The right unitor of the product of sets is the natural isomorphism
whose component
\[ \rho ^{\mathsf{Sets}}_{X} \colon X\times \text{pt}\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X \]
at $X\in \text{Obj}\webleft (\mathsf{Sets}\webright )$ is given by
\[ \rho ^{\mathsf{Sets}}_{X}\webleft (x,\star \webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x \]
for each $\webleft (x,\star \webright )\in X\times \text{pt}$.
Invertibility
The inverse of $\rho ^{\mathsf{Sets}}_{X}$ is the morphism
\[ \rho ^{\mathsf{Sets},-1}_{X}\colon X\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\times \text{pt} \]
defined by
\[ \rho ^{\mathsf{Sets},-1}_{X}\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (x,\star \webright ) \]
for each $x\in X$. Indeed:
- Invertibility I. We have
\begin{align*} \webleft [\rho ^{\mathsf{Sets},-1}_{X}\circ \rho ^{\mathsf{Sets}}_{X}\webright ]\webleft (x,\star \webright ) & = \rho ^{\mathsf{Sets},-1}_{X}\webleft (\rho ^{\mathsf{Sets}}_{X}\webleft (x,\star \webright )\webright )\\ & = \rho ^{\mathsf{Sets},-1}_{X}\webleft (x\webright )\\ & = \webleft (x,\star \webright )\\ & = \webleft [\text{id}_{X\times \text{pt}}\webright ]\webleft (x,\star \webright ) \end{align*}
for each $\webleft (x,\star \webright )\in X\times \text{pt}$, and therefore we have
\[ \rho ^{\mathsf{Sets},-1}_{X}\circ \rho ^{\mathsf{Sets}}_{X}=\text{id}_{X\times \text{pt}}. \]
- Invertibility II. We have
\begin{align*} \webleft [\rho ^{\mathsf{Sets}}_{X}\circ \rho ^{\mathsf{Sets},-1}_{X}\webright ]\webleft (x\webright ) & = \rho ^{\mathsf{Sets}}_{X}\webleft (\rho ^{\mathsf{Sets},-1}_{X}\webleft (x\webright )\webright )\\ & = \rho ^{\mathsf{Sets},-1}_{X}\webleft (x,\star \webright )\\ & = x\\ & = \webleft [\text{id}_{X}\webright ]\webleft (x\webright ) \end{align*}
for each $x\in X$, and therefore we have
\[ \rho ^{\mathsf{Sets}}_{X}\circ \rho ^{\mathsf{Sets},-1}_{X}=\text{id}_{X}. \]
Therefore $\rho ^{\mathsf{Sets}}_{X}$ is indeed an isomorphism.
Naturality
We need to show that, given a function $f\colon X\to Y$, the diagram
commutes. Indeed, this diagram acts on elements as
and hence indeed commutes. Therefore $\rho ^{\mathsf{Sets}}$ is a natural transformation.
Being a Natural Isomorphism
Since $\rho ^{\mathsf{Sets}}$ is natural and $\rho ^{\mathsf{Sets},-1}$ is a componentwise inverse to $\rho ^{\mathsf{Sets}}$, it follows from Chapter 9: Preorders, Item 2 of Proposition 9.9.7.1.2 that $\rho ^{\mathsf{Sets},-1}$ is also natural. Thus $\rho ^{\mathsf{Sets}}$ is a natural isomorphism.
