# Quail: Cl. mnfld. story

I did this around 1/5/09. Here is always an -manifold. Here orientation always refers with respect to some ring . Some generalities. is shorthand for local homology. For there’s an inclusion giving natural maps .

For me one of the big motivations for this story is the following result.

If is closed and connected then is compact. This is an easy corollary of

Orientation result:

a) orientable implies is iso.

b) non orientable implies is injective with image

c) for .

pf of compact: take a generator and define an orientation via where . Now only for in the closure of the image of some cycle representing .

If misses a point then it is contained in the complement , hence represents zero in .

By the result is always nonzero; on the other hand the prev. comment says it can only be nonzero on the image of cycle which has compact image, hence is surjective and is compact.

The orientation result is a special case of

Orientation Lemma:

A) with compact, and then s.t. for .

B) for .

proof: Step 1, Reduce to the case .

For arb. right where and . (sim. to cov. spaces, van kampfen, schemes). Now use an induction argument.

The case is covered by the reduction hypothesis. If it holds for , then adding , have that it holds for .

The induction argument is finished because the result for implies it for via the Mayer Veitoris sequence.

B) now follows easily via the l.e.s by things being squeezed by zeros. As for A), basically get classes and the l.e.s shows they glue. As for uniqueness the idea is for all implies , this is proved via it holds for and the first map in l.e.s is injective.

Step 2, , and reduce to the case is convex. In fact if it holds for convex things it holds for a countable union of convex things via a similar argument as step 1.

Assuming it holds for convex things, the idea is to treat an arb. as if its coming from slightly bigger which is a union of convex things.

with , then the boundary is compact and it has positive distance from and so each point has a closed ball nbd not intersection . The union of these balls is . The key idea, is by construction the relative cycle and it lifts .

B) can now be proved by taking to be generator of , it lifts to zero so it must have been zero to begin with. As for A), uniqueness uses for all means its zero for all , this is proved via a commutative triangle having legs

where is any ball . For existence just take the answer for any ball .

Step 3, and convex.

Here the result is evident via . B) That the higher homology is zero is clear. As for A), uniqueness is clear for similar argument as before. Existence is just taking a generator of .

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- Published:
- January 12, 2010 / 12:17 am

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- Quail

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