Adds more on the rationalization

thesis/notes/Rationalization.tex

@@ -4,26 +4,29 @@
 
 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. Again spaces (except for $S^1$) are assumed to be $1$-connected.
 
-\section{Construction of \texorpdfstring{$S^n_\Q$}{SnQ}}
+\section{Rationalization of \texorpdfstring{$S^n$}{Sn}}
-Fix $n>0$ we will construct the rationalization in stages, where at each stage we wedge a sphere and then glue a $n+1$-cell to ``invert'' some element in the $n$th homotopy group. At each stage the space will be homotopy equivalent to $S^n$.
+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 a $n+1$-cell to ``invert'' some element in the $n$th homotopy group.
 
-\todo{Put this in a lemma. And make it more readable.}
+\todo{plaatje}
-We start with $S^n(1) = S^n$. Now assume $S^n(k)$ is constructed. Let $f: S^n \to S^n(k)$ be a representative for $1 \in \Z \iso \pi_n(S^n(k))$ and $g: S^n \to S^n$ be a representative for $k+1 \in \Z \iso \pi_n(S^n)$. These maps combine into $\phi: S^n \to S^n \vee S^n \tot{f \vee g} S^n(k) \vee S^n$. We define $S^n(k+1)$ as the pushout in the following diagram.
+
+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 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.
 
 \begin{displaymath}
 	\xymatrix{
-	S^n \ar[r]^{\phi} \arcof[d] & S^n(k) \vee S^n \ar[d] \\
+	S^n \ar[r]^{\phi} \arcof[d] & S^n(r) \vee S^n \ar[d] \\
-	D^{n+1} \ar[r] & S^n(k+1)
+	D^{n+1} \ar[r] & S^n(r+1)
 	}
 \end{displaymath}
 
-Note that $S^n(k+1)$ is homotopy equivalent to $S^n$. More importantly if we identify $\pi_n(S^n(k)) \iso \Z$ and $\pi_n(S^n(k+1)) \iso \Z$, then the inclusions $S^n(k) \subset S^n(k+1)$ induces multiplication by $k+1$ on the homotopy groups.
+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 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:
+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:
 
-$$ \Z \tot{\cdots 2} \Z \tot{\cdots 3} \Z \tot{\cdots 4} \Z \tot{\cdots 5} \cdots \Q. $$
+$$ \Z \tot{\times 2} \Z \tot{\times 3} \Z \tot{\times 4} \Z \tot{\times 5} \cdots \Q. $$
 
-Moreover we note that the generator $1 \in \pi_n(S^n)$ is sent to $1 \in \pi_n(S^n_\Q$. However the other homotopy groups are harder to calculate as we have generally no idea how the induced maps will look like. But in the case of $n=1$, the other trivial homotopy groups of $S^1$ are trivial.
+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 how the induced maps will look like. But in the case of $n=1$, the other trivial homotopy groups of $S^1$ are trivial.
 
 \Corollary{rationalization-S1}{
 	The inclusion $S^1 \to S^1_\Q$ is a rationalization.
@@ -43,7 +46,14 @@ By the Serre-Hurewicz theorem (\TheoremRef{serre-hurewicz}, with $\C$ the class
 	The inclusion $S^n \to S^n_\Q$ is a rationalization.
 }
 
-The \Def{rational disk} is now defined as cone of the rational sphere: $D^{n+1}_\Q = CS^n_\Q$. This gives an inclusions of pairs of spaces $(D^{n+1}, S^n) \subset (D^{n+1}_\Q, S^n_\Q)$.
+The \Def{rational disk} is now defined as cone of the rational sphere: $D^{n+1}_\Q = CS^n_\Q$. By the natruality of the cone construction we get the following commutative diagram of inclusions.
+
+\begin{displaymath}
+	\xymatrix{
+	S^{n} \arcof[r] \arcof[d] & S^{n}_{\Q} \arcof[d] \\
+	D^{n+1} \arcof[r] & D^{n+1}_{\Q}
+	}
+\end{displaymath}
 
 \Lemma{SnQ-extension}{
 	Let $X$ be a rational space and $f : S^n \to X$ be a map. Then this map extends to a map $f' : S^n_\Q \to X$ making the following diagram commute.
@@ -53,21 +63,23 @@ The \Def{rational disk} is now defined as cone of the rational sphere: $D^{n+1}_
 		                       & X
 		}
 	\end{displaymath}
+	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 are 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 shown 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 extend on the $n+1$-cells. This defines $f'$. 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 \todo{finish proof}
 }
-\todo{Add: unique up to homotopy. A homotopy extends to a homotopy.}
 
 \section{Rationalizations of arbitrary spaces}
 Having rational cells we wish to replace the cells in a CW complex $X$ by the rational cells to obtain a rationalization.
 
 \Lemma{rationalization-CW}{
-	Any CW complex admits a rationalization.
+	Any simply connected CW complex admits a rationalization.
 }
 \Proof{
 	Let $X$ be a CW complex. We will define $X_\Q$ with induction on the dimension of the cells. Since $X$ is simply connected we can start with $X^0_\Q = X^1_\Q = \ast$. Now assume that the rationalization $X^k \tot{\phi^k} X^k_\Q$ is already defined. Let $A$ be the set of $k+1$-cells and $f_\alpha : S^k \to X^{k+1}$ be the attaching maps. Then by \LemmaRef{SnQ-extension} these extend to $g_\alpha = (\phi^k \circ f_\alpha)' : S^k_\Q \to X^k_\Q$. This defines $X^{k+1}_\Q$ as the pushout in the following diagram.
@@ -82,7 +94,15 @@ Having rational cells we wish to replace the cells in a CW complex $X$ by the ra
 	Now by the universal property of $X^{k+1}$, we get a map $\phi^{k+1} : X^{k+1} \to X^{k+1}_\Q$ which is compatible with $\phi^k$ and which is a rationalization.
 }
 
-\todo{For arbitrary spaces}
+\Lemma{rationalization}{
+	Any simply connected space admits a rationalization.
+}
+\Proof{
+	Let $Y \tot{f} X$ be a CW approximation and let $Y \tot{\phi} Y_\Q$ be the rationalization of $Y$. Now we define $X_\Q$ as the double mapping cylinder (or homotopy pushout):
+	$$ X_\Q = X \cup_f (Y \times I) \cup_{\phi} Y_\Q. $$
+
+	\todo{bewijs afmaken met excision?}
+}
 
 \Theorem{}{
 	The above construction is in fact a \Def{localization}, i.e. for any map $f : X \to Z$ to a rational space $Z$, there is an extension $f' : X_\Q \to Z$ making the following diagram commute.
@@ -94,7 +114,7 @@ Having rational cells we wish to replace the cells in a CW complex $X$ by the ra
 		}
 	\end{displaymath}
 
-	Moreover, any $f''$ making the diagram commute is homotopic to $f'$ and if $g : X \to Z$ is homotopic to $f$ then the extension $g'$ is homotopic to $f'$.
+	Moreover, $f'$ is determined up to homotopy and homotopic maps have homotopic extensions.
 }
 
 The extension property allows us to define a rationalization of maps. Given $f : X \to Y$, we can consider the composite $if : X \to Y \to Y_\Q$. Now this extends to $(if)' : X_\Q \to Y_\Q$. Note that this construction is not functorial, since there are choices of homotopies involved. When passing to the homotopy category, however, this construction \emph{is} functorial and has an universal property.