# Does a map with fibers $S^2\vee S^2$ have to be a locally trivial $S^2\vee S^2$ bundle?

Let $X\to Y$ be a proper map between pseudo-manifolds such that fibers are $S^2\vee S^2$, is it true that $X\to Y$ is locally trivial $S^2\vee S^2$ bundle?

• If the map is a submersion, proper and $X$ and $Y$ the answer is yes. See Ehresman lemma. en.wikipedia.org/wiki/Ehresmann's_lemma – Tsemo Aristide Jan 31 '16 at 1:59
• @TsemoAristide If the map is submersion, I think the fibers would also be smooth? I think the map is not a submersion. – Qixiao Jan 31 '16 at 2:02
• @Tsemo: If the map is a submersion the fibers are manifolds. – user98602 Jan 31 '16 at 2:03
• I have a hard time imagining it possible for a manifold to bundle with $S^2 \vee S^2$ fibers over another space, even locally. A simple test obstruction is that for every $x$ in $M$ the relative homology groups $H_*(M, M-x)$ should be the homology of a sphere for every point, but it seems hard to maintain this homogeneity of data between points corresponding to general points on one of the spheres and those at the wedge point. – JHance Jan 31 '16 at 2:35
• That's a good point. You can't possibly have fiber $S^2 \vee S^2$ since then you'd have an open set homeomorphic to $(S^2 \vee S^2) \times \Bbb R^k$, which is not a manifold! – user98602 Jan 31 '16 at 2:45