It looks like you're new here. If you want to get involved, click one of these buttons!
\(\newcommand{\ul}[1]{\underline{#1}}\) \(\def\da#1#2{{@V#1V#2V}}\) Consider the set \(\ul{3}=\{1,2,3\}\).
For every \(c\in\mathrm{Ob}(\mathcal{C})\) $$\require{AMScd}\begin{equation}\label{eqn.yanking}\tag{4.57}\begin{CD}c @= c\\@V{\cong}VV {}@AA{\cong}A\\c{\otimes}I @. I{\otimes}c\\@V{c\otimes\eta_c}VV {}@AA{\epsilon_c\otimes c}A\\c\otimes(c^*\otimes c) @>>\cong> (c\otimes c^*)\otimes c\end{CD}\hspace{.8in}\begin{CD}c^* @= c^*\\@V{\cong}VV {}@AA{\cong}A\\I{\otimes}c^* @. c^*{\otimes}I\\@V{\eta_c\otimes c^*}VV {}@AA{c^*\otimes\epsilon_c}A\\(c^*\otimes c)\otimes c^* @>>\cong> c^*\otimes(c\otimes c^*)\end{CD}\end{equation}$$
Comments
\[ \require{enclose} \def\uline#1#2{\enclose{updiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} \def\dline#1#2{\enclose{downdiagonalstrike}{\phantom{\Rule{#1em}{#2em}{0em}}}} % \def\place#1#2#3{\smash{\rlap{\hskip{#1em}\raise{#2em}{#3}}}} \begin{array} % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} % \hskip1em\Rule{0em}{5em}{1.5em} &\place{0}{2}{\xrightarrow{\eta_\ul{3}}}\hskip1em& % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} \place{6}{4}{\bullet} \place{6}{2}{\bullet} \place{6}{0}{\bullet} % \place{.3}{4.2}{\Rule{6em}{.1em}{0em}} \place{.3}{2.2}{\Rule{6em}{.1em}{0em}} \place{.3}{0.2}{\Rule{6em}{.1em}{0em}} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} \place{7}{4}{1} \place{7}{2}{2} \place{7}{0}{3} % \hskip8em\Rule{0em}{5em}{1.5em} \end{array} \]. By which I mean that wherever one has \(\ul{3}\), one may insert the corelation on the right using the unit natural transformation.
\[ \begin{array} % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} \place{6}{4}{\bullet} \place{6}{2}{\bullet} \place{6}{0}{\bullet} % \place{.3}{4.2}{\Rule{6em}{.1em}{0em}} \place{.3}{2.2}{\Rule{6em}{.1em}{0em}} \place{.3}{0.2}{\Rule{6em}{.1em}{0em}} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} \place{7}{4}{1} \place{7}{2}{2} \place{7}{0}{3} % \hskip8em\Rule{0em}{5em}{1.5em} &\place{0}{2}{\xrightarrow{\epsilon_\ul{3}}}\hskip1em& % \hskip1em % \place{0}{4}{\bullet} \place{0}{2}{\bullet} \place{0}{0}{\bullet} % \place{-1}{4}{1} \place{-1}{2}{2} \place{-1}{0}{3} % \hskip1em\Rule{0em}{5em}{1.5em} \end{array} \]. By which I mean that anywhere one finds the unit corelation, one may remove it, using the counit natural transformation.
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 \(\ul{3}\).
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}\\).