7.4.1 Foundations
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:
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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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$\odot \colon \mathsf{Sets}\times \mathsf{Sets}_{*}\to \mathsf{Sets}_{*}$ is the tensor functor of Item 1 of Proposition 7.2.1.1.6.
In detail, the right tensor product of $(X,x_{0})$ and $(Y,y_{0})$ is the pointed set $(X\rhd Y,[y_{0}])$ consisting of:
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The Underlying Set. The set $X\rhd Y$ defined by
\begin{align*} X\rhd Y & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\lvert X\right\rvert \odot Y\\ & \cong \bigvee _{x\in X}(Y,y_{0}), \end{align*}
where $\left\lvert X\right\rvert $ denotes the underlying set of $(X,x_{0})$.
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The Underlying Basepoint. The point $[(x_{0},y_{0})]$ of $\bigvee _{x\in X}(Y,y_{0})$, which is equal to $[(x,y_{0})]$ for any $x\in X$.
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:
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.
We write $x\rhd y$ for the element $[(x,y)]$ of
\[ X\rhd Y\cong \left\lvert X\right\rvert \odot Y. \]
Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets.
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Functoriality. The assignments $X,Y,(X,Y)\mapsto X\rhd Y$ define functors
\[ \begin{array}{ccc} X\rhd -\colon \mkern -15mu & \mathsf{Sets}_{*} \mkern -17.5mu& {}\mathbin {\to }\mathsf{Sets}_{*},\\ -\rhd Y\colon \mkern -15mu & \mathsf{Sets}_{*} \mkern -17.5mu& {}\mathbin {\to }\mathsf{Sets}_{*},\\ -_{1}\rhd -_{2}\colon \mkern -15mu & \mathsf{Sets}_{*}\times \mathsf{Sets}_{*} \mkern -17.5mu& {}\mathbin {\to }\mathsf{Sets}_{*}. \end{array} \]
In particular, given pointed maps
\begin{align*} f & \colon (X,x_{0}) \to (A,a_{0}),\\ g & \colon (Y,y_{0}) \to (B,b_{0}), \end{align*}
the induced map
\[ f\rhd g\colon X\rhd Y\to A\rhd B \]
is given by
\[ [f\rhd g](x\rhd y)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x)\rhd g(y) \]
for each $x\rhd y\in X\rhd Y$.
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Adjointness I. We have an adjunction
witnessed by a bijection of sets
\[ \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}(X\rhd Y,Z)\cong \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}(Y,[X,Z]^{\rhd }_{\mathsf{Sets}_{*}}) \]
natural in $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, where $[X,Y]^{\rhd }_{\mathsf{Sets}_{*}}$ is the pointed set of Definition 7.4.2.1.1.
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Adjointness II. The functor
\[ -\rhd Y\colon \mathsf{Sets}_{*}\to \mathsf{Sets}_{*} \]
does not admit a right adjoint.
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Adjointness III. We have a ${\text{忘}}$-relative adjunction
witnessed by a bijection of sets
\[ \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}(X\rhd Y,Z)\cong \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}(|X|,\mathsf{Sets}_{*}(Y,Z)) \]
natural in $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
This follows from the definition of $\rhd $ as a composition of functors (Definition 7.4.1.1.1).
This follows from Item 3 of Proposition 7.2.1.1.6.
For $-\rhd Y$ to admit a right adjoint would require it to preserve colimits by , of . However, we have
\begin{align*} \mathrm{pt}\rhd X & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|\mathrm{pt}|\odot X\\ & \cong X\\ & \ncong \mathrm{pt}, \end{align*}
and thus we see that $-\rhd Y$ does not have a right adjoint.
This follows from Item 2 of Proposition 7.2.1.1.6.
