From a4f31d1813d4a5f3bba203dab3f4784c84587b50 Mon Sep 17 00:00:00 2001 From: Joshua Moerman Date: Fri, 30 Jan 2015 12:07:04 +0100 Subject: [PATCH] Updates the images in the chapter on rationalization --- thesis/notes/Rationalization.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/thesis/notes/Rationalization.tex b/thesis/notes/Rationalization.tex index 828ea4d..72fdef2 100644 --- a/thesis/notes/Rationalization.tex +++ b/thesis/notes/Rationalization.tex @@ -5,7 +5,7 @@ In this section we will prove the existence of rationalizations $X \to X_\Q$. We \section{Rationalization of \texorpdfstring{$S^n$}{Sn}} We will construct $S^n_\Q$ as an infinite telescope, as depicted for $n=1$ in the following picture. -\todo{plaatje} +\cimage{telescope} The space will consist of multiple copies of $S^n$, one for each $k \in \N^{>0}$, glued together by $(n+1)$-cells. The role of the $k$th copy (together with the gluing) is to be able to ``divide by $k$''. So $S^n_\Q$ will be of the form $S^n_\Q = \bigvee_{k>0} S^n \cup_{h} \coprod_{k>0} D^{n+1}$. We will define the attaching map $h$ by doing the construction in stages. @@ -63,7 +63,7 @@ The \Def{rational disk} is now defined as cone of the rational sphere: $D^{n+1}_ } \Proof{ Note that $f$ represents a class $\alpha \in \pi_n(X)$. Since $\pi_n(X)$ is a $\Q$-vector space there exists elements $\frac{1}{2}\alpha, \frac{1}{3}\alpha, \ldots$ with representatives $\frac{1}{2}f, \frac{1}{3}f, \ldots$. Recall that $S^n_\Q$ consists of many copies of $S^n$, we can define $f'$ on the $k$th copy to be $\frac{1}{k!}f$, as depicted in the following diagram. - \cimage[scale=0.6]{SnQ_Extension} + \cimage{telescope_maps} Since $[\frac{1}{(k-1)!}f] = k[\frac{1}{k!}f] \in \pi_n(X)$ we can define $f'$ accordingly on the $n+1$-cells. Since our inclusion $i: S^n \cof S^n_\Q$ is in the first sphere, we get $f = f' \circ i$.