The diagonal of the smash product of pointed sets is the natural transformation
at $\webleft (X,x_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$ is given by the composition
for each $x\in X$.
The diagonal of the smash product of pointed sets is the natural transformation
at $\webleft (X,x_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$ is given by the composition
for each $x\in X$.
and thus $\Delta ^{\wedge }_{X}$ is a morphism of pointed sets.
the diagram
Let $\webleft (X,x_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$.
Monoidality. The diagonal
of the smash product of pointed sets is a monoidal natural transformation:
Compatibility With Strong Monoidality Constraints. For each $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$, the diagram
Compatibility With Strong Unitality Constraints. The diagrams
where we recall that the equalities
The Diagonal of the Unit. The component
of $\Delta ^{\wedge }$ at $S^{0}$ is an isomorphism.
Item 1a: Compatibility With Strong Monoidality Constraints: We need to show that the diagram
Item 1b: Compatibility With Strong Unitality Constraints: As shown in the proof of Definition 7.5.5.1.1, the inverse of the left unitor of $\mathsf{Sets}_{*}$ with respect to to the smash product of pointed sets at $\webleft (X,x_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$ is given by
for each $x\in X$, so when $X=S^{0}$, we have
But since $1\wedge 0=0\wedge 0$ and
it follows that we indeed have $\Delta ^{\wedge }_{S^{0}}=\lambda ^{\mathsf{Sets}_{*},-1}_{S^{0}}$.
This finishes the proof.