2.
Both of the following conditions are satisfied:
-
(a)
We have $x=x_{0}$ or $y=y_{0}$.
-
(b)
We have $x'=x_{0}$ or $y'=y_{0}$.
In the first case, $\sigma ^{\mathsf{Sets}_{*}}_{X}$ clearly sends both elements to the same element in $X$. Meanwhile, in the latter case both elements are equal to the basepoint $x_{0}\wedge y_{0}$ of $X\wedge Y$, which gets sent to the basepoint $y_{0}\wedge x_{0}$ of $Y\wedge X$.
Being a Morphism of Pointed Sets
As just mentioned, we have
\[ \sigma ^{\mathsf{Sets}_{*}}_{X}\webleft (x_{0}\wedge y_{0}\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}y_{0}\wedge x_{0}, \]
and thus $\sigma ^{\mathsf{Sets}_{*}}_{X}$ is a morphism of pointed sets.
Invertibility
The inverse of $\sigma ^{\mathsf{Sets}_{*}}_{X,Y}$ is given by the morphism
\[ \sigma ^{\mathsf{Sets}_{*},-1}_{X,Y} \colon Y\wedge X \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X\wedge Y \]
defined by
\[ \sigma ^{\mathsf{Sets}_{*},-1}_{X,Y}\webleft (y\wedge x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x\wedge y \]
for each $y\wedge x\in Y\wedge X$.
Naturality
We need to show that, given morphisms of pointed sets
\begin{align*} f & \colon \webleft (X,x_{0}\webright ) \to \webleft (A,a_{0}\webright ),\\ g & \colon \webleft (Y,y_{0}\webright ) \to \webleft (B,b_{0}\webright )\end{align*}
the diagram
commutes. Indeed, this diagram acts on elements as
and hence indeed commutes, showing $\sigma ^{\mathsf{Sets}_{*}}$ to be a natural transformation.
Being a Natural Isomorphism
Since $\sigma ^{\mathsf{Sets}_{*}}$ is natural and $\sigma ^{\mathsf{Sets}_{*},-1}$ is a componentwise inverse to $\sigma ^{\mathsf{Sets}_{*}}$, it follows from
Chapter 11: Categories,
Item 2 of
Proposition 11.9.7.1.2 that $\sigma ^{\mathsf{Sets}_{*},-1}$ is also natural. Thus $\sigma ^{\mathsf{Sets}_{*}}$ is a natural isomorphism.
