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Stream: learning: questions

Topic: invertible endofunctors inducing an equivalence relationship

David Egolf (Mar 03 2023 at 20:21):

[Please replace "invertible functor" with "equivalence" in your reading of this question, if it would make the question more interesting to you.]

Let $\mathsf{C}$ be a category. For any morphisms $a,b$ in $\mathsf{C}$, let $a \sim b$ if there exists an invertible functor $F: \mathsf{C} \to \mathsf{C}$ so that $F(a) = b$.

I think $\sim$ is an equivalence relationship:
I know that the identity functor $Id: \mathsf{C} \to \mathsf{C}$ is an invertible functor, acting as its own inverse. So $a \sim a$.

If $F:\mathsf{C} \to \mathsf{C}$ is an invertible functor, then its inverse $G$ is as well. If $a \sim b$, then $F(a) = b$, so $G(F(a)) = G(b)$ and so $G(b) = a$. That means $b \sim a$ if $a \sim b$.

If $a \sim b$ and $b \sim c$, then $F(a) = b$ and $H(b) = c$ for invertible functors $F,H: \mathsf{C} \to \mathsf{C}$. Composing these two to get another invertible functor, we find $(H \circ F)(a) = H(F(a)) = H(b) = c$ and so $a \sim c$.

If there is a standard name for this equivalence relationship, I'd be interested to learn it! I'm also interested in any examples of the resulting equivalence classes in different categories. For example, I think conjugacy classes in a group are one interesting example of equivalence classes generated in this way!

John Baez (Mar 03 2023 at 20:27):

This is a very nice question, but please replace "invertible functor" with "equivalence" if you want good answers. An equivalence of categories is a functor that invertible up to natural isomorphism.

John Baez (Mar 03 2023 at 20:28):

Isomorphisms of categories are too brittle to be useful: equivalences do what you really want.

John Baez (Mar 03 2023 at 20:28):

All the stuff you said would work for equivalences.

John Baez (Mar 03 2023 at 20:29):

For example, I think conjugacy classes in a group are one interesting example of equivalence classes generated in this way!

How do you want to think about that? You can think of group elements as morphisms in the one-object category corresponding to that group.

John Baez (Mar 03 2023 at 20:30):

But you seem to be wanting to think of them as objects in some category.

David Egolf (Mar 03 2023 at 20:33):

John Baez said:

This is a very nice question, but please replace "invertible functor" with "equivalence" if you want good answers. An equivalence of categories is a functor that invertible up to natural isomorphism.

I see! Unfortunately I'm out of steam for today to edit my question right now. But I would be very happy to instead consider the improved version of the question with "equivalence" replacing "invertible functor".

David Egolf (Mar 03 2023 at 20:34):

John Baez said:

For example, I think conjugacy classes in a group are one interesting example of equivalence classes generated in this way!

How do you want to think about that? You can think of group elements as morphisms in the one-object category corresponding to that group.

I was thinking of a group as an object with a single category, and with all morphisms isomorphisms. Then we have an invertible functor from the group back to itself provided by conjugation with any element. This induces an equivalence relationship on the morphisms of this category as described above, I believe.

John Baez (Mar 03 2023 at 20:40):

Oh, yes it does. For some reason I automatically assumed you were going to put an equivalence relation on objects of the category $\mathsf{C}$ such that $a \sim b$ iff there's some equivalence $F: \mathsf{C} \to \mathsf{C}$ such that $F(a) = b$.

John Baez (Mar 03 2023 at 20:40):

You can also do that.

John Baez (Mar 03 2023 at 20:41):

And by the way, if we use equivalences $F : \mathsf{C} \to \mathsf{C}$, then the equivalence relation I just mentioned is the same as this morally superior one: $a \sim b$ iff there's some equivalence $F: \mathsf{C} \to \mathsf{C}$ such that $F(a) \cong b$.

John Baez (Mar 03 2023 at 20:43):

The reason is that you can tweak an equivalence, getting a new equivalence where $F(a)$ is changed to any other isomorphic object! So if you had $F(a) \simeq b$ you can change $F$ to a new equivalence $F'$ with $F(a) = b$.

John Baez (Mar 03 2023 at 20:43):

You can't do this with isomorphisms: that's an example of how they are "brittle".

John Baez (Mar 03 2023 at 20:44):

But anyway, okay, so you were putting an equivalence relation on morphisms. You can do that too.

John Baez (Mar 03 2023 at 20:45):

So then the good news is: conjugate elements of a group will indeed give equivalent morphisms in your sense.

John Baez (Mar 03 2023 at 20:45):

And the bad news is: group elements don't need to be conjugate to be equivalent in your sense.

John Baez (Mar 03 2023 at 20:46):

It's not really "bad" news - it's just true news, nothing in math is bad. I've just heard too many "good news, bad news" jokes. :upside_down:

John Baez (Mar 03 2023 at 20:47):

So just for fun:

Puzzle: in the group $\mathbb{Z}/4$, thought of as a 1-object category, find two elements that are equivalent in your sense but not conjugate.

John Baez (Mar 03 2023 at 20:48):

And I'll say that in this puzzle it doesn't matter whether you use isomorphisms of categories, as you originally wanted to, or equivalences of categories. So you can use isomorphisms; they're a bit simpler.

David Egolf (Mar 04 2023 at 16:01):

John Baez said:

So just for fun:

Puzzle: in the group $\mathbb{Z}/4$, thought of as a 1-object category, find two elements that are equivalent in your sense but not conjugate.

I remember seeing that the group of integers together with addition has an automorphism corresponding to negation. So, in the sense described above, we have $a \sim -a$ for each integer $a$ in $\mathbb{Z}$.

Maybe this works for the group $\mathbb{Z}/4$ too (which I assume is the group of integers modulo 4 together with addition)? Now, $\mathbb{Z}/4$ becomes a ring when we add in multiplication modulo 4, and the equivalence class of the integers in $\mathbb{Z}/4$ containing -1 is 3. So, I'm wondering if multiplying by 3 (modulo 4) is an automorphism of $\mathbb{Z}/4$. Since 3 is relatively prime to 4, this map should be injective and hence (in this case) surjective. According to google, ring bijections are ring isomorphisms, and so multiplying by 3 (modulo 4) should in particular be a group isomorphism.
So, "negating" our elements (multiplying by 3, modulo 4) is a group automorphism and acts on elements as follows:

• $0 \mapsto 0$
• $1 \mapsto 3$
• $2 \mapsto 2$
• $3 \mapsto 1$
  In the sense described above, that means we again we have $a \sim -a$ for each $a \in \mathbb{Z}/4$. So, $0 \sim 0$, $1 \sim 3$, and $2 \sim 2$.

This group is commutative, so any two distinct elements are not conjugate. So, $1 \sim 3$ even though 1 and 3 are not conjugate.

David Egolf (Mar 04 2023 at 16:05):

John Baez said:

It's not really "bad" news - it's just true news, nothing in math is bad. I've just heard too many "good news, bad news" jokes. :upside_down:

Yes, I find it fun that this construction does something a bit different than relating conjugating elements!

John Baez (Mar 04 2023 at 16:09):

Yes, you got it! The group $\mathbb{Z}/3$ also has this property, that there are "equivalent" elements that are non-conjugate.

John Baez (Mar 04 2023 at 16:11):

I guess every finite abelian group has this property except $\mathbb{Z}/2$.

John Baez (Mar 04 2023 at 16:13):

In the language of group theory, what we're talking about is related to the difference between an automorphism of a group $G$, which is an invertible map $f : G \to G$ that preserves multiplication (and thus the identity and inverses too), and an [[inner automorphism]], which is an automorphism of the form

$f(g) = hgh^{-1}$

John Baez (Mar 04 2023 at 16:14):

Two group elements are "equivalent" in your sense iff there's an automorphism mapping the first to the second.

John Baez (Mar 04 2023 at 16:14):

Two group elements are conjugate iff there's an inner automorphism mapping the first two the second.

John Baez (Mar 04 2023 at 16:15):

Abelian groups never have any inner automorphisms except the identity, but they often have lots of automorphisms, and they always have the automorphism

$f(g) = g^{-1}$

which is the one you wound up using to solve the puzzle.

John Baez (Mar 04 2023 at 16:16):

(But you wrote it as $f(g) = -g$ since you were using additive notation, as people often do for abelian groups.)

John Baez (Mar 04 2023 at 16:17):

There's a lot to say about the distinction between the looser concept of "equivalent" and the stricter concept of "conjugate" - it's a really interesting issue.

John Baez (Mar 04 2023 at 16:18):

Some of this stuff is stuff group theorists talk about, and some is stuff category theorists are more likely to talk about.

David Egolf (Mar 04 2023 at 16:35):

For fun, I think here are two more examples of things that are related in the sense described above:

• Any two identity morphisms (or objects) in a discrete category
• $f$ and $g$ (or $a$ and $b$) in this little category:

a little category

John Baez (Mar 04 2023 at 21:30):

Yes, basically you've given a condition under which two morphisms are "indistinguishable unless we attach arbitrary names to the objects or morphisms in the category".

John Baez (Mar 05 2023 at 18:44):

Before I forget to do it, let me say the nice category-theoretic way to distinguish:

1) an element $g$ of a group $G$ that's mapped to an element $g'$ by some automorphism: $f(g) = g'$

from

2) an element $g$ of a group $G$ that's mapped to an element $g'$ by some inner automorphism: $hgh^{-1} = g'$

John Baez (Mar 05 2023 at 18:44):

The group elements $g$ and $g'$ are "more alike" in the second case than in the first case, but what do we mean by that?

John Baez (Mar 05 2023 at 18:45):

Think of a group as a 1-object category and think of its elements as morphisms.

John Baez (Mar 05 2023 at 18:49):

Then case 1) happens iff there's an equivalence of categories mapping $g$ to $g'$.

John Baez (Mar 05 2023 at 18:49):

But case 2) happens iff there's an equivalence of categories that's naturally isomorphic to the identity mapping $g$ to $g'$.

John Baez (Mar 05 2023 at 18:50):

I love this. Not only is $g'$ a lot like $g$, but the way they're alike is a lot like the identity.

David Egolf (Mar 05 2023 at 20:22):

That is very cool indeed! I love how category theory nicely provides a variety of levels of the concept of "same-ness".

John Baez (Mar 05 2023 at 20:34):

Yes, it's great!

I think a nice example is the group of symmetries of a square (the dihedral group $D_4$). The reflection that switches the top and bottom edges is a lot like a reflection that switches two opposite corners. But I think it's more like the reflection that switches the left and right edges.

John Baez (Mar 05 2023 at 20:35):

If I'm not confused, the first pair of symmetries are related by an an automorphism of the group, while the second pair are related by an inner automorphism - an equivalence that's naturally isomorphic to the identity.

David Egolf (Mar 05 2023 at 21:27):

That example reminds me of some of the patterns used to solve the 3x3x3 Rubik's cube! It's common to "get pieces into position, do something, and then return pieces back", which follows the form $hgh^{-1}$.

John Baez (Mar 05 2023 at 22:37):

Yes! In fact I've seen a whole book about how conjugation is used to solve problems. For example: if you drop a ball on the other side of the fence but can't reach it, you walk down the fence to a gate, go through the gate, then walk back along the fence to your ball.

John Baez (Mar 05 2023 at 22:38):

These "conjugations" are, in some perhaps vague sense, example of natural isomorphisms.

John Baez (Mar 05 2023 at 22:38):

I wish I could remember the name of that book!