4.4.1 Foundations

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

Definition 4.4.1.1.1 ▶ The Right Tensor Product of Pointed Sets

The right tensor product of pointed sets is the functor[1]

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

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 4.2.1.1.6.

Remark 4.4.1.1.2 ▶ Unwinding Definition 4.4.1.1.1: 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:

  1. Pointed maps $f\colon X\rhd Y\to Z$.
  2. 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 4.4.1.1.3 ▶ Unwinding Definition 4.4.1.1.1: 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 4.4.1.1.4 ▶ 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 $x\in X$.

Notation 4.4.1.1.5 ▶ Elements of Right Tensor Products of Pointed Sets

We write[2] $x\rhd y$ for the element $\webleft [\webleft (x,y\webright )\webright ]$ of

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

Remark 4.4.1.1.6 ▶ Basepoints of Right Tensor Products of Pointed Sets

Employing the notation introduced in Notation 4.4.1.1.5, 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 4.4.1.1.7 ▶ Properties of Right Tensor Products of Pointed Sets

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

  1. Functoriality. The assignments $X,Y,\webleft (X,Y\webright )\mapsto X\rhd Y$ define functors
    \begin{gather*} \begin{aligned} X\rhd - & \colon \mathsf{Sets}_{*} \to \mathsf{Sets}_{*},\\ -\rhd Y & \colon \mathsf{Sets}_{*} \to \mathsf{Sets}_{*},\\ \end{aligned}\\ -_{1}\rhd -_{2} \colon \mathsf{Sets}_{*}\times \mathsf{Sets}_{*} \to \mathsf{Sets}_{*}. \end{gather*}

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

    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 4.4.2.1.1.

  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 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 4.4.1.1.7.
Item 1: Functoriality
Clear.
Item 2: Adjointness I
This follows from Item 3 of Proposition 4.2.1.1.6.
Item 3: Adjointness II
For $-\rhd Y$ to admit a right adjoint would require it to preserve colimits by of . 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 4.2.1.1.6.

Remark 4.4.1.1.8 ▶ 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 4.3.1.1.7 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 by Item 3 of Proposition 4.4.1.1.7, no functor can.[3]

Footnotes

[1] Further Notation: Also written $\rhd _{\mathsf{Sets}_{*}}$.
[2] Further Notation: Also written $x\rhd _{\mathsf{Sets}_{*}}y$.
[3] The functor $\textbf{Sets}_{*}\webleft (Y,-\webright )$ is instead right adjoint to $-\wedge Y$, the smash product of pointed sets of Definition 4.5.1.1.1. See Item 2 of Proposition 4.5.1.1.9.
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