The skew associator of the right tensor product of pointed sets is the natural transformation
\[ \alpha ^{\mathsf{Sets}_{*},\rhd } \colon {\rhd }\circ {\webleft (\text{id}_{\mathsf{Sets}_{*}}\times {\rhd }\webright )} \Longrightarrow {\rhd }\circ {\webleft ({\rhd }\times \text{id}_{\mathsf{Sets}_{*}}\webright )}\circ {\mathbf{\alpha }^{\mathsf{Cats},-1}_{\mathsf{Sets}_{*},\mathsf{Sets}_{*},\mathsf{Sets}_{*}}} \]
as in the diagram
whose component
\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z} \colon X\rhd \webleft (Y\rhd Z\webright ) \to \webleft (X\rhd Y\webright )\rhd Z \]
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
\begin{align*} X\rhd \webleft (Y\rhd Z\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|X|\odot \webleft (Y\rhd Z\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|X|\odot \webleft (|Y|\odot Z\webright )\\ & \cong \bigvee _{x\in X}\webleft (|Y|\odot Z\webright )\\ & \cong \bigvee _{x\in X}\webleft (\bigvee _{y\in Y}Z\webright )\\ & \to \bigvee _{\webleft [\webleft (x,y\webright )\webright ]\in \bigvee _{x\in X}Y}Z\\ & \cong \bigvee _{\webleft [\webleft (x,y\webright )\webright ]\in |X|\odot Y}Z\\ & \cong ||X|\odot Y|\odot Z\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|X\rhd Y|\odot Z\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft (X\rhd Y\webright )\rhd Z,\end{align*}
where the map
\[ \bigvee _{x\in X}\webleft (\bigvee _{y\in Y}Z\webright )\to \bigvee _{\webleft [\webleft (x,y\webright )\webright ]\in \bigvee _{x\in X}Y}Z \]
is given by $\webleft [\webleft (x,\webleft [\webleft (y,z\webright )\webright ]\webright )\webright ]\mapsto \webleft [\webleft (\webleft [\webleft (x,y\webright )\webright ],z\webright )\webright ]$.
Firstly, note that, given $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, the map
\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z} \colon X\rhd \webleft (Y\rhd Z\webright ) \to \webleft (X\rhd Y\webright )\rhd Z \]
is indeed a morphism of pointed sets, as we have
\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}\webleft (x_{0}\rhd \webleft (y_{0}\rhd z_{0}\webright )\webright )=\webleft (x_{0}\rhd y_{0}\webright )\rhd z_{0}. \]
Next, we claim that $\alpha ^{\mathsf{Sets}_{*},\rhd }$ is a natural transformation. 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 $\alpha ^{\mathsf{Sets}_{*},\rhd }$ to be a natural transformation. This finishes the proof.
