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Assume F is null bordant. Does it imply that the total space of fiber bundle $F\hookrightarrow E\to M$ is null bordant?

in particular what if $F$ is sphere?

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In the smooth category the answer is no. This is because of the existence of exotic spheres. According to Milnor, one can construct exotix $S^7$ as a fibre bundle over $S^4$ with fibre $S^3$. However, these $S^7$'s are not the boundary of any smooth $8$-manifold! –  Somnath Basu Mar 22 '11 at 21:54
I do not think that Somnaths answer is correct; the structural group of Milnors $S^3$-bundles is $SO(4)$, hence these exotic spheres are boundaries of disc bundles. Milnors argument is that his spheres do not bound a manifold with trivial fourth Betti number. –  Johannes Ebert Mar 22 '11 at 23:01
@ Johannes - I see my mistake now. I recalled wrongly Milnor's result on exotic spheres. –  Somnath Basu Mar 22 '11 at 23:31
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1 Answer

up vote 5 down vote accepted

There exist oriented surface bundles $E \to B$ on closed surfaces such that $E$ has nonzero signature (first found by Atiyah and Hirzebruch). Hence $E$ is not (oriented) nullbordant, even though base and fibre are nullbordant.

Textbook reference: Morita, Geometry of characteristic classes.

A lot of material on characteristic numbers of total spaces of fibre bundles is contained in Hirzebruch "Manifolds and modular forms"

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Kodaira did it before Atiyah and Hirzebruch, both of whom published the results in Kodaira's festschrift. –  Ben Wieland Mar 23 '11 at 4:28
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