From 155d05c384667612e9fdd9f2556d69bd6229ab61 Mon Sep 17 00:00:00 2001 From: Joshua Moerman Date: Thu, 9 May 2013 22:33:03 +0200 Subject: [PATCH] Thesis: Added a chapter about homotopy --- thesis/2_ChainComplexes.tex | 6 ++-- thesis/5_Homotopy.tex | 60 +++++++++++++++++++++++++++++++++++++ thesis/DoldKan.tex | 4 +++ thesis/preamble.tex | 2 ++ 4 files changed, 69 insertions(+), 3 deletions(-) create mode 100644 thesis/5_Homotopy.tex diff --git a/thesis/2_ChainComplexes.tex b/thesis/2_ChainComplexes.tex index a1d377c..06e1720 100644 --- a/thesis/2_ChainComplexes.tex +++ b/thesis/2_ChainComplexes.tex @@ -12,8 +12,8 @@ Of course we can make this more general by taking for example $R$-modules instea \begin{definition} Given a chain complex $C$ we define the following subgroups: \begin{itemize} - \item $Z_n(C) = ker(\del: C_n \to C_{n-1}) \nsubgrp C_n$, and - \item $B_n(C) = im(\del: C_{n+1} \to C_n) \nsubgrp C_n$. + \item $Z_n(C) = \ker(\del: C_n \to C_{n-1}) \nsubgrp C_n$, and + \item $B_n(C) = \im(\del: C_{n+1} \to C_n) \nsubgrp C_n$. \end{itemize} \end{definition} \begin{lemma} @@ -21,7 +21,7 @@ Of course we can make this more general by taking for example $R$-modules instea $$ B_n(C) \nsubgrp Z_n(C).$$ \end{lemma} \begin{proof} - It follows from $\del_n \circ \del_{n+1} = 0$ that $im(\del: C_{n+1} \to C_n)$ is a subset of $ker(\del: C_n \to C_{n-1})$. Those are exactly the abelian groups $B_n(C)$ and $Z_n(C)$, so $ B_n(C) \nsubgrp Z_n(C) $. + It follows from $\del_n \circ \del_{n+1} = 0$ that $\im(\del: C_{n+1} \to C_n)$ is a subset of $\ker(\del: C_n \to C_{n-1})$. Those are exactly the abelian groups $B_n(C)$ and $Z_n(C)$, so $ B_n(C) \nsubgrp Z_n(C) $. \end{proof} \begin{definition} Given a chain complex $C$ we define the \emph{$n$-th homology group} $H_n(C)$: diff --git a/thesis/5_Homotopy.tex b/thesis/5_Homotopy.tex new file mode 100644 index 0000000..21fd403 --- /dev/null +++ b/thesis/5_Homotopy.tex @@ -0,0 +1,60 @@ +\section{Homotopy} +\label{sec:Homotopy} + +We've already seen homology in chain complexes. We can of course now translate this notion to simplicial abelian groups, by assigning a simplicial abelian group $X$ to $H_n(N(X))$. But there is a more general notion of homotopy for simplicial sets, which is also similar to the notion of homotopy in topology. We will define the notion of homotopy groups for simplicial sets. + +When dealing with homotopy in a topological space $X$ we always need a base-point $\ast \in X$. This is also the case for homotopy in simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast \in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots) \in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned about simplicial abelien groups, where there is an obvious choice for the base-point, namely $0$. + +\begin{definition} + Given a simplicial set $X$ with base-point $\ast$, we define $Z_n(X)$ to be the set of $n$-simplices with the base-point as boundary, i.e.: + $$ Z_n(X) = \{ x \in X_n | d_i(x) = \ast \text{ for all } i < n \}. $$ + For two $n$-simplices $x, x' \in Z_n(X)$, we define $x \sim x'$ if there exists $y \in X_{n+1}$ such that: + \begin{align} + d_{n+1}(y) &= x' \\ + d_n(y) &= x \\ + d_i(y) &= \ast \text{ for all } i < n. + \end{align} +\end{definition} + +Of course we would like $\sim$ to be an equivalence relation, however this is not true for all simplicial sets. For example there is in general no reason for symmetry, existence of a $1$-simplex $y$ from $x$ to $x'$ does not give us a $1$-simplex $y'$ from $x'$ to $x$. One can give an precise condition on when it is a equivalence relation, the so called Kan-condition. In our case of abelien groups, however, we can prove this directly. + +\todo{Htp: Insert prove $\sim$ is eq. rel. for $\sAb$.} + +\todo{Htp: Discuss/picturize Kan-condition?} + +\begin{definition} + Given a simplicial abelian group $X$, we define the $n$-th homotopy group as: + $$ \pi_n(X) = Z_n(X) / \sim. $$ +\end{definition} + +Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and $\sim$ also defines a subgroup. It is relatively straight forward to prove that this definition coincides with the $n$-th homology group of the associated normalized chain complex. + +\begin{lemma} + For any simplicial abelian group $X$: + $$ \pi_n(X) = H_n(N(X)). $$ +\end{lemma} +\begin{proof} + By writing out the definitions of the $n$-cycles and $n$-boundaries of the normalized chain complex, we see: + \begin{align*} + \ker(\del) &= \{ x \in N(X)_n \I \del(x) = 0 \} \\ + &= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i < n \text{ and } d_n(x) = 0 \} \\ + &= \{ x \in X_n \I d_i(x) = 0 \text{ forall } i \leq n \} \\ + &= Z_n(X) + \end{align*} + \begin{align*} + \im(\del) &= \{ \del(y) \I y \in N(X)_{n+1} \} \\ + &= \{ d_{n+1} \I y \in X_{n+1}, d_i(y) = 0 \text{ for all } i \leq n \} \\ + &= \{ x \in N(X)_n \I x \sim 0 \} + \end{align*} + So we see that $\pi_n(X) = Z_n(X) / \sim = \ker(\del) / \im(\del) = H_n(N(X))$. +\end{proof} + +\begin{corollary} + For a chain complex $C$ we have $H_n(C) \iso \pi_n(K(C))$ +\end{corollary} +\begin{proof} + By the established equivalence we have: + $$ \pi_n(K(C)) \iso H_n(N(K(C))) \iso H_n(C). $$ +\end{proof} + + diff --git a/thesis/DoldKan.tex b/thesis/DoldKan.tex index 249cebe..0aeeffa 100644 --- a/thesis/DoldKan.tex +++ b/thesis/DoldKan.tex @@ -13,6 +13,7 @@ \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} +\newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} @@ -48,6 +49,9 @@ In the first section some definitions from category theory are given, because we \newpage \input{../thesis/4_Constructions} +\newpage +\input{../thesis/5_Homotopy} + \newpage \listoftodos % \nocite{*} diff --git a/thesis/preamble.tex b/thesis/preamble.tex index ccc9390..5171603 100644 --- a/thesis/preamble.tex +++ b/thesis/preamble.tex @@ -5,6 +5,8 @@ \usepackage{listings} \usepackage{mathtools} +\DeclareMathOperator{\im}{Im} + \usepackage{tikz} % http://pdp7.org/blog/?p=133 \usetikzlibrary{matrix, arrows, decorations} \tikzset{node distance=3em, row sep=3em, column sep=3em, auto}