Finally got the Kreck lemma right

thesis/notes/Basics.tex

@@ -67,20 +67,24 @@ The following two theorems can be found in textbooks about homological algebra s
 \Theorem{kunneth}{
 	(Künneth Theorem)
 	For spaces $X$ and $Y$, there is a short exact sequence
-	$$ 0 \to H(X; A) \tensor H(Y; A) \to H(X \times Y; A) \to \Tor(H(X; A), H(Y; A)) \to 0, $$
-	where $H(X; A)$ and $H(X; A)$ are considered as graded modules and their tensor product and torsion groups are graded. \todo{Geef algebraische versie voor ketencomplexen}
+	\[ \footnotesize \xymatrix @C=0.3cm{
+		0 \ar[r] & H(X; A) \tensor H(Y; A) \ar[r] & H(X \times Y; A) \ar[r] & \Tor_{\ast-1}(H(X; A), H(Y; A)) \ar[r] & 0
+	},\]
+	where $H(X; A)$ and $H(X; A)$ are considered as graded modules and their tensor product and torsion groups are graded. \todo{Geef algebraische versie voor ketencomplexen? en cohomology?}
 }
 
-\section{Immediate results for rational homotopy theory}
+\section{Consequences for rational homotopy theory}
 
 The latter two theorems have a direct consequence for rational homotopy theory. By taking $A = \Q$ we see that the torsion groups vanish. We have the immediate corollary.
 
 \Corollary{rational-corollaries}{
-	We have the following natural isomorphisms in homology
-	$$ H(X) \tensor \Q \tot{\iso} H(X; \Q), $$
-	$$ H(X \times Y; \Q) \tot{\iso} H(X; \Q) \tensor H(Y; \Q). $$
-	Furthermore we can relate homology and cohomology in a natural way:
-	$$ H^n(X; \Q) \tot{\iso} \Hom(H_n(X); \Q). $$
+	We have the following natural isomorphisms in rational homology, and we can relate rational cohomolgy naturally to rational homology
+	\begin{align*}
+		H_\ast(X) \tensor \Q &\tot{\iso} H_\ast(X; \Q), \\
+		H_\ast(X; \Q) \tensor H_\ast(Y; \Q) &\tot{\iso} H_\ast(X \times Y; \Q), \\
+		H^\ast(X; \Q) &\tot{\iso} \Hom(H_\ast(X); \Q). 
+	\end{align*}
+
 }
 
 The long exact sequence for a Serre fibration also has a direct consequence for rational homotopy theory.

thesis/notes/Serre.tex

@@ -35,10 +35,10 @@ Note that the maps $0 \to C$ and $C \to 0$ are $\C$-isomorphisms for any $C \in
 In the following arguments we will consider fibrations and need to compute homology thereof. Unfortunately there is no long exact sequence for homology of a fibration, however the following lemma expresses something similar. It is usually proven with spectral sequences,  \cite[Ch. 2 Thm 1]{serre}. However in \cite{kreck} we find a more geometric proof.
 
 \Lemma{kreck}{
-	Let $\C$ be a Serre class. Let $p: E \fib B$ be a fibration between $0$-connected spaces and $F$ its fiber. If $\RH_i(F) \in \C$ for all $i < n$, then
+	Let $\C$ be a Serre class and $p: E \fib B$ be a fibration between $0$-connected spaces with a $0$-connected fiber $F$. If $\RH_i(F) \in \C$ for all $i < n$ and $B$ is $m$-connected, then
 	\begin{itemize}
-		\item $H_i(E, F) \to H_i(B, b_0)$ is a $\C$-iso for $i \leq n+1$ and
-		\item $H_i(E) \to H_i(B)$ is a $\C$-iso for all $i \leq n$.
+		\item $H_i(E, F) \to H_i(B, b_0)$ is a $\C$-iso for $i \leq n+m$ and
+		\item $H_i(E) \to H_i(B)$ is a $\C$-iso for all $i < n+m$.
 	\end{itemize}
 }
 \Proof{
@@ -61,19 +61,20 @@ In the following arguments we will consider fibrations and need to compute homol
 
 	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 $p$ was 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 and identify the cells of $B$:
-	\begin{align*}
-		H_{i+1}(E^{k+1}, E^k) &\iso (H_\ast(\coprod_\alpha D^{k+1}_\alpha, \coprod_\alpha S^k_\alpha) \tensor H_\ast(F))_{i+1} \\
-		&\iso (H_\ast(B^{k+1}, B^k) \tensor H_\ast(F)) \\
-		&= \bigoplus_{j+l=i+1} H_j(B^{k+1}, B^k) \tensor H_l(F)
-	\end{align*}
-	\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
-	$$ p_\ast : H_{i+1}(E^{k+1}, E^k) \to H_{i+1}(B^{k+1}, B^k) $$
-
-	This finished the induction on $k$.
+	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.
+	\[ \scriptsize \xymatrix @C=0.4cm {
+		0 \ar[r] & (H(B^{k+1}, B^k) \tensor H(F))_{i+1} \ar[d]^-{p'} \ar[r] & H_{i+1}(E^{k+1}, E^k) \ar[d]^-{p_\ast} \ar[r] & \Tor(H(B^{k+1}, B^k), H(F))_i \ar[d]^-{p''} \ar[r] & 0 \\
+		0 \ar[r] & H_{i+1}(B^{k+1}, B^k) \ar[r] & H_{i+1}(B^{k+1}, B^k) \ar[r] & 0 \ar[r] & 0
+	}\]
+	Now it remains to show that $p'$ and $p''$ are $\C$-iso, as it will then follow from the five lemma that $p_\ast$ is a $\C$-iso.
+
+	First note that $p'$ is surjective as it is an isomorphism on the subspace $H_{i+1}(B^{k+1}, B^k) \tensor H_0(F)$. Its kernel on the other hand is precisely given by the terms $H_{i+1-q}(B^{k+1}, B^k) \tensor H_q(F)$ for $q>0$. By assumption we have $H_q(F) \in \C$ for all $0 < q < n$ and $H_{i+1}(B^{k+1}, B^k) = 0$ for all $i+1 \leq m$. By the tensor property of a Serre class the kernel is in $\C$ for all $i < n+m$. So indeed $p'$ is a $\C$-iso for all $i < n+m$.
+
+	For $p''$ a similar reasoning holds, it is clearly surjective and we only need to prove that the kernel of $p''$ (which is the Tor group itself) is in $\C$. First notice that $\Tor(H_i(B^{k+1}, B^k), H_0(F)) = 0$ as $H_0(F) \iso \Z$. Then consider the other terms of the graded Tor group. Again we use the assumed bounds to conclude that the Tor group is in $\C$ for $i \leq n+m$. So indeed $p''$ is a $\C$-iso for all $i \leq n+m$.
+
+	Now we conclude that $p_\ast : H_{i+1}(B^{k+1}, B^k) \to H_{i+1}(E^{k+1}, E^k)$ is indeed a $\C$-iso for all $i < n+m$. And by the long exact sequence of triples shown above we get a $\C$-iso $p_\ast : H_i(E^{k}, F) \to H_i(B^{k}, b_0)$ for all $i \leq n+m$. This finished the induction on $k$.
+
+	This concludes that $H_i(E, F) \to H_i(B, b_0)$ is a $\C$-iso and by another application of the long exact sequence (of the pair $(E,F)$) and the five lemma we get the $\C$-iso $H_i(E) \to H_i(B)$.
 }
 
 \Lemma{homology-em-space}{