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Stream: learning: questions

Topic: morphisms to products

David Egolf (Feb 20 2023 at 20:05):

Let us assume we know of two morphisms $s,r: A \to B$, in a category $\mathsf{C}$ that has binary products. In this setting, there are some morphisms to $B \times B$ from $A$ that must exist, because of the fact that $B \times B$ is a product. How many of these are there?

For a specific example, in the category of sets, consider $s,r: \mathbb{R} \to \mathbb{R}$ with $s(x) = 2x$ and $r(x) = x^2$ for each $x \in \mathbb{R}$. Then we get four functions from $\mathbb{R}$ to $\mathbb{R} \times \mathbb{R}$. Namely, the ones that act like this: $x \mapsto (2x, 2x)$, $x \mapsto (x^2, x^2)$, $x \mapsto (2x, x^2)$, and $x \mapsto (x^2, 2x)$.

morphism to product

Returning to the category $\mathsf{C}$, we have a function $!: \mathsf{C}(A,B) \times \mathsf{C}(A,B) \to \mathsf{C}(A, B \times B)$ where $\mathsf{C}(A,B)$ denotes the set of morphisms from $A$ to $B$ (so I am assuming $\mathsf{C}$ is locally small). As illustrated in the "morphism to product" figure, this function takes in two morphisms from $A$ to $B$, and provides the unique morphism that makes the above diagram commute (and such a morphism exists and is unique because $B \times B$ together with $\pi_1$ and $\pi_2$ is a product of $B$ and $B$).

If we start with two morphisms from $A$ to $B$, do we always get four distinct morphisms from $A$ to $B \times B$ using $!$, as we did in the example above in the category of sets? More generally, I'm wondering if the function $!$ is injective, so that we always get different morphisms from $A$ to $B \times B$ when we specify different pairs of morphisms from $A$ to $B$.

Ralph Sarkis (Feb 20 2023 at 20:23):

The universal property of the product says precisely that this function $!$ is a bijection.

Jonas Frey (Feb 20 2023 at 20:25):

Ralph Sarkis said:

The universal property of the product says precisely that this function $!$ is a bijection.

(so in particular, it is injective, just to spell it out)

David Egolf (Feb 20 2023 at 20:43):

I thought I knew what a "universal property" was, but apparently I do not! Thank-you for pointing me in the right direction. It looks like this nlab page will be helpful.

David Egolf (Feb 20 2023 at 20:45):

I will see if I can figure this out based on this hint :smile: .

Ralph Sarkis (Feb 20 2023 at 21:02):

If you like to understand things from a more general/abstract point of view, you can try to prove that the Hom functors are continuous (the proof on the nlab can seem a bit opaque).

John Baez (Feb 20 2023 at 22:26):

Here's a common shorthand way of stating the universal property for product:

"A morphism from A to X $\times$ Y is the same as a morphism from A to X and and morphism from A to Y."

This is shorthand for saying there's a natural isomorphism

$\mathsf{C}(A,X \times Y) \cong \mathsf{C}(A,X) \times \mathsf{C}(A,Z)$

John Baez (Feb 20 2023 at 22:29):

And that, in turn, is shorthand for stating how the product has a universal property such that for any pair of morphisms $f: A \to X$, $g: A \to Y$, there exists a unique morphism $f: A \to X \times Y$ such that... blah-di-blah-di-blah.

John Baez (Feb 20 2023 at 22:29):

The "blah-di-blah-di-blah" is important but we often don't want to say it all, so we often use these shorter formulations.

John Baez (Feb 20 2023 at 22:30):

The shorter formulations are weaker, in that they follow from the precise one. But you have to do a little work to see why!

John Baez (Feb 20 2023 at 22:31):

Also, to understand the first, shortest formulation I gave, you need to understand what category theorists mean by "the same as". It doesn't always mean "equal"!

John Baez (Feb 20 2023 at 22:32):

In this case it means there's a natural bijection.

David Egolf (Feb 21 2023 at 17:13):

Thanks for your responses! I'm currently working on wrapping my head around how we get from a universal property (in terms of representability of a functor) to a limit/colimit description of a representing object. I'll try to answer my question above using this once I get a bit further in understanding this.

David Egolf (Feb 22 2023 at 19:37):

This thread has been great - I've wanted to understand some of this stuff for a while, and having a specific example has been really helpful.

I think I now understand the following example: in the category of sets, if there exists some object $Z$ so that $\mathsf{Set}(-,Z) \cong \mathsf{Set}(-,A) \times \mathsf{Set}(-,B)$ then $Z$ is forced to satisfy the diagram that means it is a product of $A$ and $B$. So if $\mathsf{Set}(-,Z) \cong \mathsf{Set}(-,A) \times \mathsf{Set}(-,B)$ then $Z \cong A \times B$. It was very cool to see how this is required by starting with the natural isomorphism diagram between $\mathsf{Set}(-,Z)$ and $\mathsf{Set}(-,A) \times \mathsf{Set}(-,B)$.

David Egolf (Feb 22 2023 at 19:38):

However, I think what I'm asking in this thread actually involves going in the other direction: In a category $\mathsf{C}$, if $\mathsf{Z} \cong A \times B$, then is $\mathsf{C}(-,Z) \cong \mathsf{C}(-,A) \times \mathsf{C}(-,B)$?
Maybe now that I have a little practice with related ideas, I can figure out how to construct a natural isomorphism from the fact that $Z$ is a product.

John Baez (Feb 23 2023 at 07:04):

David Egolf said:

However, I think what I'm asking in this thread actually involves going in the other direction: In a category $\mathsf{C}$, if $\mathsf{Z} \cong A \times B$, then is $\mathsf{C}(-,Z) \cong \mathsf{C}(-,A) \times \mathsf{C}(-,B)$?

This is true, and it feels easier to me than the implication in the other direction!

John Baez (Feb 23 2023 at 07:07):

The reason is that if $Z \cong A \times B$ I can use the definition of $A \times B$ to get a morphism from $X$ to $Z$ from a pair of morphisms $X \to A$, $X \to B$, and also vice versa. This is what the universal property definition of $A \times B$ does.

John Baez (Feb 23 2023 at 07:08):

So we just need to check that the resulting bijection $\mathsf{C}(X,Z) \cong \mathsf{C}(X,A) \times \mathsf{C}(X,B)$ is actually natural, and that feels like a "follow your nose" argument.

David Egolf (Feb 23 2023 at 17:18):

John Baez said:

The reason is that if $Z \cong Z \times B$ I can use the definition of $A \times B$ to get a morphism from $X$ to $Z$ from a pair of morphisms $X \to Z$, $Y \to Z$, and also vice versa. This is what the universal property definition of $A \times B$ does.

I think there is at least one typo here, and I'm getting confused staring at this. I'm pretty sure you mean to start by assuming that $Z \cong A \times B$, unless I'm missing something?

David Egolf (Feb 23 2023 at 17:43):

John Baez said:

So we just need to check that the resulting bijection $\mathsf{C}(X,Z) \cong \mathsf{C}(X,A) \times \mathsf{C}(X,B)$ is actually natural, and that feels like a "follow your nose" argument.

I'm a bit stuck on showing that there is a bijection at all. Here's what I've got so far:
If $Z$ is a product of $A$ and $B$ we have this diagram:
product

Given a choice of an object $X$ in $\mathsf{C}$, then because $Z$ together with $\pi_1$ and $\pi_2$ is a product of $A$ and $B$, for each pair $(f_1, f_2)$ there is a unique morphism $!(f_1, f_2)$ that makes this diagram commute. So, given a pair of morphisms in $\mathsf{C}(X,A) \times \mathsf{C}(X,B)$ we get a unique morphism in $\mathsf{C}(X,Z)$ that makes this diagram commute. That means that $!$ is a function from $\mathsf{C}(X,A) \times \mathsf{C}(X,B)$ to $\mathsf{C}(X,Z)$. We want to show this function is a bijection.

For any $h \in \mathsf{C}(X,Z)$, we can compose with $\pi_1$ and $\pi_2$ to get $\pi_1 \circ h \in \mathsf{C}(X,A)$ and $\pi_2 \circ h \in \mathsf{C}(X,B)$. If we set $f_1$ and $f_2$ to be these composed morphisms, then the diagram commutes.
So, for any $h \in \mathsf{C}(X,Z)$ there is an element $g = (\pi_1 \circ h, \pi_2 \circ h) \in \mathsf{C}(X,A) \times \mathsf{C}(X,B)$ so that $!(g) = h$. That means that $!$ is a surjective function.

It remains to show that $!$ is an injective function, and that is where I'm stuck.

Ralph Sarkis (Feb 23 2023 at 17:49):

If both $!(f_1,f_2) = {!}(g_1,g_2)$, then $!(g_1,g_2)$ makes the diagram above commute, so you can find what $g_1$ and $g_2$ are.

John Baez (Feb 23 2023 at 18:49):

David Egolf said:

John Baez said:

The reason is that if $Z \cong Z \times B$ I can use the definition of $A \times B$ to get a morphism from $X$ to $Z$ from a pair of morphisms $X \to Z$, $Y \to Z$, and also vice versa. This is what the universal property definition of $A \times B$ does.

I think there is at least one typo here, and I'm getting confused staring at this. I'm pretty sure you mean to start by assuming that $Z \cong A \times B$, unless I'm missing something?

Ugh, yes there are lots of typos there.

John Baez (Feb 23 2023 at 18:53):

But please read what I mean, not what I wrote! :upside_down:

I meant:

The reason is that if $Z \cong A \times B$ I can use the definition of $A \times B$ to get a morphism from $X$ to $Z$ from a pair of morphisms $X \to A$, $X \to B$, and also vice versa. This is what the universal property definition of $A \times B$ does.

John Baez (Feb 23 2023 at 18:56):

It remains to show that $!$ is an injective function, and that is where I'm stuck.

There are always two ways to attempt to show a function is bijective: the noble way and the ignoble way. The noble way is to construct an inverse. The ignoble way is to show it's injective and surjective. I recommend always starting with the noble way, and resorting to the ignoble way only if the noble way fails.

John Baez (Feb 23 2023 at 18:56):

The noble way is noble because it's a general fact about categories that "an isomorphism is a morphism with an inverse".

John Baez (Feb 23 2023 at 18:57):

The ignoble way is ignoble because it's a very special fact about the category Set that "an isomorphism is an epimorphism that's also a monomorphism". This fails in most categories.

John Baez (Feb 23 2023 at 18:58):

So, it's kind of shocking to me that you first tried the ignoble way.

John Baez (Feb 23 2023 at 19:00):

This is probably because instead of "isomorphism" you were saying "bijection", and thinking "one-to-one and onto".

John Baez (Feb 23 2023 at 19:00):

Here's another way to see why the noble way is nicer: it's simpler!

John Baez (Feb 23 2023 at 19:02):

Compare:

A function $f: X \to Y$ is an isomorphism if there's a function $g: Y \to X$ such that $fg = 1$ and $gf = 1$

to

A function $f: X \to Y$ is an isomorphism if for every $y \in Y$ there exists $x \in X$ such that $f(x) = y$ and for every $x, x' \in X$ if $f(x) = f(x')$ then $x = x'$.

John Baez (Feb 23 2023 at 19:03):

The second one is more complicated.

David Egolf (Feb 24 2023 at 16:51):

Ralph Sarkis said:

If both $!(f_1,f_2) = {!}(g_1,g_2)$, then $!(g_1,g_2)$ makes the diagram above commute, so you can find what $g_1$ and $g_2$ are.

This hint was a great help for figuring out how to show that $!$ is injective. Assume that $!(f_1,f_2) = !(g_1,g_2)$. Then $\pi_1 \circ !(f_1,f_2) = f_1$, and similarly $\pi_1 \circ !(g_1,g_2) = g_1$, because of the diagram that is required to commute (see above). In addition, because $!(f_1,f_2) = !(g_1,g_2)$, then $\pi_1 \circ !(f_1, f_2) = g_1$. So $f_1 = g_1$. Using $\pi_2$, we can similarly show that $f_2 = g_2$. So if $!(f_1,f_2) = !(g_1,g_2)$, then $(f_1, f_2) = (g_1,g_2)$. We conclude that $!$ is injective. We already saw that it is surjective, so it is bijective.

John Baez (Feb 24 2023 at 17:08):

(deleted)

David Egolf (Feb 24 2023 at 17:10):

John Baez said:

The noble way is noble because it's a general fact about categories that "an isomorphism is a morphism with an inverse".

This does have its appeal! The simplicity you mention is also appealing.

It sounds fun to show that $!$ is an isomorphism by finding an inverse for it. It will be interesting to see how that approach compares to showing $!$ is injective and surjective.

We recall that $!$ maps from $\mathsf{C}(X,A) \times \mathsf{C}(X,B) \to \mathsf{C}(X,Z)$, where $Z$ together with $\pi_1$ and $\pi_2$ is a product of $A$ and $B$. Let $r$ be a proposed inverse, with $r: \mathsf{C}(X,Z) \to \mathsf{C}(X,A) \times \mathsf{C}(X,B)$ acting by $r(f) = (\pi_1 \circ f, \pi_2 \circ f)$ for each $f \in \mathsf{C}(X,Z)$.

Computing $(r \circ !)(f_1, f_2)$, we get $(r \circ !)(f_1, f_2) = r(!(f_1, f_2)) = (\pi_1 \circ !(f_1, f_2), \pi_2 \circ !(f_1, f_2))$. Because the diagram commutes, $\pi_1 \circ !(f_1, f_2) = f_1$ and $\pi_2 \circ !(f_1, f_2) = f_2$. So, we find that $(r \circ !)(f_1, f_2) = (f_1, f_2)$ for each $(f_1, f_2)$. That means $r \circ !$ is the identity morphisms on $\mathsf{C}(X,A) \times \mathsf{C}(X,B)$.

Computing $(! \circ r)(f)$, we get $(! \circ r)(f)= !(( \pi_1 \circ f, \pi_2 \circ f))$. We need a morphism in $\mathsf{C}(X,Z)$ so that the diagram commutes with these morphisms to $A$ and $B$. I think $f$ does it, so that $(! \circ r)(f) = f$. That means $! \circ r$ is the identity morphisms on $\mathsf{C}(X,Z)$.

So, $r$ is an inverse of $!$, and so $!$ is an isomorphism (and hence a bijection).

David Egolf (Feb 24 2023 at 17:13):

This more general approach felt easier, but it's hard to make a fair comparison when I'd warmed up by doing the other method first. :upside_down: I do like how this approach uses an argument pattern that is less specific to $\mathsf{Set}$, though!

John Baez (Feb 24 2023 at 17:14):

Great! Actually, after further thought, this is how I'd try to make everything as simple as possible. The product $A \times B$ comes with projections

$\pi_1 : A \times B \to A$

and

$\pi_2: A \times B \to B$

so any morphism

$f: X \to A \times B$

gives a pair of morphisms

$\pi_1 \circ f : X \to A, \quad \pi_2 \circ f : X \to B$

I have just described a map

$\mathsf{C}(X, A \times B) \to \mathsf{C}(X, A) \times \mathsf{C}(X, B)$

sending $f$ to the pair $(\pi_1 \circ f, \pi_2 \circ f)$.

Now: the universal property of the product says this map is a bijection.

That is - saying the exact same thing more slowly - given any pair of maps $f_1 : X \to A, f_2: X \to B$, there exists a unique $f: X \to A \times B$ such that

$f_1 = \pi_1 \circ f, \quad f_2 = \pi_2 \circ f$

So, this bijectiveness of the map

$\mathsf{C}(X, A \times B) \to \mathsf{C}(X, A) \times \mathsf{C}(X, B)$

that I described is the universal property of the product!

David Egolf (Feb 24 2023 at 17:31):

John Baez said:

So, this bijectiveness of the map

$\mathsf{C}(X, A \times B) \to \mathsf{C}(X, A) \times \mathsf{C}(X, B)$

that I described is the universal property of the product!

Oh, that is a nice way to put it! It is interesting that the injectivity and surjectivity become easier to see from the product diagram when we work with the inverse of $!$.

John Baez (Feb 24 2023 at 17:43):

Yeah, I think it's important that the inverse of the map you're calling !, not ! itself, is the really obvious map.

This is a general fact about limits and colimits: the universal property says that some "obvious map" has an inverse.