To compare the total, base and fiber spaces 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 $B$ no need simply connected) and $F$, $F^{\prime }$, $B$, and $B^{\prime }$ are path-connected spaces. $\require{AMScd}$ \begin{CD} F @>{}>> E @>{}>> B \\ @VVV @VVV @VVV\\ F' @>{}>> E' @>{}>> B' \end{CD} If \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, then is \begin{equation*} H^{\ast }\left( F^{\prime };% %TCIMACRO{\U{211a} }% %BeginExpansion \mathbb{Q} %EndExpansion \right) \rightarrow H^{\ast }\left( F;% %TCIMACRO{\U{211a} } %BeginExpansion \mathbb{Q} %EndExpansion \right) \end{equation*} an isomorphism?

I want to use Zeeman comparison theorem but the local coefficient system of the fiber bundle $F\rightarrow E\rightarrow B$ has not simple.

Answer 1

No. Consider the map from the fibre bundle $$B\mathbb{Z} \to BD_\infty \to B\mathbb{Z}/2$$ to $* \to * \to *$. Here $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$ is the infinite dihedral group.

You can easily use the Serre spectral sequence to compute that $H^*(BD_\infty ; \mathbb{Q}) = \mathbb{Q}$ (alternatively this follows from the free product decomposition $D_\infty = \mathbb{Z}/2 * \mathbb{Z}/2$), and of course $H^*(B\mathbb{Z}/2;\mathbb{Q})=\mathbb{Q}$, so this satisfies your hypotheses.

But $B\mathbb{Z} \to *$ is not an isomorphism on rational cohomology.

Answer 2

Let $f$ be the map between the two Serre fibrations. The question then asks: if $f$ is a rational homotopy equivalence on the base and total spaces, then is it a rational homotopy equivalence on the fibers? In other words, is the induced map $f_\mathbf{Q}:F_\mathbf{Q}\to F'_\mathbf{Q}$ a homotopy equivalence? As pointed out above, the answer to this question is no in general. Conceptually, the reason for this negative answer can be boiled down to the fact that rationalization preserves (homotopy) colimits, but fiber sequences are (homotopy) limits. Concretely, this means that the fiber of the rationalized map $E_\mathbf{Q}\to B_\mathbf{Q}$ (which maps, via the homotopy equivalence $F_\mathbf{Q}$, to $E'_\mathbf{Q}\to B'_\mathbf{Q}$) is not necessarily $F_\mathbf{Q}$. (However, if $E$ and $B$ are simply connected, the map from $F_\mathbf{Q}$ to the fiber of $E_\mathbf{Q}\to B_\mathbf{Q}$ is indeed a homotopy equivalence.) Similarly, the fiber of the map $E'_\mathbf{Q}\to B'_\mathbf{Q}$ is not necessarily $F'_\mathbf{Q}$. This means that one cannot use the long exact sequence in homotopy to deduce that $f_\mathbf{Q}:F_\mathbf{Q}\to F'_\mathbf{Q}$ induces isomorphisms on rational homotopy, i.e., is a rational homotopy equivalence (hence giving a negative answer to your question in general, but a positive answer for simply connected base and total spaces).