Are there fibrations $F\_i \to X\_i \to B\_i$ $(i=1,2)$ with path-connected bases $B\_i$ and connected fibres $F\_i$ such that their corresponding Leray-Serre spectral sequences (integral coefficients) are isomorphic as abelian groups, i.e. 
$$E\_r^{p,q}(1) \cong E\_r^{p,q}(2)\;\;\;(2 \le r \le \infty,\;i,j \ge 0)$$
and such that their integral cohomology isn't isomorphic, i.e. there is $p \ge 0$ with 
$$H^p(X\_1;\mathbb{Z}) \not\cong H^p(X\_2;\mathbb{Z})$$
as abelian groups ? 



Fibrations with isomorphic Leray-Serre spectral sequences and non-isomorphic cohomology ? 
Such an example is given by the pair of fibrations
$$
K(\mathbb{Z}/2,1) \to K(\mathbb{Z},2) \to K(\mathbb{Z},2)
$$
(coming from the Bockstein exact sequence) and by
$$
K(\mathbb{Z}/2,1) \to K(\mathbb{Z}/2,1) \times K(\mathbb{Z},2) \to K(\mathbb{Z},2).
$$
(a trivial fibration).


In both cases, the (cohomological!) Leray-Serre spectral sequence is concentrated in even total degree. However, the two spaces have differing $H^2$ with integral coefficients ($\mathbb{Z}$ versus $\mathbb{Z} \times \mathbb{Z}/2$).




---


**Added (by Ralph):** Here are some more details for the spectral sequences. We know 
$K(\mathbb{Z}/2,1)=\mathbb{R}P^\infty$ and $K(\mathbb{Z},2)=\mathbb{C}P^\infty$ and 
$$H^p(\mathbb{R}P^\infty;\mathbb{Z})=
\begin{cases}
\mathbb{Z} & p=0 \newline \mathbb{Z}\_2 & p > 0 \text{ even }\;\;, \newline 0 & p > 0 \text{ odd }
\end{cases}
\hspace{10pt}
H^p(\mathbb{C}P^\infty;M)=
\begin{cases}
M & p> 0 \text{ even} \newline 0 & p > 0 \text{ odd } 
\end{cases}$$
where $\mathbb{Z}\_2 := \mathbb{Z}/2$ and $M$ are trivial coefficients. Since $\mathbb{Z}\_2$ has only two elements, the coefficient system in the LS spectral sequence of the first fibration is trivial and we obtain for $E\_2^{p,q}(1)=H^P(\mathbb{C}P^\infty;H^q(\mathbb{R}P^\infty;\mathbb{Z}))$: 
$$E\_2(1)=\;
\begin{array}{cccccccc}
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \newline 
\mathbb{Z}\_2 & 0 & \mathbb{Z}\_2 & 0 & \mathbb{Z}\_2 & 0 & \mathbb{Z}\_2 & \cdots \newline
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline 
\mathbb{Z}\_2 & 0 & \mathbb{Z}\_2 & 0 & \mathbb{Z}\_2 & 0 & \mathbb{Z}\_2 & \cdots\newline
0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots\newline 
\mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & 0 & \mathbb{Z} & \cdots\newline
\end{array}$$
Now, for positional reasons, $E\_2(1)=E\_\infty(1)$. 


As the 2nd fibration is trivial, the coefficient system in its LS spectral sequence is also trivial. Hence both spectral sequences agree (in all terms), while the cohomologies differ: $H^p(\mathbb{C}P^\infty;\mathbb{Z})$ is described above and 
$$H^p(\mathbb{C}P^\infty \times \mathbb{R}P^\infty;\mathbb{Z})=
\begin{cases}\mathbb{Z} \oplus \mathbb{Z}\_2^n & p= 2n \newline 0 & p \text{ odd.} 
\end{cases}$$


