TITLE: Does a smooth homeomorphism of closed manifolds preserve cobordism fundamental class?
QUESTION [10 upvotes]: 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?

REPLY [6 votes]: 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.