4
$\begingroup$

It is well known the following result:

Lemma: Let $F\rightarrow E\rightarrow B$ be a fibration with $B$ connected and simply connected. Suppose that $F$ is $n$-acyclic, i.e. $H^{p}\left( F;% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) =H^{p}\left( pt;% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) $ for $p\leq n$. Then $$ H^{p}\left( B; \mathbb{Q} \right) \rightarrow H^{p}\left( E;% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) $$ is an isomorphism for $p\leq n$ and a monomorphism for $q=n+1$.

The converse of this lemma is true? That is, if $$ H^{p}\left( B;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{p}\left( E;% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) $$ is an isomorphism for $p\leq n$ and a monomorphism for $q=n+1$, then $F$ is $% n$-acyclic?

$\endgroup$
1
$\begingroup$

Let $M$ be the mapping cylinder of the projection $p\colon E\to B$, so $p$ factors as an inclusion $i\colon E\to M$ followed by a homotopy equivalence $M\to B$. Let $PM$ be the path space of $M$, and let $G$ be the pullback of $E\to M$ and $PM\to M$. It is then standard that $G$ is homotopy equivalent to $F$, so it will suffice to prove that $G$ is highly connected. Note that $PM$ is contractible so $$ H^m(PM,G;\mathbb{Q})\simeq\tilde{H}^{m+1}(G;\mathbb{Q}) \simeq \tilde{H}^{m+1}(F;\mathbb{Q}) $$ The diagram $\require{AMScd}$ \begin{CD} \Omega B @>>> G @>>> E \\ @| @VVV @VViV \\ \Omega B @>>> PM @>>> M \end{CD} gives rise to a relative Serre spectral sequence $$ E_2^{ij} = H^i(M,E;H^j(\Omega B;\mathbb{Q})) \Longrightarrow H^{i+j}(PM,G;\mathbb{Q}) = \tilde{H}^{i+j+1}(F;\mathbb{Q}) $$ If the map $E\to B\simeq M$ is a cohomology isomorphism through a large range, then the $E_2$ page of the spectral sequence will be hihgly connected, so the $E_\infty$ page will be highly connected, so $H^*(F;\mathbb{Q})$ will be highly connected.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.