From: Nathaniel Wesley Filardo Date: Tue, 28 Feb 2012 23:26:32 +0000 (-0500) Subject: Add yoneda.tex X-Git-Url: https://hydra-www.ietfng.org/gitweb/?a=commitdiff_plain;h=279a0963ef660727882ec448e39b6c441c92115c;p=ctcheat Add yoneda.tex --- diff --git a/yoneda.tex b/yoneda.tex new file mode 100644 index 0000000..273d70a --- /dev/null +++ b/yoneda.tex @@ -0,0 +1,166 @@ +\documentclass[10pt,letterpaper]{article} +\DeclareSymbolFont{AMSb}{U}{msb}{m}{n} +\DeclareMathAlphabet{\mathbbm}{U}{bbm}{m}{n} + +\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} + +\section{Covariant Hom functors} + +These are defined in \S3.20.4, but a longer example never hurt anybody, right? + +Suppose $a^2 = id_A$, $b^2 = id_{B'}$, and $f' = g \circ f$ in this category $\mathbf{A}$. +\[ \xymatrix@C=1in{ + & B \ar[d]^g \ar@(dr,ur)_{id_B} \\ + A \ar@(dr,d)^{id_A} \ar@(dl,l)^{a} \ar[ru]^f \ar[r]^{f'} & B' \ar@(dl,d)_{b} \ar@(dr,r)_{id_{B'}} \\ + } \] + +If we actually draw out all the arrows, we get this diagram: +\[ \xymatrix@C=1in{ + & B \ar@<1pt>[d]_{b \circ g} \ar@<-1pt>[d]^g \ar@(dr,ur)_{id_B} \\ + A \ar@(dr,d)^{id_A} \ar@(dl,l)^{a} + \ar@<1pt>[ru]^f \ar@<-1pt>[ru]_{f \circ a} + \ar@<3pt>[r] \ar@<1pt>[r] \ar@<-1pt>[r] \ar@<-3pt>[r] + & B' \ar@(dl,d)_{b} \ar@(dr,r)_{id_{B'}} \\ + } \] +where the four arrows from $A$ to $B'$ are $\set{f', f' \circ a, b \circ f', b \circ f' \circ a}$. + +$\mbox{hom}(A,-)$ is the following full subcategory of $\mathbf{Set}$; please +note the similarities---the {\em representation} of the structure reachable from $A$: +\[ \xymatrix@C=1in{ + & \set{f, f \circ a} \ar@<1pt>[d] \ar@<-1pt>[d] \\ + \set{id_A, a} \ar@<1pt>[ru] \ar@<-1pt>[ru] + \ar@<3pt>[r] \ar@<1pt>[r] \ar@<-1pt>[r] \ar@<-3pt>[r] + & \txt{$\{f', f' \circ a,$\\ $b \circ f', b \circ f' \circ a\}$} \\ +} \] +The two arrows from $\mbox{hom}(A,A) = \set{id_A, a}$ to $\mbox{hom}(A,B) += \set{f, f \circ a}$ are obtained by pre-composition: +\begin{align*} + \mbox{hom}(A,f) &= \set{ id_A \mapsto f, a \mapsto f \circ a } \\ + \mbox{hom}(A,f \circ a) &= \set{ id_a \mapsto f \circ a, a \mapsto f } +\end{align*} +(the last entry holds because $(f \circ a) \circ a = f \circ (a \circ a) = f +\circ id_A = f$). The four arrows from $\mbox{hom}(A,A)$ to +$\mbox{hom}(A,B') = \set{f', f' \circ a, b \circ f', b \circ f' \circ a}$ are +again obtained by pre-composition: +\begin{align*} + \mbox{hom}(A,f') &= \set{id_A \mapsto f', a \mapsto f' \circ a} \\ + \mbox{hom}(A,f' \circ a) &= \set{id_a \mapsto f' \circ a, a \mapsto f'} \\ + \mbox{hom}(A,b \circ f') &= \set{id_a \mapsto b \circ f', a \mapsto b \circ f' \circ a} \\ + \mbox{hom}(A,b \circ f' \circ a) &= \set{id_a \mapsto b \circ f' \circ a, a \mapsto b \circ f'} +\end{align*} +The two vertical arrows are (again by precomposition, and recall that $f' = g \circ f$): +\begin{align*} + \mbox{hom}(A,g) &= \set{f \mapsto f', f \circ a \mapsto f' \circ a} \\ + \mbox{hom}(A,b \circ g) &= \set{f \mapsto b \circ f', f \circ a \mapsto b \circ f' \circ a} +\end{align*} +It is easy to check that, indeed, composition still holds: the four horizontal +arrows are each the result of composition of a choice of vertical and diagonal arrows, +and we haven't missed any. + +\pagebreak +\section{Proposition 6.18 and The Yoneda Lemma} + +Let's restrict our attention to this category $\mathbf{A}$ +(to truly appreciate the significance of this result, I encourage you to work out +the details in full for a slightly larger category!): +\[ \xymatrix{ A \ar[r]^f & B } \] + +The image of this in $\mathbf{Set}$ under $\mbox{hom}(A,-)$ is just +\[ \xymatrix@C=.5in{ \set{id_A} \ar[r]^{\mbox{hom}(A,f)} & \set{id_B} } \] + +Now suppose we have some other functor $F : \mathbf{A} \to \mathbf{Set}$, +whose image is +\[ \xymatrix@C=.5in{ \set{a_0, \dots} \ar[r]^{Ff} & \set{b_0, \dots} } \] +(where $FA = \set{a_0, \dots}$ and $FB = \set{b_0, \dots}$.) + +Now, the claim of Proposition 6.18 is that there exists a unique natural +transformation $\tau : \mbox{hom}(A,-) \stackrel{\cdot}{\to} F$ if we additionally +constrain $\tau_A(id_A) = a_0$. OK, so, first off: what does that mean? $\tau$ being +natural means $\forall B,C,g : B \to C$, this commutes: +\[ \xymatrix{ + \mbox{hom}(A,B) \ar[r]^{\tau_B} \ar[d]_{\mbox{hom}(A,g)} & FB \ar[d]^{Fg} \\ + \mbox{hom}(A,C) \ar[r]^{\tau_C} & FC +} \] +or more specifically, at $A,B,f$ (first generically, then expanding some computations): +\[ \xymatrix{ + \mbox{hom}(A,A) \ar[r]^{\tau_A} \ar[d]_{\mbox{hom}(A,f)} & FA \ar[d]^{Ff} \\ + \mbox{hom}(A,B) \ar[r]^{\tau_B} & FB +} \qquad \xymatrix{ + \set{id_A} \ar[r]^{\tau_A} \ar[d]_{\set{id_A \mapsto f}} & \set{a_0,\dots} \ar[d]^{Ff} \\ + \set{f} \ar[r]^{\tau_B} & \set{b_0,\dots} +} \] +and so requiring $\tau_A(id_A) = a_0$ makes sense. If this is to be natural, it +must be the case (for all $B$ and $f : A \to B$; note that this works even to +define $\tau_A$ at inputs other than $id_A$ just as well!) that +\begin{align*} + \tau_B(f) &= \tau_B(f \circ id_A) \\ + &= \tau_B(\mbox{hom}(A,f)(id_A)) & \forall_x . f \circ x = \mbox{hom}(A,f)(x) \\ + &= F(f)(\tau_A(id_A)) & \text{naturality of $\tau$} \\ + &= F(f)(a_0) & \text{requirement} +\end{align*} +So $\tau$ is fully determined by naturality and the requirement given, +precisely because $\mbox{hom}(A,-)$ on arrows captures pre-composition. +So: given a choice of $a_0 \in FA$, we can fully specify a natural transformation $\tau$. + + +Conversely, given a $\tau'$, it must pick out some $\tau_A(id_A) \in FA$. Therefore, +the Yoneda lemma: + +\begin{quote}{\em + Given a functor $F : \mathbf{A} \to \mathbf{Set}$, the set + $\set{\tau \middle\vert \tau : \mbox{hom}(A,-) \stackrel{\cdot}{\to} F}$ + is isomorphic (in $\mathbf{Set}$) to $FA$. + The isomorphism is witnessed by the function $Y(\tau) = \tau_A(id_A)$. +}\end{quote} + +\end{document}