7.4.1 Foundations
Let $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ 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.
In detail, the right tensor product of $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ is the pointed set $\webleft (X\rhd Y,\webleft [y_{0}\webright ]\webright )$ consisting of:
-
•
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}\webleft (Y,y_{0}\webright ), \end{align*}
where $\left\lvert X\right\rvert $ denotes the underlying set of $\webleft (X,x_{0}\webright )$.
-
•
The Underlying Basepoint. The point $\webleft [\webleft (x_{0},y_{0}\webright )\webright ]$ of $\bigvee _{x\in X}\webleft (Y,y_{0}\webright )$, which is equal to $\webleft [\webleft (x,y_{0}\webright )\webright ]$ for any $x\in X$.
Since $\bigvee _{y\in Y}\webleft (X,x_{0}\webright )$ is defined as the quotient of $\coprod _{x\in X}Y$ by the equivalence relation $R$ generated by declaring $\webleft (x,y\webright )\sim \webleft (x',y'\webright )$ if $y=y'=y_{0}$, we have, by Chapter 10: Conditions on Relations, 
, a natural bijection
\[ \mathsf{Sets}_{*}\webleft (X\rhd Y,Z\webright ) \cong \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (\coprod _{X\in X}Y,Z\webright ), \]
where $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )$ is the set
\[ \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (\coprod _{x\in X}Y,Z\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ f\in \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}\webleft (\coprod _{x\in X}Y,Z\webright )\ \middle |\ \begin{aligned} & \text{for each $x,y\in X$, if}\\ & \text{$\webleft (x,y\webright )\sim _{R}\webleft (x',y'\webright )$, then}\\ & \text{$f\webleft (x,y\webright )=f\webleft (x',y'\webright )$}\end{aligned} \right\} . \]
However, the condition $\webleft (x,y\webright )\sim _{R}\webleft (x',y'\webright )$ only holds when:
2.
We have $y=y'=y_{0}$.
So, given $f\in \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}\webleft (\coprod _{x\in X}Y,Z\webright )$ with a corresponding $\overline{f}\colon X\rhd Y\to Z$, the latter case above implies
\begin{align*} f\webleft (\webleft [\webleft (x,y_{0}\webright )\webright ]\webright ) & = f\webleft (\webleft [\webleft (x',y_{0}\webright )\webright ]\webright )\\ & = f\webleft (\webleft [\webleft (x_{0},y_{0}\webright )\webright ]\webright ), \end{align*}
and since $\overline{f}\colon X\rhd Y\to Z$ is a pointed map, we have
\begin{align*} f\webleft (\webleft [\webleft (x_{0},y_{0}\webright )\webright ]\webright ) & = \overline{f}\webleft (\webleft [\webleft (x_{0},y_{0}\webright )\webright ]\webright )\\ & = z_{0}. \end{align*}
Thus the elements $f$ in $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )$ are precisely those functions $f\colon X\times Y\to Z$ satisfying the equality
\[ f\webleft (x,y_{0}\webright )=z_{0} \]
for each $y\in Y$, giving an equality
\[ \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )=\operatorname {\mathrm{Hom}}^{\otimes ,\mathrm{R}}_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright ) \]
of sets, which when composed with our earlier isomorphism
\[ \mathsf{Sets}_{*}\webleft (X\rhd Y,Z\webright ) \cong \operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright ), \]
gives our desired natural bijection, finishing the proof.
We write $x\rhd y$ for the element $\webleft [\webleft (x,y\webright )\webright ]$ of
\[ X\rhd Y\cong \left\lvert X\right\rvert \odot Y. \]
Let $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ be pointed sets.
-
1.
Functoriality. The assignments $X,Y,\webleft (X,Y\webright )\mapsto X\rhd Y$ 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 \webleft (X,x_{0}\webright ) \to \webleft (A,a_{0}\webright ),\\ g & \colon \webleft (Y,y_{0}\webright ) \to \webleft (B,b_{0}\webright ), \end{align*}
the induced map
\[ f\rhd g\colon X\rhd Y\to A\rhd B \]
is given by
\[ \webleft [f\rhd g\webright ]\webleft (x\rhd y\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (x\webright )\rhd g\webleft (y\webright ) \]
for each $x\rhd y\in X\rhd Y$.
-
2.
Adjointness I. We have an adjunction
witnessed by a bijection of sets
\[ \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}\webleft (X\rhd Y,Z\webright )\cong \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}\webleft (Y,\webleft [X,Z\webright ]^{\rhd }_{\mathsf{Sets}_{*}}\webright ) \]
natural in $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$, where $\webleft [X,Y\webright ]^{\rhd }_{\mathsf{Sets}_{*}}$ is the pointed set of Definition 7.4.2.1.1.
-
3.
Adjointness II. The functor
\[ -\rhd Y\colon \mathsf{Sets}_{*}\to \mathsf{Sets}_{*} \]
does not admit a right adjoint.
-
4.
Adjointness III. We have a ${\text{忘}}$-relative adjunction
witnessed by a bijection of sets
\[ \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}\webleft (X\rhd Y,Z\webright )\cong \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}\webleft (|X|,\mathsf{Sets}_{*}\webleft (Y,Z\webright )\webright ) \]
natural in $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \operatorname {\mathrm{Obj}}\webleft (\mathsf{Sets}_{*}\webright )$.
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.
