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?

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 cobordisminvariant 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. 


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 builtin 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 AtiyahHirzebruch 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 wellversed 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 $$ 0\to \Omega_n(M^{n1})\to \Omega_n(M)\to E^\infty_{n,0}\to 0, $$ where the first map is induced by inclusion of the $n1$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? 

