The Clowder Project

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

for each $(x,y),(x^{\prime },y^{\prime })\in X\times Y$ with $(x,y){\sim }_R(x^{\prime },y^{\prime })$, where ${\sim }_R$ is the relation constructing $X\wedge Y$ as

in Construction 5.5.1.1.4. The condition defining $\sim$ is that at least one of the following conditions is satisfied:

1. 1. We have $x=x^{\prime }$ and $y=y^{\prime }$;
2. 2. Both of the following conditions are satisfied:
   1. (a) We have $x=x_0$ or $y=y_0$.
   2. (b) We have $x^{\prime }=x_0$ or $y^{\prime }=y_0$.

We have five cases:

1. 3. In the first case, we clearly have
   $$[f\wedge g](x,y)=[f\wedge g](x^{\prime },y^{\prime })$$

   since $x=x^{\prime }$ and $y=y^{\prime }$.

2. 4. If $x=x_0$ and $x^{\prime }=x_0$, we have
   $$\begin{array}{rl}[f\wedge g](x_0,y)& \stackrel{\text{def}}{=}f\left(x_0\right)\wedge g\left(y\right)\\ & =a_0\wedge g\left(y\right)\\ & =a_0\wedge g\left(y^{\prime }\right)\\ & =f\left(x_0\right)\wedge g\left(y^{\prime }\right)\\ & \stackrel{\text{def}}{=}[f\wedge g](x_0,y^{\prime }).\end{array}$$
3. 5. If $x=x_0$ and $y^{\prime }=y_0$, we have
   $$\begin{array}{rl}[f\wedge g](x_0,y)& \stackrel{\text{def}}{=}f\left(x_0\right)\wedge g\left(y\right)\\ & =a_0\wedge g\left(y\right)\\ & =a_0\wedge b_0\\ & =f\left(x^{\prime }\right)\wedge b_0\\ & =f\left(x^{\prime }\right)\wedge g\left(y_0\right)\\ & \stackrel{\text{def}}{=}[f\wedge g](x^{\prime },y_0).\end{array}$$
4. 6. If $y=y_0$ and $x^{\prime }=x_0$, we have
   $$\begin{array}{rl}[f\wedge g](x,y_0)& \stackrel{\text{def}}{=}f\left(x\right)\wedge g\left(y_0\right)\\ & =f\left(x\right)\wedge b_0\\ & =a_0\wedge b_0\\ & =a_0\wedge g\left(y^{\prime }\right)\\ & =f\left(x_0\right)\wedge g\left(y^{\prime }\right)\\ & \stackrel{\text{def}}{=}[f\wedge g](x_0,y^{\prime }).\end{array}$$
5. 7. If $y=y_0$ and $y^{\prime }=y_0$, we have
   $$\begin{array}{rl}[f\wedge g](x,y_0)& \stackrel{\text{def}}{=}f\left(x\right)\wedge g\left(y_0\right)\\ & =f\left(x\right)\wedge b_0\\ & =f\left(x^{\prime }\right)\wedge b_0\\ & =f\left(x\right)\wedge g\left(y_0\right)\\ & \stackrel{\text{def}}{=}[f\wedge g](x^{\prime },y_0).\end{array}$$

Thus $f\wedge g$ is well-defined. Next, we claim that $\wedge$ preserves identities and composition:

• Preservation of Identities. We have
  $$\begin{array}{rl}[{\text{id}}_X\wedge {\text{id}}_Y](x\wedge y)& \stackrel{\text{def}}{=}{\text{id}}_X\left(x\right)\wedge {\text{id}}_Y\left(y\right)\\ & =x\wedge y\\ & =\left[{\text{id}}_{X\wedge Y}\right](x\wedge y)\end{array}$$

  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
  $$\begin{array}{rl}f& :(X,x_0)\to (X^{\prime },x_0^{\prime }),\\ h& :(X^{\prime },x_0^{\prime })\to (X^{″},x_0^{″}),\\ g& :(Y,y_0)\to (Y^{\prime },y_0^{\prime }),\\ k& :(Y^{\prime },y_0^{\prime })\to (Y”,y{”}_0),\end{array}$$

  we have

  $$\begin{array}{rl}[(h\circ f)\wedge (k\circ g)](x\wedge y)& \stackrel{\text{def}}{=}h\left(f\left(x\right)\right)\wedge k\left(g\left(y\right)\right)\\ & \stackrel{\text{def}}{=}[h\wedge k](f\left(x\right)\wedge g\left(y\right))\\ & \stackrel{\text{def}}{=}[h\wedge k]\left([f\wedge g](x\wedge y)\right)\\ & \stackrel{\text{def}}{=}[(h\wedge k)\circ (f\wedge g)](x\wedge y)\end{array}$$

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

  $$(h\circ f)\wedge (k\circ g)=(h\wedge k)\circ (f\wedge g).$$

as the proof of the adjunction $X\wedge -⊣{\mathbf{\text{Sets}}}_{\ast }(X,-)$ is similar. We claim we have a bijection

natural in $(X,x_0),(Y,y_0),(Z,z_0)\in \text{Obj}\left({\mathsf{Sets}}_{\ast }\right)$, impliying the desired adjunction. Indeed, this bijection is a restriction of the bijection

of Chapter 2: Constructions With Sets, Item 2 of Proposition 2.1.3.1.3:

• A map
  $$\xi :X\times Y\to Z$$

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

  
  where ${\xi }_x^{†}:Y\to Z$ is the map defined by
  $${\xi }_x^{†}\left(y\right)\stackrel{\text{def}}{=}\xi (x,y)$$

  for each $y\in Y$, where:

  • The map ${\xi }^{†}$ is indeed pointed, as we have
    $$\begin{array}{rl}{\xi }_{x_0}^{†}\left(y\right)& \stackrel{\text{def}}{=}\xi (x_0,y)\\ & \stackrel{\text{def}}{=}{z}_0\end{array}$$

    for each $y\in Y$. Thus ${\xi }_{x_0}^{†}={\mathrm{\Delta }}_{z_0}$ and ${\xi }^{†}$ is pointed.

  • The map ${\xi }_x^{†}$ indeed lies in ${\mathbf{\text{Sets}}}_{\ast }(Y,Z)$, as we have
    $$\begin{array}{rl}{\xi }_x^{†}\left(y_0\right)& \stackrel{\text{def}}{=}\xi (x,y_0)\\ & \stackrel{\text{def}}{=}{z}_0.\end{array}$$

• Conversely, a map
  
  in ${\text{Hom}}_{{\mathsf{Sets}}_{\ast }}(X,{\mathbf{\text{Sets}}}_{\ast }(Y,Z))$ gets sent to the map
  $${\xi }^{†}:X\times Y\to Z$$

  defined by

  $${\xi }^{†}(x,y)\stackrel{\text{def}}{=}{\xi }_x\left(y\right)$$

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

  • Left Bilinearity. We have
    $$\begin{array}{rl}{\xi }^{†}(x_0,y)& \stackrel{\text{def}}{=}{\xi }_{x_0}\left(y\right)\\ & \stackrel{\text{def}}{=}{\mathrm{\Delta }}_{z_0}\left(y\right)\\ & \stackrel{\text{def}}{=}{z}_0\end{array}$$

    for each $y\in Y$, since ${\xi }_{x_0}={\mathrm{\Delta }}_{z_0}$ as $\xi$ is assumed to be a pointed map.

  • Right Bilinearity. We have
    $$\begin{array}{rl}{\xi }^{†}(x,y_0)& \stackrel{\text{def}}{=}{\xi }_x\left(y_0\right)\\ & \stackrel{\text{def}}{=}{z}_0\end{array}$$

    for each $x\in X$, since ${\xi }_x\in {\mathbf{\text{Sets}}}_{\ast }(Y,Z)$ is a morphism of pointed sets.

where $\sim$ identifies the elemenet $\star$ in $\text{pt}$ with all elements of the form $(x_0,y)$ and $(x,y_0)$ in $X\times Y$. Thus , of coupled with Remark 5.5.1.1.8 then gives us a well-defined map

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

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