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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?

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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 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.

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  • $\begingroup$ So this does indeed work for non-compact manifolds! Dear Igor, thank you so much for this extremely helpful reply. I greatly appreciate it! $\endgroup$
    – Stefan Kebekus
    Commented Mar 3, 2023 at 15:22
  • $\begingroup$ One place the statement appears in print is Proposition 10.2 of the final Kirby-Siebenmann essay. $\endgroup$
    – Connor Malin
    Commented Mar 3, 2023 at 15:29
  • $\begingroup$ @ConnorMalin: as far as I can see Proposition 10.2 says that $BSO\to BSTOP$ is a rational homotopy equivalence, which uses that the spaces are simply-connected. The non-orientable case is a bit more hassle. One can reduce to the orientable case by taking Whitney sum of tangent bundles with line bundles corresponding to $w_1$, which is a homeomorphism invariant, and then use the Whitney sum formula for Pontryagin classes to show that $p_i$ is a topological invariant. $\endgroup$
    – Igor Belegradek
    Commented Mar 3, 2023 at 15:40
  • $\begingroup$ Right, I should have mentioned Kirby proved $\pi_0(\mathrm{TOP})=\mathbb{Z}/2$ which then immediately gives the nonorientable version. $\endgroup$
    – Connor Malin
    Commented Mar 3, 2023 at 15:45
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    $\begingroup$ By the LES of a fibration, the map $\mathrm{STOP}/\mathrm{SO} \rightarrow \mathrm{TOP}/\mathrm{O}$ is an equivalence on homotopy groups if $B\mathrm{SO} \rightarrow B\mathrm{O}$ and $B\mathrm{STOP}\rightarrow B\mathrm{TOP}$ are equivalences on homotopy groups. For higher homotopy groups the latter two statements are automatic since this is an inclusion of path components. For the fundamental group, this follows from Kirby's result that $\pi_0(\mathrm{TOP})=\mathbb{Z}/2$. $\endgroup$
    – Connor Malin
    Commented Mar 3, 2023 at 16:23

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