to compare cohomologies of fibers of two fiber bundles

Question

Consider the following commutative diagram of the fiber bundles $% F\rightarrow E\rightarrow B$ and $F^{\prime }\rightarrow E^{\prime }\rightarrow B^{\prime }$ where $B^{\prime }$ is simply connected space (but $B$ is not simply connected space) and all spaces are path-connected spaces. $\require{AMScd}$ \begin{CD} F @>{}>> E @>{}>> B \\ @VVV @VVV @VVV\\ F' @>{}>> E' @>{}>> B' \end{CD} Suppose that \begin{equation*} H^{\ast }\left( B^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( B;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} and \begin{equation*} H^{\ast }\left( E^{\prime };% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( E;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} are isomorphisms.

If \begin{equation*} H^{i }\left( F;% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right)=0 \end{equation*} for all $i \geq n$ ($n$ fixed), then \begin{equation*} H^{i }\left( F^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right)=0 \end{equation*} for all $i \geq n$?

Answer 1

No. Let $B'$ be any space, and take $E'=PB'$ and $F'=\Omega B$. The Kan-Thurston theorem gives a map $f\colon B\to B'$ such that $H^*(f;\mathbb{Q})$ is an isomorphism but $\Omega B$ is discrete, so $H^i(\Omega B;\mathbb{Q})=0$ for $i>0$. The diagram $\require{AMScd}$ \begin{CD} \Omega B @>{}>> PB @>{}>> B \\ @VVV @VVV @VVV\\ \Omega B' @>{}>> PB' @>{}>> B' \end{CD} satisfies most of your hypotheses, but $H^*(\Omega B')$ need not be bounded above.

One problem with the above example is that we have fibrations, but these need not be fibre bundles. If necessary this can be fixed by a detour into simplicial sets and simplicial groups.

Another problem with the above example is that the space $F=\Omega B$ is disconnected. If you want an example where absolutely everything is connected, we can proceed as follows. We can assume that we have actual fibre bundles, and then let $\Sigma_BE$ denote the fibrewise unreduced suspension of $E$, which is a fibre bundle over $B$ with fibre $\Sigma F$, which is always connected. We can describe $\Sigma_BE$ as the homotopy pushout of $B\xleftarrow{}E\xrightarrow{}B$, and from this we see that the map $\Sigma_BE\to\Sigma_{B'}E'$ is a homology equivalence provided that $B\to B'$ and $E\to E'$ are homology equivalences. Thus, we can apply this procedure to the previous counterexample to obtain a new counterexample in which everything is connected.

In general, if you know that a result is true for simply connected spaces, and you want to check whether that assumption can be relaxed, you should ask yourself whether the Kan-Thurston theorem gives counterexamples.