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}}
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 $S^n$.
\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 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.}
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.
\todo{plaatje}
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.
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:
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.
Joshua Moerman
10 years ago