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 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.
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:
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}.
}\]
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 $S^n_\Q=\colim_{r} S^n(r)$.
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 $r$.
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:
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 $i \neq0, n$ 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 $i \neq0, n$ 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 $f''$.
Joshua Moerman
10 years ago