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}
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.
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}{
\Corollary{rational-corollaries}{
We have the following natural isomorphisms in homology
We have the following natural isomorphisms in rational homology, and we can relate rational cohomolgy naturally to rational homology
@ -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.
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}{
\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}
\begin{itemize}
\item$H_i(E, F)\to H_i(B, b_0)$ is a $\C$-iso for $i \leq n+1$ and
\item$H_i(E, F)\to H_i(B, b_0)$ is a $\C$-iso for $i \leq n+m$ and
\item$H_i(E)\to H_i(B)$ is a $\C$-iso for all $i \leq n$.
\item$H_i(E)\to H_i(B)$ is a $\C$-iso for all $i < n+m$.
\end{itemize}
\end{itemize}
}
}
\Proof{
\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:
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:
Now we can apply the Künneth theorem and identify the cells of $B$:
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.
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.
\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 $1\leq l \leq i+1$:
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 $i < n+m$.
$$ H_{i+1-l}(B^{k+1}, B^k)\tensor H_l(F). $$
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
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 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 $H_i(E)\to H_i(B)$.
Joshua Moerman
10 years ago