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Let $f:M\to N$ be a smooth map of closed oriented smooth manifolds which is also a homeomorphism. Let $[M]\in H_\bullet(M;\mathbb Z)$ denote the fundamental class (and similarly for $N$). It is clear that $f_\ast[M]=N$. Is the same true if $[M],[N]$ instead denote the fundamental classes in oriented bordism?

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The statement after "it is clear that" is false (take an orientation-reversing diffeomorphism, e.g.). Other than that, the statement is true, and also in cobordism where it is even trivially true if you define cobordism theory geometrically. –  Dylan Wilson Jul 4 '13 at 6:59
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@Dylan: please read questions carefully before you vote to close. User35353 wrote ''smooth map which is also a homeomorphism'' and not ''diffeomorphism'' (in the latter case, the answer is indeed trivial). It is a nontrivial question and I would like to see a qualified answer (which I cannot provide). –  Johannes Ebert Jul 4 '13 at 7:42
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@Fernando : For a diffeomorphism, the inverse is also smooth. Not so with a smooth homeomorphism. –  Andy Putman Jul 4 '13 at 10:10
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I think the question is difficult. Here are my reasons: 1. it is true that homeomorphic manifolds are oriented cobordant, even if they are not diffeomorphic. This requires, to my knowledge, two fields medal theorems: Thom and Wall did prove that two oriented smooth manifolds are cobordant if their Pontrjagin numbers and Stiefl-Whitney numbers agree. The Stiefel-Whitney numbers are homotopy invariants, because you can express then in terms of the Spivak normal fibration and the Steenrod operations. The Pontrjagin numbers are invariant under homeomorphisms by Novikov's famous result. –  Johannes Ebert Jul 4 '13 at 14:28
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Last but not least: any proof attempt that does not refer to special structural properties of oriented cobordism is doomed to fail. There exists exotic spheres that are not spin cobordant to the standard sphere. This means that the result is wrong for spin bordism. –  Johannes Ebert Jul 4 '13 at 14:37

1 Answer 1

Let $X$ be an oriented $d$-manifold, and consider the homomorphism $$\rho_i : \Omega_d(X) \to H^{4i}(X;\mathbb{Z})$$ which sends $f : M \to X$ to $f_!(p_i(TM))$, the pushforward along $f$ of the $i$th Pontrjagin class of $M$. This is cobordism-invariant by the usual argument.

Now $\rho_i(id_M) = p_i(TM)$, and if $f : N \to M$ is a homeomorphism then $\rho_i(f) = (f^{-1})^*(p_i(TN))$.

So if the cobordism fundamental class were homeomorphism invariant, the integral Pontrjagin classes would be. This is false cf. chapter 4.4 of the Novikov conjecrure book by Kreck and Lueck.

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Thanks! If two smooth manifolds are homeomorphic are they necessarily related by a smooth homeomorphism? (maybe this is hard in general, but maybe one can at least say something for one of the counterexamples to the homeomorphism invariance of integral Pontryagin classes). –  John Pardon Jul 5 '13 at 18:00

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