The symmetry of the coproduct of sets is the natural isomorphism
at $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ is defined by
for each $(x,y)\in X\times Y$.
The symmetry of the coproduct of sets is the natural isomorphism
at $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ is defined by
for each $(x,y)\in X\times Y$.
where $S=\left\{ 0,1\right\} \times (X\cup Y)$ and
where $S'=\left\{ 0,1\right\} \times (Y\cup X)=\left\{ 0,1\right\} \times (X\cup Y)=S$.
defined by
and hence given by
for each $z\in Y\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}X$. Indeed:
Invertibility I. We have
for each $(0,x)\in X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y$ and
for each $(1,y)\in X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y$, and therefore we have
Invertibility II. We have
for each $(0,y)\in Y\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}X$ and
for each $(1,x)\in Y\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}X$, and therefore we have
Therefore $\sigma ^{\mathsf{Sets},\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}}_{X,Y}$ is indeed an isomorphism.