Manually typesets first half

thesis/chapters/CDGA_As_Algebraic_Model_For_Rational_Homotopy_Theory.tex

@@ -2,10 +2,10 @@
 \Chapter{Homotopy Theory For cdga's}{HomotopyTheoryCDGA}
 
 Recall that a cdga $A$ is a commutative differential graded algebra, meaning that
-\begin{itemize}
+\begin{itemize}\itemsep0em
 	\item it has a grading: $A = \bigoplus_{n\in\N} A^n$,
 	\item it has a differential: $d: A \to A$ with $d^2 = 0$,
-	\item it has an associative and unital multiplication: $\mu: A \tensor A \to A$ and
+	\item it has a multiplication: $\mu: A \tensor A \to A$ which is associative and unital and
 	\item it is commutative: $x y = (-1)^{\deg{x}\cdot\deg{y}} y x$.
 \end{itemize}
 And all of the above structure is compatible with each other (e.g. the differential is a derivation of degree $1$, the maps are graded, \dots). The exact requirements are stated in the appendix on algebra. An algebra $A$ is augmented if it has a specified map (of algebras) $A \tot{\counit} \k$. Furthermore we adopt the notation $A^{\leq n} = \bigoplus_{k \leq n} A^k$ and similarly for $\geq n$.
@@ -15,7 +15,7 @@ There is a left adjoint $\Lambda$ to the forgetful functor $U$ which assigns the
 In homological algebra we are especially interested in \emph{quasi isomorphisms}, i.e. maps $f: A \to B$ inducing an isomorphism on cohomology: $H(f): HA \iso HB$. This notions makes sense for any object with a differential.
 
 We furthermore have the following categorical properties of cdga's:
-\begin{itemize}
+\begin{itemize}\itemsep0em
 	\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.

thesis/notes/Basics.tex

@@ -13,7 +13,7 @@ In the following definition \emph{space} is to be understood as a topological sp
 }
 
 \Definition{rational-homotopy-groups}{
-	We define the \emph{rational homotopy groups} of a $0$-connected space $X$ with abelian fundamental group as:
+	We define the \emph{rational homotopy groups} of a \linebreak $0$-connected space $X$ with abelian fundamental group as:
 	$$ \pi_i(X) \tensor \Q \quad \forall i > 0.$$
 }
 
@@ -45,7 +45,7 @@ We will now recall known results from algebraic topology, without proof. One can
 }
 
 \Theorem{serre-les}{
-	(Long Exact Sequence of Homotopy Groups) Let $f: X \to Y$ be a Serre fibration, then there is a long exact sequence:
+	(Long Exact Sequence of Homotopy Groups) Let \linebreak $f: X \to Y$ be a Serre fibration, then there is a long exact sequence:
 	$$ \cdots \tot{\del} \pi_i(F) \tot{i_\ast} \pi_i(X) \tot{f_\ast} \pi_i(Y) \tot{\del} \cdots \to \pi_0(Y) \to \ast, $$
 	where $F$ is the fiber of $f$.
 }

thesis/notes/CDGA_Basic_Examples.tex

@@ -3,7 +3,7 @@ We will first define some basic cochain complexes which model the $n$-disk and $
 $$ D(n) = ... \to 0 \to \k \to \k \to 0 \to ... $$
 $$ S(n) = ... \to 0 \to \k \to 0 \to 0 \to ... $$
 
-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, where weak equivalences are quasi isomorphisms, fibrations are degreewise surjective and cofibrations are degreewise injective for positive degrees \cite[Example 1.6]{goerss2}.
+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 \linebreak $H^n(S(n)) = \k \cdot [a]$. 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, where weak equivalences are quasi isomorphisms, fibrations are degreewise surjective and cofibrations are degreewise injective for positive degrees \cite[Example 1.6]{goerss2}.
 
 \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.
@@ -14,11 +14,12 @@ As we do not directly need this claim, we omit the proof. However, in the next s
 $S(n)$ 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 $b^k$ and $c b^k$ when $n$ is even, and it consists of linear combinations of $c^k b$ and $c^k$ when $n$ is odd. In both cases we can compute the differentials using the Leibniz rule:
-$$ d(b^k) = k \cdot c b^{k-1} $$
-$$ d(c b^k) = 0 $$
 
-$$ d(c^k b) = c^{k+1} $$
+\[\xymatrix @R=0cm{
-$$ d(c^k) = 0 $$
+	d(b^k) = k \cdot c b^{k-1}   &   d(c^k b) = c^{k+1}\\
+	d(c b^k) = 0   &   d(c^k) = 0
+	}
+\]
 
 Those cocycles are in fact coboundaries (using that $\k$ is a field of characteristic $0$):
 $$ c b^k = \frac{1}{k} d(b^{k+1}) $$
@@ -40,6 +41,6 @@ The situation for $\Lambda S(n)$ is easier as it has only one generator (as alge
 We will prove this theorem in the next section. Note that the functors $\Lambda$ and $U$ thus form a Quillen pair with this model structure.
 
 \subsection{Why we need $\Char{\k} = 0$ for algebras}
-The above Quillen pair $(\Lambda, U)$ fails to be a Quillen pair if $\Char{\k} = p \neq 0$. We will show this by proving that the maps $\Lambda(j_n)$ are not weak equivalences for even $n$. Consider $b^p \in \Lambda D(n)$, then by the Leibniz rule:
+The above Quillen pair $(\Lambda, U)$ fails to be a Quillen pair if \linebreak $\Char{\k} = p \neq 0$. We will show this by proving that the maps $\Lambda(j_n)$ are not weak equivalences for even $n$. Consider $b^p \in \Lambda D(n)$, then by the Leibniz rule:
 $$ d(b^p) = p \cdot c b^{p-1} = 0. $$
 So $b^p$ is a cocycle. Now assume $b^p = d x$ for some $x$ of degree $p n - 1$, then $x$ contains a factor $c$ for degree reasons. By the calculations above we see that any element containing $c$ has a trivial differential or has a factor $c$ in its differential, contradicting $b^p =  d x$. So this cocycle is not a coboundary and $\Lambda D(n)$ is not acyclic.

thesis/notes/Homotopy_Relations_CDGA.tex

@@ -1,7 +1,7 @@
 
 Although the abstract theory of model categories gives us tools to construct a homotopy relation (\DefinitionRef{homotopy}), it is useful to have a concrete notion of homotopic maps.
 
-Consider the free cdga on one generator $\Lambda(t, dt)$, where $\deg{t} = 0$, this can be thought of as the (dual) unit interval with endpoints $1$ and $t$. Notice that this cdga is isomorphic to $\Lambda(D(0))$ as defined in the previous section. We define two \emph{endpoint maps} as follows:
+Consider the free cdga on one generator $\Lambda(t, dt)$, where \linebreak $\deg{t} = 0$, this can be thought of as the (dual) unit interval with endpoints $1$ and $t$. Notice that this cdga is isomorphic to \linebreak $\Lambda(D(0))$ as defined in the previous section. We define two \emph{endpoint maps} as follows:
 $$ d_0, d_1 : \Lambda(t, dt) \to \k $$
 $$ d_0(t) = 1, \qquad d_1(t) = 0, $$
 this extends linearly and multiplicatively. Note that it follows that we have $d_0(1-t) = 0$ and $d_1(1-t) = 1$. These two functions extend to tensor products as $d_0, d_1: \Lambda(t, dt) \tensor X \to \k \tensor X \tot{\iso} X$.

thesis/notes/Model_Of_CDGA.tex

@@ -76,7 +76,7 @@ Next we will prove the factorization property [MC5]. We will prove one part dire
 	The maps $i_n$ are trivial cofibrations and the maps $j_n$ are cofibrations.
 \end{lemma}
 \begin{proof}
-	Since $H(\Lambda D(n)) = \k$ (as stated earlier this uses $\Char{\k} = 0$) we see that indeed $H(i_n)$ is an isomorphism. For the lifting property of $i_n$ and $j_n$ simply use surjectivity of the fibrations and the freeness of $\Lambda D(n)$ and $\Lambda S(n)$.
+	Since $H(\Lambda D(n)) = \k$ (as stated earlier this uses \linebreak $\Char{\k} = 0$) we see that indeed $H(i_n)$ is an isomorphism. For the lifting property of $i_n$ and $j_n$ simply use surjectivity of the fibrations and the freeness of $\Lambda D(n)$ and $\Lambda S(n)$.
 \end{proof}
 
 \begin{lemma}

thesis/notes/Rationalization.tex

@@ -17,7 +17,7 @@ We start the construction with $S^n(1) = S^n$. Now assume $S^n(r) = \bigvee_{i=1
 } \]
 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 $S^n_\Q = \colim_{r} S^n(r)$.
 
-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 $r+1$. This shows 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 $r$.
+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, let $i_r: S^n \to S^n(r+1)$ be the inclusion of the $r$th sphere and let $i_{r+1} : S^n \to S^n(r+1)$ be 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 \linebreak the class $[i_r]$ by $r+1$. This shows 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:
 \[ \xymatrix{ \Z \ar[r]^-{\times 2} & \Z \ar[r]^-{\times 3} & \Z \ar[r]^-{\times 4} & \Z \ar@{-->}[rr] & & \Q } \]

thesis/notes/Serre.tex

@@ -13,7 +13,7 @@ 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 a Serre class as a class of groups we want to \emph{ignore}.
+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 \linebreak (stronger) requirements. One should think of a Serre class as a class of groups we want to \emph{ignore}.
 
 \Example{serre-classes}{
 	We give three Serre classes without proof.
@@ -58,7 +58,7 @@ In the following arguments we will consider fibrations and need to compute homol
 
 	The morphism in the middle is a $\C$-iso by induction. We will prove that the left morphism is a $\C$-iso which implies by the five lemma that the right morphism is one as well.
 
-	As we are working with relative homology $H_{i+1}(E^{k+1}, E^k)$, we only have to consider the interiors of the $k+1$-cells (by excision). Each interior of a $k+1$-cell is a product, as $p$ is a fiber bundle. So we note that we have an isomorphism:
+	As we are working with relative homology $H_{i+1}(E^{k+1}, E^k)$, we only have to consider the interiors of the $(k+1)$-cells (by excision). Each interior of a $(k+1)$-cell is a product, as $p$ is a fiber bundle. So we note that we have an isomorphism:
 	$$ H_{i+1}(E^{k+1}, E^k) \iso H_{i+1}(\coprod_\alpha D^{k+1}_\alpha \times F, \coprod_\alpha S^k_\alpha \times F). $$
 	Now we can apply the Künneth theorem for this product to obtain a natural short exact sequence, furthermore we apply the Künneth theorem for $(B^{k+1}, B^k) \times \ast$ to obtain a second short exact sequence as follows.
 	\[ \scriptsize \xymatrix @C=0.4cm {
@@ -95,8 +95,8 @@ For the main theorem we need the following decomposition of spaces. The construc
 	$$ \cdots \fib X(n+1) \fib X(n) \fib X(n-1) \fib \cdots \fib X(1) = X, $$
 	such that:
 	\begin{itemize}
-		\item $K(\pi_n(X), n-1) \cof X(n+1) \fib X(n)$ is a fiber sequence,
+		\item $K(\pi_n(X), n-1) \cof X(n+1) \fib X(n)$ is a fiber sequence.
-		\item There is a space $X'_n$ weakly equivalent to $X(n)$ such that $X(n+1) \cof X'_n \fib K(\pi_n(X), n)$ is a fiber sequence, and
+		\item There is a space $X'_n$ weakly equivalent to $X(n)$ such that \linebreak $X(n+1) \cof X'_n \fib K(\pi_n(X), n)$ is a fiber sequence.
 		\item $X(n)$ is $(n-1)$-connected and $\pi_i(X(n)) \to \pi_i(X)$ is an isomorphism for all $i \geq n$.
 	\end{itemize}
 }
@@ -111,9 +111,9 @@ For the main theorem we need the following decomposition of spaces. The construc
 
 	For the induction step we may 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 by induction hypothesis. Furthermore the theorem assumes that $\pi_{n-1}(X) \in \C$ and hence we conclude $H_{n-1}(X) \in \C$.
 
-	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 $X(2) = X(1) = X$.
+	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 $1$-connected and that $X(2) = X(1) = X$.
 
-	\Claim{}{For all $j < n$ and $i \leq n$ the induced map $H_i(X(j+1)) \to H_i(X(j))$ is a $\C$-iso.}
+	\Claim{}{For all $j < n$ and $i \leq n$ the induced map \linebreak $H_i(X(j+1)) \to H_i(X(j))$ is a $\C$-iso.}
 	Note that $X(j+1) \fib X(j)$ is a fibration with $F = K(\pi_j(X), j-1)$ as its fiber. So by \LemmaRef{homology-em-space} we know $H_i(F) \in \C$ for all $i$. Apply \LemmaRef{kreck} to obtain a $\C$-iso $H_i(X(j+1)) \to H_i(X(j))$ for all $j < n$ and all $i > 0$. This proves the claim.
 
 	Considering this claim for all $j < n$ gives a chain of $\C$-isos $H_i(X(n)) \to H_i(X(n-1)) \to \cdot \to H_i(X(2)) = H_i(X)$ for all $i \leq n$. Consider the following diagram: