The associator of the product of sets is the natural isomorphism
\[ \alpha ^{\mathsf{Sets}} \colon \mathord {\times }\circ {\webleft (\mathord {\times }\times \text{id}_{\mathsf{Sets}}\webright )} \mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}\mathord {\times }\circ {\webleft (\text{id}_{\mathsf{Sets}}\times \mathord {\times }\webright )}\circ {\mathbf{\alpha }^{\mathsf{Cats}_{\mathsf{2}}}_{\mathsf{Sets},\mathsf{Sets},\mathsf{Sets}}}, \]
as in the diagram
whose component
\[ \alpha ^{\mathsf{Sets}}_{X,Y,Z} \colon \webleft (X\times Y\webright )\times Z \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\times \webleft (Y\times Z\webright ) \]
at $\webleft (X,Y,Z\webright )$ is given by
\[ \alpha ^{\mathsf{Sets}}_{X,Y,Z}\webleft (\webleft (x,y\webright ),z\webright ) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (x,\webleft (y,z\webright )\webright ) \]
for each $\webleft (\webleft (x,y\webright ),z\webright )\in \webleft (X\times Y\webright )\times Z$.
Invertibility
The inverse of $\alpha ^{\mathsf{Sets}}_{X,Y,Z}$ is the morphism
\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z} \colon X\times \webleft (Y\times Z\webright ) \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }\webleft (X\times Y\webright )\times Z \]
defined by
\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\webleft (x,\webleft (y,z\webright )\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (\webleft (x,y\webright ),z\webright ) \]
for each $\webleft (x,\webleft (y,z\webright )\webright )\in X\times \webleft (Y\times Z\webright )$. Indeed:
- Invertibility I. We have
\begin{align*} \webleft [\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets}}_{X,Y,Z}\webright ]\webleft (\webleft (x,y\webright ),z\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\webleft (\alpha ^{\mathsf{Sets}}_{X,Y,Z}\webleft (\webleft (x,y\webright ),z\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\webleft (x,\webleft (y,z\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (\webleft (x,y\webright ),z\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{id}_{\webleft (X\times Y\webright )\times Z}\webright ]\webleft (\webleft (x,y\webright ),z\webright ) \end{align*}
for each $\webleft (\webleft (x,y\webright ),z\webright )\in \webleft (X\times Y\webright )\times Z$, and therefore we have
\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets}}_{X,Y,Z}=\text{id}_{\webleft (X\times Y\webright )\times Z}. \]
- Invertibility II. We have
\begin{align*} \webleft [\alpha ^{\mathsf{Sets}}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\webright ]\webleft (x,\webleft (y,z\webright )\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets}}_{X,Y,Z}\webleft (\alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\webleft (x,\webleft (y,z\webright )\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\alpha ^{\mathsf{Sets}}_{X,Y,Z}\webleft (\webleft (x,y\webright ),z\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (x,\webleft (y,z\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\text{id}_{\webleft (X\times Y\webright )\times Z}\webright ]\webleft (x,\webleft (y,z\webright )\webright ) \end{align*}
for each $\webleft (x,\webleft (y,z\webright )\webright )\in X\times \webleft (Y\times Z\webright )$, and therefore we have
\[ \alpha ^{\mathsf{Sets},-1}_{X,Y,Z}\circ \alpha ^{\mathsf{Sets}}_{X,Y,Z}=\text{id}_{X\times \webleft (Y\times Z\webright )}. \]
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 9: Preorders, Item 2 of Proposition 9.9.7.1.2 that $\alpha ^{\mathsf{Sets},-1}$ is also natural. Thus $\alpha ^{\mathsf{Sets}}$ is a natural isomorphism.
