John, thanks for the guideline to better understanding!

>Is there something else you would prefer?

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

>Are you happy with thinking about this as a functor?

Actually, I was thinking why not a functor LOL. But associativity is built off of the product functor so it has to be a natural transformation! So if we set the cartesian product as a functor and all equations using the cartesian product will be a natural transformation.

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

>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

Yes, they are not equal because the tuple is structured differently ie wearing different clothes. More specifically, if they are equal then \\((X \times Y) = X \text{ and } Z = (Y \times Z)\\) which not true generally (this can be true for monoidal products when \\( Y=I\\)). But they are isomorphic since the cardinalities are the same.

>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?

If I am not mistaken, the "lots of isomorphisms" is referring to the fact that there are many instances of tuples that are isomorphic? The best one is the one that can describe all of those isomorphisms. As you pointed out, it is \\(\alpha_{X,Y,Z}\\) which needs to satisfy \\(\alpha_{XYZ}^{-1} \alpha_{XYZ} = 1_{(XY)Z} \text{ and } \alpha_{XYZ} \alpha_{XYZ}^{-1} = 1_{X(YZ)}\\) and \\(\alpha_{X',Y',Z'} \circ F(f,f',f'') = G(f,f',f'') \circ \alpha_{X,Y,Z}\\)?

>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]

![associativity](http://aether.co.kr/images/universal_property_associativity.svg)

This is probably wrong but the best I could do... We need the diagram above to commute giving us the following equations:

\[f_x=\rho_x f_{x(yz)}=\pi_x \rho_{xy} f_{(xy)z}\]

\[f_y=\pi_y \rho_{yz} f_{x(yz)}=\pi_y \rho_{xy} f_{(xy)z}\]

\[f_z=\rho_z f_{(xy)z}=\pi_z \rho_{yz} f_{x(yz)}\]

\[\alpha_{xyz} f_{(xy)z} = f_{x(yz)}\]

\[\alpha_{xyz}^{-1} f_{x(yz)} = f_{(xy)z}\]

The last two equations when factored out will extract each component x,y,z from both sides. For example:

\[\rho_x = \pi_x \rho_{xy} \alpha_{xyz}^{-1}\]

Ultimately, it seems that the associativity isomorphism stems from the fact that these are just tuples in disguise and the cardinality is the same no matter where you put the parentheses since it always factors down the triangle? The last two questions are still unclear to me...

>Is there something else you would prefer?

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

>Are you happy with thinking about this as a functor?

Actually, I was thinking why not a functor LOL. But associativity is built off of the product functor so it has to be a natural transformation! So if we set the cartesian product as a functor and all equations using the cartesian product will be a natural transformation.

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

>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

Yes, they are not equal because the tuple is structured differently ie wearing different clothes. More specifically, if they are equal then \\((X \times Y) = X \text{ and } Z = (Y \times Z)\\) which not true generally (this can be true for monoidal products when \\( Y=I\\)). But they are isomorphic since the cardinalities are the same.

>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?

If I am not mistaken, the "lots of isomorphisms" is referring to the fact that there are many instances of tuples that are isomorphic? The best one is the one that can describe all of those isomorphisms. As you pointed out, it is \\(\alpha_{X,Y,Z}\\) which needs to satisfy \\(\alpha_{XYZ}^{-1} \alpha_{XYZ} = 1_{(XY)Z} \text{ and } \alpha_{XYZ} \alpha_{XYZ}^{-1} = 1_{X(YZ)}\\) and \\(\alpha_{X',Y',Z'} \circ F(f,f',f'') = G(f,f',f'') \circ \alpha_{X,Y,Z}\\)?

>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]

![associativity](http://aether.co.kr/images/universal_property_associativity.svg)

This is probably wrong but the best I could do... We need the diagram above to commute giving us the following equations:

\[f_x=\rho_x f_{x(yz)}=\pi_x \rho_{xy} f_{(xy)z}\]

\[f_y=\pi_y \rho_{yz} f_{x(yz)}=\pi_y \rho_{xy} f_{(xy)z}\]

\[f_z=\rho_z f_{(xy)z}=\pi_z \rho_{yz} f_{x(yz)}\]

\[\alpha_{xyz} f_{(xy)z} = f_{x(yz)}\]

\[\alpha_{xyz}^{-1} f_{x(yz)} = f_{(xy)z}\]

The last two equations when factored out will extract each component x,y,z from both sides. For example:

\[\rho_x = \pi_x \rho_{xy} \alpha_{xyz}^{-1}\]

Ultimately, it seems that the associativity isomorphism stems from the fact that these are just tuples in disguise and the cardinality is the same no matter where you put the parentheses since it always factors down the triangle? The last two questions are still unclear to me...