Top Homology of Manifolds via Formal Homology Theory

In the lecture on algebraic topology, which I am currently attending, we tried to proof as much as possible using just the Eilenberg-Steenrod axioms for homology theory before we constructed singular homology. In particular we introduced the topic of homological orientation of manifolds by saying that a topological manifold is orientable, if it admits a fundamental class. We were not told however, how $$H_n(M)$$ can be calculated, so it is not clear to me that fundamental classes should exist at all.

Researching homological orientation I found the definition of orientability via local orientations (noting that $$H_n(M,M\setminus \{x\}) \cong H_n(\mathbb S^n) \cong R$$ for $$H_\bullet$$ being a ordinary additive homology theory with coefficient ring $$R$$).

Is there a proof of the fact that for a connected closed manifold $$M$$ we have $$H_n(M) = R$$ if and only if $$M$$ is $$R$$-orientable, which avoids details of the construction of singular homology, i.e. uses tools from formal homology theory only (e.g. Mayer Vietoris sequence, the triple exact sequence or excission)?

The proofs of Hatcher or Greenberg-Harper use a rather technical lemma, which seems to rely on the explicit representation of cycles. For me this is a problem in the sense that we worked quite alot with fundamental classes before even constructing singular homology. As I prefer to learn things in order, I would have to reorder the whole lecture when preparing for the exam, so I am keen to have at least the implication $$R$$-orientable $$\Rightarrow$$ top homology is nonzero cyclic $$R$$-module.

PS: I would be interested in a formal (i.e. following from the Eilenberg-Steenrod axioms and not involving the construction of singular homology) way to proof any of the following facts:

• For an $$n$$-dimensional topological manifold $$M$$ it holds that $$H_k(M) = 0$$ for $$k>n$$
• For a connected closed manifold $$M$$ it holds that $$H_n(M) = 0$$ if and only if $$M$$ is not $$R$$-orientable
• For a connected noncompact manifold $$M$$ it holds that $$H_n(M)=0$$

Update
I want to give an update on what I have tried so far. Please be aware, that I am currently attending the lecture, so I have no advanced understanding in the tools involved and what turned out to fail for me may actually work.

• I tried my best to apply any of the long exact sequences onto canonical choices of subsets of the manifold, but to no avail.
• Another approach was to cover the compact manifold $$M$$ with finitely many open subsets homeomorphic to $$\mathbb R^n$$ and try to use Mayer Vietoris to calculate the top homology group. However I could not find a version of Mayer Vietoris using a cover with more that $$2$$ sets and I could not come up with a way to inductively circumvent this.
• My last approach was to find a CW-structure on the manifold. Besides being circular in the sense that we used degrees (and in particular orientations) to show that cellular homology is isomorphic to a given ordinary homology theory, this would have been nice to at least get the result that the homology in degrees greater than $$n$$ vanish. It turns out though, that one does not actually know, whether each topological manifold admits a CW-structure. The best I could find was some kind of approximation by CW-structures, but this seems to be linked to homotopy groups and hence (as far as I know) to singular homology. Even assuming that higher homology groups vanish did not enable me to show my original question, so it would not have helped anyway.
• I doubt that it is possible. Usually the proof is given in the context of "duality on manifolds". This involves cohomology and products (cap-, cup- and slant-products). I do not see how you get that stuff only via the homology axioms. But perhaps I am too pessimistic ... – Paul Frost Jan 25 at 16:02
• I start to think that too. There seem to be alot of properties inherent to singular homology, which require the explicit description of cycles and boundaries. Moreover the treatment and notation of homology theories seems to vary quite alot, so it is hard to see, whether e.g. an exercise is meant to be made using explicit descriptions or axiomatic methods... – PrudiiArca Jan 25 at 16:27
• You should have a look at "Switzer, Robert M. Algebraic topology-homotopy and homology. Springer, 2017." He does everything for homology theories defined by spectra. – Paul Frost Jan 25 at 16:44
• Thank you for the reference! I could not find the requested result there yet, but I will double check. – PrudiiArca Jan 25 at 16:47
• Look at Theorem 14.13. If you take the Eilenberg-MacLane-spectrum, then you get the desired result. But all that is again not axiom-based. – Paul Frost Jan 25 at 16:53

Have a look at the book written by Bredon; "Topology and Geometry". In the chapter about the orientation bundle he proves this statement and if I'm not mistaken he only uses properties of homology you get out of the axioms. (I'm sorry i cannot rephrase the whole proof here since it's long and contains many elaborate ideas but if you'd wish I could provide you with some more details.)

• This looks very promising. I am a little concerned about one detail though. In theorem 7.8 (iii) there is the remark „where the top isom comes from any chain being contained in compact subset, the isom already holds on chain level“. Doesnt this use the explicit description of chains? – PrudiiArca Jan 27 at 23:21
• @PrudiiArca I haven't access to Bredon's book, I only saw the contents via Amazon preview. It seems again to be in the context of "duality on manifolds" involving cohomology and products. – Paul Frost Jan 31 at 13:09
• Afaik Bredon, Hatcher and Greenberg-Harper all use an orientation sheaf/bundle and relate this to the homology groups. But this relating morphism uses cycles having compact support (which definitely relies on the explicit description of singular homology) and even if one can circumvent this, the main theorem uses an exhaustion of the manifold by compact sets, which I cannot handle just with the axioms of a homology theory – PrudiiArca Jan 31 at 15:03