# Topological cobordisms between smooth manifolds

Wall has calculated enough about the cobordism ring of oriented smooth manifolds that we know that two oriented smooth manifolds are oriented cobordant if and only if they have the same Stiefel--Whitney and Pontrjagin numbers.

Novikov has shown that rational Pontrjagin classes may be defined for topological manifolds; thus smooth manifolds which are topologically cobordant have equal Pontrjagin numbers. It is also easy to see that they have the same Stiefel--Whitney numbers (for this they only need to be Poincare cobordant).

It follows that smooth manifolds which are topologically cobordant are in fact smoothly cobordant. Is there a direct geometric proof of this fact?

• Is this analogous, or equivalent through Thom-Pontrjagin construction, to asking whether or not within the homotopy class of a continuous map between two smooth manifolds, there is a choice of a smooth map, which vaguely I would imagine would follow from Weierstrass theorem?! – user51223 Jun 12 '15 at 12:05
• No: it is not the quality of the map that affects the outcome of the Pontrjagin--Thom construction, but the quality of the bundle. – Oscar Randal-Williams Jun 23 '15 at 14:35
• What about the unoriented case? Is there a direct proof that if a smooth manifold bounds a topological manifold (or even a Poincare complex) then it bounds a smooth manifold? – Tom Goodwillie Mar 3 '17 at 20:47
• I can see that the Pontryagin classes are defined for non-smooth manifolds (more generally for any oriented topological $R^n$-bundle) because the map $BSO\to BSTOP$ is known to be a rational homotopy equivalence. But why Stiefel-Whitney numbers (or classes?) can be defined for non-smooth manifolds? – Victor Jan 4 '18 at 1:58
• @Victor: on a closed smooth manifold, the Wu formula describes how to recover the Stiefel-Whitney classes from Poincaré duality and the Steenrod module structure of mod 2 cohomology. The Steenrod module structure and Poincaré duality are present on closed topological manifolds, so one can use them in the same way to define Stiefel-Whitney classes. Then Stiefel-Whitney numbers can be obtained by evaluating on the fundamental class as usual. – Arun Debray Jun 26 '18 at 19:53