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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
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2 Answers

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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Edit: The proof below is circular. Thanks to everyone who commented, especially Eric, for pointing this out. I will leave it here in case it is of use to somebody.


The answer is yes, although as Dylan points out I think the question should be modified so that $f: M\to N$ is a smooth orientation preserving homeomorphism. Then it is built-in that $f_\ast[M]=[N]\in H_n(N;\mathbb{Z})$.

Consider the Thom map $\mu: \Omega_n(M)\to H_n(M;\mathbb{Z})$ which sends a bordism class $[g: X^n\to M]$ to $g_\ast[X]$. This is a split surjection; we simply send the generator $[M]$ of $H_n(M;\mathbb{Z})=\mathbb{Z}$ to $[M]=[\operatorname{Id}:M\to M]\in \Omega_n(M)$. So we have a split short exact sequence of abelian groups $$ 0 \to \ker \mu \to \Omega_n(M)\to H_n(M;\mathbb{Z})\to 0.$$ Now the key point is this splitting is natural with respect to orientation preserving homeomorphisms of $n$-manifolds. This follows from naturality of the Atiyah--Hirzebruch spectral sequence $$ H_p(M;\Omega_q(\ast))\implies \Omega_\ast(M). $$

Added: I realize this was a bit of a smoke and mirrors answer! But perhaps it holds up. Here is the argument I had in mind, those more well-versed in spectral sequences can tell me if it works.

We have $H_n(M;\mathbb{Z})=E^2_{n,0}=E^\infty_{n,0}$, and the split short exact sequence described above can be identified with $$ 0\to \Omega_n(M^{n-1})\to \Omega_n(M)\to E^\infty_{n,0}\to 0, $$ where the first map is induced by inclusion of the $n-1$-skeleton. The fact that this splits means that the fundamental class $[M]_\Omega \in \Omega_n(M)$ does not involve terms of lower filtration. So it's represented by the permanent cycle $[M]\in H_n(M;\mathbb{Z})=E^2_{n,0}$, Likewise $[N]_\Omega\in\Omega_n(N)$ is represented by $[N]\in H_n(N;\mathbb{Z})$.

Now naturality says that the map $f_\ast: \Omega_\ast(M)\to \Omega_\ast(N)$ is induced by the map on the $E^2$-page, which by assumption sends $[M]$ to $[N]$. Doesn't this imply the result?

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How does naturality of the AHSS give naturality of the splitting? –  Eric Wofsey Jul 4 '13 at 14:16
    
Why is this splitting ''natural with respect to homeomorphisms''? –  Johannes Ebert Jul 4 '13 at 14:24
    
I do not think your updated version works. Global objection: if it were correct, the argument worked for spin bordism, where the conclusion is known to be false. Local objection: in order to make the argument with the exact sequence work, you need to know that the splitting is natural, which amounts to a circular conclusion. Without the naturality of the splitting, you only get that the two fundamental classes have the same image in homology. –  Johannes Ebert Jul 4 '13 at 16:53
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The problem is that the phrase "does not involve terms of lower filtration" is meaningless unless you've already chosen a splitting. $\Omega_n(M)$ is only filtered by the $E^\infty$ page, not graded: you can't split an element of $\Omega_n(M)$ into its "parts" in $E^{\infty}_{n-k,k}$; you can only identify the smallest $k$ such that it has a nonzero image in $E^\infty_{n-k,k}$. –  Eric Wofsey Jul 4 '13 at 16:53
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If you don't already know that your splitting is natural, then you don't know that your way of decomposing an element of $\Omega_n(M)$ into parts is preserved by $f_*$. –  Eric Wofsey Jul 4 '13 at 18:23
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