The Clowder Project

By Chapter 10: Conditions on Relations, , we have a natural bijection

where $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(X\times Y,Z)$ is the set

1. 1.

   We have $x=x'$ and $y=y'$.

2. 2.

   The following conditions are satisfied:

   1. (a)

      We have $x=x_{0}$ or $y=y_{0}$.

   2. (b)

      We have $x'=x_{0}$ or $y'=y_{0}$.

So, given $f\in \operatorname {\mathrm{Hom}}_{\mathsf{Sets}}(X\times Y,Z)$ with a corresponding $\overline{f}\colon X\wedge Y\to Z$, the latter case above implies

and since $\overline{f}\colon X\wedge Y\to Z$ is a pointed map, we have

Thus the elements $f$ in $\operatorname {\mathrm{Hom}}^{R}_{\mathsf{Sets}}(X\times Y,Z)$ are precisely those functions $f\colon X\times Y\to Z$ satisfying the equalities

for each $x\in X$ and each $y\in Y$, giving an equality

of sets, which when composed with our earlier isomorphism

gives our desired natural bijection, finishing the proof.

Indeed, since $X\cong \bigvee _{x\in X^{-}}S^{0}$, we have

where we have used that $\wedge $ preserves colimits in both variables via for the second isomorphism above, since it has right adjoints in both variables by Item 2.

A similar proof applies to the isomorphism $X\wedge Y\cong \bigvee _{y\in Y^{-}}X$.

sending $(x,y)$ to $f(x)\wedge g(y)$, which we need to show satisfies

for each $(x,y),(x',y')\in X\times Y$ with $(x,y)\sim _{R}(x',y')$, where $\mathord {\sim }_{R}$ is the relation constructing $X\wedge Y$ as

in Construction 7.5.1.1.4. The condition defining $\mathord {\sim }$ is that at least one of the following conditions is satisfied:

1. 1.

   We have $x=x'$ and $y=y'$;

2. 2.

   Both of the following conditions are satisfied:

   1. (a)

      We have $x=x_{0}$ or $y=y_{0}$.

   2. (b)

      We have $x'=x_{0}$ or $y'=y_{0}$.

We have five cases:

1. 1.

   In the first case, we clearly have

   \[ [f\wedge g](x,y)=[f\wedge g](x',y') \]

   since $x=x'$ and $y=y'$.

2. 2.

   If $x=x_{0}$ and $x'=x_{0}$, we have

   \begin{align*} [f\wedge g](x_{0},y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x_{0})\wedge g(y)\\ & = a_{0}\wedge g(y)\\ & = a_{0}\wedge g(y')\\ & = f(x_{0})\wedge g(y')\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[f\wedge g](x_{0},y’).\end{align*}
3. 3.

   If $x=x_{0}$ and $y'=y_{0}$, we have

   \begin{align*} [f\wedge g](x_{0},y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x_{0})\wedge g(y)\\ & = a_{0}\wedge g(y)\\ & = a_{0}\wedge b_{0}\\ & = f(x')\wedge b_{0}\\ & = f(x')\wedge g(y_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[f\wedge g](x’,y_{0}).\end{align*}
4. 4.

   If $y=y_{0}$ and $x'=x_{0}$, we have

   \begin{align*} [f\wedge g](x,y_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x)\wedge g(y_{0})\\ & = f(x)\wedge b_{0}\\ & = a_{0}\wedge b_{0}\\ & = a_{0}\wedge g(y')\\ & = f(x_{0})\wedge g(y')\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[f\wedge g](x_{0},y’).\end{align*}
5. 5.

   If $y=y_{0}$ and $y'=y_{0}$, we have

   \begin{align*} [f\wedge g](x,y_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}f(x)\wedge g(y_{0})\\ & = f(x)\wedge b_{0}\\ & = f(x')\wedge b_{0}\\ & = f(x)\wedge g(y_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[f\wedge g](x’,y_{0}).\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*} [\operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}_{Y}](x\wedge y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\operatorname {\mathrm{id}}_{X}(x)\wedge \operatorname {\mathrm{id}}_{Y}(y)\\ & = x\wedge y\\ & = [\operatorname {\mathrm{id}}_{X\wedge Y}](x\wedge y) \end{align*}

  for each $x\wedge y\in X\wedge Y$, and thus

  \[ \operatorname {\mathrm{id}}_{X}\wedge \operatorname {\mathrm{id}}_{Y}=\operatorname {\mathrm{id}}_{X\wedge Y}. \]
• •

  Preservation of Composition. Given pointed maps

  \begin{align*} f & \colon (X,x_{0}) \to (X',x'_{0}),\\ h & \colon (X',x'_{0}) \to (X'',x''_{0}),\\ g & \colon (Y,y_{0}) \to (Y',y'_{0}),\\ k & \colon (Y’,y’_{0}) \to (Y^{\prime \prime },y^{\prime \prime }_{0}), \end{align*}

  we have

  \begin{align*} [(h\circ f)\wedge (k\circ g)](x\wedge y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}h(f(x))\wedge k(g(y))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[h\wedge k](f(x)\wedge g(y))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[h\wedge k]([f\wedge g](x\wedge y))\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}[(h\wedge k)\circ (f\wedge g)](x\wedge y) \end{align*}

  for each $x\wedge y\in X\wedge Y$, and thus

  \[ (h\circ f)\wedge (k\circ g)=(h\wedge k)\circ (f\wedge g). \]

This finishes the proof.

as the proof of the adjunction $X\wedge -\dashv \boldsymbol {\mathsf{Sets}}_{*}(X,-)$ is similar. We claim we have a bijection

natural in $(X,x_{0}),(Y,y_{0}),(Z,z_{0})\in \operatorname {\mathrm{Obj}}(\mathsf{Sets}_{*})$, impliying the desired adjunction. Indeed, this bijection is a restriction of the bijection

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

• •

  A map

  \[ \xi \colon X\times Y\to Z \]

  in $\operatorname {\mathrm{Hom}}^{\otimes }_{\mathsf{Sets}_{*}}(X\times Y,Z)$ gets sent to the pointed map

  
  where $\xi ^{\dagger }_{x}\colon Y\to Z$ is the map defined by
  \[ \xi ^{\dagger }_{x}(y)\mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (x,y) \]

  for each $y\in Y$, where:

  • •

    The map $\xi ^{\dagger }$ is indeed pointed, as we have

    \begin{align*} \xi ^{\dagger }_{x_{0}}(y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (x_{0},y)\\ & \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 $\boldsymbol {\mathsf{Sets}}_{*}(Y,Z)$, as we have

    \begin{align*} \xi ^{\dagger }_{x}(y_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi (x,y_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}z_{0}.\end{align*}
• •

  Conversely, a map

  
  in $\operatorname {\mathrm{Hom}}_{\mathsf{Sets}_{*}}(X,\boldsymbol {\mathsf{Sets}}_{*}(Y,Z))$ gets sent to the map
  \[ \xi ^{\dagger }\colon X\times Y\to Z \]

  defined by

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

  for each $(x,y)\in X\times Y$, which indeed lies in $\operatorname {\mathrm{Hom}}^{\otimes }_{\mathsf{Sets}_{*}}(X\times Y,Z)$, as:

  • •

    Left Bilinearity. We have

    \begin{align*} \xi ^{\dagger }(x_{0},y) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{x_{0}}(y)\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\Delta _{z_{0}}(y)\\ & \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 }(x,y_{0}) & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\xi _{x}(y_{0})\\ & \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}z_{0}\end{align*}

    for each $x\in X$, since $\xi _{x}\in \boldsymbol {\mathsf{Sets}}_{*}(Y,Z)$ is a morphism of pointed sets.

This finishes the proof.

where $\mathord {\sim }$ identifies the elemenet $\star $ in $\mathrm{pt}$ with all elements of the form $(x_{0},y)$ and $(x,y_{0})$ in $X\times Y$. Thus Chapter 10: Conditions on Relations, Item 4 of Proposition 10.6.2.1.3 coupled with Remark 7.5.1.1.8 then gives us a well-defined map

via $[(\star ,(x,y))]\mapsto x\wedge y$, with inverse

given by $x\wedge y\mapsto [(\star ,(x,y))]$.