Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets.
The right tensor product of pointed sets is the functor
\[ \rhd \colon \mathsf{Sets}_{*}\times \mathsf{Sets}_{*}\to \mathsf{Sets}_{*} \]
defined as the composition
\[ \mathsf{Sets}_{*}\times \mathsf{Sets}_{*}\overset {{\text{忘}}\times \mathsf{id}}{\to }\mathsf{Sets}\times \mathsf{Sets}_{*}\overset {\odot }{\to }\mathsf{Sets}_{*}, \]
where:
-
•
${\text{忘}}\colon \mathsf{Sets}_{*}\to \mathsf{Sets}$ is the forgetful functor from pointed sets to sets.
-
•
$\odot \colon \mathsf{Sets}\times \mathsf{Sets}_{*}\to \mathsf{Sets}_{*}$ is the tensor functor of Item 1 of Proposition 7.2.1.1.6.
Since $\bigvee _{y\in Y}(X,x_{0})$ is defined as the quotient of $\coprod _{x\in X}Y$ by the equivalence relation $R$ generated by declaring $(x,y)\sim (x',y')$ if $y=y'=y_{0}$, we have, by Chapter 10: Conditions on Relations, 
, a natural bijection
\[ \mathsf{Sets}_{*}(X\rhd Y,Z) \cong \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(\coprod _{X\in X}Y,Z), \]
where $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(X\times Y,Z)$ is the set
\[ \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(\coprod _{x\in X}Y,Z)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ f\in \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}(\coprod _{x\in X}Y,Z)\ \middle |\ \begin{aligned} & \text{for each $x,y\in X$, if}\\ & \text{$(x,y)\sim _{R}(x',y')$, then}\\ & \text{$f(x,y)=f(x',y')$}\end{aligned} \right\} . \]
However, the condition $(x,y)\sim _{R}(x',y')$ only holds when:
-
1.
We have $x=x'$ and $y=y'$.
-
2.
We have $y=y'=y_{0}$.
So, given $f\in \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}(\coprod _{x\in X}Y,Z)$ with a corresponding $\overline{f}\colon X\rhd Y\to Z$, the latter case above implies
\begin{align*} f([(x,y_{0})]) & = f([(x',y_{0})])\\ & = f([(x_{0},y_{0})]), \end{align*}
and since $\overline{f}\colon X\rhd Y\to Z$ is a pointed map, we have
\begin{align*} f([(x_{0},y_{0})]) & = \overline{f}([(x_{0},y_{0})])\\ & = z_{0}. \end{align*}
Thus the elements $f$ in $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(X\times Y,Z)$ are precisely those functions $f\colon X\times Y\to Z$ satisfying the equality
\[ f(x,y_{0})=z_{0} \]
for each $y\in Y$, giving an equality
\[ \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(X\times Y,Z)=\operatorname {\mathrm{Hom}}^{\otimes ,\mathrm{R}}_{\mathsf{Sets}_{*}}(X\times Y,Z) \]
of sets, which when composed with our earlier isomorphism
\[ \mathsf{Sets}_{*}(X\rhd Y,Z) \cong \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(X\times Y,Z), \]
gives our desired natural bijection, finishing the proof.
