The converse of Vietoris-Begle theorem 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?
 A: 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.  
