Subsection 5.4.1 (00EK): Foundations

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5.4.1 Foundations

Let $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ be pointed sets.

Definition 5.4.1.1.1 ▶ The Right Tensor Product of Pointed Sets

The right tensor product of pointed sets is the functor¹

\[ \rhd \colon \mathsf{Sets}_{*}\times \mathsf{Sets}_{*}\to \mathsf{Sets}_{*} \]

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 5.2.1.1.6.

¹Further Notation: Also written $\rhd _{\mathsf{Sets}_{*}}$.

Remark 5.4.1.1.2 ▶ Unwinding Definition 5.4.1.1.1: Universal Property I

The right tensor product of pointed sets satisfies the following natural bijection:

\[ \mathsf{Sets}_{*}\webleft (X\rhd Y,Z\webright )\cong \textup{Hom}^{\otimes ,\mathrm{R}}_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright ). \]

That is to say, the following data are in natural bijection:

Pointed maps $f\colon X\rhd Y\to Z$.
Maps of sets $f\colon X\times Y\to Z$ satisfying $f\webleft (x,y_{0}\webright )=z_{0}$ for each $x\in X$.

Remark 5.4.1.1.3 ▶ Unwinding Definition 5.4.1.1.1: Universal Property II

The right tensor product of pointed sets may be described as follows:

The right tensor product of $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ is the pair $\webleft (\webleft (X\rhd Y,x_{0}\rhd y_{0}\webright ),\iota \webright )$ consisting of
- A pointed set $\webleft (X\rhd Y,x_{0}\rhd y_{0}\webright )$;
- A right bilinear morphism of pointed sets $\iota \colon \webleft (X\times Y,\webleft (x_{0},y_{0}\webright )\webright )\to X\rhd Y$;
satisfying the following universal property:
- Given another such pair $\webleft (\webleft (Z,z_{0}\webright ),f\webright )$ consisting of
  - A pointed set $\webleft (Z,z_{0}\webright )$;
  - A right bilinear morphism of pointed sets $f\colon \webleft (X\times Y,\webleft (x_{0},y_{0}\webright )\webright )\to X\rhd Y$;
  there exists a unique morphism of pointed sets $X\rhd Y\overset {\exists !}{\to }Z$ making the diagram
  
  commute.

Construction 5.4.1.1.4 ▶ The Right Tensor Product of Pointed Sets

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 $x\in X$.

Proof of Construction 5.4.1.1.4.

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 , , a natural bijection

\[ \mathsf{Sets}_{*}\webleft (X\rhd Y,Z\webright ) \cong \textup{Hom}^{R}_{\mathsf{Sets}}\webleft (\coprod _{X\in X}Y,Z\webright ), \]

where $\textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )$ is the set

\[ \textup{Hom}^{R}_{\mathsf{Sets}}\webleft (\coprod _{x\in X}Y,Z\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft\{ f\in \textup{Hom}_{\mathsf{Sets}}\webleft (\coprod _{x\in X}Y,Z\webright )\ \middle |\ \begin{aligned} & \text{for each $x,y\in X$, if}\\ & \text{$\webleft (x,y\webright )\sim _{R}\webleft (x',y'\webright )$, then}\\ & \text{$f\webleft (x,y\webright )=f\webleft (x',y'\webright )$}\end{aligned} \webright\} . \]

However, the condition $\webleft (x,y\webright )\sim _{R}\webleft (x',y'\webright )$ only holds when:

We have $x=x'$ and $y=y'$.
We have $y=y'=y_{0}$.

So, given $f\in \textup{Hom}_{\mathsf{Sets}}\webleft (\coprod _{x\in X}Y,Z\webright )$ 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 $\textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )$ 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

\[ \textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )=\textup{Hom}^{\otimes ,\mathrm{R}}_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright ) \]

of sets, which when composed with our earlier isomorphism

\[ \mathsf{Sets}_{*}\webleft (X\rhd Y,Z\webright ) \cong \textup{Hom}^{R}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright ), \]

gives our desired natural bijection, finishing the proof.

Notation 5.4.1.1.5 ▶ Elements of Right Tensor Products of Pointed Sets

We write¹ $x\rhd y$ for the element $\webleft [\webleft (x,y\webright )\webright ]$ of

\[ X\rhd Y\cong \left\lvert X\right\rvert \odot Y. \]

¹Further Notation: Also written $x\rhd _{\mathsf{Sets}_{*}}y$.

Remark 5.4.1.1.6 ▶ Basepoints of Right Tensor Products of Pointed Sets

Employing the notation introduced in Notation 5.4.1.1.5, we have

\[ x_{0}\rhd y_{0}=x\rhd y_{0} \]

for each $x\in X$, and

\[ x\rhd y_{0}=x'\rhd y_{0} \]

for each $x,x'\in X$.

Proposition 5.4.1.1.7 ▶ Properties of Right Tensor Products of Pointed Sets

Let $\webleft (X,x_{0}\webright )$ and $\webleft (Y,y_{0}\webright )$ be pointed sets.

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 $x\rhd y\in X\rhd Y$.
Adjointness I. We have an adjunction
witnessed by a bijection of sets
\[ \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X\rhd Y,Z\webright )\cong \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (Y,\webleft [X,Z\webright ]^{\rhd }_{\mathsf{Sets}_{*}}\webright ) \]

natural in $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, where $\webleft [X,Y\webright ]^{\rhd }_{\mathsf{Sets}_{*}}$ is the pointed set of Definition 5.4.2.1.1.
Adjointness II. The functor
\[ -\rhd Y\colon \mathsf{Sets}_{*}\to \mathsf{Sets}_{*} \]

does not admit a right adjoint.
Adjointness III. We have a ${\text{忘}}$-relative adjunction
witnessed by a bijection of sets
\[ \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X\rhd Y,Z\webright )\cong \textup{Hom}_{\mathsf{Sets}}\webleft (|X|,\mathsf{Sets}_{*}\webleft (Y,Z\webright )\webright ) \]

natural in $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$.

Proof of Proposition 5.4.1.1.7.

Item 1: Functoriality

This follows from the definition of $\rhd $ as a composition of functors (Definition 5.4.1.1.1).

Item 2: Adjointness I

This follows from Item 3 of Proposition 5.2.1.1.6.

Item 3: Adjointness II

For $-\rhd Y$ to admit a right adjoint would require it to preserve colimits by

. However, we have

\begin{align*} \text{pt}\rhd X & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}|\text{pt}|\odot X\\ & \cong X\\ & \ncong \text{pt}, \end{align*}

and thus we see that $-\rhd Y$ does not have a right adjoint.

Item 4: Adjointness III

This follows from Item 2 of Proposition 5.2.1.1.6.

Remark 5.4.1.1.8 ▶ On the Failure of $-\rhd Y$ To Be a Left Adjoint

Here is some intuition on why $-\rhd Y$ fails to be a left adjoint. Item 4 of Proposition 5.3.1.1.7 states that we have a natural bijection

\[ \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X\rhd Y,Z\webright )\cong \textup{Hom}_{\mathsf{Sets}}\webleft (|X|,\mathsf{Sets}_{*}\webleft (Y,Z\webright )\webright ), \]

so it would be reasonable to wonder whether a natural bijection of the form

\[ \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X\rhd Y,Z\webright )\cong \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X,\textbf{Sets}_{*}\webleft (Y,Z\webright )\webright ), \]

also holds, which would give $-\rhd Y\dashv \textbf{Sets}_{*}\webleft (Y,-\webright )$. However, such a bijection would require every map

\[ f\colon X\rhd Y\to Z \]

to satisfy

\[ f\webleft (x_{0}\rhd y\webright )=z_{0} \]

for each $x\in X$, whereas we are imposing such a basepoint preservation condition only for elements of the form $x\rhd y_{0}$. Thus $\textbf{Sets}_{*}\webleft (Y,-\webright )$ can’t be a right adjoint for $-\rhd Y$, and as shown byItem 3 of Proposition 5.4.1.1.7, no functor can.¹

¹The functor $\textbf{Sets}_{*}\webleft (Y,-\webright )$ is instead right adjoint to $-\wedge Y$, the smash product of pointed sets of Definition 5.5.1.1.1. See Item 2 of Proposition 5.5.1.1.10.

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