The converse of Vietoris-Begle theorem

Question

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?

Answer 1

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.