In ordinary homology, the classical results give the following situation:
for a compact, connected, topological manifold $M$ of dimension $n$ we have, for each ring $R$, that $H_n(M,M \setminus \{x\};R) \simeq R$. The orientation of $M$ is a section $o$ such that $o(x) \in H_n(M,M \setminus\{x\};R)$ in a generator of this homology module, for each $x \in M$. This definition heavily uses the 'machinery' of this particular homology theory: in other words, it is highly nonaxiomatic. However, I met the sentence that "for each (co)homology theory there is a notion of orientability". So here is my (possibly a little bit vaque):
Question 1 Is the notion of orientability in any (co)homology theory axiomatic? If so, how to interpret the above, particular definition in axiomatic terms?
Recently I've got interested in $K$theory and $K$homology. These are homology theories for $C^*$algebras: however, the construction doesn't remind the standard constructions in homological algebra. To be more precise, it doesn't involve chain complexes and typical cycle in the theory is not an element in kernel of some (boundary) operator modulo boudary.
What happens, is that one have only two $K$theory groups, in contrast to ordinary homology. I know that orientability in $K$theory is the existence of $spin^c$ structure but let me ask:
Question 2 What is the definition of $K$orientable space?
It happens, that the notion of orientability (via singular homology), when restricted to smooth manifolds, is the same as ordinary orientability. So it is also very natural to ask:
Question 3 For what class of spaces $K$orientability is still well defined?
Finally, there is a notion of a fundamental class in singular cohomology: it is an image of the orientation under the isomorphism $\Gamma\Big(M,\bigcup_{x \in M}H_n(M,M \setminus \{x\};R)\Big) \simeq H_n(M;R)$. As before, this is very specific, so again:
Question 4 is it possible to define the notion of a fundamental class axiomaticly?
I apologize for a rather long question: I would be grateful if anybody could clarify this issue for me.

2$\begingroup$ Orientability of a vector bundle is the esistence of a Thom class. Orientability of a manifold = that of its tangent bundle. $\endgroup$– Alex DegtyarevOct 14, 2014 at 16:58

6$\begingroup$ Every (co)homology theory is represented by a spectrum, and orientability, orientations, and fundamental classes can all be expressed as structures associated to a spectrum. This is all worked out in Rudyak's book "On Thom Spectra, Orientability, and Cobordism". In particular chapter 5 works out the general theory (yielding the axiomatic description you want) and chapter 6 specializes to Ktheory. $\endgroup$– Paul SiegelOct 14, 2014 at 17:13

$\begingroup$ This paper, may be of some use to you. $\endgroup$– Nerses AramianOct 14, 2014 at 22:06

$\begingroup$ @Paul Siegel: don't you need a ring spectrum? $\endgroup$– John KleinOct 15, 2014 at 16:55
1 Answer
Ad question 1: Let $E$ be any ring spectrum, with associated cohomology and homology theory. An $E$orientation of a rank $n$ vector bundle $V \to X$ is a class $u \in E^n (V,V0)$ such that the restriction to any point $x \in X$, i.e. the image in $E^n (V_x,V_x0) \cong E^n (R^n,R^{n}0) \cong E^0$ is a generator. For a closed manifold $M^n$; an orientation is a class $a \in E_n (M)$, such that for all $x \in M$, the image of $a$ in $E_n (M,Mx) \cong E_n (R^n, R^n0)=E_0$ is a generator.
This is an axiomatic definition. For smooth manifolds, orientations of $TM$ and of $M$ are in bijection by the AtiyahMilnor duality theorem.
Proving that a vector bundle has an $E$orientation in practice is a different question. This is \it{almost never} a tautology. In principle, there is a homotopy theoretic strategy; but the computations quickly become intractable. Look in Rudyak's book ''Thom spectra...'', chapter VI, to see what I mean (there he discusses KTheory).
If there is a concrete geometric model of $E$Theory, the problem is much easier. For example, if $E$ is real $K$Theory, then a vector bundle is $KO$orientable iff it is spin. This depends heavily on the description of $K$theory in terms of vector bundles, i.e. it relies on linear algebra related to Clifford algebras. There might exist a purely homotopytheoretic proof, but I bet that it is quite hard and Rudyak's exposition does not give a homotopytheoretic proof of that fact.

1$\begingroup$ For a more concise version of Rudyak's book one may look at map.mpimbonn.mpg.de/… $\endgroup$– ThiKuOct 15, 2014 at 15:30