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Massey Singular Homology theory states at IX.2

Let $M$ be an n-dimensional manifold with orientation $\mu$; it would be advantageous if there were a global homology class $\mu_M$ \in $H_n(M,\mathbb{Z})$ such that for any $x \in M$ $\mu_x = \rho_X(\mu_M)$. Unfortunately this can not be true if M is non compact.

Here $\rho_x$ is induced by the inclusion map. $(M, M-M) \to (M, M - \{x\})$ It suggests also to use the following proposition

Let (X,A) be a pair consisting of a topological space X and subspace A. (a) Given any homology class $u \in H_n(X,A)$ there exists a compact pair $(C,D) \subset$. $(X,A)$ and a homology class $u'$ e $H_n(C,D)$ such that $i_*(u') = u$, where $i:(C,D) \to (X,A)$ is the inclusion map. (b) Let $(C,D)$ be any compact pair such that $(C,D) \subset (Χ,Α)$, and $v\in H_m(C.D)$ a homology class such that $i_*(v) = 0$· Then there exists a compact pair $(C',D')$ such that $(C,D) \subset (C',D') \subset (Χ,Α)$ and $j_*(v) = 0$, where $j:(C,D) \to (C',D')$ is the inclusion map.

I can't see how the latter implies the non existence of a global orientation for a non compact manifold

namenick
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  • If $X$ is an open $n$-manifold, then $H_n(X)=0$. You can find a proof of this in any algebraic topology textbook, say, Hatcher or Massey. – Moishe Kohan Nov 27 '22 at 01:26
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    Assume such a $\mu_M$ existed. The proposition tells you there is a compact $C$, such that $\mu_M$ is in the image of $H_n(C)\rightarrow H_n(M)$. Since $M$ is non-compact, there is an $x\in M-C$. Now, what happens if you look at the image $\mu_M$ under $H_n(M)\rightarrow H_n(M,M-x)$? – Thorgott Nov 27 '22 at 03:14
  • @Thorgott. Let's suppose that $\rho_x(\mu_M) = \mu_x$ then, considering the composition of the two inclusions, we have that the inclusion map $(C, \emptyset) \to (M, M - x)$ is surjective in homology. But since $C \subset M - x$ such a map is the zero map. This is a contradiction because $H_n(M, M - {x} ) \neq 0$ – namenick Nov 27 '22 at 17:29

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