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Exercise 60 - Chapter 4

David Lambert
in Exercises

\(\newcommand{\ul}[1]{\underline{#1}}\) \(\def\da#1#2{{@V#1V#2V}}\) Consider the set \(\ul{3}=\{1,2,3\}\).

  1. Draw a picture of the unit corelation \(\varnothing\to\ul{3}\sqcup\ul{3}\).
  2. Draw a picture of the counit corelation \(\ul{3}\sqcup\ul{3}\to\varnothing\).
  3. Check that the snake equations \eqref{eqn.yanking} hold. (Since every object is its own dual, you only need to check one of them.)

For every \(c\in\mathrm{Ob}(\mathcal{C})\) $$\require{AMScd}\begin{equation}\label{eqn.yanking}\tag{4.57}\begin{CD}c @= c\\@V{\cong}VV {}@AA{\cong}A\\c{\otimes}I @. I{\otimes}c\\@V{c\otimes\eta_c}VV {}@AA{\epsilon_c\otimes c}A\\c\otimes(c^*\otimes c) @>>\cong> (c\otimes c^*)\otimes c\end{CD}\hspace{.8in}\begin{CD}c^* @= c^*\\@V{\cong}VV {}@AA{\cong}A\\I{\otimes}c^* @. c^*{\otimes}I\\@V{\eta_c\otimes c^*}VV {}@AA{c^*\otimes\epsilon_c}A\\(c^*\otimes c)\otimes c^* @>>\cong> c^*\otimes(c\otimes c^*)\end{CD}\end{equation}$$

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    1.
    David Lambert

    1. \[ \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} \begin{array} % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} % \hskip1em\Rule{0em}{5em}{1.5em} &\place{0}{2}{\xrightarrow{\eta_\ul{3}}}\hskip1em& % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} \place{6}{4}{\bullet} \place{6}{2}{\bullet} \place{6}{0}{\bullet} % \place{.3}{4.2}{\Rule{6em}{.1em}{0em}} \place{.3}{2.2}{\Rule{6em}{.1em}{0em}} \place{.3}{0.2}{\Rule{6em}{.1em}{0em}} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} \place{7}{4}{1} \place{7}{2}{2} \place{7}{0}{3} % \hskip8em\Rule{0em}{5em}{1.5em} \end{array} \]. By which I mean that wherever one has \(\ul{3}\), one may insert the corelation on the right using the unit natural transformation.

    2. \[ \begin{array} % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} \place{6}{4}{\bullet} \place{6}{2}{\bullet} \place{6}{0}{\bullet} % \place{.3}{4.2}{\Rule{6em}{.1em}{0em}} \place{.3}{2.2}{\Rule{6em}{.1em}{0em}} \place{.3}{0.2}{\Rule{6em}{.1em}{0em}} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} \place{7}{4}{1} \place{7}{2}{2} \place{7}{0}{3} % \hskip8em\Rule{0em}{5em}{1.5em} &\place{0}{2}{\xrightarrow{\epsilon_\ul{3}}}\hskip1em& % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} % \hskip1em\Rule{0em}{5em}{1.5em} \end{array} \]. By which I mean that anywhere one finds the unit corelation, one may remove it, using the counit natural transformation.

    3. Here the snake equations correspond to the fact that inserting and then removing the unit corelation leaves one with the same sets and corelations with which one started. In terms of diagrams, one composes the two figures above, so that \(\ul{3}\) is mapped to \(\ul{3}\).

    Comment Source:1. \\[ \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} \begin{array} % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} % \hskip1em\Rule{0em}{5em}{1.5em} &\place{0}{2}{\xrightarrow{\eta\_\ul{3}}}\hskip1em& % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} \place{6}{4}{\bullet} \place{6}{2}{\bullet} \place{6}{0}{\bullet} % \place{.3}{4.2}{\Rule{6em}{.1em}{0em}} \place{.3}{2.2}{\Rule{6em}{.1em}{0em}} \place{.3}{0.2}{\Rule{6em}{.1em}{0em}} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} \place{7}{4}{1} \place{7}{2}{2} \place{7}{0}{3} % \hskip8em\Rule{0em}{5em}{1.5em} \end{array} \\]. By which I mean that wherever one has \\(\ul{3}\\), one may insert the corelation on the right using the unit natural transformation. 2. \\[ \begin{array} % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} \place{6}{4}{\bullet} \place{6}{2}{\bullet} \place{6}{0}{\bullet} % \place{.3}{4.2}{\Rule{6em}{.1em}{0em}} \place{.3}{2.2}{\Rule{6em}{.1em}{0em}} \place{.3}{0.2}{\Rule{6em}{.1em}{0em}} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} \place{7}{4}{1} \place{7}{2}{2} \place{7}{0}{3} % \hskip8em\Rule{0em}{5em}{1.5em} &\place{0}{2}{\xrightarrow{\epsilon\_\ul{3}}}\hskip1em& % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} % \hskip1em\Rule{0em}{5em}{1.5em} \end{array} \\]. By which I mean that anywhere one finds the unit corelation, one may remove it, using the counit natural transformation. 3. Here the snake equations correspond to the fact that inserting and then removing the unit corelation leaves one with the same sets and corelations with which one started. In terms of diagrams, one composes the two figures above, so that \\(\ul{3}\\) is mapped to \\(\ul{3}\\).
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