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Oct 4, 2017 at 16:17 comment added მამუკა ჯიბლაძე @EvgenyShinder It is defined for any spaces afaik. Pushforward requires manifolds, I believe. Isomorphism needs K-orientability, which is stronger than ordinary orientability (approximately - stable spin$^c$-structure).
Oct 4, 2017 at 16:01 answer added Dmitry Vaintrob timeline score: 2
Oct 2, 2017 at 8:55 comment added Evgeny Shinder vap: My understanding is that wrong way (pushforward) maps on topological K-theory will only be defined under additional assumptions such as spaces being smooth and compact (or at least the space on the top is smooth, and the map is proper). Otherwise pushforward of a vector bundle does not have to be a finite complex of vector bundles. At least that's how it is in algebraic K-theory: K-theory is contravariantly functorial for all morphisms, and G-theory is covariantly functorial for proper morphisms.
Oct 2, 2017 at 8:51 comment added Evgeny Shinder მამუკა ჯიბლაძე: for the Baum-Douglas cycles, is the basic assumption that $X$ is a manifold? Would then K-cohomology and K-homology canonically isomorphic (in the compact manifold case)?
Oct 2, 2017 at 8:49 comment added Evgeny Shinder user36931: I looked at the paper of Thomason, and I don't quite yet see how it's relevant. Do you mean his approach of inverting the Bott element in the Algebraic K-theory to relate to topological K-theory?
Sep 30, 2017 at 7:14 comment added vap Topological K-theory can be defined by looking at complexes of vector bundles (or locally free sheaves) and can be endowed with covariant functoriality by using wrong way maps, as in the definition of topological index, or à la Grothendieck.
Sep 29, 2017 at 18:27 comment added მამუკა ჯიბლაძე Closely related objects, providing a link between $G$-like homology classes and "true" cycles are Baum-Douglas cycles. Very briefly, such cycles on a manifold $X$ consist of a submanifold $Q\hookrightarrow X$ together with a spin$^c$ structure on $Q$; with appropriate bordism relation they indeed give $K$-homology. To obtain a "$G$-like" class out of it, take pushforward to $X$ of the tangent bundle of $Q$; under certain conditions it is represented by a finite complex of vector bundles. See the link for details and references.
Sep 29, 2017 at 15:04 comment added user36931 Your guess in 2) is basically correct. You can have a look at papers of Thomason for precise details. See for example projecteuclid.org/download/pdffirstpage_1/euclid.dmj/1077306718 and references there-in
Sep 29, 2017 at 12:39 history asked Evgeny Shinder CC BY-SA 3.0