Let $X$ be a set.
Omitted.
We claim there’s an adjunction $\webleft (-\webright )^{+}\dashv {\text{忘}}$, witnessed by a bijection of sets
\begin{align*} \mathsf{Sets}_{*}\webleft (\webleft (X^{+},\star _{X}\webright ),\webleft (Y,y_{0}\webright )\webright )\cong \mathsf{Sets}\webleft (X,Y\webright ),\end{align*}
natural in $X\in \text{Obj}\webleft (\mathsf{Sets}\webright )$ and $\webleft (Y,y_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$.
- Map I. We define a map
\[ \Phi _{X,Y}\colon \mathsf{Sets}_{*}\webleft (\webleft (X^{+},\star _{X}\webright ),\webleft (Y,y_{0}\webright )\webright )\to \mathsf{Sets}\webleft (X,Y\webright ) \]
by sending a pointed function
\[ \xi \colon \webleft (X^{+},\star _{X}\webright )\to \webleft (Y,y_{0}\webright ) \]
to the function
\[ \xi ^{\dagger }\colon X\to Y \]
given by
\[ \xi ^{\dagger }\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi \webleft (x\webright ) \]
for each $x\in X$.
- Map II. We define a map
\[ \Psi _{X,Y}\colon \mathsf{Sets}\webleft (X,Y\webright )\to \mathsf{Sets}_{*}\webleft (\webleft (X^{+},\star _{X}\webright ),\webleft (Y,y_{0}\webright )\webright ) \]
given by sending a function $\xi \colon X\to Y$ to the pointed function
\[ \xi ^{\dagger }\colon \webleft (X^{+},\star _{X}\webright )\to \webleft (Y,y_{0}\webright ) \]
defined by
\[ \xi ^{\dagger }\webleft (x\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} \xi \webleft (x\webright ) & \text{if $x\in X$,}\\ y_{0} & \text{if $x=\star _{X}$} \end{cases} \]
for each $x\in X^{+}$.
- Invertibility I. We claim that
\[ \Psi _{X,Y}\circ \Phi _{X,Y}=\text{id}_{\mathsf{Sets}_{*}\webleft (\webleft (X^{+},\star _{X}\webright ),\webleft (Y,y_{0}\webright )\webright )}, \]
which is clear.
- Invertibility II. We claim that
\[ \Phi _{X,Y}\circ \Psi _{X,Y}=\text{id}_{\mathsf{Sets}\webleft (X,Y\webright )}, \]
which is clear.
- Naturality for $\Phi $, Part I. We need to show that, given a pointed function $g\colon \webleft (Y,y_{0}\webright )\to \webleft (Y',y'_{0}\webright )$, the diagram
commutes. Indeed, given a pointed function
\[ \xi ^{\dagger }\colon \webleft (X^{+},\star _{X}\webright )\to \webleft (Y,y_{0}\webright ) \]
we have
\begin{align*} \webleft [\Phi _{X,Y'}\circ g_{*}\webright ]\webleft (\xi \webright ) & = \Phi _{X,Y'}\webleft (g_{*}\webleft (\xi \webright )\webright )\\ & = \Phi _{X,Y'}\webleft (g\circ \xi \webright )\\ & = g\circ \xi \\ & = g\circ \Phi _{X,Y'}\webleft (\xi \webright )\\ & = g_{*}\webleft (\Phi _{X,Y'}\webleft (\xi \webright )\webright )\\ & = \webleft [g_{*}\circ \Phi _{X,Y'}\webright ]\webleft (\xi \webright ). \end{align*}
- Naturality for $\Phi $, Part II. We need to show that, given a pointed function $f\colon \webleft (X,x_{0}\webright )\to \webleft (X',x'_{0}\webright )$, the diagram
commutes. Indeed, given a function
\[ \xi \colon X'\to Y, \]
we have
\begin{align*} \webleft [\Phi _{X,Y}\circ f^{*}\webright ]\webleft (\xi \webright ) & = \Phi _{X,Y}\webleft (f^{*}\webleft (\xi \webright )\webright )\\ & = \Phi _{X,Y}\webleft (\xi \circ f\webright )\\ & = \xi \circ f\\ & = \Phi _{X',Y}\webleft (\xi \webright )\circ f\\ & = f^{*}\webleft (\Phi _{X',Y}\webleft (\xi \webright )\webright )\\ & = f^{*}\webleft (\Phi _{X',Y}\webleft (\xi \webright )\webright )\\ & = \webleft [f^{*}\circ \Phi _{X',Y}\webright ]\webleft (\xi \webright ). \end{align*}
- Naturality for $\Psi $. Since $\Phi $ is natural in each argument and $\Phi $ is a componentwise inverse to $\Psi $ in each argument, it follows from Chapter 9: Categories, Item 2 of Proposition 9.9.7.1.2 that $\Psi $ is also natural in each argument.
Item 3: Symmetric Strong Monoidality With Respect to Wedge Sums
The isomorphism
\[ \phi \colon X^{+}\vee Y^{+}\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }\webleft (X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y\webright )^{+} \]
is given by
\[ \phi \webleft (z\webright )=\begin{cases} x & \text{if $z=\webleft [\webleft (0,x\webright )\webright ]$ with $x\in X$,}\\ y & \text{if $z=\webleft [\webleft (1,y\webright )\webright ]$ with $y\in Y$,}\\ \star _{X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y} & \text{if $z=\webleft [\webleft (0,\star _{X}\webright )\webright ]$,}\\ \star _{X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y} & \text{if $z=\webleft [\webleft (1,\star _{Y}\webright )\webright ]$}\end{cases} \]
for each $z\in X^{+}\vee Y^{+}$, with inverse
\[ \phi ^{-1} \colon \webleft (X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y\webright )^{+} \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X^{+}\vee Y^{+} \]
given by
\[ \phi ^{-1}\webleft (z\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} \webleft [\webleft (0,x\webright )\webright ] & \text{if $z=\webleft [\webleft (0,x\webright )\webright ]$,}\\ \webleft [\webleft (0,y\webright )\webright ] & \text{if $z=\webleft [\webleft (1,y\webright )\webright ]$,}\\ p_{0} & \text{if $z=\star _{X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y}$} \end{cases} \]
for each $z\in \webleft (X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y\webright )^{+}$.
Meanwhile, the isomorphism $\text{pt}\cong \text{Ø}^{+}$ is given by sending $\star _{X}$ to $\star _{\text{Ø}}$.
That these isomorphisms satisfy the coherence conditions making the functor $\webleft (-\webright )^{+}$ symmetric strong monoidal can be directly checked element by element.
Item 4: Symmetric Strong Monoidality With Respect to Smash Products
The isomorphism
\[ \phi \colon X^{+}\wedge Y^{+}\overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }\webleft (X\times Y\webright )^{+} \]
is given by
\[ \phi \webleft (x\wedge y\webright )=\begin{cases} \webleft (x,y\webright ) & \text{if $x\neq \star _{X}$ and $y\neq \star _{Y}$}\\ \star _{X\times Y} & \text{otherwise}\end{cases} \]
for each $x\wedge y\in X^{+}\wedge Y^{+}$, with inverse
\[ \phi ^{-1} \colon \webleft (X\times Y\webright )^{+} \overset {\scriptstyle \mathord {\sim }}{\dashrightarrow }X^{+}\wedge Y^{+} \]
given by
\[ \phi ^{-1}\webleft (z\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\begin{cases} x\wedge y & \text{if $z=\webleft (x,y\webright )$ with $\webleft (x,y\webright )\in X\times Y$,}\\ \star _{X}\wedge \star _{Y} & \text{if $z=\star _{X\times Y}$,}\\ \end{cases} \]
for each $z\in \webleft (X\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}Y\webright )^{+}$.
Meanwhile, the isomorphism $S^{0}\cong \text{pt}^{+}$ is given by sending $\star $ to $1\in S^{0}=\webleft\{ 0,1\webright\} $ and $\star _{\text{pt}}$ to $0\in S^{0}$.
That these isomorphisms satisfy the coherence conditions making the functor $\webleft (-\webright )^{+}$ symmetric strong monoidal can be directly checked element by element.
