The morphism $\smash {\lambda ^{\mathsf{Sets}_{*},\rhd }_{X}}$ is non-invertible, as it is non-surjective when viewed as a map of sets, since the elements $0\rhd x$ of $S^{0}\rhd X$ with $x\neq x_{0}$ are outside the image of $\lambda ^{\mathsf{Sets}_{*},\rhd }_{X}$, which sends $x$ to $1\rhd x$.
Proof of Definition 4.4.5.1.1.
Firstly, note that, given $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, the map
\[ \lambda ^{\mathsf{Sets}_{*},\rhd }_{X} \colon X \to S^{0}\rhd X \]
is indeed a morphism of pointed sets, as we have
\begin{align*} \lambda ^{\mathsf{Sets}_{*},\rhd }_{X}\webleft (x_{0}\webright ) & = 1\rhd x_{0}\\ & = 0\rhd x_{0}.\end{align*}
Next, we claim that $\lambda ^{\mathsf{Sets}_{*},\rhd }$ 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 $\lambda ^{\mathsf{Sets}_{*},\rhd }$ to be a natural transformation. This finishes the proof.
