Subsection 5.1.7 (01NX): The Symmetry

The symmetry of the product of sets is the natural isomorphism

whose component

\[ \sigma ^{\mathsf{Sets}}_{X,Y} \colon X\times Y \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }Y\times X \]

at $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ is defined by

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

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

Invertibility

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

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

defined by

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

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

•
Invertibility I. We have

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

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

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

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

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

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

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

Naturality

We need to show that, given functions

\begin{align*} f & \colon X \to A,\\ g & \colon Y \to B\end{align*}

the diagram

commutes. Indeed, this diagram acts on elements as

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

Being a Natural Isomorphism

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

The Clowder Project

5.1.7 The Symmetry