The map $f\wedge g$ comes from , of via the map
\[ f\wedge g\colon X\times Y\to A\wedge B \]
sending $\webleft (x,y\webright )$ to $f\webleft (x\webright )\wedge g\webleft (y\webright )$, which we need to show satisfies
\[ \webleft [f\wedge g\webright ]\webleft (x,y\webright )=\webleft [f\wedge g\webright ]\webleft (x',y'\webright ) \]
for each $\webleft (x,y\webright ),\webleft (x',y'\webright )\in X\times Y$ with $\webleft (x,y\webright )\sim _{R}\webleft (x',y'\webright )$, where $\mathord {\sim }_{R}$ is the relation constructing $X\wedge Y$ as
\[ X\wedge Y\cong \webleft (X\times Y\webright )/\mathord {\sim }_{R} \]
in Construction 5.5.1.1.4. The condition defining $\mathord {\sim }$ is that at least one of the following conditions is satisfied:
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We have $x=x'$ and $y=y'$;
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Both of the following conditions are satisfied:
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We have $x=x_{0}$ or $y=y_{0}$.
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We have $x'=x_{0}$ or $y'=y_{0}$.
We have five cases:
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In the first case, we clearly have
\[ \webleft [f\wedge g\webright ]\webleft (x,y\webright )=\webleft [f\wedge g\webright ]\webleft (x',y'\webright ) \]
since $x=x'$ and $y=y'$.
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If $x=x_{0}$ and $x'=x_{0}$, we have
\begin{align*} \webleft [f\wedge g\webright ]\webleft (x_{0},y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (x_{0}\webright )\wedge g\webleft (y\webright )\\ & = a_{0}\wedge g\webleft (y\webright )\\ & = a_{0}\wedge g\webleft (y'\webright )\\ & = f\webleft (x_{0}\webright )\wedge g\webleft (y'\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [f\wedge g\webright ]\webleft (x_{0},y’\webright ).\end{align*}
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If $x=x_{0}$ and $y'=y_{0}$, we have
\begin{align*} \webleft [f\wedge g\webright ]\webleft (x_{0},y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (x_{0}\webright )\wedge g\webleft (y\webright )\\ & = a_{0}\wedge g\webleft (y\webright )\\ & = a_{0}\wedge b_{0}\\ & = f\webleft (x'\webright )\wedge b_{0}\\ & = f\webleft (x'\webright )\wedge g\webleft (y_{0}\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [f\wedge g\webright ]\webleft (x’,y_{0}\webright ).\end{align*}
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If $y=y_{0}$ and $x'=x_{0}$, we have
\begin{align*} \webleft [f\wedge g\webright ]\webleft (x,y_{0}\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (x\webright )\wedge g\webleft (y_{0}\webright )\\ & = f\webleft (x\webright )\wedge b_{0}\\ & = a_{0}\wedge b_{0}\\ & = a_{0}\wedge g\webleft (y'\webright )\\ & = f\webleft (x_{0}\webright )\wedge g\webleft (y'\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [f\wedge g\webright ]\webleft (x_{0},y’\webright ).\end{align*}
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If $y=y_{0}$ and $y'=y_{0}$, we have
\begin{align*} \webleft [f\wedge g\webright ]\webleft (x,y_{0}\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f\webleft (x\webright )\wedge g\webleft (y_{0}\webright )\\ & = f\webleft (x\webright )\wedge b_{0}\\ & = f\webleft (x'\webright )\wedge b_{0}\\ & = f\webleft (x\webright )\wedge g\webleft (y_{0}\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [f\wedge g\webright ]\webleft (x’,y_{0}\webright ).\end{align*}
Thus $f\wedge g$ is well-defined. Next, we claim that $\wedge $ preserves identities and composition:
- Preservation of Identities. We have
\begin{align*} \webleft [\text{id}_{X}\wedge \text{id}_{Y}\webright ]\webleft (x\wedge y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\text{id}_{X}\webleft (x\webright )\wedge \text{id}_{Y}\webleft (y\webright )\\ & = x\wedge y\\ & = \webleft [\text{id}_{X\wedge Y}\webright ]\webleft (x\wedge y\webright ) \end{align*}
for each $x\wedge y\in X\wedge Y$, and thus
\[ \text{id}_{X}\wedge \text{id}_{Y}=\text{id}_{X\wedge Y}. \]
- Preservation of Composition. Given pointed maps
\begin{align*} f & \colon \webleft (X,x_{0}\webright ) \to \webleft (X',x'_{0}\webright ),\\ h & \colon \webleft (X',x'_{0}\webright ) \to \webleft (X'',x''_{0}\webright ),\\ g & \colon \webleft (Y,y_{0}\webright ) \to \webleft (Y',y'_{0}\webright ),\\ k & \colon \webleft (Y’,y’_{0}\webright ) \to \webleft (Y”,y”_{0}\webright ), \end{align*}
we have
\begin{align*} \webleft [\webleft (h\circ f\webright )\wedge \webleft (k\circ g\webright )\webright ]\webleft (x\wedge y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}h\webleft (f\webleft (x\webright )\webright )\wedge k\webleft (g\webleft (y\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [h\wedge k\webright ]\webleft (f\webleft (x\webright )\wedge g\webleft (y\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [h\wedge k\webright ]\webleft (\webleft [f\wedge g\webright ]\webleft (x\wedge y\webright )\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft [\webleft (h\wedge k\webright )\circ \webleft (f\wedge g\webright )\webright ]\webleft (x\wedge y\webright ) \end{align*}
for each $x\wedge y\in X\wedge Y$, and thus
\[ \webleft (h\circ f\webright )\wedge \webleft (k\circ g\webright )=\webleft (h\wedge k\webright )\circ \webleft (f\wedge g\webright ). \]
This finishes the proof.
We prove only the adjunction $-\wedge Y\dashv \textbf{Sets}_{*}\webleft (Y,-\webright )$, witnessed by a natural bijection
\[ \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X\wedge Y,Z\webright )\cong \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X,\textbf{Sets}_{*}\webleft (Y,Z\webright )\webright ), \]
as the proof of the adjunction $X\wedge -\dashv \textbf{Sets}_{*}\webleft (X,-\webright )$ is similar. We claim we have a bijection
\[ \textup{Hom}^{\otimes }_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright )\cong \textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X,\textbf{Sets}_{*}\webleft (Y,Z\webright )\webright ) \]
natural in $\webleft (X,x_{0}\webright ),\webleft (Y,y_{0}\webright ),\webleft (Z,z_{0}\webright )\in \text{Obj}\webleft (\mathsf{Sets}_{*}\webright )$, impliying the desired adjunction. Indeed, this bijection is a restriction of the bijection
\[ \mathsf{Sets}\webleft (X\times Y,Z\webright )\cong \mathsf{Sets}\webleft (X,\mathsf{Sets}\webleft (Y,Z\webright )\webright ) \]
of Chapter 2: Constructions With Sets, Item 2 of Proposition 2.1.3.1.3:
- A map
\[ \xi \colon X\times Y\to Z \]
in $\textup{Hom}^{\otimes }_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright )$ gets sent to the pointed map
where $\xi ^{\dagger }_{x}\colon Y\to Z$ is the map defined by
\[ \xi ^{\dagger }_{x}\webleft (y\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi \webleft (x,y\webright ) \]
for each $y\in Y$, where:
- The map $\xi ^{\dagger }$ is indeed pointed, as we have
\begin{align*} \xi ^{\dagger }_{x_{0}}\webleft (y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi \webleft (x_{0},y\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}z_{0}\end{align*}
for each $y\in Y$. Thus $\xi ^{\dagger }_{x_{0}}=\Delta _{z_{0}}$ and $\xi ^{\dagger }$ is pointed.
- The map $\xi ^{\dagger }_{x}$ indeed lies in $\textbf{Sets}_{*}\webleft (Y,Z\webright )$, as we have
\begin{align*} \xi ^{\dagger }_{x}\webleft (y_{0}\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi \webleft (x,y_{0}\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}z_{0}.\end{align*}
- Conversely, a map in $\textup{Hom}_{\mathsf{Sets}_{*}}\webleft (X,\textbf{Sets}_{*}\webleft (Y,Z\webright )\webright )$ gets sent to the map
\[ \xi ^{\dagger }\colon X\times Y\to Z \]
defined by
\[ \xi ^{\dagger }\webleft (x,y\webright )\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{x}\webleft (y\webright ) \]
for each $\webleft (x,y\webright )\in X\times Y$, which indeed lies in $\textup{Hom}^{\otimes }_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright )$, as:
- Left Bilinearity. We have
\begin{align*} \xi ^{\dagger }\webleft (x_{0},y\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{x_{0}}\webleft (y\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Delta _{z_{0}}\webleft (y\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}z_{0}\end{align*}
for each $y\in Y$, since $\xi _{x_{0}}=\Delta _{z_{0}}$ as $\xi $ is assumed to be a pointed map.
- Right Bilinearity. We have
\begin{align*} \xi ^{\dagger }\webleft (x,y_{0}\webright ) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{x}\webleft (y_{0}\webright )\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}z_{0}\end{align*}
for each $x\in X$, since $\xi _{x}\in \textbf{Sets}_{*}\webleft (Y,Z\webright )$ is a morphism of pointed sets.
This finishes the proof.
This follows from Item 2 and , of .
Following the description of Chapter 2: Constructions With Sets, Remark 2.2.4.1.3, we have
\[ \text{pt}\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}_{X\vee Y}\webleft (X\times Y\webright )\cong \webleft (\text{pt}\times \webleft (X\times Y\webright )\webright )/\mathord {\sim }, \]
where $\mathord {\sim }$ identifies the elemenet $\star $ in $\text{pt}$ with all elements of the form $\webleft (x_{0},y\webright )$ and $\webleft (x,y_{0}\webright )$ in $X\times Y$. Thus , of coupled with Remark 5.5.1.1.8 then gives us a well-defined map
\[ \text{pt}\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}_{X\vee Y}\webleft (X\times Y\webright )\to X\wedge Y \]
via $\webleft [\webleft (\star ,\webleft (x,y\webright )\webright )\webright ]\mapsto x\wedge y$, with inverse
\[ X\wedge Y\to \text{pt}\mathchoice {\mathbin {\textstyle \coprod }}{\mathbin {\textstyle \coprod }}{\mathbin {\scriptstyle \textstyle \coprod }}{\mathbin {\scriptscriptstyle \textstyle \coprod }}_{X\vee Y}\webleft (X\times Y\webright ) \]
given by $x\wedge y\mapsto \webleft [\webleft (\star ,\webleft (x,y\webright )\webright )\webright ]$.
Item 5: Distributivity Over Wedge Sums
This follows from Proposition 5.5.9.1.1, , of , and the fact that $\vee $ is the coproduct in $\mathsf{Sets}_{*}$ (Chapter 4: Pointed Sets, Definition 4.3.3.1.1).