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Stream: learning: questions

Topic: Exponentiating two morphisms

Nick Smith (Jul 23 2021 at 08:13):

I have a question related to my quest to understand closed categories:

If there is a monoidal structure on a category, you can use this to "multiply" two morphisms. For example, if I have $A \xrightarrow f C$ and $B \xrightarrow g D$ I can obtain $A \times B \xrightarrow {f \times g} C \times D$. The morphism $f \times g$ is guaranteed to exist by the bifunctor $\times: C \times C \to C$.

But there's also a bifunctor corresponding to the property of being monoidal closed. It's called the "internal hom functor", and has the form $[-,-] : C^{op} \times C \to C$. I want to understand how this acts on pairs of morphisms, akin to above, but it's melting my brain. What does it mean to "raise one morphism to the power of another"? Are there any examples of how this works for simple categories? For Set perhaps? And is it actually a useful concept, or is it just a curiosity?

Matteo Capucci (he/him) (Jul 23 2021 at 09:46):

It is analogous to the hom-functor, so it acts by precomposition on the contravariant slot and by postcomposition on the covariant slot

Matteo Capucci (he/him) (Jul 23 2021 at 09:47):

i.e. $[f, g] : [C, B] \to [A,D]$ sends a map $h : C \to B$ to the map $A \overset{f}\to C \overset{h}\to B \overset{g}\to D$

Fawzi Hreiki (Jul 23 2021 at 10:34):

Just a note: the notation $(f, g)$ usually refers to the pairing of two morphisms with the same domain. What you're describing is usually denoted $f \times g$.

John Baez (Jul 23 2021 at 15:14):

Nick Smith:

And is it actually a useful concept, or is it just a curiosity?

I don't use it much myself, but it's theoretically very important because it shows the functor $[-,-]: C^{\mathrm op} \times C \to C$ knows how morphisms are composed.

John Baez (Jul 23 2021 at 15:15):

The formula Matteo gave illustrates this.

John Baez (Jul 23 2021 at 15:17):

But be a bit careful with this formula! It makes perfect sense when $C = \mathrm{Set}$, because it seems to be treating $[c,b]$ as a set of morphisms from $c$ to $b$, with $h$ as an element of this set. (Matteo used upper-case letters for objects of $C$ but I'll use lower-case ones so I can say $c \in C$ without making the universe collapse.)

John Baez (Jul 23 2021 at 15:21):

But the concept of cartesian closed category is the most exciting precisely when the objects are not sets.

John Baez (Jul 23 2021 at 15:22):

Nonetheless we can still define an 'element' of an object of $C$, and any morphism $h: c \to b$ gives an 'element' of $[c,b]$. Do you know how this works?

John Baez (Jul 23 2021 at 15:23):

Using this, Matteo's remark makes sense for arbitrary cartesian closed categories. Thus, the internal hom functor of a cartesian closed category always keeps track of how morphisms are composed. But it does more, too.

Nick Smith (Jul 24 2021 at 01:58):

Matteo Capucci (he/him) said:

i.e. $[f, g] : [C, B] \to [A,D]$ sends a map $h : C \to B$ to the map $A \overset{f}\to C \overset{h}\to B \overset{g}\to D$

Wow, this is exactly the intuitive explanation I was looking for. That's so simple! I'm surprised all the resources on cartesian closed categories begin by waffling on about evaluation maps and adjunctions. None of the ones I've seen even bother to talk about this simple bifunctor. I think this example makes clear why closed categories are interesting.

So the composite $1 \xrightarrow {\text{my choice}} [C, B] \xrightarrow {[f,g]} [A, D]$ allows you to "select" how you'd like to connect f to g (at least in $\bold{Set}$), and tells you what the end result is. It's like playing with lego!

Nick Smith (Jul 24 2021 at 02:32):

John Baez said:

the concept of cartesian closed category is the most exciting precisely when the objects are not sets.

Yes, it seems the monoidal closed structure of $\bold{Poly}$ is more interesting than $\bold{Set}$! I understand the internal hom functor for $\otimes$ better than the one for $\times$ (Poly is closed in two ways). It firstly tells you how a morphism $h$ constructs a composite $f ; h ; g$ (as above), but it also seems to tell you which "wire" ($\mapsto$) between directions in $h$ induced each wire in $f ; h ; g$. It's an explanation: "here is why I'm connected the way I am". Of course, most people here probably aren't familar with $\bold{Poly}$...

Nonetheless we can still define an 'element' of an object of $C$, and any morphism $h: c \to b$ gives an 'element' of $[c,b]$. Do you know how this works?

I suppose I don't really know how $h: c \to b$ gets encoded as an "element" of $[c, b]$ when we're talking about arbitrary categories, but I think I know how to point to these elements: you can use a map from the unit of the monoidal structure? For $\times$, this is the terminal object $1$ (thus $1 \xrightarrow {el} [c,b]$), and in the case of $\otimes$ in $\bold{Poly}$, this is the monomial $y$ (thus $y \xrightarrow {el} [c,b]$). But the objects may have additional properties that aren't captured by these maps alone, right?

John Baez (Jul 24 2021 at 14:56):

Nick Smith said:

Matteo Capucci (he/him) said:

i.e. $[f, g] : [c, b] \to [a,d]$ sends a map $h : c \to b$ to the map $a \overset{f}\to c \overset{h}\to b \overset{g}\to d$

Wow, this is exactly the intuitive explanation I was looking for. That's so simple! I'm surprised all the resources on cartesian closed categories begin by waffling on about evaluation maps and adjunctions. None of the ones I've seen even bother to talk about this simple bifunctor.

It's not "waffling", since this fact follows from the more terse definition: a cartesian closed category is a cartesian category where each functor $a \times -$ has a right adjoint.

John Baez (Jul 24 2021 at 14:59):

Also, note that this fact is not sufficient to define the bifunctor $[-,-]: C^{\rm op} \times C \to C$, except in cases like $\mathrm{Set}$ where the object $[c,b]$ is a mere set, and $[f,g]$ is a mere function.

John Baez (Jul 24 2021 at 15:02):

Here's where you should have been told this formula: when someone told you about the functor

$\mathrm{hom} : C^{\rm op} \times C \to \mathrm{Set}$

which any category $C$ has. You need to know this functor to know what it means when people say

$L$ and $R$ are adjoint functors if there's a natural isomorphism $\mathrm{hom}(La,b) \cong \mathrm{hom}(a,Rb)$

After all, you can't know what it means for this isomorphism to be natural until you know how $\mathrm{hom}$ is a functor - including what it does to morphisms!

Nick Smith (Jul 25 2021 at 00:38):

John Baez said:

[...] the functor

$\mathrm{hom} : C^{\rm op} \times C \to \mathrm{Set}$

which any category $C$ has. You need to know this functor to know what it means when people say

$L$ and $R$ are adjoint functors if there's a natural isomorphism $\mathrm{hom}(La,b) \cong \mathrm{hom}(a,Rb)$

After all, you can't know what it means for this isomorphism to be natural until you know how $\mathrm{hom}$ is a functor - including what it does to morphisms!

When I first encountered adjunctions (a few months ago), they were explained in terms of "an isomorphism between sets", so the whole functor thing was swept under a rug.

Nick Smith (Jul 25 2021 at 00:44):

Anyway, thank you for the clarifications. I'll continue studying these things. I really do appreciate being able to get help from experts. This Zulip is magical!

John Baez (Jul 25 2021 at 15:42):

Nick Smith said:

John Baez said:

[...] the functor

$\mathrm{hom} : C^{\rm op} \times C \to \mathrm{Set}$

which any category $C$ has. You need to know this functor to know what it means when people say

$L$ and $R$ are adjoint functors if there's a natural isomorphism $\mathrm{hom}(La,b) \cong \mathrm{hom}(a,Rb)$

After all, you can't know what it means for this isomorphism to be natural until you know how $\mathrm{hom}$ is a functor - including what it does to morphisms!

When I first encountered adjunctions (a few months ago), they were explained in terms of "an isomorphism between sets", so the whole functor thing was swept under a rug.

It's good to think about why merely having an isomorphism of sets

$\mathrm{hom}(La,b) \cong \mathrm{hom}(a,Rb)$

is not enough for $L$ and $R$ to be adjoint functors. It really needs to be a natural isomorphism. Without naturality a lot of things go wrong when you try to actually use adjoint functors.

But to even state the naturality condition we need to explain how $\mathrm{hom}(L-,-)$ and $\mathrm{hom}(-,R-)$ are functors. And for that we need to explain how $\mathrm{hom}(-,-)$ is a functor. So we need to understand what it does to morphisms.

John Baez (Jul 25 2021 at 15:46):

A while back Jade and I were trying to construct the right adjoint to a functor $L$, and we found a functor $R$ and an isomorphism of sets

$\mathrm{hom}(La,b) \cong \mathrm{hom}(a,Rb)$

So we thought we had found the right adjoint! But it turned out this isomorphism wasn't natural! So we had found the wrong adjoint.