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Stream: learning: questions

Topic: isomorphic objects and functors

David Egolf (May 01 2023 at 22:25):

Let $F: A \to D$ be a functor, with $F(a) = x$ for some object $a \in A$ and some object $x \in D$. Assume that there is a $y$ different from $x$ with $x \cong y$ in $D$.

I'm wondering if we can modify $F$ to get a new functor $F'$ so that $F'(a) = y$. I want to require that $F'(a') = F(a')$ for all $a' \neq a$. I'm curious if it's possible to do this so that $F'$ and $F$ are naturally isomorphic.

To do this, we have to figure out how $F'$ acts on morphisms that go to or from $a$. If $f: a \to b$ is a morphism from $a$ to $b \neq a$, then $F(f): x \to F(b)$ and $F'(f): y \to F(b)$. We have a triangular diagram formed by these morphisms and the isomorphism $\phi: x \to y$. It's tempting to set $F'(f)$ so that this diagram commutes, so $F'(f) = F(f) \circ \phi^{-1}$.

Similarly, if $g: b \to a$ is a morphism to $a$ from $b \neq a$, then $F(g): F(b) \to x$ and $F'(g): F(b) \to y$. We again have a triangular diagram formed by these morphisms and the isomorphism $\phi: x \to y$. In this case, I want to set $F'(g) = \phi \circ F(g)$, so the diagram commutes.

Finally, if $h: a \to a$ is a morphism from $a$ to $a$, then $F(h): x \to x$ and $F'(h): y \to y$. Using the isomorphism $\phi: x \to y$, we get a square diagram that we would like to commute, so that $\phi \circ F(f) = F'(f) \circ \phi$, implying that $F'(f) = \phi \circ F(f) \circ \phi^{-1}$.

The next order of business would be to check that $F'$ is actually a functor. There are bunch of cases to check, when investigating whether $F'(n) \circ F'(m) = F'(n \circ m)$ for composable morphisms $n$ and $m$ in $A$. For example, the easiest case is when neither $n$ nor $m$ go to or from $a$, and in this case we can use the fact that $F$ is a functor to show $F'(n) \circ F'(m) = F'(n \circ m)$. However, there are a lot more cases to check depending on whether $n$ or $m$ or both go to and/or from $a$!

Does $F'$ really form a functor? And is there some way to find this out without checking a bunch of cases (I'm currently counting 9 cases to check)? Finally, if $F'$ does form a functor, is it naturally isomorphic to $F$?

Martti Karvonen (May 01 2023 at 22:30):

Yes, this works! Note that your $F'$ and the natural isomorphism depend on the chosen isomorphism $x\cong y$.

Martti Karvonen (May 01 2023 at 22:32):

More generally, given an arbitrary family of isomorphisms $\phi_x\colon F(x)\to x'=:F'(x)$, there is a unique way of making $F'$ into a functor so that the maps $\phi_x$ define a natural isomorphism $\phi\colon F\to F'$.

Martti Karvonen (May 01 2023 at 22:36):

So one can arbitrarily tweak the values of a functor within isomorphism and get a naturally isomorphic functor.

John Baez (May 01 2023 at 23:15):

David Egolf said:

Let $F: A \to D$ be a functor, with $F(a) = x$ for some object $a \in A$ and some object $x \in D$. Assume that there is a $y$ different from $x$ with $x \cong y$ in $D$.

I'm wondering if we can modify $F$ to get a new functor $F'$ so that $F'(a) = y$. I want to require that $F'(a') = F(a')$ for all $a' \neq a$. I'm curious if it's possible to do this so that $F'$ and $F$ are naturally isomorphic.

Yes, you can. I didn't look carefully at your approach, since it's less work to say what I'd do - you can tell me if that's what you were doing. The basic idea is to take $F$ and tweak it minimally to achieve the desired effect.

So, on objects, let $F' = F$ except for one thing: make $F'(a) = y$. On morphisms you need to do a bit more work since whenever $F$ sent a morphism to one with source or target (or both) equal to $x$, you need $F'$ to send it to one where the source or target is $y$. To make sure $F'$ is a functor you had better not do this in a random way! You'd better choose an isomorphism $\alpha: x \to y$ and use this and its inverse to "reroute" morphisms that start or end in $x$. That is:

• if $F(f) : z \to z'$ with $z,z' \ne x$ you let $F'(f) = F(f)$
• if $F(f) : z \to x$ with $z \ne x$ you let $F'(f) = \alpha \circ F(f)$
• if $F(f) : x \to z$ with $z \ne x$ you let $F'(f) = F(f) \circ \alpha^{-1}$
• if $F(f): x \to x$ you let $F'(f) = \alpha \circ F(f) \circ \alpha^{-1}$

And yes, there had better be some more intelligent way to say all this, check that $F'$ is a functor, and check that $F'$ is naturally isomorphic to $F$!

John Baez (May 01 2023 at 23:17):

The basic principle here is that "a chosen isomorphism is as good as equality".

John Baez (May 01 2023 at 23:22):

One reason I've never thought about how to do the above process intelligently is that I've never needed to make sure a functor doesn't hit some particular object. And it's not like I ever say "ooh, that object is disgusting, I'm going to make sure my functor avoids it". :upside_down:

John Baez (May 01 2023 at 23:28):

But it's nice to know we could do it if we had to. Like if Elon Musk starting charging us a fee each time we used $\mathbb{R}^3$, we could switch to using some other 3-dimensional real vector space.

David Egolf (May 01 2023 at 23:39):

John Baez said:

Yes, you can. I didn't look carefully at your approach, since it's less work to say what I'd do - you can tell me if that's what you were doing.

I believe your approach is the same as the one I described above. So that's encouraging!

Figuring out how to check that the approach works without a lot of tedious work still interests me. I've been thinking about the process of moving "structure" or "properties" between isomorphic objects a little bit the last few days. I'm interested in understanding when statements about one object can be transformed into statements about an isomorphic object, especially when these statements involve interactions with other categories. I'm also interested in understanding if there is some process that can take a property that doesn't "transform" properly between isomorphic objects, and extends it to one that does.

By the way, these musings are inspired by the earlier discussion on essential fibers, which I think is like an upgraded version of "fiber" that doesn't change in an undesired way as we move between isomorphic objects.

John Baez (May 02 2023 at 00:15):

I'm interested in understanding when statements about one object can be transformed into statements about an isomorphic object.

The short answer is: "always".

However, making sure the answer is "always" requires that we doing things right, like using essential fibers instead of fibers, etc. Any construction that goes wrong - that makes it impossible to replace an object by an isomorphic object - is said to violate the [[principle of equivalence]], and we avoid it. For short, we say it is "evil".

One main advantage of modern foundations of mathematics, like homotopy type theory / univalent mathematics, is that it becomes essentially impossible to formulate constructions that violate the principle of equivalence.

John Baez (May 02 2023 at 00:17):

As you'll see at the link, Michael Makkai proposed the Principle of Isomorphism:

“all grammatically correct properties of objects of a fixed category are to be invariant under isomorphism”.

John Baez (May 02 2023 at 00:19):

The principle of equivalence is a generalization of this.

But you and I have been doing category theory the old way, where there aren't built-in guardrails, and it's up to the user to use only constructions that don't violate the principle of equivalence.

John Baez (May 02 2023 at 00:55):

David Egolf said:

Figuring out how to check that the approach works without a lot of tedious work still interests me.

Yes, there should be something better than what I said. Here's something I tried but discarded, still perhaps worth thinking about. Forget the functor $F: A \to D$ for a minute and think about the category $D$.

If $x$ and $y$ are objects of $D$ for which there's an isomorphism $f: x \to y$, I claim there's a functor

$G: D \to D$

that

1) is the identity on all objects except perhaps $x$,
2) has $G(x) = y$,
3) is naturally isomorphic to $1_D : D \to D$, via some natural isomorphism $\alpha: 1_D \Rightarrow G$ with $\alpha_x = f$

John Baez (May 02 2023 at 00:59):

Then, given any functor $F: A \to D$, the functor $G \circ F: A \to D$ has these properties:

A) Whenever $F(a) = x$, $G(a) = y$,
B) $G \circ F$ is naturally isomorphic to $F$.

Item B) follows from item 3) and a bit of category theory that you might enjoy learning!

Mike Shulman (May 02 2023 at 01:12):

Even in old-fashioned mathematics, there are abstract notions that characterize "transportable" structure and properties, and general theorems about them. For instance, a (usually forgetful) functor $U:B\to X$ is called an isofibration if for any object $b\in B$ and isomorphism $\phi:U(b) \cong x$ in $X$, there exists an object $b'\in B$ such that $U(b') = x$ and an isomorphism $\psi:b\cong b'$ in $B$ such that $U(\psi)=\phi$.

This includes your example of functors: let $B$ be your category of functors $A\to D$, and $X$ the category $D^{{\rm ob}(A)}$ of families of objects of $D$ indexed by the set of objects of $A$, and let $U$ forget the action of a functor on morphisms but remember its action on objects. Then what you want is the fact that $U$ is an isofibration: given any functor $F$, and for each object $a\in A$ an object $F'(a)\in D$ and an isomorphism $F(a) \cong F'(a)$ (this is the isomorphism $\phi$), there exists a functor $F'$ with this action on objects and a natural isomorphism $F\cong F'$ (this is the isomorphism $\psi$).

Of course, one still has to prove that this functor is an isofibration. It's instructive to do this by hand, as the others in this thread have done; but once you get familiar with this sort of stuff there are general theorems you can appeal to. For instance, this functor $U$ happens to be (strictly) monadic, and any (strictly) monadic functor is an isofibration.

Jason Erbele (May 02 2023 at 01:32):

John Baez said:

Any construction that goes wrong - that makes it impossible to replace an object by an isomorphic object - is said to violate the [[principle of equivalence]], and we avoid it. For short, we say it is "evil".

I was about to mention that there are some interesting "evil" categorical gadgets, like [[dagger categories]], but apparently the equivalence elves have been working on that particular example, so there is now a non-"evil" formulation for the dagger. I guess the moral of the story is that if something "evil" is useful enough, it can (always?) find "salvation" with an appropriate change in perspective.

David Egolf (May 02 2023 at 21:32):

Thank-you everyone for your interesting responses! I will plan to read them properly and respond once I can (unfortunately today I have a nasty headache!).

David Egolf (May 03 2023 at 18:29):

John Baez said:

As you'll see at the link, Michael Makkai proposed the Principle of Isomorphism:

“all grammatically correct properties of objects of a fixed category are to be invariant under isomorphism”.

This is thought provoking for me. Without having thought deeply about any of this, I would have expected that we would instead want the properties of isomorphic objects to be invariant "up to isomorphism", not completely invariant. So, if $a \cong b$ then $P(a) \cong P(b)$ for any property $P$. We would need a category that the property values live in as objects though, for us to assess when two property values are isomorphic.

I would probably need to read the original context to better understand what Makkai meant.

David Egolf (May 03 2023 at 18:35):

John Baez said:

Then, given any functor $F: A \to D$, the functor $G \circ F: A \to D$ has these properties:

A) Whenever $F(a) = x$, $G(a) = y$,
B) $G \circ F$ is naturally isomorphic to $F$.

Item B) follows from item 3) and a bit of category theory that you might enjoy learning!

If I understand, the "bit of category theory" being referred to would involve showing the following:

• if a functor $G: D \to D$ satisfies $G \cong 1_\mathsf{D}$,
• then for any functor $F: A \to D$, we have $G \circ F \cong F$

Hmm. Let me see if I can prove this.

David Egolf (May 03 2023 at 19:09):

I think I got it! In sketch form:

• We first write out the commutative square associated with $G \cong 1_D$
• This square commutes for any $f: d \to d'$ in $D$
• In particular, it commutes for any $F(g): F(a) \to F(a')$ in $D$, for any $g: a \to a'$ in $A$.
• Expanding this out in the commutative square corresponding to $G \cong 1_D$, we get a commutative square that provides a natural isomorphism from $G \circ F$ to $F$

I suppose it could still be a lot of work to show that there is a $G$ that satisfies properties (1), (2) and (3), though.

John Baez (May 03 2023 at 19:13):

David Egolf said:

John Baez said:

As you'll see at the link, Michael Makkai proposed the Principle of Isomorphism:

“all grammatically correct properties of objects of a fixed category are to be invariant under isomorphism”.

This is thought provoking for me. Without having thought deeply about any of this, I would have expected that we would instead want the properties of isomorphic objects to be invariant "up to isomorphism", not completely invariant. So, if $a \cong b$ then $P(a) \cong P(b)$ for any property $P$. We would need a category that the property values live in as objects though, for us to assess when two property values are isomorphic.

Right. Ordinarily when people say "property" they mean something that can be true or false, so that $P(a)$ takes values in the 2-element poset $T \le F$. This poset has only identity morphisms, so two properties are isomorphic iff they are equal.

John Baez (May 03 2023 at 19:15):

That's in classical logic. In intuitionistic logic, which Makkai is deeply familiar with, we often replace the 2-element poset $T \le F$ by a more general poset of truth values. But since it's a poset, properties values in this will still be isomorphic iff they are equal.

John Baez (May 03 2023 at 19:15):

I imagine this might be what Makkai meant when he required properties to be invariant under isomorphism.

John Baez (May 03 2023 at 19:18):

A more modern attitude, which Makkai understood long before most, would be that for things valued in a category, "invariant" means "invariant up to coherent isomorphism". And more generally, for things valued in an n-category, "invariant" means "invariant up to coherent equivalence".

John Baez (May 03 2023 at 19:18):

(I could explain "coherent" but I don't feel like it right this second.)

John Baez (May 03 2023 at 19:19):

Still, it would be a bit unusual to use the word "property" to mean something valued in a category or n-category that's not just a poset (or preorder).

John Baez (May 03 2023 at 19:29):

"Property" is basically the same as "predicate", to hark back to an earlier conversation.

John Baez (May 03 2023 at 19:30):

David Egolf said:

If I understand, the "bit of category theory" being referred to would involve showing the following:

• if a functor $G: D \to D$ satisfies $G \cong 1_\mathsf{D}$,
• then for any functor $F: A \to D$, we have $G \circ F \cong F$

Exactly.

John Baez (May 03 2023 at 19:34):

David Egolf said:

I think I got it!

I think you did! And in the process you've started learning about "whiskering".

Given a functor $F: A \to B$ and functors $G, G': B \to C$ and a natural transformation $\alpha: G \Rightarrow G'$, there is a natural transformation called $F$ left whiskered by $\alpha$, and sometimes written $\alpha \circ F$, which does this:

$\alpha \circ F: G \circ F \Rightarrow G' \circ F$

John Baez (May 03 2023 at 19:36):

There is also "right whiskering", and these operations satisfy a bunch of nice rules.

John Baez (May 03 2023 at 19:38):

Using these rules you can easily show that if $\alpha$ has an inverse $\beta$ (so it's a natural isomorphism) then $\alpha \circ F$ has an inverse, namely $\beta \circ G$.

You can also see this directly, and it sounds like you already have.

David Egolf (May 03 2023 at 20:06):

John Baez said:

Still, it would be a bit unusual to use the word "property" to mean something valued in a category or n-category that's not just a poset (or preorder).

I guess I still have essential fibers on my mind here. I would be tempted to use the word "property" to refer to the essential fiber of an object with respect to a given functor ("a property of this object is its essential fiber with respect to this functor"). But an essential fiber is a category, which doesn't necessarily live in a preorder or partial order, so it sounds like "property" probably isn't the word I'm looking for here.

Hmm. Maybe even in this example there is a property like "induces this specific category as an essential fiber", which would lead to a statement about the object that can be true or false. Maybe I'm looking for a phrase like "induced structure" instead of "property", to refer to the essential fiber.

Spencer Breiner (May 03 2023 at 20:07):

John Baez said:

...$F$ left whiskered by $\alpha$...

I think this should this be $\alpha$ whiskered by $F$; the functors are the "whiskers" because they are one-dimensional.

Discussion of right vs left is an invitation to argument, but in "Invitation to Higher Gauge Theory" you called this a right whiskering (I guess to agree with applicative notation $\alpha\circ F$). I'd prefer to stick with diagrammatic notation (e.g., $F;\alpha$), and call it a left whiskering of $\alpha$ by $F$.

John Baez (May 03 2023 at 20:14):

David Egolf said:

I would be tempted to use the word "property" to refer to the essential fiber of an object with respect to a given functor ("a property of this object is its essential fiber with respect to this functor").

Don't. Just say no.

John Baez (May 03 2023 at 20:16):

For example, suppose you have the forgetful functor from groups to sets. Then you'd be calling a group a "property of" its underlying set. That's just not how we mathematicians talk. We'd call the group a structure on its underlying set.

John Baez (May 03 2023 at 20:18):

Someday you'll learn about how forgetful functors can forget stuff, or just structure, or just properties. When this sinks in it really clarifies a lot about mathematics.

John Baez (May 03 2023 at 20:19):

Briefly: properties take values in {T,F}. Structures take values in a set. Stuff takes values in a category. 2-stuff takes values in a 2-category. Etc.

John Baez (May 03 2023 at 20:26):

In our earlier conversation, we saw that the homotopy fiber of a functor between categories is a category, in general. So we say that in a general a functor forgets stuff. But we came up with conditions under which the homotopy fiber is just a set - or more precisely, equivalent to the discrete category on a set. If all the homotopy fibers are just sets, it makes sense to say our functor is just forgetting structure.

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

If we look at, say, the forgetful functor from abelian groups to groups, the homotopy fibers are all sets with either 0 or 1 elements, depending on whether our group is abelian or not. So in this case our functor is just forgetting a property: the property of being abelian.

John Baez (May 03 2023 at 20:36):

Spencer Breiner said:

John Baez said:

...$F$ left whiskered by $\alpha$...

I think this should this be $\alpha$ whiskered by $F$; the functors are the "whiskers" because they are one-dimensional.

You're right! I got mixed up. For the non-expert: when you draw this sort of thing as a diagram, the natural transformation $\alpha$ looks like a mouth, and $F$ looks like a whisker poking out. If you have $\alpha \circ F$ it looks like a mouth with a whisker poking out to the right... at least in one convention for drawing these things. Then this is "right whiskering". If you draw it the other way then it's called "left whiskering".

Here I hadn't mixed up left and right; I'd mixed up whiskers and mouths, which is much worse.

Discussion of right vs left is an invitation to argument, but in "Invitation to Higher Gauge Theory" you called this a right whiskering (I guess to agree with applicative notation $\alpha\circ F$).

I'm glad someone is keeping track of these things. I don't really care much what's left and what's right, though in any given text I try to be consistent (and often fail).

David Egolf (May 04 2023 at 20:48):

Mike Shulman said:

Even in old-fashioned mathematics, there are abstract notions that characterize "transportable" structure and properties, and general theorems about them. For instance, a (usually forgetful) functor $U:B\to X$ is called an isofibration if for any object $b\in B$ and isomorphism $\phi:U(b) \cong x$ in $X$, there exists an object $b'\in B$ such that $U(b') = x$ and an isomorphism $\psi:b\cong b'$ in $B$ such that $U(\psi)=\phi$.

This includes your example of functors: let $B$ be your category of functors $A\to D$, and $X$ the category $D^{{\rm ob}(A)}$ of families of objects of $D$ indexed by the set of objects of $A$, and let $U$ forget the action of a functor on morphisms but remember its action on objects. Then what you want is the fact that $U$ is an isofibration: given any functor $F$, and for each object $a\in A$ an object $F'(a)\in D$ and an isomorphism $F(a) \cong F'(a)$ (this is the isomorphism $\phi$), there exists a functor $F'$ with this action on objects and a natural isomorphism $F\cong F'$ (this is the isomorphism $\psi$).

Thanks for introducing me to the concept of "isofibration"!

I don't yet understand your second paragraph, though. In the category $D^{{\rm ob}(A)}$, what are the objects, and what are the morphisms? My first guess from your description is that an object is a function to the objects of $D$ from the objects of $A$, and that a morphism is a function between objects of $A$ that makes the resulting triangular diagram commute. However, I'm unsure if this is correct - I'm still learning the terminology involved here.

Mike Shulman (May 04 2023 at 21:21):

An object of $D^{ob(A)}$ is a function from the objects of $A$ to the objects of $D$, and a morphism from $F:ob(A) \to ob(D)$ to $G:ob(A) \to ob(D)$ is a function sending each $a\in ob(A)$ to a morphism $F(a) \to G(a)$ in $D$.

There are two ways to see this as an instance of something more general. First, it is actually the ordinary functor category from $ob(A)$, regarded as a discrete category, to $D$. Second, it is the product (in Cat) of $ob(A)$ copies of the category $D$.

David Egolf (May 04 2023 at 23:36):

Awesome, that makes a lot of sense! Thanks for explaining.

John Baez (May 04 2023 at 23:43):

This is another nice example of how category theorists freely think of sets as their corresponding discrete categories (we also saw that in the homotopy fiber story). The exponential of sets $X^Y$ and the exponential of a category by a set $C^Y$ become commonly used special cases of the exponential of categories $C^D$.

David Egolf (May 05 2023 at 00:00):

I see! So $X^Y$ for sets $X$ and $Y$ can be viewed as the set of functions from $Y$ to $X$, but it can also be thought of as a category of functors between discrete categories. The objects are functors - which are just functions - and the morphisms are natural transformations, which are sort of "translations" between functions. One cool thing this approach enables: we get a notion of "isomorphic functions" for functions from $Y$ to $X$. That is, we can say two functions are isomorphic if they are naturally isomorphic when viewed as functors between discrete categories.

I think I'd thought about this example before, but not the case where one of the two categories in the exponential is discrete and the other isn't necessarily discrete. Very interesting to learn about!

John Baez (May 05 2023 at 00:16):

That is, we can say two functions are isomorphic if they are naturally isomorphic when viewed as functors between discrete categories.

Yes, good point! But when exactly are two functions naturally isomorphic according to this definition? There's a very simple criterion.

David Egolf (May 05 2023 at 00:24):

Oh, I think they are naturally isomorphic exactly when they are equal. If $f \cong g$ as functors between discrete categories, then for any morphism $1_y: y \to y$ in $Y$ we have $f(y) \cong g(y)$. But the only morphisms in the target category are identity morphisms, so $f(y) = g(y)$. So if $f \cong g$, then $f = g$.

John Baez (May 05 2023 at 00:31):

Right! Moral: when working with categories that are secretly just sets, "isomorphic" secretly just means "equal".

David Egolf (May 05 2023 at 00:46):

Oh, I think the functor category $X^Y$ is actually a discrete category when $X$ and $Y$ are discrete categories: the only morphisms are identity morphisms. It makes sense, then, that it can be reasonably viewed as a set!

David Egolf (May 05 2023 at 01:45):

Mike Shulman said:

This includes your example of functors: let $B$ be your category of functors $A\to D$, and $X$ the category $D^{{\rm ob}(A)}$ of families of objects of $D$ indexed by the set of objects of $A$, and let $U$ forget the action of a functor on morphisms but remember its action on objects. Then what you want is the fact that $U$ is an isofibration: given any functor $F$, and for each object $a\in A$ an object $F'(a)\in D$ and an isomorphism $F(a) \cong F'(a)$ (this is the isomorphism $\phi$), there exists a functor $F'$ with this action on objects and a natural isomorphism $F\cong F'$ (this is the isomorphism $\psi$).

I think I understand the idea of this now! Let $F: A \to D$ be a functor that sends an object $a$ to $F(a)$. Let $U(F)$ be the corresponding functor from $ob(A)$ to $D$, that remembers only how $F$ maps on objects. Now let $U(F)'$ be a modified version of $U(F)$, that instead maps $a$ to some $d \cong F(a)$, with $d \neq F(a)$.

I think $U(F)' \cong U(F)$ in $D^{{\rm ob}(A)}$. That hopefully isn't too hard to show, since we've forgotten about the morphisms in $A$ for the moment. Then, if $U$ is an isofibration, since $U(F)$ is an object in the image of $U$ and $U(F)' \cong U(F)$, that means that there is some $F'$ so that $U(F') = U(F)'$. Further, this $F'$ is isomorphic to $F$ in $D^A$.

I'd have to check the details still to make sure I didn't make a mistake, but I suspect this is the basic idea!

John Baez (May 05 2023 at 14:17):

David Egolf said:

Oh, I think the functor category $X^Y$ is actually a discrete category when $X$ and $Y$ are discrete categories: the only morphisms are identity morphisms. It makes sense, then, that it can be reasonably viewed as a set!

True! In fact $X^Y$ is a discrete category whenever $X$ is discrete; there's no need for $Y$ to be discrete. After all, a morphism in $X^Y$ is a natural transformation between functors $F, G: Y \to X$, and such a natural transformation must be the identity if $X$ has only identity morphisms.

John Baez (May 05 2023 at 14:22):

You may have noticed here that I mentioned 3 cases but not the 4th:

John Baez said:

This is another nice example of how category theorists freely think of sets as their corresponding discrete categories (we also saw that in the homotopy fiber story). The exponential of sets $X^Y$ and the exponential of a category by a set $C^Y$ become commonly used special cases of the exponential of categories $C^D$.

John Baez (May 05 2023 at 14:24):

Raising a set to the power of a category gives a set, by what I just pointed out... but this case seems less useful than the other 3.