The symmetry of the coproduct of sets is the natural isomorphism
at $X,Y\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}\webright )$ is defined by
for each $\webleft (x,y\webright )\in X\times Y$.
The symmetry of the coproduct of sets is the natural isomorphism
at $X,Y\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}\webright )$ is defined by
for each $\webleft (x,y\webright )\in X\times Y$.
where $S=\left\{ 0,1\right\} \times \webleft (X\cup Y\webright )$ and
where $S'=\left\{ 0,1\right\} \times \webleft (Y\cup X\webright )=\left\{ 0,1\right\} \times \webleft (X\cup Y\webright )=S$.
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 $\webleft (0,x\webright )\in X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y$ and
for each $\webleft (1,y\webright )\in X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y$, and therefore we have
Invertibility II. We have
for each $\webleft (0,y\webright )\in Y\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}X$ and
for each $\webleft (1,x\webright )\in Y\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}X$, 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.