Options

Exercise 35 - Chapter 4

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

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

Comments

  • 1.

    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)\\).
Sign In or Register to comment.