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# Exercise 35 - Chapter 4

edited August 2018

Explain why the companion $$\widehat{\mathrm{id}}$$ of $$\mathrm{id}:\mathcal{P}\to\mathcal{P}$$ really has the formula given in Eq. \eqref{eq1}.

$$\label{eq1}\tag{4.25}U_{\mathcal{X}}(x,y):=\mathcal{X}(x,y).$$

By definition of the identity $$\newcommand{\idprof}[1]{\mathrm{id}(#1)}\newcommand{\cat}[1]{\mathcal{#1}}\newcommand{\companion}[1]{\widehat{#1}}\cat{V}-$$functor, $$\cat{P}(\idprof{p},\idprof{q})=\cat{P}(p,q)$$. By the definitions of companion (, conjoint) and unit profunctor, $$\companion{\mathrm{id}}(p,q)=\cat{P}(\idprof{p},q)=\cat{P}(p,q)=\cat{P}(p,\idprof{q})=\check{\mathrm{id}}(p,q)=U_\cat{P}(p,q)$$.
Comment Source:By definition of the identity \$$\newcommand{\idprof}[1]{\mathrm{id}(#1)}\newcommand{\cat}[1]{\mathcal{#1}}\newcommand{\companion}[1]{\widehat{#1}}\cat{V}-\$$functor, \$$\cat{P}(\idprof{p},\idprof{q})=\cat{P}(p,q)\$$. By the definitions of companion (, conjoint) and unit profunctor, \$$\companion{\mathrm{id}}(p,q)=\cat{P}(\idprof{p},q)=\cat{P}(p,q)=\cat{P}(p,\idprof{q})=\check{\mathrm{id}}(p,q)=U\_\cat{P}(p,q)\$$.