@ -28,7 +28,7 @@ One can check that $\Apl \in \simplicial{\CDGA_\k}$. We will denote the subspace
$\Apl^k$ is contractible.
}
\Proof{
\todo{Note geometric interpretation}We will prove this by defining an extra degeneracy $s: \Apl_n \to\Apl_{n+1}$. Define for $i =1, \ldots, n$:
We will prove this by defining an extra degeneracy $s: \Apl_n \to\Apl_{n+1}$. In the more geometric context of topological $n$-simplices we would achieve this by dividing by $1-x_0$. However, since this algebra consists of polynomials only, this cannot be done. Instead, we will multiply everything by $(1-x_0)^2$, so that we can divide by $1-x_0$. Define for $i =1, \ldots, n$:
\begin{align*}
s(1) &= (1-x_0)^2 \\
s(x_i) &= (1-x_0) \cdot x_{i+1}
@ -53,7 +53,7 @@ One can check that $\Apl \in \simplicial{\CDGA_\k}$. We will denote the subspace
}
\Lemma{apl-kan-complex}{
$\Apl^k$ is a Kan complex.\todo{also without $k$?}
$\Apl^k$ is a Kan complex.
}
\Proof{
By the simple fact that $\Apl^k$ is a simplicial group, it is a Kan complex \cite{goerss}.
@ -136,5 +136,3 @@ This allows us to directly relate the rational homotopy groups to the cohomology
}
In fact we have that $S^n_\Q$ is an H-space if and only if $n$ is odd. The only if part is precisely the corollary above, the if part follows from the fact that $S^n_\Q$ is the loop space $K(\Q^\ast, n)$ for odd $n$.
@ -46,13 +46,13 @@ The second observation is that $Q$ is nicely behaved on tensor products and coke
\[ Q(\coker(f))\iso\coker(Qf). \]
}
\Proof{
First note that the cokernel of $f$ in the category of augmented cdga's is $\coker(f)= B / f(\overline{A})$ and that its augmentation ideal is $\overline{B}/ f(\overline{A})$\todo{$B /f(\overline{A})B$}. Just as above we make a simple calculation, where $p: \overline{B}\to Q(B)$ is the projection map:
First note that the cokernel of $f$ in the category of augmented cdga's is $\coker(f)= B /{f(\overline{A})B}$ and that its augmentation ideal is $\overline{B}/{f(\overline{A})B}$, where $f(\overline{A})B$ is the ideal generated by $f(\overline{A})$. Just as above we make a simple calculation, where $p: \overline{B}\to Q(B)$ is the projection map:
@ -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\todo{Refereer}. 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.
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 see as follows that it is a very good path object (\DefinitionRef{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.
@ -30,7 +30,7 @@ We will often say \Def{minimal model} or \Def{minimal algebra} to mean minimal S
Let $(A, d)$ be a cdga which is $1$-reduced, such that $A =\Lambda V$ is free as cga. Then the differential $d$ is decomposable if and only if $(A, d)$ is a Sullivan algebra filtered by degree.
}
\Proof{
Let $V$ be filtered by degree: $V(k)= V^{\leq k}$\todo{is this notation introduced?}. Now $d(v)\in\Lambda V^{< k}$ for any $v \in V^k$. For degree reasons $d(v)$ is a product, so $d$ is decomposable.
Let $V$ be filtered by degree: $V(k)= V^{\leq k}$. Now $d(v)\in\Lambda V^{< k}$ for any $v \in V^k$. For degree reasons $d(v)$ is a product, so $d$ is decomposable.
For the converse take $V(n)= V^{\leq n}$ (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:
$$\deg{x}+\deg{y}=\deg{xy}=\deg{dv}=\deg{v}+1= n +1. $$
@ -10,19 +10,26 @@ As this thesis considers different categories, each with its own homotopy theory
\begin{definition}
A \Def{model category} is a category $\cat{C}$ together with three subcategories:
\begin{itemize}
\begin{itemize}\itemsep0em
\item a class of \Def{weak equivalences}$\W$,
\item a class of \Def{fibrations}$\Fib$ and
\item a class of \Def{cofibrations}$\Cof$,
\end{itemize}
such that the following five axioms hold:
\begin{itemize}
\begin{itemize}\itemsep0em
\item[MC1] All finite limits and colimits exist in $\cat{C}$.
\item[MC2] If $f$, $g$ and $fg$ are maps such that two of them are weak equivalences, then so it the third. This is called the \emph{2-out-of-3} property.
\item[MC3] All three classes of maps are closed under retracts\todo{Either draw the diagram or define a retract earlier}.
\item[MC4] In any commuting square as follows where $i \in\Cof$ and $p \in\Fib$,
\cdiagram{Model_Liftproblem}
\item[MC2] The \emph{2-out-of-3} property: if $f$, $g$ and $fg$ are maps such that two of them are weak equivalences, then so it the third.
\item[MC3] All three classes of maps are closed under retracts. A class $\mathfrak{K}$ is closed under retracts if, when given a diagram
\[\xymatrix{
A' \ar[r]^-i \ar[d]^-g & A \ar[r]^-r \ar[d]^-f & A' \ar[d]^-g \\
X' \ar[r]^-j & X \ar[r]^-s & X'
}\]
with $r \circ i =\id$ and $s \circ j =\id$, then $f \in\mathfrak{K}$ implies $g \in\mathfrak{K}$.
\item[MC4] In any commuting square with $i \in\Cof$ and $p \in\Fib$
\[\xymatrix{
A \ar[d]^i \ar[r]& X \ar[d]^p \\
B \ar[r]& Y
}\]
there exist a lift $h: B \to Y$ if either (a) $i \in\W$ or (b) $p \in\W$.
\item[MC5] Any map $f : A \to B$ can be factored in two ways:
\begin{itemize}
@ -96,7 +103,8 @@ Of course the most important model category is the one of topological spaces. We
\item Cofibrations: all monomorphisms.
\end{itemize}
}
\todo{small object arg?}
Both of these examples are often proven to be model categories by using \emph{Quillen's small object arguments}. This technique can be found in \cite{goerss2, dwyer, may2}.
In this thesis we often restrict to $1$-connected spaces. The full subcategory $\Top_1$ of $1$-connected spaces satisfies MC2-MC5: the 2-out-of-3 property, retract property and lifting properties hold as we take the \emph{full} subcategory, factorizations exist as the middle space is $1$-connected as well. Both products and coporducts exist. However $\Top_1$ does not have all limits and colimits.
@ -107,7 +107,7 @@ As a consequence of the above two lemmas, the class generated by $J$ is containe
Now if $[y]\in H(Y)$ is some class, then $dy =0$, and so by the above we find a preimage $x$ of $y$ such that $dx =0$, proving that $H(f)$ is surjective. Now let $[x]\in H(X)$ such that $[f(x)]=0$, then there is an element $\beta$ such that $f(x)= d\beta$, again by the above we can lift $\beta$ to get $x = d\alpha$., hence $H(f)$ is injective. Conclude that $f$ is a trivial fibration.
\end{proof}
We can use Quillen's small object argument with the set $J$. The argument directly proves the following lemma. Together with the above lemmas this translates to the required factorization.\todo{Definieer wat ``small'' betkent en geef een referentie}
We can use Quillen's small object argument with the set $J$. The argument directly proves the following lemma. Together with the above lemmas this translates to the required factorization.
\Lemma{cdga-mc5b}{
A map $f: A \to X$ can be factorized as $f = pi$ where $i$ is in the class generated by $J$ and $p$ has the RLP w.r.t. $J$.
@ -46,7 +46,17 @@ Another way to model the $n$-simplex is by the singular cochain complex associat
$$ C_n = C^\ast(\Delta^n; \k), $$
where $C^\ast(\Delta^n; \k)$ is the (normalized) singular cochain complex of $\Delta^n$ with coefficients in $\k$. The inclusion maps $d^i : \Delta^n \to\Delta^{n+1}$ and the maps $s^i: \Delta^n \to\Delta^{n-1}$ induce face and degeneracy maps on the dga's $C_n$, turning $C$ into a simplicial dga. Again we can extend this to functors by Kan extensions
\cdiagram{C_Extension}
\todo{show that $C^\ast$ really is sing cohom} where the left adjoint is precisely the functor $C^\ast$ as noted in \cite{felix}. We will relate $\Apl$ and $C$ in order to obtain a natural quasi isomorphism $A(X)\we C^\ast(X)$ for every $X \in\sSet$. Furthermore this map preserves multiplication on the homology algebras.
This left adjoint functor $C^\ast : \sSet\to\opCat{\DGA_\k}$ is (just as above) defined as $C^\ast(X)=\Hom_\sSet(X, C^\ast(\Delta[-]; \k))$. To see that this is precisely the classical definition of the singular cochain complex, we make the following calculation.
\begin{align*}
C^\ast(X) &= \Hom(X, C^\ast(\Delta[n])) \\
&= \Hom(X, \Hom(N \Z\Delta[n], \k)) \\
&\ison{1}\Hom(X, \Gamma(\k)) \\
&\iso\Hom(N \Z (X), \k)
\end{align*}
where $\Z$ is the free simplicial abelian group, $N$ is the normalized chain complex (this is the Dold-Kan equivalence) and $\Gamma$ its right adjoint. At \refison{1} we used the definition $\Gamma(C)=\Hom(N \Z\Delta[n], C)$. Now the conclusion of this calculation is that $C^\ast(X)$ is precisely the dual complex of $N \Z(X)$, which is the singular (normalized) chain complex.
We will relate $\Apl$ and $C$ in order to obtain a natural quasi isomorphism $A(X)\we C^\ast(X)$ for every $X \in\sSet$. Furthermore this map preserves multiplication on the homology algebras.
\subsection{Integration and Stokes' theorem for polynomial forms}
@ -56,11 +66,12 @@ In this section we will prove that the singular cochain complex is quasi isomorp
For any $v \in\Apl_n^n$, we can write $v$ 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):
\todo{rewrite?}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
Let $x$ be a $k$-simplex of $\Delta[n]$. Then $x$ induces a linear map $x^\ast: \Apl_n \to\Apl_k$. Now if we have an element $v \in\Apl_n^k$, then observe that$x^\ast(v)\in\Apl_k^k$ is an element 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 $\oint_n(v)\in C_n$\todo{normalised}. By linearity of $\int_n$ and $x^\ast$, we have a linear map $\oint_n: \Apl_n \to C_n$.
Note that $\oint_n(v): \Delta[n]\to\k$ is just a map of sets, so we can extend this linearly to chains on $\Delta[n]$ to obtain a linear map $\oint_n(v): \Z\Delta[n]\to\k$.
If $x$ is a degenerate simplex $x = s_j x'$, then $x^\ast(v)= s_j^\ast(x'^\ast(v))$. Now $x'^\ast(v)\in\Apl_{k-1}^k =0$ and so the integral vanishes on degenerate simplices. In other words we get $\oint_n(v)\in C_n$. 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 $\oint : \Apl\to C$ 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:
\begin{align*}
@ -70,7 +81,8 @@ Next we will show that $\oint = \{\oint_n\}_n$ is a simplicial map and that each
For it to be a chain map, we need to prove $d \circ\oint_n =\oint_n \circ d$. This is precisely the same calculation as \emph{Stokes' theorem}. \todo{Reference}
For it to be a chain map, we need to prove $d \circ\oint_n =\oint_n \circ d$. This is precisely \emph{Stokes' theorem} and any prove will apply here as well \cite{bousfield}.
We now proved that $\oint$ is indeed a simplicial chain map. Note that $\oint_n$ need not to preserve multiplication, so it fails to be a map of cochain algebras. However $\oint(1)=1$ and so the induced map on homology sends the class of $1$ in $H(\Apl_n)=\k\cdot[1]$ to the class of $1$ in $H(C_n)=\k\cdot[1]$. We have proven the following lemma.
@ -95,7 +107,7 @@ We will now prove that the map $\oint: A(X) \to C^\ast(X)$ is a quasi isomorphis
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\todo{see 7.0.2, what about $C$?}, 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}) to conclude that also $A(X')\we C^\ast(X')$.
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}) to conclude that also $A(X')\we C^\ast(X')$.
This proves $A(X)\we C^\ast(X)$ for any simplicial set with finitely many non-degenerate simplices. We can extend this to simplicial sets of finite dimension by attaching many simplices at once. For this we observe that both $A$ and $C^\ast$ send coproducts to products and that cohomology commutes with products:
@ -139,5 +151,3 @@ We will now prove that the map $\oint: A(X) \to C^\ast(X)$ is a quasi isomorphis
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 $X$.
@ -88,7 +88,7 @@ In the following arguments we will consider fibrations and need to compute homol
Now $\Omega K(G, n+1)= K(G, n)$, and we can apply \LemmaRef{kreck} as the reduced homology of the fiber is in $\C$ by induction hypothesis. Conclude that the homology of $P K(G, n+1)$ is $\C$-isomorphic to the homology of $K(G, n)$. Since $\RH_\ast(P K(G, n+1))=0$, we get $\RH_\ast(K(G, n+1))\in\C$.
}
For the main theorem we need the following construction. \todo{referentie}
For the main theorem we need the following decomposition of spaces. The construction can be found in \cite{serre} or \cite{kreck}.
\Lemma{whitehead-tower}{
(Whitehead tower)
We can decompose a $0$-connected space $X$ into fibrations:
@ -216,7 +216,7 @@ By using the class of $\Q$-vector spaces we get a dual theorem.
Y \arwe[rr]^-\lambda\ar[dr]_-q && Y \times_K MK \arfib[dl]^-{\overline{q}}\\
& K &
}\]
Where $MK$ is the Moore path space and $\overline{q}$ is induced by the map sending a path to its endpoint.\todo{read} Now $\overline{q}\lambda f$ is homotopic to the constant map, so there is a homotopy $h: \overline{q}\lambda f \eq\ast$ which we can lift against the fibration $\overline{q}$ to a homotopy $h' : \lambda f \eq f_1$ with $\overline{q} f_1=\ast$. In other words $f_1$ lands in the fiber of $\overline{q}$.
Where $MK$ is the Moore path space and $\overline{q}$ is induced by the map sending a path to its endpoint. Now $\overline{q}\lambda f$ is homotopic to the constant map, so there is a homotopy $h: \overline{q}\lambda f \eq\ast$ which we can lift against the fibration $\overline{q}$ to a homotopy $h' : \lambda f \eq f_1$ with $\overline{q} f_1=\ast$. In other words $f_1$ lands in the fiber of $\overline{q}$.
Joshua Moerman
9 years ago