\chapter[A and K form a Quillen pair]{$A$ and $K$ form a Quillen pair}
\label{sec:a-k-quillen-pair}
\chapter{The main equivalence}
In this section we aim to prove that the homotopy theory of rational spaces is the same as the homootopy theory of cdga's over $\Q$. Before we prove the equivalence, we will show that $A$ and $K$ form a Quillen pair. This already provides an adjunction between the homotopy categories. Besides the equivalence of the homotopy categories we will also investigate homotopy groups on a cdga directly. The homotopy groups of a space will be dual to the homotopy groups of the associated cdga.
We will prove that $A$ preserves cofibrations and trivial cofibrations. We only have to check this fact for the generating (trivial) cofibrations in $\sSet$. Note that the contravariance of $A$ means that a (trivial) cofibrations should be sent to a (trivial) fibration.
@ -9,17 +10,25 @@ We will prove that $A$ preserves cofibrations and trivial cofibrations. We only
\end{lemma}
\begin{proof}
Let $\phi\in A(\del\Delta[n])$ be an element of degree $k$, hence it is a map $\del\Delta[n]\to\Apl^k$. We want to extend this to the whole simplex. By the fact that $\Apl^k$ is Kan and contractible we can find a lift $\overline{\phi}$ in the following diagram showing the surjectivity.
\cimage[scale=0.5]{Extend_Boundary_Form}
\begin{displaymath}
\xymatrix{
\del\Delta[n]\ar[r]^\phi\arcof[d]^i &\Apl^k \\
\Delta[n]\ar@{-->}[ru]_{\overline{\phi}}
}
\end{displaymath}
\end{proof}
\begin{lemma}
$A(j) : A(\Delta[n])\to A(\Lambda^n_k)$ is surjective and a quasi isomorphism.
$A(j) : A(\Delta[n])\to A(\Lambda^k_n)$ is surjective and a quasi isomorphism.
\end{lemma}
\begin{proof}
As above we get surjectivity from the Kan condition. To prove that $A(j)$ is a quasi isomorphism we pass to the singular cochain complex and use that $C^\ast(j) : C^\ast(\Delta[n])\we C^\ast(\Lambda^n_k)$ is a quasi isomorphism. Consider the following diagram and conclude that $A(j)$ is surjective and a quasi isomorphism.
Since $A$ is a left adjoint, it preserves all colimits and by functoriality it preserves retracts. From this we can conclude the following corollary.
@ -42,7 +51,7 @@ The induced adjunction in the previous corollary is given by $LA(X) = A(X)$ for
}
\section{Homotopy groups of \texorpdfstring{$K(A)$}{K(A)}}
\section{Homotopy groups of cdga's}
We are after an equivalence of homotopy categories, so it is natural to ask what the homotopy groups of $K(A)$ are for a cdga $A$. In order to do so, we will define homotopy groups of cdga's directly and compare the two notions.
Recall that an augmented cdga is a cdga $A$ with an algebra map $A \tot{\counit}\k$ such that $\counit\unit=\id$.
@ -60,6 +69,7 @@ Note that for a free cdga $\Lambda C$ there is a natural augmentation and the ch
Let $A$ be an augmented cdga, then
$$[A, V(n)]\tot{\iso}\Hom_\k(\pi^n(A), \k). $$
}
\todo{prove}
We will denote the dual of a vector space as $V^\ast=\Hom_\k(V, \k)$.
@ -24,10 +24,10 @@ In this section we will discuss the so called minimal models. These are cdga's w
$$(M, d)\we(A, d). $$
\end{definition}
The requirement that there exists a filtration can be replaced by a stronger statement.
In the following lemma we see that the filtration is sometimes naturally there for $1$-reduced cdga's.
\begin{lemma}
Let $(A, d)$ be a cdga which is $1$-reduced, quasi-free and with a decomposable differential. Then $(A, d)$ is a minimal algebra.
Let $(A, d)$ be a cdga which is $1$-reduced, such that $A$ is free as cga and $d$ is decomposable. Then $(A, d)$ is a minimal algebra.
\end{lemma}
\begin{proof}
Let $V$ generate $A$. Take $V(n)=\bigoplus_{k=0}^n V^k$ (note that $V^0= V^1=0$). Since $d$ is decomposable we see that for $v \in V^n$: $d(v)= x \cdot y$ for some $x, y \in A$. Assuming $dv$ to be non-zero we can compute the degrees:
@ -35,6 +35,18 @@ The requirement that there exists a filtration can be replaced by a stronger sta
As $A$ is $1$-reduced we have $\deg{x}, \deg{y}\geq2$ and so by the above $\deg{x}, \deg{y}\leq n-1$. Conclude that $d(V(k))\subset\Lambda(V(n-1))$.
\end{proof}
The above definition is the same as in \cite{felix} without assuming connectivity. We find some different definitions of (minimal) Sullivan algebras in the literature. For example we find a definition using well orderings in \cite{hess}. The decomposability of $d$ also admits a different characterization (at least in the connected case). The equivalence of the definitions is expressed in the following two lemmas.
\Lemma{}{
A cdga $(\Lambda V, d)$ is a Sullivan algebra if and only if there exists a well order $J$ such that $V$ is generated by $v_j$ for $j \in J$ and $d v_j \in\Lambda V_{<j}$.
}
\Lemma{}{
Let $(\Lambda V, d)$ be a Sullivan algebra with $V^0=0$, then $d$ is decomposable if and only if there is a well order $J$ as above such that $i < j$ implies $\deg{v_i}\leq\deg{v_j}$.
}
It is clear that induction will be an important technique when proving things about (minimal) Sullivan algebras. We will first prove that minimal models always exist for $1$-connected cdga's and afterwards prove uniqueness.
\section{Existence}
@ -58,15 +70,19 @@ The requirement that there exists a filtration can be replaced by a stronger sta
\section{Uniqueness}
Before we state the uniqueness theorem we need some more properties of minimal models.
Before we state the uniqueness theorem we need some more properties of minimal models. In this section we will use some general facts about model categories.
\begin{lemma}
Sullivan algebras are cofibrant and the inclusions $(\Lambda V(k), d)\to(\Lambda V(k+1), d)$ are cofibrations.
\end{lemma}
\begin{proof}
Consider the following lifting problem, where $\Lambda V$ is a Sullivan algebra.
\cimage[scale=0.5]{Sullivan_Lifting}
\begin{displaymath}
\xymatrix{
\k\ar[r]^\unit\ar[d]^\unit& X \artfib[d]^p \\
\Lambda V \ar[r]^g & Y
}
\end{displaymath}
By the left adjointness of $\Lambda$ we only have to specify a map $\phi: V \to X$ such that $p \circ\phi= g$. We will do this by induction. Note that the induction step proves precisely that $(\Lambda V(k), d)\to(\Lambda V(k+1), d)$ is a cofibrations.
@ -98,13 +98,10 @@ 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:
\cdiagram{Apl_C_Quasi_Iso_Pushout}
We can apply our two functors to it, and use the natural transformation $\oint$ to obtain the following cube:
\cdiagram{Apl_C_Quasi_Iso_Cube}
Note that $A(\Delta[n])\we C^\ast(\Delta[n])$ by \CorollaryRef{apl-c-quasi-iso}, $A(X)\we C^\ast(X)$ by assumption and $A(\del\Delta[n])\we C^\ast(\del\Delta[n])$ by induction. Secondly note that both $A$ and $C^\ast$ send injective maps to surjective maps, so we get fibrations on the right side of the diagram. Finally note that the front square and back square are pullbacks, by adjointness of $A$ and $C^\ast$. Apply the cube lemma (\LemmaRef{cube-lemma}, \cite[Lemma 5.2.6]{hovey}) to conclude that also $A(X')\we C^\ast(X')$.
@ -114,25 +111,41 @@ We will now prove that the map $\oint: A(X) \to C^\ast(X)$ is a quasi isomorphis
This means that we can extend the previous argument to pushout of this form:
\cimage[scale=0.5]{Apl_C_Quasi_Iso_Pushout2}
\begin{displaymath}
\xymatrix{
\coprod_{\alpha\in A}\del\Delta[n]\arcof[d]\ar[r]\xypo& X \ar[d]\\
\coprod_{\alpha\in A}\Delta[n]\ar[r]& X'
}
\end{displaymath}
Finally we can write any simplicial set $X$ as a colimit of finite dimensional ones as:
$$ sk_0 X \cof sk_1 X \cof sk_2\cof\dots\colim sk_n X = X, $$
where $sk_i X$ has no non-degenerate simplices in dimensions $n > i$. By the above $\oint$ gives a quasi isomorphism on all the terms $sk_i X$. So we are in the following situation:
We will define long exact sequences for both sequences in the following way. Consider cochain algebras $B_i$ as follows:
$$ B =\lim_i B_i \dots\fib B_2\fib^{b_1} B_1\fib^{b_0} B_0. $$
We will define long exact sequences for both sequences in the following way. As the following construction is quite general, consider arbitrary cochain algebras $B_i$ as follows:
Define a map $t: \prod_i B_i \to\prod_i B_i$ defined by $t(x_0, x_1, \dots)=(x_0+ b_0(x_1), x_1+ b_1(x_2), \dots)$. Note that $t$ is surjective and that $B \iso\ker(t)$. So we get the following natural short exact sequence and its associated natural long exact sequence in homology:
$$0\to B \tot{i}\prod_i B_i \tot{t}\prod_i B_i \to0, $$
In our case we get two such long exact sequences with $\oint$ connecting them. As cohomology commutes with products we get isomorphisms on the left and right in the following diagram.
So by the five lemma we can conclude that the middle morphism is an isomorphism as well, proving $H^n(A(X))\tot{\iso} H^n(C^\ast(X))$ for all $n$. This proves the statement for all $X$.
So by the five lemma we can conclude that the middle morphism is an isomorphism as well, proving the isomorphism $H^n(A(X))\tot{\iso} H^n(C^\ast(X))$ for all $n$. This proves the statement for all $X$.
@ -82,7 +82,7 @@ Having rational cells we wish to replace the cells in a CW complex $X$ by the ra
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.
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 pullback in the following diagram.
\todo{Write down actual kunneth theorem with torsion (still works though)}
Note that this is the graded tensor product, and that the term $H_{i+1}(B^{k+1}, B^k)\tensor H_0(F)= H_{i+1}(B^{k+1}, B^k)$ and that this identification is compatible with the induced map $p_\ast : H_{i+1}(E^{k+1}, E^k)\to H_{i+1}(B^{k+1}, B^k)$ (hence the map is surjective). To prove that the map is a $\C$-iso, we need to prove that the kernel is in $\C$. The kernel is the sum of the following terms, with $1\leq l \leq i+1$:
$$ H_{i+1-l}(B^{k+1}, B^k)\tensor H_l(F). $$
Now we can use the assumption that $H_l(F)\in\C$ for $1\leq l < n$ and that for $B \in\C$ we have $A \tensor B \in\C$ for all $A$ (by \LemmaRef{Serre-properties}). This concludes that the kernel $H_{i+1-l}(B^{k+1}, B^k)\tensor H_l(F)$ is indeed in $\C$. And hence the induced map is a $\C$-iso for all
Joshua Moerman
10 years ago