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