The Clowder Project

7.5.1 Foundations

Let $(X,x_{0})$ and $(Y,y_{0})$ be pointed sets.

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