Proposition 5.5.8.1.2 (00GV): Properties of the Diagonal of $\wedge $

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Let $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$.

Monoidality. The diagonal
\[ \Delta ^{\wedge }\colon \text{id}_{\mathsf{Sets}_{*}}\Longrightarrow {\wedge }\circ {\Delta ^{\mathsf{Cats}_{\mathsf{2}}}_{\mathsf{Sets}_{*}}}, \]

of the smash product of pointed sets is a monoidal natural transformation:
1. Compatibility With Strong Monoidality Constraints. For each $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, the diagram
  
  commutes.
2. Compatibility With Strong Unitality Constraints. The diagrams
  commute, i.e. we have
  \begin{align*} \Delta ^{\wedge }_{S^{0}} & = \lambda ^{\mathsf{Sets}_{*},-1}_{S^{0}}\\ & = \rho ^{\mathsf{Sets}_{*},-1}_{S^{0}}, \end{align*}
  
  where we recall that the equalities
  
  \begin{align*} \lambda ^{\mathsf{Sets}_{*}}_{S^{0}} & = \rho ^{\mathsf{Sets}_{*}}_{S^{0}},\\ \lambda ^{\mathsf{Sets}_{*},-1}_{S^{0}} & = \rho ^{\mathsf{Sets}_{*},-1}_{S^{0}}\end{align*}
  
  are always true in any monoidal category by , of .
The Diagonal of the Unit. The component
\[ \Delta ^{\wedge }_{S^{0}} \colon S^{0} \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }S^{0}\wedge S^{0} \]

of $\Delta ^{\wedge }$ at $S^{0}$ is an isomorphism.

Item 1: Monoidality

We claim that $\Delta ^{\wedge }$ is indeed monoidal:

Item (a): Compatibility With Strong Monoidality Constraints: We need to show that the diagram

commutes. Indeed, this diagram acts on elements as

and hence indeed commutes.
Item (b): Compatibility With Strong Unitality Constraints: As shown in the proof of Definition 5.5.5.1.1, the inverse of the left unitor of $\mathsf{Sets}_{*}$ with respect to to the smash product of pointed sets at $\webleft (X,x_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$ is given by
\[ \lambda ^{\mathsf{Sets}_{*},-1}_{X}\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}1\wedge x \]

for each $x\in X$, so when $X=S^{0}$, we have

\begin{align*} \lambda ^{\mathsf{Sets}_{*},-1}_{S^{0}}\webleft (0\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}1\wedge 0,\\ \lambda ^{\mathsf{Sets}_{*},-1}_{S^{0}}\webleft (1\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}1\wedge 1. \end{align*}

But since $1\wedge 0=0\wedge 0$ and

\begin{align*} \Delta ^{\wedge }_{S^{0}}\webleft (0\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}0\wedge 0,\\ \Delta ^{\wedge }_{S^{0}}\webleft (1\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}1\wedge 1, \end{align*}

it follows that we indeed have $\Delta ^{\wedge }_{S^{0}}=\lambda ^{\mathsf{Sets}_{*},-1}_{S^{0}}$.

This finishes the proof.

Item 2: The Diagonal of the Unit

This follows from Item 1 and the invertibility of the left/right unitor of $\mathsf{Sets}_{*}$ with respect to $\wedge $, proved in the proof of Definition 5.5.5.1.1 for the left unitor or the proof of Definition 5.5.6.1.1 for the right unitor.