From: Nathaniel Wesley Filardo Date: Mon, 4 Nov 2013 04:58:32 +0000 (-0500) Subject: Add example: adjoints of the diagonal functor X-Git-Url: https://hydra-www.ietfng.org/gitweb/?a=commitdiff_plain;h=668b2731d3117c9f9f08b1435132a7ba960865fe;p=ctcheat Add example: adjoints of the diagonal functor --- diff --git a/adjoints-diag.tex b/adjoints-diag.tex new file mode 100644 index 0000000..e57097a --- /dev/null +++ b/adjoints-diag.tex @@ -0,0 +1,222 @@ +\documentclass[letterpaper]{article} +\DeclareSymbolFont{AMSb}{U}{msb}{m}{n} +\DeclareMathAlphabet{\mathbbm}{U}{bbm}{m}{n} + +\title{$\Sigma \dashv \Delta \dashv \Pi$} + +\usepackage{amsmath,amssymb,amsthm,latexsym} +\usepackage{fancyhdr} +\usepackage[tiny,center,compact,sc]{titlesec} +\usepackage[cm]{fullpage} +\usepackage{pstricks} +\usepackage{graphicx} +\usepackage{verbatim} +\usepackage{bm} +\usepackage{ifthen} +\usepackage{epsfig} +\usepackage[all]{xypic} +\usepackage{textcomp} +\usepackage{url} +\usepackage{multirow} +\usepackage{hyperref} +\usepackage{breakurl} + +\renewcommand{\baselinestretch}{0.9} + +%\newtheorem{thm}{Thm}[section] +%\newtheorem{dfn}{Def}[section] + +\setlength{\parindent}{0pt} +\setlength{\parskip}{3pt} + +%Scalable bracket-like +\newcommand{\paren}[1]{\left({#1}\right)} +\newcommand{\brak}[1]{\left[{#1}\right]} +\newcommand{\abs}[1]{\left\lvert{#1}\right\rvert} +\newcommand{\ang}[1]{\left\langle{#1}\right\rangle} +\newcommand{\set}[1]{\left\{#1\right\}} + +%Mathematics +\newcommand{\condexp}[1]{\ifthenelse{\equal{#1}{false}}{}{^{#1}}} +\newcommand{\dd}[3][false]{\frac{d\condexp{#1}{#2}}{d{#3}\condexp{#1}}} +\newcommand{\pd}[3][false]{\frac{\partial\condexp{#1}{#2}}{\partial{#3}\condexp{#1}}} + +\newcommand{\ifrac}[2]{{#1}/{#2}} + +%Quantum Mechanics +\newcommand{\ket}[1]{\left\lvert{#1}\right\rangle} +\newcommand{\bra}[1]{\left\langle{#1}\right\rvert} +\newcommand{\braket}[2]{\left\langle{#1}\middle\vert{#2}\right\rangle} +\newcommand{\Braket}[3]{\left\langle{#1}\middle\vert{#2}\middle\vert{#3}\right\rangle} +\newcommand{\dyad}[2]{\left\lvert{#1}\middle\rangle\middle\langle{#2}\right\rvert} + +\DeclareMathOperator{\mm}{\mid\mid} + +\newcommand{\defn}[1]{{\bf #1}} + +\begin{document} +\maketitle + +\section{Getting Started} + +The ``diagonal functor'' $\Delta : \mathbf{C} \to \mathbf{C}^2$ is +charmingly degenerate: +$\Delta X = (X,X), \Delta (f : A \to B) = (f,f) : (A,A) \to (B,B)$. +(In the ${}^2$ category, morphisms are such that $(f,g)(a,b) = (f a,g b)$.) +For notational clarity, we'll use $A \times B$ for products within +$\mathbf{C}$ and $(A,B)$ for objects in $\mathbf{C}^2$. + +Define $\Pi_1 : \mathcal{C}^2 \to \mathcal{C}$ to be the first +projector, and $\Pi_2$ the second. If $\mathcal{C}$ has +products, then we may define $\Pi : \mathcal{C}^2 \to \mathcal{C}$ as +$(A,B) \mapsto A \times B$, and $\Pi_1 = \pi_1 \circ \Pi$. +If $\mathcal{C}$ has coproducts, define $\Sigma : \mathcal{C}^2 \to +\mathcal{C}$ by $(A,B) \mapsto A + B$. + +While on the topic of notation, recall that, if $\mathcal{C}$ is so +equipped, we may form arrows involving products and coproducts from others: +$\ang{f : A \to B,g : A \to C}(a) = (f a) \times (g a) \in B \times C$ and +$\brak{f : A \to C,g : B \to C}((a : A) + (b : B)) \in C$ is case analysis. + +\section{Left Adjoint} +\subsection{Unit} + +What would a left adjunction to $\Delta$ be? It would be a functor $F : +\mathcal{C}^2 \to \mathcal{C}$ and natural transformation $\eta$ where, in +the category $\mathcal{C}^2$, +\[ \xymatrix{ + X \ar[r]^{\eta_X} \ar[rd]_{f} & \Delta (F X) \ar[d]^{\Delta (f^\#)} \\ + & \Delta Y +} \equiv \xymatrix { + X \ar[r]^(.4){\eta_X} \ar[rd]_{f} & (F X, F X) \ar[d]^{(f^\#, f^\#)} \\ + & (Y, Y) +} \equiv \xymatrix { + (A,B) \ar[r]^(.4){\eta_X} \ar[rd]_{f} & (F (A,B), F (A,B)) \ar[d]^{(f^\#, f^\#)} \\ + & (Y, Y) +} \] + + +If this diagram is to commute for all $f$, then: $\Pi_1 \circ f = \Pi_1 +\circ (f^\#, f^\#) \circ \eta_X = f^\# \circ \Pi_1 \circ \eta_X$, and +similarly for the $\Pi_2$ component. Intuitively, this can only work in the +case where $f^\#$ is able to discriminate whether it has been handed the +$\Pi_1$ or $\Pi_2$ projection of $\eta_X$'s output. That sounds like a +perfect use of coproducts! If we take $\eta_X = (i_1, i_2): X +\to \Delta (\Sigma X)$ (i.e. $\eta_X (x) = (i_1 x, i_2 x) : (A,B) \to (A + +B, A + B)$) and define $f^\# = \brak{\Pi_1 \circ f, \Pi_2 \circ f}$, then we +see that +$f^\# \circ \Pi_1 \circ \eta_X = + \brak{\Pi_1 \circ f, \Pi_2 \circ f} \circ \Pi_1 \circ (i_1, i_2) = + \brak{\Pi_1 \circ f, \Pi_2 \circ f} \circ i_1 = + \Pi_1 \circ f +$ as required. Any such $f^\#$ is clearly unique. + +All that remains is to check that $\eta_X$ is natural from +$I$ to $\Delta \Sigma$. That is, does this commute for all $f : A \to +B$? +\[ \xymatrix{ + A \ar[r]^{\eta_A} \ar[d]^f & \Delta \Sigma A \ar[d]^{\Delta \Sigma f} \\ + B \ar[r]^{\eta_B} & \Delta \Sigma B \\ +} \equiv \xymatrix{ + (A_1, A_2) \ar[r]^(.35){\eta_A} \ar[d]^{(f_1,f_2)} & (A_1 + A_2, A_1 + A_2) + \ar[d]^{(f_1 + f_2, f_1 + f_2)} \\ + (B_1, B_2) \ar[r]^(.35){\eta_B} & (B_1 + B_2, B_1 + B_2) \\ +} \] + +Well: +\begin{align*} + \Delta \Sigma f \circ \eta_A + &= \paren{(f_1 + f_2), (f_1 + f_2)} \circ (i_1, i_2) & \text{defn}~\eta, \Delta, \Sigma \\ + &= ((f_1 + f_2) \circ i_1, (f_1 + f_2) \circ i_2) & \circ \\ + &= (i_1 \circ f_1, i_2 \circ f_2) & (f + g) \circ i_1 = i_1 \circ f \\ + &= (i_1, i_2) \circ (f_1,f_2) & \circ \\ + &= \eta_B \circ f & \text{defn}~\eta,f +\end{align*} + +So we have: $\Sigma \dashv \Delta$. + +\subsection{Counit} + +Looking at this the other way, we have, in $\mathcal{C}$, +\[ \xymatrix{ + \Sigma (\Delta X) \ar[r]^{\epsilon_X} & X \\ + \Sigma Y \ar[u]^{\Sigma f'} \ar[ur]_{f} +} \equiv \xymatrix{ + X + X \ar[r]^(.75){\epsilon_X} & X \\ + Y_1 + Y_2 \ar[u]^{f'_1 + f'_2} \ar[ur]_{f} +} \] +Then if we take $\eta_X = [id,id]$ we can define $f' = (f \circ i_1) + (f +\circ i_2)$. This is unique and $\eta_X$ is natural by inspection. + +\section{Right Adjoint} +\subsection{Unit} + +What about the other way around? Now we have, in $\mathcal{C}$ this time, +\[ \xymatrix{ + X \ar[r]^{\eta_X} \ar[rd]_{f} & G (X, X) \ar[d]^{G (f^\#)} \\ + & G Y +} \equiv \xymatrix { + X \ar[r]^{\eta_X} \ar[rd]_{f} & G (X, X) \ar[d]^{G (f^\#_1, f^\#_2)} \\ + & G (Y_1, Y_2) +} \] + +Let's speculate that $G = \Pi_1$ and see what goes wrong. That would mean +that, for each $f : X \to Y$, there is some unique $f^\# : (X,X) \to (Y,Y')$ +such that $f = \Pi_1 f^\# \circ \eta_X$. But that can't possibly be true, because +given such a $f^\#$, one that differs only in its second component will also +work, so we've violated the ``exists unique'' part of the definition. + +But if we take $G = \Pi$, then +\[ \xymatrix{ + X \ar[r]^{\eta_X} \ar[rd]_{f} & G (X, X) \ar[d]^{G (f^\#)} \\ + & G Y +} \equiv \xymatrix { + X \ar[r]^{\eta_X} \ar[rd]_{f} & G (X, X) \ar[d]^{G (f^\#_1, f^\#_2)} \\ + & G (Y_1, Y_2) +} \equiv \xymatrix { + X \ar[r]^{\eta_X} \ar[rd]_{f} & X \times X \ar[d]^{f^\#_1 \times f^\#_2} \\ + & Y_1 \times Y_2 +} \] +And we can see that taking $\eta_X = \ang{id,id}$ and $f^\# = (\pi_1 \circ f) +\times (\pi_2 \circ f)$ makes this commute with a unique $f^\#$ for each +$f$. $\eta_X$ is clearly natural. Thus we have that $\Delta \dashv \Pi$. + +\subsection{Counit} + +Here the counit diagram takes place in $\mathcal{C}^2$: +\[ \xymatrix{ + \Delta (\Pi X) \ar[r]^{\epsilon_X} & X \\ + \Delta Y \ar[u]^{\Delta f'} \ar[ur]_{f} +} \equiv \xymatrix { + \Delta (\Pi (X_1,X_2)) \ar[r]^(.65){\epsilon_X} & (X_1, X_2) \\ + \Delta Y \ar[u]^{\Delta f'} \ar[ur]_{f} +} \equiv \xymatrix { + (X_1 \times X_2, X_1 \times X_2) \ar[r]^(.65){\epsilon_X} & (X_1, X_2) \\ + (Y,Y) \ar[u]^{(f',f')} \ar[ur]_{(f_1,f_2)} +} \] + +Take $\eta_X = (\pi_1, \pi_2)$, then $f' = \ang{f_1, f_2}$. Uniqueness of +$f'$ is immediate. Naturality of $\eta_X$ is immediate from the action of +$\Delta\Pi$ on arrows: +\[ \xymatrix{ + (A \times B, A \times B) \ar[r] \ar[d]^{\Delta \Pi f} & (A, B) \ar[d]^{f} \\ + (A' \times B', A' \times B') \ar[r] & (A', B') +} \] +$\Delta \Pi f = \Delta \Pi (f_1,f_2) = \Delta (f_1 \times f_2) = (f_1 \times +f_2, f_1 \times f_2)$, and so +$(\pi_1, \pi_2) \circ \Delta \Pi f = (\pi_1, \pi_2) \circ (f_1 \times +f_2, f_1 \times f_2) = (f_1, f_2) = (f_1,f_2) \circ (\pi_1, \pi_2)$. + +\section{Notes} + +Note that for the unit of the left adjunction and the counit of the right +adjunction, we had to choose ``non-obvious'' natural transformations, +whereas for the other two we had things ``built from identities''. In the +former two cases, there are actually other functions which would work, +notably $\eta_X = \ang{i_2, i_1}$ and $\epsilon_X = (\pi_2,\pi_1)$. + +$\Delta \dashv \Pi$ has some reading as ``diagonals are free products'' +though I do not find that terribly informative; I have yet to find +``coproducts are free diagonals'' a useful statement at all. + +\end{document}