Item 2 (025W) — The Clowder Project

7.2.1 Tensors of Pointed Sets by Sets

Let $(X,x_{0})$ be a pointed set and let $A$ be a set.

Definition 7.2.1.1.1 ▶ Tensors of Pointed Sets by Sets

The tensor of $(X,x_{0})$ by $A$¹ is the tensor $A\odot (X,x_{0})$² of $(X,x_{0})$ by $A$ as in , .

¹Further Terminology: Also called the copower of $(X,x_{0})$ by $A$.
²Further Notation: Often written $A\odot X$ for simplicity.

Remark 7.2.1.1.2 ▶ Unwinding Definition 7.2.1.1.1

In detail, the tensor of $(X,x_{0})$ by $A$ is the pointed set $A\odot (X,x_{0})$ satisfying the following universal property:

(★)

\[ \mathsf{Sets}_{*}(A\odot X,K)\cong \mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K)), \]

This universal property is in turn equivalent to the following one:

(★)

\[ \mathsf{Sets}_{*}(A\odot X,K) \cong \mathsf{Sets}^{\otimes }_{\mathbb {E}_{0}}(A\times X,K), \]

\[ \mathsf{Sets}^{\otimes }_{\mathbb {E}_{0}}(A\times X,K) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\left\{ f\in \mathsf{Sets}(A\times X,K)\ \middle |\ \begin{aligned} & \text{for each $a\in A$, we}\\ & \text{have $f(a,x_{0})=k_{0}$}\end{aligned} \right\} . \]

Proof of the Equivalence in Remark 7.2.1.1.2.

We claim that we have a bijection

\[ \mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K))\cong \mathsf{Sets}^{\otimes }_{\mathbb {E}_{0}}(A\times X,K) \]

natural in $(K,k_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$. Indeed, this bijection is a restriction of the bijection

\[ \mathsf{Sets}(A,\mathsf{Sets}(X,K))\cong \mathsf{Sets}(A\times X,K) \]

of Chapter 4: Constructions With Sets, Item 2 of Proposition 4.1.3.1.3:

•
A map
in $\mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K))$ gets sent to the map

\[ \xi ^{\dagger }\colon A\times X\to K \]

defined by

\[ \xi ^{\dagger }(a,x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{a}(x) \]

for each $(a,x)\in A\times X$, which indeed lies in $\mathsf{Sets}^{\otimes }_{\mathbb {E}_{0}}(A\times X,K)$, as we have

\begin{align*} \xi ^{\dagger }(a,x_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{a}(x_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}k_{0}\end{align*}

for each $a\in A$, where we have used that $\xi _{a}\in \mathsf{Sets}_{*}(X,K)$ is a morphism of pointed sets.
•
Conversely, a map

\[ \xi \colon A\times X\to K \]

in $\mathsf{Sets}^{\otimes }_{\mathbb {E}_{0}}(A\times X,K)$ gets sent to the map
where

\[ \xi ^{\dagger }_{a}\colon X \to K \]

is the map defined by

\[ \xi ^{\dagger }_{a}(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (a,x) \]

for each $x\in X$, and indeed lies in $\mathsf{Sets}_{*}(X,K)$, as we have

\begin{align*} \xi ^{\dagger }_{a}(x_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (a,x_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}k_{0}.\end{align*}

This finishes the proof.

Construction 7.2.1.1.3 ▶ Construction of Tensors of Pointed Sets by Sets

Concretely, the tensor of $(X,x_{0})$ by $A$ is the pointed set $A\odot (X,x_{0})$ consisting of:

•
The Underlying Set. The set $A\odot X$ given by

\[ A\odot X\cong \bigvee _{a\in A}(X,x_{0}), \]

where $\bigvee _{a\in A}(X,x_{0})$ is the wedge product of the $A$-indexed family $((X,x_{0}))_{a\in A}$ of Chapter 6: Pointed Sets, Definition 6.3.2.1.1.
•
The Basepoint. The point $[(a,x_{0})]=[(a',x_{0})]$ of $\bigvee _{a\in A}(X,x_{0})$.

Proof of Construction 7.2.1.1.3.

(Proven below in a bit.)

Notation 7.2.1.1.4 ▶ Elements of Tensors of Pointed Sets by Sets

We write $a\odot x$ for the element $[(a,x)]$ of

\begin{align*} A\odot X & \cong \bigvee _{a\in A}(X,x_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}(\coprod _{i\in I}X_{i})/\mathord {\sim }.\end{align*}

Remark 7.2.1.1.5 ▶ Basepoints of Tensors of Pointed Sets by Sets

Taking the tensor of any element of $A$ with the basepoint $x_{0}$ of $X$ leads to the same element in $A\odot X$, i.e. we have

\[ a\odot x_{0}=a'\odot x_{0}, \]

for each $a,a'\in A$. This is due to the equivalence relation $\mathord {\sim }$ on

\[ \bigvee _{a\in A}(X,x_{0})\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\coprod _{a\in A}X/\mathord {\sim } \]

identifying $(a,x_{0})$ with $(a',x_{0})$, so that the equivalence class $a\odot x_{0}$ is independent from the choice of $a\in A$.

Proof of Construction 7.2.1.1.3.

We claim we have a bijection

\[ \mathsf{Sets}_{*}(A\odot X,K)\cong \mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K)) \]

natural in $(K,k_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.

1.
Map I. We define a map

\[ \Phi _{K}\colon \mathsf{Sets}_{*}(A\odot X,K) \to \mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K)) \]

by sending a morphism of pointed sets

\[ \xi \colon (A\odot X,a\odot x_{0})\to (K,k_{0}) \]

to the map of sets
where

\[ \xi _{a}\colon (X,x_{0})\to (K,k_{0}) \]

is the morphism of pointed sets defined by

\[ \xi _{a}(x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (a\odot x) \]

for each $x\in X$. Note that we have

\begin{align*} \xi _{a}(x_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (a\odot x_{0})\\ & = k_{0},\end{align*}

so that $\xi _{a}$ is indeed a morphism of pointed sets, where we have used that $\xi $ is a morphism of pointed sets.

2.
Map II. We define a map

\[ \Psi _{K}\colon \mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K))\to \mathsf{Sets}_{*}(A\odot X,K) \]

given by sending a map
to the morphism of pointed sets

\[ \xi ^{\dagger }\colon (A\odot X,a\odot x_{0})\to (K,k_{0}) \]

defined by

\[ \xi ^{\dagger }(a\odot x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{a}(x) \]

for each $a\odot x\in A\odot X$. Note that $\xi ^{\dagger }$ is indeed a morphism of pointed sets, as we have

\begin{align*} \xi ^{\dagger }(a\odot x_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{a}(x_{0})\\ & = k_{0}, \end{align*}

where we have used that $\xi (a)\in \mathsf{Sets}_{*}(X,K)$ is a morphism of pointed sets.

Invertibility I. We claim that

\[ \Psi _{K}\circ \Phi _{K}=\operatorname {\mathrm{id}}_{\mathsf{Sets}_{*}(A\odot X,K)}. \]

Indeed, given a morphism of pointed sets

\[ \xi \colon (A\odot X,a\odot x_{0})\to (K,k_{0}), \]

we have

\begin{align*} [\Psi _{K}\circ \Phi _{K}](\xi ) & = \Psi _{K}(\Phi _{K}(\xi ))\\ & = \Psi _{K}([\mspace {-3mu}[a\mapsto [\mspace {-3mu}[x\mapsto \xi (a\odot x)]\mspace {-3mu}]]\mspace {-3mu}])\\ & = \Psi _{K}([\mspace {-3mu}[a'\mapsto [\mspace {-3mu}[x'\mapsto \xi (a'\odot x')]\mspace {-3mu}]]\mspace {-3mu}])\\ & = [\mspace {-3mu}[a\odot x\mapsto \mathrm{ev}_{x}(\mathrm{ev}_{a}([\mspace {-3mu}[a'\mapsto [\mspace {-3mu}[x'\mapsto \xi (a'\odot x')]\mspace {-3mu}]]\mspace {-3mu}]))]\mspace {-3mu}]\\ & = [\mspace {-3mu}[a\odot x\mapsto \mathrm{ev}_{x}([\mspace {-3mu}[x'\mapsto \xi (a\odot x')]\mspace {-3mu}])]\mspace {-3mu}]\\ & = [\mspace {-3mu}[a\odot x\mapsto \xi (a\odot x)]\mspace {-3mu}]\\ & = \xi .\end{align*}

Invertibility II. We claim that

\[ \Phi _{K}\circ \Psi _{K}=\operatorname {\mathrm{id}}_{\mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K))}. \]

Indeed, given a morphism $\xi \colon A\to \mathsf{Sets}_{*}(X,K)$, we have

\begin{align*} [\Phi _{K}\circ \Psi _{K}](\xi ) & = \Phi _{K}(\Psi _{K}(\xi ))\\ & = \Phi _{K}([\mspace {-3mu}[a\odot x\mapsto \xi _{a}(x)]\mspace {-3mu}])\\ & = [\mspace {-3mu}[a\mapsto [\mspace {-3mu}[x\mapsto \xi _{a}(x)]\mspace {-3mu}]]\mspace {-3mu}]\\ & = [\mspace {-3mu}[a\mapsto \xi (a)]\mspace {-3mu}]\\ & = \xi .\end{align*}

Naturality of $\Phi $. We need to show that, given a morphism of pointed sets

\[ \phi \colon (K,k_{0})\to (K',k'_{0}), \]

the diagram

commutes. Indeed, given a morphism of pointed sets

\[ \xi \colon (A\odot X,a\odot x_{0})\to (K,k_{0}), \]

we have

\begin{align*} [\Phi _{K'}\circ \phi _{*}](\xi ) & = \Phi _{K'}(\phi _{*}(\xi ))\\ & = \Phi _{K'}(\phi \circ \xi )\\ & = (\phi \circ \xi )^{\dagger }\\ & = [\mspace {-3mu}[a\mapsto \phi \circ \xi (a\odot -)]\mspace {-3mu}]\\ & = [\mspace {-3mu}[a\mapsto \phi _{*}(\xi (a\odot -))]\mspace {-3mu}]\\ & = (\phi _{*})_{*}([\mspace {-3mu}[a\mapsto \xi (a\odot -]\mspace {-3mu}]))\\ & = (\phi _{*})_{*}(\Phi _{K}(\xi ))\\ & = [(\phi _{*})_{*}\circ \Phi _{K}](\xi ). \end{align*}

Naturality of $\Psi $. Since $\Phi $ is natural and $\Phi $ is a componentwise inverse to $\Psi $, it follows from Chapter 11: Categories, Item 2 of Proposition 11.9.7.1.2 that $\Psi $ is also natural.

This finishes the proof.

Proposition 7.2.1.1.6 ▶ Properties of Tensors of Pointed Sets by Sets

Let $(X,x_{0})$ be a pointed set and let $A$ be a set.

1.
Functoriality. The assignments $A,(X,x_{0}),(A,(X,x_{0}))$ define functors

\[ \begin{array}{ccc} A\odot -\colon \mkern -15mu & \mathsf{Sets}\mathrlap {{}_{*}} \mkern -17.5mu& {}\mathbin {\to }\mathsf{Sets}_{*},\\ -\odot X\colon \mkern -15mu & \mathsf{Sets} \mkern -17.5mu& {}\mathbin {\to }\mathsf{Sets}_{*},\\ -_{1}\odot -_{2}\colon \mkern -15mu & \mathsf{Sets}\times \mathsf{Sets}_{*} \mkern -17.5mu& {}\mathbin {\to }\mathsf{Sets}_{*}. \end{array} \]

In particular, given:
- •
  A map of sets $f\colon A\to B$;
- •
  A pointed map $\phi \colon (X,x_{0})\to (Y,y_{0})$;
the induced map

\[ f\odot \phi \colon A\odot X\to B\odot Y \]

is given by

\[ [f\odot \phi ](a\odot x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(a)\odot \phi (x) \]

for each $a\odot x\in A\odot X$.
2.
Adjointness I. We have an adjunction
witnessed by a bijection

\[ \mathsf{Sets}_{*}(A\odot X,K)\cong \mathsf{Sets}(A,\mathsf{Sets}_{*}(X,K)), \]

natural in $A\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ and $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
3.
Adjointness II. We have an adjunctions
witnessed by a bijection

\[ \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}(A\odot X,Y)\cong \operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}(X,A\pitchfork Y), \]

natural in $A\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ and $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
4.
As a Weighted Colimit. We have

\[ A\odot X\cong \operatorname {\mathrm{colim}}^{[A]}(X), \]

where in the right hand side we write:
- •
  $A$ for the functor $A\colon \mathrm{pt}\to \mathsf{Sets}$ picking $A\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$;
- •
  $X$ for the functor $X\colon \mathrm{pt}\to \mathsf{Sets}_{*}$ picking $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
5.
Iterated Tensors. We have an isomorphism of pointed sets

\[ A\odot (B\odot X)\cong (A\times B)\odot X, \]

natural in $A,B\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ and $(X,x_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$.
6.
Interaction With Homs. We have a natural isomorphism

\[ \mathsf{Sets}_{*}(A\odot X,-)\cong A\pitchfork \mathsf{Sets}_{*}(X,-). \]
7.
The Tensor Evaluation Map. For each $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, we have a map

\[ \mathrm{ev}^{\odot }_{X,Y}\colon \mathsf{Sets}_{*}(X,Y)\odot X\to Y, \]

natural in $X,Y\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, and given by

\[ \mathrm{ev}^{\odot }_{X,Y}(f\odot x)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x) \]

for each $f\odot x\in \mathsf{Sets}_{*}(X,Y)\odot X$.
8.
The Tensor Coevaluation Map. For each $A\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ and each $X\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, we have a map

\[ \mathrm{coev}^{\odot }_{A,X}\colon A\to \mathsf{Sets}_{*}(X,A\odot X), \]

natural in $A\in \operatorname {\mathrm{Obj}}(\mathsf{Sets})$ and $X\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, and given by

\[ \mathrm{coev}^{\odot }_{A,X}(a)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[\mspace {-3mu}[x\mapsto a\odot x]\mspace {-3mu}] \]

for each $a\in A$.