Definition 5.1.4.1.1 (01NS): The Associator of $\times $

The associator of the product of sets is the natural isomorphism

\[ \alpha ^{\mathsf{Sets}} \colon \mathord {\times }\circ {(\mathord {\times }\times \operatorname {\mathrm{id}}_{\mathsf{Sets}})} \mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}\mathord {\times }\circ {(\operatorname {\mathrm{id}}_{\mathsf{Sets}}\times \mathord {\times })}\circ {\mathbf{\alpha }^{\mathsf{Cats}_{\mathsf{2}}}_{\mathsf{Sets},\mathsf{Sets},\mathsf{Sets}}}, \]

as in the diagram

whose component

\[ \alpha ^{\mathsf{Sets}}_{X,Y,Z} \colon (X\times Y)\times Z \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\times (Y\times Z) \]

at $(X,Y,Z)$ is given by

\[ \alpha ^{\mathsf{Sets}}_{X,Y,Z}((x,y),z) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(x,(y,z)) \]

for each $((x,y),z)\in (X\times Y)\times Z$.

Invertibility

The inverse of $\alpha ^{\mathsf{Sets}}_{X,Y,Z}$ is the morphism

\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z} \colon X\times (Y\times Z) \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }(X\times Y)\times Z \]

defined by

\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}(x,(y,z))\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}((x,y),z) \]

for each $(x,(y,z))\in X\times (Y\times Z)$. Indeed:

•
Invertibility I. We have

\begin{align*} [\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets}}_{X,Y,Z}]((x,y),z) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}(\alpha ^{\mathsf{Sets}}_{X,Y,Z}((x,y),z))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}(x,(y,z))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}((x,y),z)\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[\operatorname {\mathrm{id}}_{(X\times Y)\times Z}]((x,y),z) \end{align*}

for each $((x,y),z)\in (X\times Y)\times Z$, and therefore we have

\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets}}_{X,Y,Z}=\operatorname {\mathrm{id}}_{(X\times Y)\times Z}. \]
•
Invertibility II. We have

\begin{align*} [\alpha ^{\mathsf{Sets}}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}](x,(y,z)) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets}}_{X,Y,Z}(\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}(x,(y,z)))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets}}_{X,Y,Z}((x,y),z)\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(x,(y,z))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[\operatorname {\mathrm{id}}_{(X\times Y)\times Z}](x,(y,z)) \end{align*}

for each $(x,(y,z))\in X\times (Y\times Z)$, and therefore we have

\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets}}_{X,Y,Z}=\operatorname {\mathrm{id}}_{X\times (Y\times Z)}. \]

Therefore $\alpha ^{\mathsf{Sets}}_{X,Y,Z}$ is indeed an isomorphism.

Naturality

We need to show that, given functions

\begin{align*} f & \colon X \to X',\\ g & \colon Y \to Y',\\ h & \colon Z \to Z’ \end{align*}

the diagram

commutes. Indeed, this diagram acts on elements as

and hence indeed commutes, showing $\alpha ^{\mathsf{Sets}}$ to be a natural transformation.

Being a Natural Isomorphism

Since $\alpha ^{\mathsf{Sets}}$ is natural and $\alpha ^{\mathsf{Sets},-1}$ is a componentwise inverse to $\alpha ^{\mathsf{Sets}}$, it follows from Chapter 11: Categories, Item 2 of Proposition 11.9.7.1.2 that $\alpha ^{\mathsf{Sets},-1}$ is also natural. Thus $\alpha ^{\mathsf{Sets}}$ is a natural isomorphism.

The Clowder Project

5.1.4 The Associator