The morphism $\smash {\rho ^{\mathsf{Sets}_{*},\lhd }_{X}}$ is non-invertible, as it is non-surjective when viewed as a map of sets, since the elements $x\lhd 0$ of $X\lhd S^{0}$ with $x\neq x_{0}$ are outside the image of $\rho ^{\mathsf{Sets}_{*},\lhd }_{X}$, which sends $x$ to $x\lhd 1$.
Proof of Definition 4.3.6.1.1.
Firstly, note that, given $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, the map
\[ \rho ^{\mathsf{Sets}_{*},\lhd }_{X} \colon X \to X\lhd S^{0} \]
is indeed a morphism of pointed sets as we have
\begin{align*} \rho ^{\mathsf{Sets}_{*},\lhd }_{X}\webleft (x_{0}\webright ) & = x_{0}\lhd 1\\ & = x_{0}\lhd 0.\end{align*}
Next, we claim that $\rho ^{\mathsf{Sets}_{*},\lhd }$ is a natural transformation. We need to show that, given a morphism of pointed sets
\[ f\colon \webleft (X,x_{0}\webright )\to \webleft (Y,y_{0}\webright ), \]
the diagram
commutes. Indeed, this diagram acts on elements as
and hence indeed commutes, showing $\rho ^{\mathsf{Sets}_{*},\lhd }$ to be a natural transformation. This finishes the proof.
