2.
Both of the following conditions are satisfied:
-
(a)
We have $x=x_{0}$ or $y=0$.
-
(b)
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:
-
1.
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
\[ \rho ^{\mathsf{Sets}_{*},-1}_{X}\circ \rho ^{\mathsf{Sets}_{*}}_{X}=\operatorname {\mathrm{id}}_{X\wedge S^{0}}. \]
-
2.
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
\[ \rho ^{\mathsf{Sets}_{*}}_{X}\circ \rho ^{\mathsf{Sets}_{*},-1}_{X}=\operatorname {\mathrm{id}}_{X}. \]
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 11: Categories,
Item 2 of
Proposition 11.9.7.1.2 that $\rho ^{\mathsf{Sets}_{*},-1}$ is also natural. Thus $\rho ^{\mathsf{Sets}_{*}}$ is a natural isomorphism.