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# Exercise 40 - Chapter 4

$$\def\cat#1{{\mathcal{#1}}}$$ $$\def\comp#1{{\widehat{#1}}}$$ $$\def\conj#1{{\check{#1}}}$$ $$\def\id{{\mathrm{id}}}$$ Let $$\cat{V}$$ be a skeletal quantale, let $$\cat{P}$$ and $$\cat{Q}$$ be $$\cat{V}$$-categories, and let $$F\colon\cat{P}\to\cat{Q}$$ and $$G\colon\cat{Q}\to\cat{P}$$ be $$\cat{V}$$-functors.

1. Show that $$F$$ and $$G$$ are $$\cat{V}$$-adjoints (as in \ref{eq1}) if and only if the companion of the former equals the conjoint of the latter: $$\comp{F}=\conj{G}$$.

2. Use this to prove that $$\comp{\id}=\conj{\id}$$, as was stated in 4.34.

$$\label{eq1}\tag{4.39}\cat{P}(p,G(q))\cong\cat{Q}(F(p),q)$$

1. Suppose $$\def\cat#1{{\mathcal{#1}}}$$ $$\def\comp#1{{\widehat{#1}}}$$ $$\def\conj#1{{\check{#1}}}$$ $$\def\id{{\mathrm{id}}}\comp{F}=\conj{G}$$. By the definitions of companion and conjoint, $$\cat{Q}(F(p),q)=\comp{F}(p,q)=\conj{G}(p,q)=\cat{P}(p,G(q))$$ for every $$p\in\cat{P}$$ and $$q\in\cat{Q}$$. Suppose $$F$$ and $$G$$ are $$\cat{V}$$-adjoints. Since $$\cat{V}$$ is a skeletal quantale, \eqref{eq1} holds with equality. By the definitions of companion and conjoint $$\conj{G}(p,q)=\cat{P}(p,G(q))=\cat{Q}(F(p),q)=\comp{F}(p,q)$$ holds for every $$p\in\cat{P}$$ and $$q\in\cat{Q}$$. Thus, $$\comp{F}=\conj{G}$$.
2. $$\cat{P}(p,\id(q))=\cat{P}(p,q)=\cat{P}(\id(p),q)$$, so $$\comp{\id}=\conj{\id}$$.
Comment Source:1. Suppose \$$\def\cat#1{{\mathcal{#1}}}\$$ \$$\def\comp#1{{\widehat{#1}}}\$$ \$$\def\conj#1{{\check{#1}}}\$$ \$$\def\id{{\mathrm{id}}}\comp{F}=\conj{G}\$$. By the definitions of companion and conjoint, \$$\cat{Q}(F(p),q)=\comp{F}(p,q)=\conj{G}(p,q)=\cat{P}(p,G(q))\$$ for every \$$p\in\cat{P}\$$ and \$$q\in\cat{Q}\$$. Suppose \$$F\$$ and \$$G\$$ are \$$\cat{V}\$$-adjoints. Since \$$\cat{V}\$$ is a skeletal quantale, \eqref{eq1} holds with equality. By the definitions of companion and conjoint \$$\conj{G}(p,q)=\cat{P}(p,G(q))=\cat{Q}(F(p),q)=\comp{F}(p,q)\$$ holds for every \$$p\in\cat{P}\$$ and \$$q\in\cat{Q}\$$. Thus, \$$\comp{F}=\conj{G}\$$. 2. \$$\cat{P}(p,\id(q))=\cat{P}(p,q)=\cat{P}(\id(p),q)\$$, so \$$\comp{\id}=\conj{\id}\$$.