>\[ X \times Y = \lbrace (x,y) : x \in X, y \in Y \rbrace .\]

>Given this definition, we can work out \\( (X \times Y) \times Z \\) and \\( X \times (Y \times Z) \\) and see that they are different sets. (Can you or someone else do it?)

\[( X \times Y) \times Z = \lbrace (a,b) : a \in X \times Y, b \in Z \rbrace .\]

\[ X \times (Y \times Z )= \lbrace (c,d) : c \in X , d \in Y \times Z \rbrace .\]

>However, we can also see an obvious one-to-one correspondence between these sets! We can write down a formula for this bijection and call it the associator

>\[ \alpha_{X,Y,Z} \colon (X \times Y) \times Z \to X \times (Y \times Z). \]

>(Can you or someone else do it?)

Let \\(\pi_1 \text{ and } \pi_2\\) be functions that extract out the first and second components respectively. Then a formula for the bijection would be :

\[ \pi_1 a \mapsto c\]

\[ \pi_2 a \mapsto \pi_1 d \]

\[ b \mapsto \pi_2 d \]

>To be precise, there will be 402,387,260 ... 000 bijections between them. But there will be one "best" bijection.

This sounded like Dr. Strange from Infinity Wars LOL. I get utterly confused whenever I hear the words "best" or "canonical" in category theory. I think it means the one that can best represent them all in which in this case it would be the associator. If I have to choose from actual bijections, I honestly don't know.

>Given this definition, we can work out \\( (X \times Y) \times Z \\) and \\( X \times (Y \times Z) \\) and see that they are different sets. (Can you or someone else do it?)

\[( X \times Y) \times Z = \lbrace (a,b) : a \in X \times Y, b \in Z \rbrace .\]

\[ X \times (Y \times Z )= \lbrace (c,d) : c \in X , d \in Y \times Z \rbrace .\]

>However, we can also see an obvious one-to-one correspondence between these sets! We can write down a formula for this bijection and call it the associator

>\[ \alpha_{X,Y,Z} \colon (X \times Y) \times Z \to X \times (Y \times Z). \]

>(Can you or someone else do it?)

Let \\(\pi_1 \text{ and } \pi_2\\) be functions that extract out the first and second components respectively. Then a formula for the bijection would be :

\[ \pi_1 a \mapsto c\]

\[ \pi_2 a \mapsto \pi_1 d \]

\[ b \mapsto \pi_2 d \]

>To be precise, there will be 402,387,260 ... 000 bijections between them. But there will be one "best" bijection.

This sounded like Dr. Strange from Infinity Wars LOL. I get utterly confused whenever I hear the words "best" or "canonical" in category theory. I think it means the one that can best represent them all in which in this case it would be the associator. If I have to choose from actual bijections, I honestly don't know.