Is there a counterexample to the following assertion?:

Let $p_1:E_1\to B_1$ and $p_2:E_2\to B_2$ be fibrations with the same fiber $\mathbb S ^1$ such that $E_1$ and $E_2$ are homeomorphic (and both connected). Then $B_1$ and $B_2$ are homotopic.

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    $\begingroup$ Did you mean that you have a map of fibrations, i.e. a commuting square involving $p_1$ and $p_2$? $\endgroup$ – David Roberts Jun 22 '15 at 3:50

$S^1\times S^3$ fibers both over $S^3$ (obvious) and $S^1\times S^2$ (identity cross Hopf). BTW, you don't say "homotopic" about spaces.

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    $\begingroup$ Or more generally, if $F\to E\to B$ is a bundle with $E$ not a product, then $F\times E \to E$, $F\times E \to F\times B$ works. $\endgroup$ – Allen Knutson Jun 21 '15 at 22:53

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