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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.

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

Comments

  • 1.
    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}\\).
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