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 $i \geq n$?

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.

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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