The generators $e$ and $f$ in the last proof are related by the so callend \Def{Whitehead product}. The whitehead product is a bilinear map $\pi_p(X)\times\pi_q(X)\to\pi_{p+q-1}(X)$ satisfying a graded commutativity relation and a graded Jacobi relation, see \cite{felix}. If we define a \Def{Whitehead algebra} to be a graded vector space with such a map satisfying these relations, we can summarize the above two propositions as follows \cite{berglund}.
The generators $e$ and $f$ in the last proof are related by the so called \Def{Whitehead product}. The whitehead product is a bilinear map $\pi_p(X)\times\pi_q(X)\to\pi_{p+q-1}(X)$ satisfying a graded commutativity relation and a graded Jacobi relation, see \cite{felix}. If we define a \Def{Whitehead algebra} to be a graded vector space with such a map satisfying these relations, we can summarize the above two propositions as follows \cite{berglund}.
\Corollary{}{
The rational homotopy groups of $S^n$ are given by
Recall the following facts about cdga's over a ring $\k$:
\begin{itemize}
\item A map $f: A \to B$ in $\CDGA_\k$ is a \emph{quasi isomorphism} if it induces isomorphisms in cohomology.
\item The finite coproduct in $\CDGA_\k$ is the (graded) tensor products.
\item The finite coproduct in $\CDGA_\k$ is the (graded) tensor product.
\item The finite product in $\CDGA_\k$ is the cartesian product (with pointwise operations).
\item The equalizer (resp. coequalizer) of $f$ and $g$ is given by the kernel (resp. cokernel) of $f - g$. Together with the (co)products this defines pullbacks and pushouts.
\item$\k$ and $0$ are the initial and final object.
We assume the reader is familiar with category theory, basics from algebraic topology and the basics of simplicial sets. Some knowledge about differential graded algebra (or homological algebra) and model categories is assumed, but the reader may review this in the appendices.
\begin{itemize}
\item$\k$ will denote an arbitrary commutative ring (or field, if indicated at the start of a section). Modules, tensor products, \dots are understood as $\k$-modules, tensor products over $\k$, \dots. If ambiguitity can occur notation will be explicit.
\item$\k$ will denote an arbitrary commutative ring (or field, if indicated at the start of a section). Modules, tensor products, \dots are understood as $\k$-modules, tensor products over $\k$, \dots. If ambiguity can occur notation will be explicit.
\item$\cat{C}$ will denote an arbitrary category.
\item$\cat{0}$ (resp. $\cat{1}$) will denote the initial (resp. final) objects in a category $\cat{C}$.
\item$\Hom_\cat{C}(A, B)$ will denote the set of maps from $A$ to $B$ in the category $\cat{C}$. The subscript $\cat{C}$ is occasionally left out if the category is clear from the context.
In this section we will state the aim of rational homotopy theory. Moreover we will recall classical theorems from algebraic topology and deduce rational versions of them.
In the following definition \emph{space} is to be understood as a topological space or a simplicial set. We will restrict to simply connected spaces.
In the following definition \emph{space} is to be understood as a topological space or a simplicial set. We will restrict ourselves to simply connected spaces.\todo{Per definitie/stelling samenhangendheid aangeven}
\Definition{rational-space}{
A space $X$ is a \emph{rational space} if
@ -28,7 +28,7 @@ Note that for a rational space $X$, the homotopy groups are isomorphic to the ra
Note that a weak equivalence (and hence also a homotopy equivalence) is always a rational homotopy theory. Furthermore if $f: X \to Y$ is a map between rational spaces, then $f$ is a rational homotopy equivalence if and only if $f$ is a weak equivalence.
We will later see that any space admits a rationalization. The theory of rational homotopy theory is then the study of the homotopy category $\Ho_\Q(\Top)\iso\Ho(\Top_\Q)$, which is on its own turn equivalent to $\Ho(\sSet_\Q)\iso\Ho_\Q(\sSet)$.
We will later see that any space admits a rationalization. The theory of rational homotopy theory is then the study of the homotopy category $\Ho_\Q(\Top)\iso\Ho(\Top_\Q)$, which is on its own turn equivalent to $\Ho(\sSet_\Q)\iso\Ho_\Q(\sSet)$.\todo{Notatie}
\section{Classical results from algebraic topology}
@ -36,7 +36,7 @@ We will now recall known results from algebraic topology, without proof. One can
\Theorem{relative-hurewicz}{
(Relative Hurewicz) For any inclusion of spaces $A \subset X$ and all $i > 0$, there is a natural map
$$ h_i : \pi_i(X, A)\to H_i(X, A). $$
$$ h_i : \pi_i(X, A)\to H_i(X, A). $$\todo{Andere letter dan $A$}
If furthermore $(X,A)$ is $n$-connected ($n > 0$), then the map $h_i$ is an isomorphism for all $i \leq n +1$.
}
@ -84,7 +84,7 @@ The long exact sequence for a Serre fibration also has a direct consequence for
\Corollary{rational-les}{
Let $f: X \to Y$ be a Serre fibration of $1$-connected spaces, then there is a natural long exact sequence of rational homotopy groups:
We will first define some basic cochain complexes which model the $n$-disk and $n$-sphere. $D(n)$ is the cochain complex generated by one element $b \in D(n)^n$ and its differential $c = d(b)\in D(n)^{n+1}$. $S(n)$ is the cochain complex generated by one element $a \in S(n)^n$ which differential vanishes (i.e. $d a =0$). In other words:
We will first define some basic cochain complexes which model the $n$-disk and $n$-sphere. $D(n)$ is the cochain complex generated by one element $b \in D(n)^n$ and its differential $c = d(b)\in D(n)^{n+1}$. \todo{Herschrijf}$S(n)$ is the cochain complex generated by one element $a \in S(n)^n$ with trivial differential (i.e. $d a =0$). In other words:
$$ D(n)= ... \to0\to\k\to\k\to0\to ... $$
$$ S(n)= ... \to0\to\k\to0\to0\to ... $$
Note that $D(n)$ is acyclic for all $n$, or put in different words: $j_n : 0\to D(n)$ induces an isomorphism in cohomology. The sphere $S(n)$ has exactly one non-trivial cohomology group $H^n(S(n))=\k\cdot[a]$. There is an injective function $i_n : S(n+1)\to D(n)$, sending $a$ to $c$. The maps $j_n$ and $i_n$ play the following important role in the model structure of cochain complexes:
\todo{Introduceer de model structuur}
\begin{claim}
The set $I =\{i_n : S(n+1)\to D(n)\I n \in\N\}$ generates all cofibrations and the set $J =\{j_n : 0\to D(n)\I n \in\N\}$ generates all trivial cofibrations.
\end{claim}
As we do not directly need this claim, we omit the proof. However, in the next section we will prove a similar result for cdga's in detail.
$S(n)$ plays a another special role: maps from $S(n)$ to some cochain complex $X$ correspond directly to elements in the kernel of $\restr{d}{X^n}$. Any such map is null-homotopic precisely when the corresponding elements in the kernel is a coboundary. So there is a natural isomorphism: $\Hom(S(n), X)/\simeq\iso H^n(X)$.
$S(n)$ plays a another special role: maps from $S(n)$ to some cochain complex $X$ correspond directly to elements in the kernel of $\restr{d}{X^n}$. Any such map is null-homotopic precisely when the corresponding elements in the kernel is a coboundary. So there is a natural isomorphism: $\Hom(S(n), X)/{\simeq}\iso H^n(X)$.
By using the free cdga functor we can turn these cochain complexes into cdga's $\Lambda(D(n))$ and $\Lambda(S(n))$. So $\Lambda(D(n))$ consists of linear combinations of $b^n$ and $c b^n$ when $n$ is even, and it consists of linear combinations of $c^n b$ and $c^n$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule:
By using the free cdga functor we can turn these cochain complexes into cdga's $\Lambda(D(n))$ and $\Lambda(S(n))$. So $\Lambda(D(n))$ consists of linear combinations of $b^n$\todo{gebruik niet weer $n$}and $c b^n$ when $n$ is even, and it consists of linear combinations of $c^n b$ and $c^n$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule:
$$ d(b^n)= n \cdot c b^{n-1}$$
$$ d(c b^n)=0$$
@ -27,7 +28,7 @@ $$ c^n = d(b c^{n-1}) $$
There are no additional cocycles in $\Lambda(D(n))$ besides the constants and $c$. So we conclude that $\Lambda(D(n))$ is acyclic as an algebra. In other words $\Lambda(j_n): \k\to\Lambda D(n)$ is a quasi isomorphism.
The situation for $\Lambda S(n)$ is easier: when $n$ is even it is given by polynomials in $a$, if $n$ is odd it is an exterior algebra (i.e. $a^2=0$). Again the sets $\Lambda(I)=\{\Lambda(i_n) : \Lambda S(n+1)\to\Lambda D(n)\I n \in\N\}$ and $\Lambda(J)=\{\Lambda(j_n) : \k\to\Lambda D(n)\I n \in\N\}$ play an important role.
The situation for $\Lambda S(n)$ is easier: when $n$ is even it is given by polynomials in $a$, if $n$ is odd it is an exterior algebra \todo{?}(i.e. $a^2=0$). Again the sets $\Lambda(I)=\{\Lambda(i_n) : \Lambda S(n+1)\to\Lambda D(n)\I n \in\N\}$ and $\Lambda(J)=\{\Lambda(j_n) : \k\to\Lambda D(n)\I n \in\N\}$ play an important role.
\begin{theorem}
The sets $\Lambda(I)$ and $\Lambda(J)$ generate a model structure on $\CDGA_\k$ where:
@ -12,7 +12,7 @@ this extends linearly and multiplicatively. Note that it follows that we have $d
such that $d_0 h = g$ and $d_1 h = f$.
}
In terms of model categories, such a homotopy is a right homotopy and the object $\Lambda(t, dt)\tensor X$ is a path object for $X$. We can easily see that it is a very good path object. First note that $\Lambda(t, dt)\tensor X \tot{(d_0, d_1)} X \oplus X$ is surjective (for $(x, y)\in X \oplus X$ take $t \tensor x +(1-t)\tensor y$). Secondly we note that $\Lambda(t, dt)=\Lambda(D(0))$ and hence $\k\to\Lambda(t, dt)$ is a cofibration, by \LemmaRef{model-cats-coproducts} we have that $X \to\Lambda(t, dt)\tensor X$ is a cofibration.
In terms of model categories, such a homotopy is a right homotopy and the object $\Lambda(t, dt)\tensor X$ is a path object for $X$. We can easily see that it is a very good path object. First note that $\Lambda(t, dt)\tensor X \tot{(d_0, d_1)} X \oplus X$ is surjective (for $(x, y)\in X \oplus X$ take $t \tensor x +(1-t)\tensor y$). Secondly we note that $\Lambda(t, dt)=\Lambda(D(0))$ and hence $\k\to\Lambda(t, dt)$ is a cofibration, by \LemmaRef{model-cats-coproducts} we have that $X \to\Lambda(t, dt)\tensor X$ is a (necessarily trivial) cofibration.
Clearly we have that $f \simeq g$ implies $f \simeq^r g$ (see \DefinitionRef{right_homotopy}), however the converse need not be true.
@ -40,6 +40,7 @@ The results from model categories immediately imply the following results. \todo
\item Let $A$ and $X$ both be cofibrant, then $f: A \we X$ is a weak equivalence if and only if $f$ is a strong homotopy equivalence. Moreover, the two induced maps are bijections:
$$ f_\ast: [Z, A]\tot{\iso}[Z, X], $$
$$ f^\ast: [X, Z]\tot{\iso}[A, X]. $$
\todo{De eerste werkt ook als $i$ gewoon een w.e. is. (Gebruik factorizatie.)}
@ -41,6 +41,7 @@ Next we will prove the factorization property [MC5]. We will do this by Quillen'
Define the following objects and sets of maps:
\begin{itemize}
\item$S(n)$ is the CDGA generated by one element $a$ of degree $n$ such that $da =0$.
\todo{Andere letters, of $\Lambda$}
\item$T(n)$ is the CDGA generated by two element $b$ and $c$ of degree $n$ and $n+1$ respectively, such that $db = c$ (and necessarily $dc =0$).
\item$I =\{ i_n: \k\to T(n)\I n \in\N\}$ is the set of units of $T(n)$.
\item$J =\{ j_n: S(n+1)\to T(n)\I n \in\N\}$ is the set of inclusions $j_n$ defined by $j_n(a)= b$.
@ -67,7 +68,7 @@ As a consequence of the above two lemmas, the class generated by $I$ is containe
If $p: X \to Y$ has the RLP w.r.t. $I$ then $p$ is a fibration.
\end{lemma}
\begin{proof}
Let $y \in Y^n$ an element of degree $n$, then we have the following commuting diagram:
Let $y \in Y^n$be an element of degree $n$, then we have the following commuting diagram:
\cdiagram{CDGA_Model_I_Fib}
where $g$ sends the generator $b$ to $y$ and $c$ to $dy$. By assumption there exists a lift $h$. Now $h(b)\in X^n$ is a preimage for $y$, proving that $p$ is surjective.
@ -58,13 +58,13 @@ In this section we will prove that the singular cochain complex is quasi isomorp
Let $v \in\Apl_n^n$, then we can always write it as $v = p(x_1, \dots, x_n)dx_1\dots dx_n$ where $p$ is a polynomial in $n$ variables. If $\Q\subset\k\subset\mathbb{C}$ we can integrate geometrically on the $n$-simplex:
which defines a well-defined linear map $\int_n : \Apl_n^n \to\k$. For general fields of characteristic zero we can define it formally on the generators of $\Apl_n^n$ (as vector space):
Let $x$ be a $k$-simplex of $\Delta[n]$, i.e. $x: \Delta[k]\to\Delta[n]$. Then $x$ induces a linear map $x^\ast: \Apl_n \to\Apl_k$. Let $v \in\Apl_n^k$, then $x^\ast(v)\in\Apl_k^k$, which we can integrate. Now define
Note that $\oint_n(v): \Delta[n]\to\k$ is just a map, we can extend this linearly to chains on $\Delta[n]$ to obtain $\oint_n(v): \Z\Delta[n]\to\k$, in other words $\oint_n(v)\in C_n$. By linearity of $\int_n$ and $x^\ast$, we have a linear map $\oint_n: \Apl_n \to C_n$.
Next we will show that $\oint=\{\oint_n\}_n$ is a simplicial map and that each $\oint_n$ is a chain map, in other words $\oint : \Apl\to C_n$ is a simplicial chain map (of complexes). Let $\sigma: \Delta[n]\to\Delta[k]$, and $\sigma^\ast: \Apl_k \to\Apl_n$ its induced map. We need to prove $\oint_n \circ\sigma^\ast=\sigma^\ast\circ\oint_k$. We show this as follow:
Next we will show that $\oint=\{\oint_n\}_n$ is a simplicial map and that each $\oint_n$ is a chain map, in other words $\oint : \Apl\to C_n$ is a simplicial chain map (of complexes). Let $\sigma: \Delta[n]\to\Delta[k]$, and $\sigma^\ast: \Apl_k \to\Apl_n$ its induced map. We need to prove $\oint_n \circ\sigma^\ast=\sigma^\ast\circ\oint_k$. We show this as follows:
@ -90,6 +90,7 @@ We will now prove that the map $\oint: A(X) \to C^\ast(X)$ is a quasi isomorphis
The induced map $\oint: A(X)\to C^\ast(X)$ is a natural quasi isomorphism.
}
\Proof{
\todo{Diagrammen typesetten}
Assume we have a simplicial set $X$ such that $\oint: A(X)\to C^\ast(X)$ is a quasi isomorphism. We can add a simplex by considering pushouts of the following form:
@ -12,13 +12,13 @@ In this section we will prove the Whitehead and Hurewicz theorems in a rational
\end{itemize}
}
Serre gave weaker axioms for his classes and proves some of the following lemmas only using these weaker axioms. However the classes we are interested in do satisfy the above (stronger) requirements. One should think of such Serre class as a class of groups we want to \emph{ignore}. We will be interested in the first two of the following examples.
Serre gave weaker axioms for his classes and proves some of the following lemmas only using these weaker axioms. However the classes we are interested in do satisfy the above (stronger) requirements. One should think of a Serre class as a class of groups we want to \emph{ignore}. We will be interested in the first two of the following examples.
\Example{serre-classes}{
We give three Serre classes without proof.
\begin{itemize}
\item The class $\C=\{0\}$. With this class the following Hurewicz and Whitehead theorem will simply be the classical statements.
\item The class $\C$ of all torsion group. Using this class we can prove the rational version of the Hurewicz and Whitehead theorems.
\item The class $\C$ of all torsion groups. Using this class we can prove the rational version of the Hurewicz and Whitehead theorems.
\item Let $P$ be a set of primes, then define a class $\C$ of torsion groups for which all $p$-subgroups are trivial for all $p \in P$. This can be used to \emph{localize} at $P$.
\end{itemize}
}
@ -51,7 +51,7 @@ In the following arguments we will consider fibrations and need to compute homol
\end{itemize}
}
\Proof{
We will assume $B$ is a CW complex and prove this by induction on its skeleton $B^k$. By connectedness we can assume $B^0=\{ b_0\}$. Restrict $E$ to $E^k$ and note $E^0= F$. Now the base case is clear: $H_i(E^0, F)\to H_i(B^0, b_0)$ is a $\C$-iso.
We will assume $B$ is a CW complex and prove this by induction on its skeleton $B^k$. By connectedness we can assume $B^0=\{ b_0\}$. Restrict $E$ to $B^k$ and note $E^0= F$. Now the base case is clear: $H_i(E^0, F)\to H_i(B^0, b_0)$ is a $\C$-iso.
For the induction step, consider the long exact sequence in homology for the triples $(E^{k+1}, E^k, F)$ and $(B^{k+1}, B^k, b_0)$:
@ -76,7 +76,7 @@ In the following arguments we will consider fibrations and need to compute homol
For the main theorem we need the following construction. \todo{Geef de constructie}
\Lemma{whitehead-tower}{
(Whitehead tower)
We can decompose a space $X$ into fibrations:
We can decompose a $0$-connected space $X$ into fibrations:
@ -92,8 +92,7 @@ For the main theorem we need the following construction. \todo{Geef de construct
If $\pi_i(X)\in C$ for all $i<n$, then $H_i(X)\in C$ for all $i<n$ and the Hurewicz map $h: \pi_i(X)\to H_i(X)$ is a $\C$-isomorphism for all $i \leq n$.
}
\Proof{
We will prove the lemma by induction on $n$. Note that the base case follows from the $1$-connectedness.
For the induction step assume that $H_i(X)\in\C$ for all $i<n-1$ and that $h_{n-1}: \pi_{n-1}(X)\to H_{n-1}(X)$ is a $\C$-iso. Now given is that $\pi_{n-1}(X)\in\C$ and hence $H_{n-1}(X)\in\C$.
We will prove the lemma by induction on $n$. Note that the base case ($n =1$) follows from the $1$-connectedness. For the induction step assume that $H_i(X)\in\C$ for all $i<n-1$ and that $h_{n-1}: \pi_{n-1}(X)\to H_{n-1}(X)$ is a $\C$-iso. Now given is that $\pi_{n-1}(X)\in\C$ and hence $H_{n-1}(X)\in\C$. \todo{kromme zin}
It remains to show that $h_n$ is a $\C$-iso. Use the Whitehead tower from \LemmaRef{whitehead-tower} to obtain $\cdots\fib X(3)\fib X(2)= X$. Note that each $X(j)$ is also $1$-connected and that $X(2)= X(1)= X$.
Joshua Moerman
10 years ago