Michael wrote:

> It seems like once we start talking about categories, all laws or equations are presented as natural transformations like the associator and unitors. I don't get why we present them as natural transformations.

Is there something else you would prefer?

That's usually the best way to think about questions like this. If you're wondering why mathematicians do something in some way, try to do it some other way and see what happens.

For example, consider the tensor product \\( \otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}\\). In fact, let's think about a specific example, the cartesian product

\[ \times \colon \mathbf{Set} \times \mathbf{Set} \to \mathbf{Set} .\]

Are you happy with thinking about this as a functor? Maybe you haven't thought about it much. Treating cartesian product as a functor automatically lets us take cartesian products of morphisms (functions) as well as objects (sets). Do you see how given functions

\[ f \colon X \to Y \]

and

\[ f' \colon X' \to Y' \]

we get a function

\[ f \times f' \colon X \times X' \to Y \times Y' \; ? \]

That's nice. Also, the _fact_ that cartesian product is a functor summarizes useful facts like

\[ (g \times g') \circ (f \times f') = (gf \times g'f') .\]

Now consider the associative law. You might hope to express it as an _equation_:

\[ (X \times Y) \times Z = X \times (Y \times Z) \]

but in fact this equation isn't true! Do you see why not? If you don't see why, maybe someone else can explain why

\[ (X \times Y) \times Z \ne X \times (Y \times Z) \]

What we really have is an isomorphism

\[ (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z). \]

However, while there are usually _lots_ of isomorphisms

\[ (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z) \]

for any particular choice of sets \\(X,Y,Z\\), there's a particular 'best' one which we call

\[ \alpha_{X,Y,Z} \colon (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z) .\]

Do you see which one I'm talking about? Can you write down a formula for it?

But even once we have the 'best' isomorphism, how can we make precise the idea that it's the 'best', or at least 'good'? Here's how: the best isomorphism

\[ \alpha_{X,Y,Z} \colon (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z) .\]

is a _natural_ isomorphism. Do you see why it's natural? To show it's natural, you need to check that certain 'naturality squares' commute - remember [Lecture 43](https://forum.azimuthproject.org/discussion/2244/lecture-43-chapter-3-natural-transformations/p1).

It turns out that requiring naturality picks out the one 'best' isomorphism. Showing this is a bit harder than the other things I've asked you to show, so I'll let you off the hook here. But as you can see, I've asked you to do a lot of little calculations. I think these are required to see what's really going on here.

I could do them for you, but it's much better to do these things oneself and build ones mathematical muscles!

> It seems like once we start talking about categories, all laws or equations are presented as natural transformations like the associator and unitors. I don't get why we present them as natural transformations.

Is there something else you would prefer?

That's usually the best way to think about questions like this. If you're wondering why mathematicians do something in some way, try to do it some other way and see what happens.

For example, consider the tensor product \\( \otimes \colon \mathcal{C} \times \mathcal{C} \to \mathcal{C}\\). In fact, let's think about a specific example, the cartesian product

\[ \times \colon \mathbf{Set} \times \mathbf{Set} \to \mathbf{Set} .\]

Are you happy with thinking about this as a functor? Maybe you haven't thought about it much. Treating cartesian product as a functor automatically lets us take cartesian products of morphisms (functions) as well as objects (sets). Do you see how given functions

\[ f \colon X \to Y \]

and

\[ f' \colon X' \to Y' \]

we get a function

\[ f \times f' \colon X \times X' \to Y \times Y' \; ? \]

That's nice. Also, the _fact_ that cartesian product is a functor summarizes useful facts like

\[ (g \times g') \circ (f \times f') = (gf \times g'f') .\]

Now consider the associative law. You might hope to express it as an _equation_:

\[ (X \times Y) \times Z = X \times (Y \times Z) \]

but in fact this equation isn't true! Do you see why not? If you don't see why, maybe someone else can explain why

\[ (X \times Y) \times Z \ne X \times (Y \times Z) \]

What we really have is an isomorphism

\[ (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z). \]

However, while there are usually _lots_ of isomorphisms

\[ (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z) \]

for any particular choice of sets \\(X,Y,Z\\), there's a particular 'best' one which we call

\[ \alpha_{X,Y,Z} \colon (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z) .\]

Do you see which one I'm talking about? Can you write down a formula for it?

But even once we have the 'best' isomorphism, how can we make precise the idea that it's the 'best', or at least 'good'? Here's how: the best isomorphism

\[ \alpha_{X,Y,Z} \colon (X \times Y) \times Z \stackrel{\sim}{\rightarrow} X \times (Y \times Z) .\]

is a _natural_ isomorphism. Do you see why it's natural? To show it's natural, you need to check that certain 'naturality squares' commute - remember [Lecture 43](https://forum.azimuthproject.org/discussion/2244/lecture-43-chapter-3-natural-transformations/p1).

It turns out that requiring naturality picks out the one 'best' isomorphism. Showing this is a bit harder than the other things I've asked you to show, so I'll let you off the hook here. But as you can see, I've asked you to do a lot of little calculations. I think these are required to see what's really going on here.

I could do them for you, but it's much better to do these things oneself and build ones mathematical muscles!