Rewrites pieces of the rationalization

thesis/notes/Rationalization.tex

@@ -5,33 +5,31 @@
 In this section we will prove the existence of rationalizations $X \to X_\Q$. We will do this in a cellular way. The $n$-spheres play an important role here, so their rationalizations will be discussed first. In this section $1$-connectedness of spaces will play an important role.
 
 \section{Rationalization of \texorpdfstring{$S^n$}{Sn}}
-In this section we fix $n>0$. We will construct $S^n_\Q$ in stages $S^n(1), S^n(2), \ldots$, where at each stage we wedge a sphere and then glue an $n+1$-cell to ``invert'' some element in the $n$th homotopy group.
+We will construct $S^n_\Q$ as an infinite telescope, as depicted for $n=1$ in the following picture.
+\todo{plaatje}
+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 $k$''.
 
-\todo{plaatje en tekst leesbaarder maken}
+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.
 
-We start the construction with $S^n(1) = S^n$. Assume we constructed $S^n(r) = \bigvee_{i=1}^{r} S^{n} \cup_{h} \coprod_{i=1}^{r-1} D^{n+1}$, where $h$ is a specific attaching map. Assume furthermore the following two properties. Firstly, the inclusion $i_r : S^n \to S^n(r)$ of the terminal sphere is a weak equivalence. Secondly, the inclusion $i_1 : S^n \to S^n(r)$ of the initial sphere induces the multiplication $\pi_n(S^n) \tot{\times r!} \pi_n(S^n(r))$ under the identification of $\pi_n(S^n) = \pi_n(S^n(r)) = \Z$.
+We start the construction with $S^n(1) = S^n$. Now assume $S^n(r) = \bigvee_{i=1}^r S^n \cup_{h(r)} \coprod_{i=1}^{r-1} D^{n+1}$ is constructed. Let $i: S^n \to S^n(r)$ be the inclusion into the last (i.e. $r$th) sphere, and let $g : S^n \to S^n$ be a representative for the class $(r+1)[\id] \in \pi_n(S^n)$. Combine the two maps to obtain $\phi : S^n \to S^n \vee S^n \tot{i \vee g} S^n(r) \vee S^n$. We define $S^n(r+1)$ as the pushout:
-
+\[ \xymatrix{
-We will construct $S^n(r+1)$ with similar properties as follows. Let $f: S^n \to S^n(r)$ be a representative for $1 \in \Z \iso \pi_n(S^n(r))$ and $g: S^n \to S^n$ be a representative for $r+1 \in \Z \iso \pi_n(S^n)$. These maps combine into $\phi: S^n \to S^n \vee S^n \tot{f \vee g} S^n(r) \vee S^n$. We define $S^n(r+1)$ as the pushout in the following diagram.
+	S^n \ar[r]^-\phi \arcof[d] \xypo & S^n(r) \vee S^n \ar[d] \\
-\begin{displaymath}
-	\xymatrix{
-	S^n \ar[r]^{\phi} \arcof[d] & S^n(r) \vee S^n \ar[d] \\
 	D^{n+1} \ar[r] & S^n(r+1)
-	}
+} \]
-\end{displaymath}
+So that $S^n(r+1) = \bigvee_{i=1}^{r+1} S^n \cup_{h(r+1)} \coprod_{i=1}^{r} D^{n+1}$. To finish the construction we define $S^n_\Q = \colim_{r} S^n(r)$.
-
-So $S^n(r+1) = \bigvee_{i=1}^{r+1} S^{n} \cup_{h'} \coprod_{i=1}^{r} D^{n+1}$. \todo{Prove the two properties}.
 
-Now to finish the construction we define the \Def{rational sphere} as $S^n_\Q = \colim_r S^n(r)$. Note that the homotopy groups commute with filtered colimits \cite[9.4]{may}, so that we can compute $\pi_n(S^n_\Q)$ as the colimit of the terms $\pi_n(S^n(r)) \iso \Z$ and the induced maps as depicted in the following diagram:
+We note two things here. First, at any stage, the inclusion $i : S^n \to S^n(r)$ into the $r$th sphere is a weak equivalence, as we can collapse the (finite) telescope to the last sphere. This identifies $\pi_n(S^n(r)) = \Z$ for all $r$. Secondly, if $i_r: S^n \to S^n(r+1)$ is the inclusion of the $r$th sphere and $i_{r+1} : S^n \to S^n(r+1)$ the inclusion of the last sphere, then $[i_r] = (r+1)[i_{r+1}] \in \pi^n(S^n(r+1))$, by construction. This means that we can divide $[i_r]$ by $r+1$. Note that the inclusion $S^n(r) \to S^n(r+1)$ induces a multiplication by $r+1$ under the identification $\pi^n(S^n(r)) = \Z$ for all $r$.
 
-$$ \Z \tot{\times 2} \Z \tot{\times 3} \Z \tot{\times 4} \Z \tot{\times 5} \cdots \Q. $$
+The $n$th homotopy group of $S^n_\Q$ can be calculated as follows.  We use the fact that the homotopy groups commute with filtered colimits \cite[9.4]{may} to compute $\pi_n(S^n_\Q)$ as the colimit of the terms $\pi_n(S^n(r)) \iso \Z$ and the induced maps as depicted in the following diagram:
+\[ \xymatrix{ \Z \ar[r]^-{\times 2} & \Z \ar[r]^-{\times 3} & \Z \ar[r]^-{\times 4} & \Z \ar@{-->}[rr] & & \Q } \]
 
-Moreover we note that the generator $1 \in \pi_n(S^n)$ is sent to $1 \in \pi_n(S^n_\Q)$ via the inclusion $S^n \to S^n_\Q$ of the initial sphere. However the other homotopy groups are harder to calculate as we have generally no idea what the induced maps are. But in the case of $n=1$, the other homotopy groups of $S^1$ are trivial.
+Moreover we note that the generator $1 \in \Z = \pi_n(S^n)$ is sent to $1 \in \Q = \pi_n(S^n_\Q)$ via the inclusion $S^n \to S^n_\Q$ of the initial sphere. However the other homotopy groups are harder to calculate as we have generally no idea what the induced maps are. But in the case of $n=1$, the other homotopy groups of $S^1$ are trivial.
 
 \Corollary{rationalization-S1}{
 	The inclusion $S^1 \to S^1_\Q$ is a rationalization.
 }
 
-For $n>1$ we can resort to homology, which also commutes with filtered colimits \cite[14.6]{may}. By connectedness we have $H_0(S^n_\Q) = \Z$ and for $i \neq 0, n$ we have $H_i(S^n) = 0$, so in these cases the homology of the colimit is also $\Z$ and resp. $0$ \todo{leesbaarder}. For $i = n$ we can use the same sequence as above (or use the Hurewicz theorem) to conclude:
+For $n>1$ we can resort to homology, which also commutes with filtered colimits \cite[14.6]{may}. By connectedness we have $H_0(S^n_\Q) = \Z$ and for $i \neq 0, n$ we have $H_i(S^n) = 0$, so the colimit is also trivial. For $i = n$ we can use the same sequence as above (or use the Hurewicz theorem) to conclude:
 
 $$ H_i(S^n_\Q) = \begin{cases}
 	\Z, &\text{ if } i = 0 \\
@@ -65,11 +63,10 @@ The \Def{rational disk} is now defined as cone of the rational sphere: $D^{n+1}_
 	Furthermore $f'$ is determined up to homotopy (i.e. any map $f''$ with $f''i = f$ is homotopic to $f'$) and homotopic maps have homotopic extensions (i.e. if $f \simeq g$, then $f' \simeq g'$).
 }
 \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 shown in the following diagram.
+	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}
 
-	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 $f = f' \circ i$.
+	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 $f = f' \circ i$.
 
 	Let $f''$ be any map such that $f''i = f$. Then $f''$ also represents $\alpha$ and all the functions $\frac{1}{2}f''$, $\frac{1}{6}f''$,\dots are hence homotopic to $\frac{1}{2}f$, $\frac{1}{6}f$,\dots. So indeed $f$ is homotopic to $f''$.