Master thesis on Rational Homotopy Theory
https://github.com/Jaxan/Rational-Homotopy-Theory
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141 lines
8.0 KiB
141 lines
8.0 KiB
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\chapter{Rationalizations}
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\label{sec:rationalizations}
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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.
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\section{Rationalization of \texorpdfstring{$S^n$}{Sn}}
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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.
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\todo{plaatje}
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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 $\pi_n(S^n) = \pi_n(S^n(r)) = \Z$.
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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.
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\begin{displaymath}
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\xymatrix{
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S^n \ar[r]^{\phi} \arcof[d] & S^n(r) \vee S^n \ar[d] \\
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D^{n+1} \ar[r] & S^n(r+1)
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}
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\end{displaymath}
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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}.
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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:
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$$ \Z \tot{\times 2} \Z \tot{\times 3} \Z \tot{\times 4} \Z \tot{\times 5} \cdots \Q. $$
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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.
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\Corollary{rationalization-S1}{
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The inclusion $S^1 \to S^1_\Q$ is a rationalization.
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}
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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. $0$. For $i = n$ we can use the same sequence as above (or use the Hurewicz theorem) to conclude:
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$$ H_i(S^n_\Q) = \begin{cases}
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\Z, &\text{ if } i = 0 \\
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\Q, &\text{ if } i = n \\
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0, &\text{ otherwise.}
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\end{cases} $$
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By the Serre-Hurewicz theorem (\TheoremRef{serre-hurewicz}, with $\C$ the class of uniquely divisible groups) we see that $S^n_\Q$ is indeed rational. Then by the Serre-Whitehead theorem (\TheoremRef{serre-whitehead}, with $\C$ the class of torsion groups) the inclusion map $S^n \to S^n_\Q$ is a rationalization.
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\Corollary{rationalization-Sn}{
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The inclusion $S^n \to S^n_\Q$ is a rationalization.
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}
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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.
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\begin{displaymath}
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\xymatrix{
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S^{n} \arcof[r] \arcof[d] & S^{n}_{\Q} \arcof[d] \\
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D^{n+1} \arcof[r] & D^{n+1}_{\Q}
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}
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\end{displaymath}
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\Lemma{SnQ-extension}{
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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.
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\begin{displaymath}
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\xymatrix{
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S^n \ar[r]^i \ar[rd]^f & S^n_\Q \ar@{-->}[d]^{f'} \\
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& X
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}
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\end{displaymath}
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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 $f' \simeq g'$).
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}
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\Proof{
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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.
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\cimage[scale=0.6]{SnQ_Extension}
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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 $f = f' \circ i$.
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Let $f''$ be any map such that $f''i = f$. Then \todo{finish proof}
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}
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\section{Rationalizations of arbitrary spaces}
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Having rational cells we wish to replace the cells in a CW complex $X$ by the rational cells to obtain a rationalization.
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\Lemma{rationalization-CW}{
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Any simply connected CW complex admits a rationalization.
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}
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\Proof{
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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.
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\begin{displaymath}
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\xymatrix{
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\coprod_A S^n_\Q \ar[r]^{(g_\alpha)} \arcof[d] \xypo & X^k_\Q \ar@{-->}[d] \\
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\coprod_A D^{n+1}_\Q \ar@{-->}[r] & X^{k+1}_\Q
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}
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\end{displaymath}
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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.
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}
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\Lemma{rationalization}{
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Any simply connected space admits a rationalization.
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}
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\Proof{
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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):
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$$ X_\Q = X \cup_f (Y \times I) \cup_{\phi} Y_\Q. $$
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\todo{bewijs afmaken met excision?}
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}
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\Theorem{}{
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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.
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\begin{displaymath}
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\xymatrix{
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X \ar[r]^i \ar[rd]^f & X_\Q \ar@{-->}[d]^{f'} \\
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& Z
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}
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\end{displaymath}
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Moreover, $f'$ is determined up to homotopy and homotopic maps have homotopic extensions.
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}
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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.
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We already mentioned in the first section that for rational spaces the notions of weak equivalence and rational equivalence coincide. Now that we always have a rationalization we have:
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\Corollary{}{
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Let $f: X \to Y$ be a map, then $f$ is a rational equivalence if and only if $f_\Q : X_\Q \to Y_\Q$ is a weak equivalence.
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}
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\Corollary{}{
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The homotopy category of $1$-connected rational spaces is equivalent to the rational homotopy category of $1$-connected spaces.
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}
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\section{Other constructions}
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There are others ways to obtain a rationalization. One of them relies on the observations that it is easy to rationalize Eilenberg-MacLane spaces.
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\Lemma{rationalization-em-space}{
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Let $A$ be an abelian group and $n \geq 1$. Then
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$$ K(A, n) \to K(A \tensor \Q, n) $$
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is a rationalization
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}
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Postnikov
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