabla (S)$ using this "equation"? Ultimately everything. But let's try to tease out that information by making clever choices of $B$, and maybe choosing an easy $S$ to start with.

By the way, right now I have no intuition at all for what $\nabla (S)$, so I'm not "cheating" and pretending I'm clueless - I'm actually clueless!

John Baez (Aug 08 2024 at 09:27):

First something really stupid. Take $S = 1$. Then the left-hand side has just one element, so the right-hand side must as well, so $\Delta(S)$ must be the terminal presheaf, which happens to be the presheaf that assigns a one-element set to every open $U \subseteq X$. So we know $\Delta(S)$ in this one case.

John Baez (Aug 08 2024 at 09:28):

But this is no big surprise: we've just reproved that a right adjoint must send a terminal object to a terminal object.

John Baez (Aug 08 2024 at 09:29):

It's also known that a right adjoint preserves products and equalizers, so we instantly know $\Delta(S)$ for every set built from one-element sets by taking products and equalizers. Unfortunately the only sets we get this way are one-element sets. :sad: So, this is not useful... not yet, anyway.

John Baez (Aug 08 2024 at 09:32):

What about $\Delta(S)$ when $S$ is the empty set? Then the left side of

$\mathrm{hom}(B(X), \emptyset) \cong \mathrm{hom}(B, \nabla(\emptyset))$

is empty unless $B(X) = \emptyset$, in which case it has just one element, the identity morphism.

What presheaf might $\nabla(\emptyset)$ be?

Ryan Schwiebert (Aug 12 2024 at 03:07):

John Baez said:

What are some things we can figure out about ∇(S) using this "equation"? Ultimately everything. But let's try to tease out that information by making clever choices of B, and maybe choosing an easy S to start with.

I'm aware of this, of course, but to me it has all the promise of working as well as guessing what a map $\mathbb R^2\to \mathbb R$ based on how it behaves above the axes. But it's a better guide than nothing!

I haven't had any inspiration for new presheaves so I'm forced to look at the ones I mentioned before. For a set $C$, the idea of having $\nabla C(S)=Set[S,C]$ seems to promisingly go right when $C$ has one element, since I think it means there's only one natural transformation $B\to Set[S,C]$. But the same doesn't seem to be true for $C=\emptyset$, since I think it amounts to saying there aren't any natural transformations from $B$ to $Set[S,\emptyset]$ ever, and that can't be ruled out if $B(X)=\emptyset$ can happen.

OTOH with $\nabla C(S)=Set[C,S]$, when $C$ is empty, there is always only one natural transformation $B\to \nabla C$, but on the other side, $B(X)\to C$ won't happen if $B(X)\neq \emptyset$.

So... I'm still basically at square one. On this matter my brainpan has boiled dry and now simply smoking and blackening.

Ryan Schwiebert (Aug 12 2024 at 03:12):

I should say I did give some time to other ideas like $\nabla C(S)=S$, $\nabla C(S)=X\setminus S$, but they seemed to fail right away with the checks on $C=\emptyset$.

Ryan Schwiebert (Aug 22 2024 at 02:49):

Exercise 2.2.11: Given an adjunction $F\dashv G$ between categories $\mathcal A$ and $\mathcal B$ with unit and counit $\eta, \epsilon$, let $Fix(GF)$ be the set of objects $A\in\mathcal A$ such that $\eta_A$ is an isomorphism, and $Fix(FG)$ be the set of objects $B\in\mathcal B$ such that $\epsilon_B$ is an isomorphism.

Show the restriction of $F$ to $Fix(GF)$ and the restriction of $G$ to $Fix(FG)$ is an equivalence of categories. Then describe what this does for a few specific examples of adjunctions.

The proof that there is an equivalence is actually refreshingly straightforward. I picked the free-forgetful adjunction $F\dashv U$ between $Set$ and $Vect_\mathbb R$ to examine as suggested above, first.

My set theory feels a little rusty, but doesn't $Fix(UF)$ consist of the emptyset along with sets with cardinality at least $\mathfrak c$? I think the other half is basically $\mathbb R$ vector spaces of dimension $0$ and dimension at least $\mathfrak c$.

If so, it feels like a bit of an oddball equivalence of two categories. I'm trying to find a heuristic to make sense of it.

I can't recall the details but I'm guessing the Galois connection between fixed fields and subgroups of automorphism groups interpreted this way results in the fundamental theorem of Galois theory.

Chris Grossack (they/them) (Aug 22 2024 at 08:07):

The emptyset shouldn't be fixed, since the free vector space on it still has a 0 vector!

Chris Grossack (they/them) (Aug 22 2024 at 08:11):

As a hint remember you're not trying to show that the objects are abstractly isomorphic. It's important that the unit/counit be the isomorphism! For instance, take a set $X$ of size continuum. Can you explicitly describe the unit map $\eta_X : X \to UFX$? Why will it never be an isomorphism?

Morgan Rogers (he/him) (Aug 22 2024 at 08:13):

Following Chris' reply, I think you should check a different type of adjunction to get a meaningful equivalence out: free-forgetful adjunctions will almost always add elements to every set that will not be in the image of the unit.

Ryan Schwiebert (Aug 22 2024 at 22:25):

Chris Grossack (they/them) said:

The emptyset shouldn't be fixed, since the free vector space on it still has a 0 vector!

Ugh, yes, how stupid of me. And on the other end, $FU(\{0\})$ is a freely generated on a single element (who cares if it used to be called "0") and so it has dimension $1$, and after applying $\epsilon$ you just get the zero vector space, so it isn't part of $Fix(FU)$.

It probably has something to do with school starting again this week and doing all this after 10:45 P.M.

But on the upside, that makes the equivalence a lot less weird than I initially thought.

Ryan Schwiebert (Aug 22 2024 at 22:30):

Working my way gradually with responses to existing comments. Nobody need pressed to comment more anytime soon...

Ryan Schwiebert (Aug 23 2024 at 02:54):

Chris Grossack (they/them) said:

As a hint remember you're not trying to show that the objects are abstractly isomorphic. It's important that the unit/counit be the isomorphism!

Oh, of course. I think the transpose of the identity is going to send each element $a\in A$ to the linear combination in $F(A)$ which is $1$ for the basis element "a", and $0$ elsewhere. Obviously not onto the full span, ever. So the canonical equivalence is empty on both sides for this example?

Ryan Schwiebert (Aug 23 2024 at 03:01):

Morgan Rogers (he/him) said:

I think you should check a different type of adjunction to get a meaningful equivalence out

Like the dude at the end of Indiana Jones and the Last Crusade, I chose poorly.

Building a list of candidates to try again on.

Ryan Schwiebert (Aug 23 2024 at 03:05):

Incidentally, the discrete-forgetful-indiscrete adjunctions between $Top$ and $Set$ are denoted with $D,U,I$ in Leinster's book, and now it lives rent-free in my head as "the drunk adjunctions."

Chris Grossack (they/them) (Aug 23 2024 at 07:50):

Those adjunctions will be more exciting!

You can also look at the inclusion map $\iota : (\mathbb{Z},\leq) \hookrightarrow (\mathbb{R},\leq)$, viewing these posets as categories. Do you see what the left/right adjoints are? What is the equivalence? (It's a bit silly here)

For another option, remember if $f:X \to Y$ we have an adjunction $f_* \dashv f^{-1}$ on the powersets of $X$ and $Y$. What equivalence do you get in this case?

Ryan Schwiebert (Aug 24 2024 at 13:04):

Chris Grossack (they/them) said:

You can also look at the inclusion map ι:(Z,≤)↪(R,≤), viewing these posets as categories. Do you see what the left/right adjoints are? What is the equivalence? (It's a bit silly here)

That's a nice stepping stone. $Ceil\dashv \iota \dashv Floor$ right? And in both cases, it gives an equivalence between the domain and codomain of $\iota$.

I think the situation is similar for closure and interior operators on a topological space $X$, $Cl\dashv \iota$ and $\iota\dashv Int$, abusing $\iota$ as representing two different inclusions, one of the closed sets into the powerset, and one of the open sets into the powerset. For each $\iota$, the equivalence is between its domain and codomain.

Ryan Schwiebert (Aug 24 2024 at 22:01):

Chris Grossack (they/them) said:

For another option, remember if f:X→Y we have an adjunction f∗​⊣f−1 on the powersets of X and Y. What equivalence do you get in this case?

The equivalence is between subsets $A$ of $X$ such that $f^{-1}f_\ast(A)=A$ and the subsets $B$ of $Y$ such that $f_\ast f^{-1}(B)=B$, I think. I don't recall ever learning any special adjective for either type... is there one?

Chris Grossack (they/them) (Aug 24 2024 at 22:35):

Subsets $A \subseteq X$ so that $f^{-1} f_* A = A$ are often called "saturated". And a subset $$B \sibseteq Y$$ satisfies the dual property exactly when it's contained in the image of $f$. So this tells you that subsets of $\text{Im}(f)$ are in natural bijection with saturated subsets of $X$. In fact, it tells you the lattice structures on these collections are the same!

Ryan Schwiebert (Aug 24 2024 at 22:57):

Chris Grossack (they/them) said:

saturated subsets

Ah yes, a very natural choice of terminology. Thank you for the suggestion.

Ryan Schwiebert (Aug 31 2024 at 00:43):

Exercise 2.2.12 delivers equivalent conditions for an adjunction to be a reflection. I was able to prove the equivalence, although the hypotheses that contributed to each piece of the proof seemed surprisingly different from each other.

Anyhow, the second part asks you to go back and look at adjunctions to see which ones are reflections. I could use some sanity checks on what I think I found. I think the free/forgetful adjunction between Set and Vect, and the adjunction between $-\times B$ and $(-)^B$ are not reflections.

On the other hand,
The free-inclusion and inclusion-(subset of units) between Mon and Grp both seem to be reflections.
The D-U and U-I adjunctions between Set and Top seem to be reflective.
The Abelianize-inclusion functor between Grp and Ab seem reflections.

Secondly, I'm also trying to reconcile this notion of "reflective subcategory" with "adjunction that is a reflection." I can't quite tell if the same thing is being described in two different ways.

Thirdly, no mention is made of the dual thing with the adjunction, where one could ask the unit is an isomorphism. Does anyone get any mileage out of an "adjunction that is a coreflection"?

John Baez (Aug 31 2024 at 01:06):

Ryan Schwiebert said:

I could use some sanity checks on what I think I found. I think the free/forgetful adjunction between Set and Vect, and the adjunction between $-\times B$ and $(-)^B$ are not reflections.

That's right. I find it helps to think of a reflection as what you get when you have a category $C$ of things, a subcategory $N$ of "nice" things that have extra properties, but where maps are defined the same way, and a functor called the reflector $L: C \to N$ that "makes things be nice" by imposing those extra properties. Part of the deal here is that if you make a thing be nice, and then make it be nice again, it doesn't change the second time, since it's already nice.

The example I always remember is abelian groups as a reflective subcategory of groups. Abelian groups are just groups with an extra property, namely being abelian. So a homomorphism between abelian group is just a group homomorphism between groups that happen to be abelian - that's what I mean by "maps are defined the same way". And the reflector is abelianization. If you abelianize a group that's already abelian, you're not doing anything to it.

This example also helps me remember that the reflector is a left adjoint. In this example, it's left adjoint to the forgetful functor that forgets the abelianness of a group.

None of this is like what's going on between sets and vector spaces! Nobody ever says sets are vector spaces with a nice property, or vector spaces are sets with a nice property. Sure, vector spaces are sets with extra structure - but the maps between them have to preserve that extra structure, they are not just arbitrary functions between sets that happen to be vector spaces.

On the other hand, the free-inclusion and inclusion-(subset of units) between Mon and Grp both seem to be reflections.

That doesn't sound quite right. It's true that groups are just monoids with an extra property. A group homomorphism is just a monoid homomorphism between monoids that happen to be groups. (We don't need to separately say that inverses are preserved - that follows automatically.)

But it sounds like you're claiming the reflector sends any monoid to its group of units: the submonoid consisting of invertible elements. That can't be. Taking the group of units defines a functor from Mon to Grp, but I don't think this is the left adjoint to the forgetful functor Grp $\to$ Mon. I think this is the right adjoint. A reflector needs to be a left adjoint!

Secondly, I'm also trying to reconcile this notion of "reflective subcategory" with "adjunction that is a reflection." I can't quite tell if the same thing is being described in two different ways.

Yes, they are equivalent concepts - two ways of talking about the same phenomenon.

Thirdly, no mention is made of the dual thing with the adjunction, where one could ask the unit is an isomorphism. Does anyone get any mileage out of an "adjunction that is a coreflection"?

It sounds like you're talking about the concept of [[coreflective subcategory]]. There are some examples of those at the link. And in fact one of them is this:

the inclusion of groups into monoids, where the right adjoint takes a monoid to its group of units.

John Baez (Aug 31 2024 at 01:49):

(Btw, @Ryan Schwiebert, if you instantly read my reply when I first wrote it, I have now massively rewritten it.)

Ryan Schwiebert (Sep 01 2024 at 02:21):

@John Baez OK, I'm revisiting the trio "free-inclusion-(subgroup of units)" between Mon and Grp. I probably went too fast.

I get where you're coming from with the "nice" thing. Forgetting that a group is abelian and then abelianizing it does not cause it to grow. That was on my mind as I looked at $FU(G)$ for a group $G$, temporarily forgetting it had inverses in $U(G)$, but then not needing to add any in $FU(G)$.

I see now why the other pair is suspect. Take any finite monoid that isn't a group, its group of units will be proper, and upon inclusion back into Mon it will have to have shrunken: isomorphism will be impossible.

Ryan Schwiebert (Sep 01 2024 at 02:26):

John Baez said:

A reflector needs to be a left adjoint!

I have to ask, since Leinster never uses the term "reflector." He classifies an adjunct pair as a "reflection" if the right adjoint is full and faithful. Is that condition equivalent to a set of conditions on the left adjoint, so that one can identify a reflection from the other member of the adjoint pair?

John Baez (Sep 01 2024 at 04:32):

Sorry: as you probably guessed, given a reflective subcategory, we call the left adjoint to the inclusion of the subcategory in the larger category the reflector.

John Baez (Sep 01 2024 at 04:35):

The nLab article on reflective subcategories gives 6 equivalent characterizations of reflective subcategories. Condition 6 only involves a property of the reflector, but I'm not very familiar with it: I'm more used to conditions 1, 2 and 5.

Ryan Schwiebert (Sep 03 2024 at 01:54):

John Baez said:

6 equivalent characterizations of reflective subcategories

Excellent: I hadn't started reading syntopically yet but when I do, I should remember to use nLab.

Ryan Schwiebert (Sep 05 2024 at 02:51):

Exercise 2.2.13 is in part the example suggested earlier here about $f_\ast\dashv f^{-1}$. The kicker is it asks for a right adjoint for $f^{-1}$ too!

Again, I'm at a loss for a candidate for a right adjoint. If $\hat f$ is the right adjoint, I think it has to send the terminal object of $\mathcal P(X)$ to the terminal object of $\mathcal P(Y)$, so $\hat f(X)=Y$. Then I did a bit of thinking about covers of the initial elements of both categories and their dual counterparts along with $f^{-1}(S)\subseteq T \iff S\subseteq \hat f(T)$, without anything productive arising.

I'm certain I will find the answer if I search for it, but before I throw in the towel, I'm hoping someone can provoke a eureka moment and get me to see how one can figure this sort of thing out. I feel like I've plied everything I've learned and it's frustrating to still come up empty handed for something this simple sounding.

Any takers?

David Egolf (Sep 05 2024 at 03:18):

Looking at $f^{-1}(S) \subseteq T \iff S \subseteq \hat{f}(T)$, I am intrigued by the fact that this lets us translate $S \subseteq \hat{f}(T)$ to some equivalent condition. This catches my attention because I want to figure out what $\hat{f}(T)$ is, and $S \subseteq \hat{f}(T)$ is a partial description of $\hat{f}(T)$ .

What if we let $S$ be a set with a single element, so $S = \{y\}$? Does that help us figure out which elements are in $\hat{f}(T)$?

David Egolf (Sep 05 2024 at 03:23):

As a side note, because I keep getting confused about this, I think we have:

• $f:X \to Y$
• $f^{-1}:\mathcal{P}(Y) \to \mathcal{P}(X)$
• $\hat{f}:\mathcal{P}(X) \to \mathcal{P}(Y)$

So, $S \subseteq Y$ and $T \subseteq X$.

Josselin Poiret (Sep 05 2024 at 09:35):

Ryan Schwiebert said:

Exercise 2.2.13 is in part the example suggested earlier here about $f_\ast\dashv f^{-1}$. The kicker is it asks for a right adjoint for $f^{-1}$ too!

Again, I'm at a loss for a candidate for a right adjoint. If $\hat f$ is the right adjoint, I think it has to send the terminal object of $\mathcal P(X)$ to the terminal object of $\mathcal P(Y)$, so $\hat f(X)=Y$. Then I did a bit of thinking about covers of the initial elements of both categories and their dual counterparts along with $f^{-1}(S)\subseteq T \iff S\subseteq \hat f(T)$, without anything productive arising.

I'm certain I will find the answer if I search for it, but before I throw in the towel, I'm hoping someone can provoke a eureka moment and get me to see how one can figure this sort of thing out. I feel like I've plied everything I've learned and it's frustrating to still come up empty handed for something this simple sounding.

Any takers?

very set-theoretically, you can massage the logical statement you get from $f^*S \to T$ (I'm taking the exercise's notation) to get something that looks like $S \subset \dots$.
$\forall x \in X, (\exists y \in S, f(x)=y) \Rightarrow x \in T$
$\forall x \in X, \forall y \in S, f(x)=y \Rightarrow x \in T$
...

John Baez (Sep 05 2024 at 15:53):

All this notation is confusing me, but when you have a function $f: X \to Y$ any subset $S \subset X$ has an image in $Y$, but also another thing, less widely discussed, which I'll call the 'dual image'. So we get two maps $P X \to P Y$, and one of these is the left adjoint of the inverse image map $f^{-1}: P Y \to P X$ while the other is the right adjoint.

• The image of $S \subseteq X$ is the subset $\{f(x) \, \vert \; x \in S \}$

and

• The dual image of $S\subseteq X$ is the complement of the image of the complement of $S$.

John Baez (Sep 05 2024 at 15:57):

A more systematic way to say it, less likely to offend intuitionistic logicians:

• $y \in Y$ is in the image of $S \subseteq X$ if some element $x \in X$ with $f(x) = y$ is an element of $S$.

• $y \in Y$ is in the dual image of $S \subseteq X$ if every element $x \in X$ with $f(x) = y$ is an element of $S$.

Ryan Schwiebert (Sep 06 2024 at 03:00):

(Working these in the order they came in.)

Josselin Poiret said:

very set-theoretically, you can massage the logical statement you get from f∗S→T (I'm taking the exercise's notation) to get something that looks like S⊂….
∀x∈X,(∃y∈S,f(x)=y)⇒x∈T
∀x∈X,∀y∈S,f(x)=y⇒x∈T

Huh! The passage to the second line is something that I missed entirely. But I continued with $\forall y\in S,(\exists x\in X, f(x)=y)\implies x\in T$ which I think suggests $\hat f(T):=f(T)\cup (Y\setminus f(X))$. Using that I can get $f^{-1}(S)\subseteq T\implies S\subseteq \hat f(T)$, but for whatever reason I can't get the other direction.

Supposing the right hand side holds ($S\subseteq \hat f(T)$). We try to show that $f^{-1}(S)\subseteq T$. Now $f(f^{-1}(S))\subseteq S\subseteq f(T)$ since it is obviously not in the $Y\setminus f(X)$ piece. I guess that also nets us $f^{-1}(S)\subseteq f^{-1}f(T)$

Then what? I keep juggling things with the adjunction $f_\ast\dashv f^{-1}$, but nothing works out :( It tells us that $ff^{-1}(T)\subseteq T$, but I'm stuck with $f^{-1}f(T)$.

Have I gone wrong somewhere here?

Josselin Poiret (Sep 06 2024 at 09:32):

Ryan Schwiebert said:

Huh! The passage to the second line is something that I missed entirely.

this part is more second nature to people used to functional programming, it's just [[currying]]!

Ryan Schwiebert said:

But I continued with ∀y∈S,(∃x∈X,f(x)=y)⟹x∈T

be careful here, the $x$ you're using in the conclusion of the implication does not refer to a bound variable! you can't curry here! rather, you end up with $\forall y \in S, \forall x \in X, f(x)=y \Rightarrow x \in T$, ie. $\forall y \in S, P(y)$ and that proposition $P(y)$ with free variable $y$ defines a subset $\{y \in Y, S(y)\}$.

Josselin Poiret (Sep 06 2024 at 09:34):

(i'm intentionally being a bit cryptic here so that you can finish figuring it out)

Ryan Schwiebert (Sep 06 2024 at 11:19):

Josselin Poiret said:

it's just currying!

It's interesting because I restrained myself from asking if it had a special name because I thought "nah that's stupid to ask." I'm glad you mention it! I have encountered currying before, and TBH I don't recognize it yet in this situation, or at least it doesn't line up with my mental model yet...

be careful here, the x you're using in the conclusion of the implication does not refer to a bound variable! you can't curry here!

Apparently I have to be more careful! And the candidate I had felt so plausible... at least it satisfies the $\hat f(X)=Y$ condition. I will keep pursuing the hint tonight...

Ryan Schwiebert (Sep 07 2024 at 02:15):

@Josselin Poiret Huh! So that's it then: $\hat f (T):=\{y\in Y\mid f(x)=y \implies x\in T\}$. The elements with fibers contained in $T$?

It doesn't seem to reduce to anything more familiar... Is it known by a better name?

Ryan Schwiebert (Sep 07 2024 at 02:27):

John Baez said:

but also another thing, less widely discussed, which I'll call the 'dual image'.

I've circled back to this. Yes, indeed, I've never run across it before. I guess most of the time we're focusing on the "forward" direction of maps but this makes sense as some sort of backward looking version.

Josselin Poiret (Sep 07 2024 at 10:02):

Ryan Schwiebert said:

Josselin Poiret Huh! So that's it then: $\hat f (T):=\{y\in Y\mid f(x)=y \implies x\in T\}$. The elements with fibers contained in $T$?

It doesn't seem to reduce to anything more familiar... Is it known by a better name?

no, no, that's exactly it (rewording it in terms of fibers is a very good observation and how I would describe it)! the second question of the exercise will give you some intuition about what it is.

John Baez (Sep 07 2024 at 12:59):

Ryan Schwiebert said:

John Baez said:

• $y \in Y$ is in the image of $S \subseteq X$ if some element $x \in X$ with $f(x) = y$ is an element of $S$.

• $y \in Y$ is in the dual image of $S \subseteq X$ if every element $x \in X$ with $f(x) = y$ is an element of $S$.

I've circled back to this. Yes, indeed, I've never run across it before.

I don't know a really standard name for this 'dual image' thing: it seems sadly unused. You might use it in computer security or other subjects where you're trying to make things safe, like food or drug manufacture and distribution.

Suppose $X$ is some set of things, and $S \subseteq X$ is the set of things of type $X$ that are 'safe'. Suppose $f: X \to Y$ converts things of type $X$ to things of type $Y$. Then you might say that a thing of a type $Y$ is safe if it can only come from things of type $X$ that are safe.

John Baez (Sep 07 2024 at 13:06):

For example a company might only want to buy some food if they're sure all ingredients in that food come from safe sources.

But this application works just as well, or even better, if we generalize $f$ to a relation from $X$ to $Y$. A relation from $X$ to $Y$ is a subset

$R \subseteq X \times Y$

Any relation from $X$ to $Y$ gives various maps $PX \to PY$ and $PY \to PX$. In particular, we get two maps $PY \to PX$ which we might call 'image' and 'dual image':

John Baez (Sep 07 2024 at 13:07):

• $y \in Y$ is in the image of $S \subseteq X$ if some element $x \in X$ with $(x,y) \in R$ is an element of $S$.

• $y \in Y$ is in the dual image of $S \subseteq X$ if every element $x \in X$ with $(x,y) \in R$ is an element of $S$.