Firstly, note that, given $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, the map
\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z} \colon X\rhd (Y\rhd Z) \to (X\rhd Y)\rhd Z \]
is indeed a morphism of pointed sets, as we have
\[ \alpha ^{\mathsf{Sets}_{*},\rhd }_{X,Y,Z}(x_{0}\rhd (y_{0}\rhd z_{0}))=(x_{0}\rhd y_{0})\rhd z_{0}. \]
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 (X,x_{0}) \to (X',x'_{0}),\\ g & \colon (Y,y_{0}) \to (Y',y'_{0}),\\ h & \colon (Z,z_{0}) \to (Z’,z’_{0}) \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.