The right unitor of the smash product of pointed sets is the natural isomorphism
whose component
\[ \rho ^{\mathsf{Sets}_{*}}_{X} \colon X\wedge S^{0} \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X \]
at $X\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$ is given by
\begin{align*} x\wedge 0 & \mapsto x_{0},\\ x\wedge 1 & \mapsto x \end{align*}
for each $x\in X$.
Well-Definedness
Let $\webleft [\webleft (x,y\webright )\webright ]=\webleft [\webleft (x',y'\webright )\webright ]$ be an element in $X\wedge S^{0}$. Then either:
-
We have $x=x'$ and $y=y'$.
-
Both of the following conditions are satisfied:
-
We have $x=x_{0}$ or $y=0$.
-
We have $x'=x_{0}$ or $y'=0$.
In the first case, $\rho ^{\mathsf{Sets}_{*}}_{X}$ clearly sends both elements to the same element in $X$. Meanwhile, in the latter case both elements are equal to the basepoint $x_{0}\wedge 0$ of $X\wedge S^{0}$, which gets sent to the basepoint $x_{0}$ of $X$.
Being a Morphism of Pointed Sets
As just mentioned, we have
\[ \rho ^{\mathsf{Sets}_{*}}_{X}\webleft (x_{0}\wedge 0\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x_{0}, \]
and thus $\rho ^{\mathsf{Sets}_{*}}_{X}$ is a morphism of pointed sets.
Invertibility
The inverse of $\rho ^{\mathsf{Sets}_{*}}_{X}$ is the morphism
\[ \rho ^{\mathsf{Sets}_{*},-1}_{X}\colon X\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\wedge S^{0} \]
defined by
\[ \rho ^{\mathsf{Sets}_{*},-1}_{X}\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x\wedge 1 \]
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\wedge 0\webright ) & = \rho ^{\mathsf{Sets}_{*},-1}_{X}\webleft (\rho ^{\mathsf{Sets}_{*}}_{X}\webleft (x\wedge 0\webright )\webright )\\ & = \rho ^{\mathsf{Sets}_{*},-1}_{X}\webleft (x_{0}\webright )\\ & = x_{0}\wedge 1\\ & = x\wedge 0, \end{align*}
and
\begin{align*} \webleft [\rho ^{\mathsf{Sets}_{*},-1}_{X}\circ \rho ^{\mathsf{Sets}_{*}}_{X}\webright ]\webleft (x\wedge 1\webright ) & = \rho ^{\mathsf{Sets}_{*},-1}_{X}\webleft (\rho ^{\mathsf{Sets}_{*}}_{X}\webleft (x\wedge 1\webright )\webright )\\ & = \rho ^{\mathsf{Sets}_{*},-1}_{X}\webleft (x\webright )\\ & = x\wedge 1 \end{align*}
for each $x\in X$, and thus we have
\[ \rho ^{\mathsf{Sets}_{*},-1}_{X}\circ \rho ^{\mathsf{Sets}_{*}}_{X}=\text{id}_{X\wedge S^{0}}. \]
- 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\wedge 1\webright )\\ & = x \end{align*}
for each $x\in X$, and thus we have
\[ \rho ^{\mathsf{Sets}_{*}}_{X}\circ \rho ^{\mathsf{Sets}_{*},-1}_{X}=\text{id}_{X}. \]
This shows $\rho ^{\mathsf{Sets}_{*}}_{X}$ to be invertible.
Naturality
We need to show that, given a morphism of pointed sets
\[ f\colon \webleft (X,x_{0}\webright )\to \webleft (Y,y_{0}\webright ), \]
the diagram
commutes. Indeed, this diagram acts on elements as
and
and hence indeed commutes, showing $\rho ^{\mathsf{Sets}_{*}}$ to be 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.
