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
\[ \alpha ^{\mathsf{Sets}_{*},\lhd }_{X,Y,Z} \colon \webleft (X\lhd Y\webright )\lhd Z \to X\lhd \webleft (Y\lhd Z\webright ) \]
is indeed a morphism of pointed sets, as we have
\[ \alpha ^{\mathsf{Sets}_{*},\lhd }_{X,Y,Z}\webleft (\webleft (x_{0}\lhd y_{0}\webright )\lhd z_{0}\webright )=x_{0}\lhd \webleft (y_{0}\lhd z_{0}\webright ). \]
Next, we claim that $\alpha ^{\mathsf{Sets}_{*},\lhd }$ 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 $\alpha ^{\mathsf{Sets}_{*},\lhd }$ to be a natural transformation. This finishes the proof.
