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

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

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

We claim we have a bijection

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

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

  • Map I. We define a map

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

    by sending a morphism of pointed sets

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

    to the map of sets

    where

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

    is the morphism of pointed sets defined by

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

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

    \begin{align*} \xi _{a}\webleft (x_{0}\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi \webleft (a\odot x_{0}\webright )\\ & = 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.

  • Map II. We define a map

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

    given by sending a map

    to the morphism of pointed sets

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

    defined by

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

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

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

  • Invertibility I. We claim that

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

    Indeed, given a morphism of pointed sets

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

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

  • Invertibility II. We claim that

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

    Indeed, given a morphism $\xi \colon A\to \mathsf{Sets}_{*}\webleft (X,K\webright )$, 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}[a\odot x\mapsto \xi _{a}\webleft (x\webright )]\mspace {-3mu}]\webright )\\ & = [\mspace {-3mu}[a\mapsto [\mspace {-3mu}[x\mapsto \xi _{a}\webleft (x\webright )]\mspace {-3mu}]]\mspace {-3mu}]\\ & = [\mspace {-3mu}[a\mapsto \xi \webleft (a\webright )]\mspace {-3mu}]\\ & = \xi .\end{align*}

  • Naturality of $\Phi $. 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 morphism of pointed sets

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

    we have

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

  • Naturality of $\Psi $. Since $\Phi $ is natural and $\Phi $ is a componentwise inverse to $\Psi $, it follows from Chapter 9: Categories, Item 2 of Proposition 9.9.7.1.2 that $\Psi $ is also natural.
This finishes the proof.


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