The associator of the smash product of pointed sets is the natural isomorphism
\[ \alpha ^{\mathsf{Sets}_{*}} \colon {\wedge }\circ {\webleft ({\wedge }\times \text{id}_{\mathsf{Sets}_{*}}\webright )} \mathbin {\overset {\mathord {\sim }}{\Longrightarrow }}{\wedge }\circ {\webleft (\text{id}_{\mathsf{Sets}_{*}}\times {\wedge }\webright )}\circ {\mathbf{\alpha }^{\mathsf{Cats}}_{\mathsf{Sets}_{*},\mathsf{Sets}_{*},\mathsf{Sets}_{*}}}, \]
as in the diagram
whose component
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z} \colon \webleft (X\wedge Y\webright )\wedge Z \xrightarrow {\cong }X\wedge \webleft (Y\wedge Z\webright ) \]
at $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$ is given by
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}\webleft (\webleft (x\wedge y\webright )\wedge z\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x\wedge \webleft (y\wedge z\webright ) \]
for each $\webleft (x\wedge y\webright )\wedge z\in \webleft (X\wedge Y\webright )\wedge Z$.
Well-Definedness
Let $\webleft [\webleft (\webleft (x,y\webright ),z\webright )\webright ]=\webleft [\webleft (\webleft (x',y'\webright ),z'\webright )\webright ]$ be an element in $\webleft (X\wedge Y\webright )\wedge Z$. Then either:
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We have $x=x'$, $y=y'$, and $z=z'$.
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Both of the following conditions are satisfied:
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We have $x=x_{0}$ or $y=y_{0}$ or $z=z_{0}$.
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We have $x'=x_{0}$ or $y'=y_{0}$ or $z'=z_{0}$.
In the first case, $\alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}$ clearly sends both elements to the same element in $X\wedge \webleft (Y\wedge Z\webright )$. Meanwhile, in the latter case both elements are equal to the basepoint $\webleft (x_{0}\wedge y_{0}\webright )\wedge z_{0}$ of $\webleft (X\wedge Y\webright )\wedge Z$, which gets sent to the basepoint $x_{0}\wedge \webleft (y_{0}\wedge z_{0}\webright )$ of $X\wedge \webleft (Y\wedge Z\webright )$.
Being a Morphism of Pointed Sets
As just mentioned, we have
\[ \alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}\webleft (\webleft (x_{0}\wedge y_{0}\webright )\wedge z_{0}\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}x_{0}\wedge \webleft (y_{0}\wedge z_{0}\webright ), \]
and thus $\alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}$ is a morphism of pointed sets.
Invertibility
Clearly, the inverse of $\alpha ^{\mathsf{Sets}_{*}}_{X,Y,Z}$ is given by the morphism
\[ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z} \colon X\wedge \webleft (Y\wedge Z\webright ) \xrightarrow {\cong }\webleft (X\wedge Y\webright )\wedge Z \]
defined by
\[ \alpha ^{\mathsf{Sets}_{*},-1}_{X,Y,Z}\webleft (x\wedge \webleft (y\wedge z\webright )\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (x\wedge y\webright )\wedge z \]
for each $x\wedge \webleft (y\wedge z\webright )\in X\wedge \webleft (Y\wedge Z\webright )$.
Naturality
We need to show that, given morphisms of pointed sets
\begin{align*} f & \colon \webleft (X,x_{0}\webright ) \to \webleft (X',x'_{0}\webright ),\\ g & \colon \webleft (Y,y_{0}\webright ) \to \webleft (Y',y'_{0}\webright ),\\ h & \colon \webleft (Z,z_{0}\webright ) \to \webleft (Z’,z’_{0}\webright ) \end{align*}
the diagram
commutes. Indeed, this diagram acts on elements as
and hence indeed commutes, showing $\alpha ^{\mathsf{Sets}_{*}}$ to be a natural transformation.
Being a Natural Isomorphism
Since $\alpha ^{\mathsf{Sets}_{*}}$ is natural and $\alpha ^{\mathsf{Sets}_{*},-1}$ is a componentwise inverse to $\alpha ^{\mathsf{Sets}_{*}}$, it follows from Chapter 8: Categories, Item 2 of Proposition 8.8.6.1.2 that $\alpha ^{\mathsf{Sets}_{*},-1}$ is also natural. Thus $\alpha ^{\mathsf{Sets}_{*}}$ is a natural isomorphism.
