Concretely, the cotensor of $\webleft (X,x_{0}\webright )$ by $A$ is the pointed set $A\pitchfork \webleft (X,x_{0}\webright )$ consisting of:

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

    \[ A\pitchfork X\cong \bigwedge _{a\in A}\webleft (X,x_{0}\webright ), \]

    where $\bigwedge _{a\in A}\webleft (X,x_{0}\webright )$ is the smash product of the $A$-indexed family $\webleft (\webleft (X,x_{0}\webright )\webright )_{a\in A}$ of Definition 4.6.1.1.1.

  • The Basepoint. The point $\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]=\webleft [\webleft (x_{0},x_{0},x_{0},\ldots \webright )\webright ]$ of $\bigwedge _{a\in A}\webleft (X,x_{0}\webright )$.

We claim we have a bijection

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

natural in $\webleft (K,k_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$.

  • Map I. We define a map

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

    by sending a morphism of pointed sets

    \[ \xi \colon \webleft (K,k_{0}\webright )\to \webleft (A\pitchfork X,\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]\webright ) \]

    to the map of sets

    where

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

    is the morphism of pointed sets defined by

    \[ \xi _{a}\webleft (k\webright )=\begin{cases} x^{k}_{a} & \text{if $\xi \webleft (k\webright )\neq \webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]$,}\\ x_{0} & \text{if $\xi \webleft (k\webright )=\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]$}\end{cases} \]

    for each $k\in K$, where $x^{k}_{a}$ is the $a$th component of $\xi \webleft (k\webright )=\webleft [\webleft (x^{k}_{a}\webright )_{a\in A}\webright ]$. Note that:

    1. The definition of $\xi _{a}\webleft (k\webright )$ is independent of the choice of equivalence class. Indeed, suppose we have
      \begin{align*} \xi \webleft (k\webright ) & = \webleft [\webleft (x^{k}_{a}\webright )_{a\in A}\webright ]\\ & = \webleft [\webleft (y^{k}_{a}\webright )_{a\in A}\webright ] \end{align*}

      with $x^{k}_{a}\neq y^{k}_{a}$ for some $a\in A$. Then there exist $a_{x},a_{y}\in A$ such that $x^{k}_{a_{x}}=y^{k}_{a_{y}}=x_{0}$. The equivalence relation $\mathord {\sim }$ on $\prod _{a\in A}X$ then forces

      \begin{align*} \webleft [\webleft (x^{k}_{a}\webright )_{a\in A}\webright ] & = \webleft [\webleft (x_{0}\webright )_{a\in A}\webright ],\\ \webleft [\webleft (y^{k}_{a}\webright )_{a\in A}\webright ] & = \webleft [\webleft (x_{0}\webright )_{a\in A}\webright ], \end{align*}

      however, and $\xi _{a}\webleft (k\webright )$ is defined to be $x_{0}$ in this case.

    2. The map $\xi _{a}$ is indeed a morphism of pointed sets, as we have
      \[ \xi _{a}\webleft (k_{0}\webright )=x_{0} \]

      since $\xi \webleft (k_{0}\webright )=\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]$ as $\xi $ is a morphism of pointed sets and $\xi _{a}\webleft (k_{0}\webright )$, defined to be the $a$th component of $\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]$, is equal to $x_{0}$.

  • Map II. We define a map

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

    given by sending a map

    to the morphism of pointed sets

    \[ \xi ^{\dagger }\colon \webleft (K,k_{0}\webright )\to \webleft (A\pitchfork X,\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]\webright ) \]

    defined by

    \[ \xi ^{\dagger }\webleft (k\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\webleft (\xi _{a}\webleft (k\webright )\webright )_{a\in A}\webright ] \]

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

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

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

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

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

    the diagram

    commutes. Indeed, given a map of sets

    we have

    \begin{align*} \webleft [\Psi _{K}\circ \webleft (\phi ^{*}\webright )_{*}\webright ]\webleft (\xi \webright ) & = \Psi _{K}\webleft (\webleft (\phi ^{*}\webright )_{*}\webleft (\xi \webright )\webright )\\ & = \Psi _{K}\webleft (\webleft (\phi ^{*}\webright )_{*}\webleft ([\mspace {-3mu}[a\mapsto \xi _{a}]\mspace {-3mu}]\webright )\webright )\\ & = \Psi _{K}\webleft (\webleft ([\mspace {-3mu}[a\mapsto \phi ^{*}\webleft (\xi _{a}\webright )]\mspace {-3mu}]\webright )\webright )\\ & = \Psi _{K}\webleft (\webleft ([\mspace {-3mu}[a\mapsto [\mspace {-3mu}[k\mapsto \xi _{a}\webleft (\phi \webleft (k\webright )\webright )]\mspace {-3mu}]]\mspace {-3mu}]\webright )\webright )\\ & = [\mspace {-3mu}[k\mapsto \webleft [\webleft (\xi _{a}\webleft (\phi \webleft (k\webright )\webright )\webright )_{a\in A}\webright ]]\mspace {-3mu}]\\ & = \phi ^{*}\webleft ([\mspace {-3mu}[k'\mapsto \webleft [\webleft (\xi _{a}\webleft (k'\webright )\webright )_{a\in A}\webright ]]\mspace {-3mu}]\webright )\\ & = \phi ^{*}\webleft (\Psi _{K'}\webleft (\xi \webright )\webright )\\ & = \webleft [\phi ^{*}\circ \Psi _{K'}\webright ]\webleft (\xi \webright ). \end{align*}

  • Naturality of $\Phi $. Since $\Psi $ is natural and $\Psi $ is a componentwise inverse to $\Phi $, it follows from Chapter 8: Categories, Item 2 of Proposition 8.8.6.1.2 that $\Phi $ is also natural.
  • Invertibility I. We claim that

    \[ \Psi _{K}\circ \Phi _{K}=\text{id}_{\mathsf{Sets}_{*}\webleft (K,A\pitchfork X\webright )}. \]

    Indeed, given a morphism of pointed sets

    \[ \xi \colon \webleft (K,k_{0}\webright )\to \webleft (A\pitchfork X,\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]\webright ) \]

    we have

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

    Now, we have two cases:

    1. If $\xi \webleft (k\webright )=\webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]$, we have
      \begin{align*} \webleft [\Psi _{K}\circ \Phi _{K}\webright ]\webleft (\xi \webright ) & = \cdots \\ & = [\mspace {-3mu}[k\mapsto \webleft [\webleft (\xi _{a}\webleft (k\webright )\webright )_{a\in A}\webright ]]\mspace {-3mu}]\\ & = [\mspace {-3mu}[k\mapsto \webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]]\mspace {-3mu}]\\ & = [\mspace {-3mu}[k\mapsto \xi \webleft (k\webright )]\mspace {-3mu}]\\ & = \xi .\end{align*}
    2. If $\xi \webleft (k\webright )\neq \webleft [\webleft (x_{0}\webright )_{a\in A}\webright ]$ and $\xi \webleft (k\webright )=\webleft [\webleft (x^{k}_{a}\webright )_{a\in A}\webright ]$ instead, we have
      \begin{align*} \webleft [\Psi _{K}\circ \Phi _{K}\webright ]\webleft (\xi \webright ) & = \cdots \\ & = [\mspace {-3mu}[k\mapsto \webleft [\webleft (\xi _{a}\webleft (k\webright )\webright )_{a\in A}\webright ]]\mspace {-3mu}]\\ & = [\mspace {-3mu}[k\mapsto \webleft [\webleft (x^{k}_{a}\webright )_{a\in A}\webright ]]\mspace {-3mu}]\\ & = [\mspace {-3mu}[k\mapsto \xi \webleft (k\webright )]\mspace {-3mu}]\\ & = \xi .\end{align*}

    In both cases, we have $\webleft [\Psi _{K}\circ \Phi _{K}\webright ]\webleft (\xi \webright )=\xi $, and thus we are done.

  • Invertibility II. We claim that

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

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

    \begin{align*} \webleft [\Phi _{K}\circ \Psi _{K}\webright ]\webleft (\xi \webright ) & = \Phi _{K}\webleft (\Psi _{K}\webleft (\xi \webright )\webright )\\ & = \Phi _{K}\webleft ([\mspace {-3mu}[k\mapsto \webleft [\webleft (\xi _{a}\webleft (k\webright )\webright )_{a\in A}\webright ]]\mspace {-3mu}]\webright )\\ & = [\mspace {-3mu}[a\mapsto [\mspace {-3mu}[k\mapsto \xi _{a}\webleft (k\webright )]\mspace {-3mu}]]\mspace {-3mu}]\\ & = \xi \end{align*}

This finishes the proof.


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