Given a homeomorphism between complex manifolds, $f : X → Y$, is it then true that the rational Pontrjagin class $p\_1(X) \in H^4(X,\mathbb Q)$ equals the pull-back $f^\* p\_1(Y)$?


If $X$ and $Y$ are compact, then I understand that this is the famous Novikov result. I am, however, unsure if the result holds in the non-compact setting. I am aware of papers that claim the result for "smooth manifolds" -- but I have not been able to find out if "manifolds" are meant to be compact by the authors.


The spaces $X$ and $Y$ that I have in mind are Zariski-open subsets of complex-algebraic varieties, and therefore topologically harmless. Would that be of any help?



Topological invariance of rational Pontrjagin classes for non-compact spaces
I am not sure about the reference, but here is an argument: the map $BO\to BTOP$ induces an isomorphism on rational cohomology, as mentioned, e.g., on p.2 of [Dalian notes on rational Pontryagin classes](https://arxiv.org/pdf/1507.00153.pdf) by Weiss. The topological rational Pontryagin class $p\_i$ is an element in $H^{4i}(BTOP;\mathbb Q)$ that corresponds to the usual Pontryagin class under the isomorphism. For a topological manifold $M$ its tangent microbundle is a homotopy class of maps $\tau\_M: M\to BTOP$. By definition, the Pontryagin class of $M$ is $\tau^\*\_M p\_i$, the $\tau\_M$-image of $p\_i$ under the map $\tau\_M$ in rational cohomology. If $h: N\to M$ is a homeomorphism, then $\tau\_M\circ h$ and $\tau\_N$ are homotopic as maps from $N$ to $BTOP$. Since homotopic maps induce the same map on cohomology, $h^\*$ sends $\tau^\*\_M p\_i$ to $\tau^\*\_N p\_i$. Thus the real work is in showing that $BO\to BTOP$ is an isomorphism on rational cohomology.


