Given a connected $n$ - manifold $M$ and assume that for any compact set $K \subset M$ we have $H_{n}(M , M - K , \mathbb{Z})$ isn't trivial . Prove that $M$ is orientable .
As an obvious corollary if $M$ is compact , connected and nonorientable we have $H_{n}(M, \mathbb{Z}) = 0$
And a general version :
If $M$ is a compact , connected , nonorientable , $n$ - dimensional manifold and $G$ is an abelian group then
$$H_{n}(M , G ) =\, _{2}G$$
Where $_{2}G$ is subgroup of $G$ contains all elements $g$ satisfy $2g = 0$
A manifold $M$ always satisfy the following :
$$H_{n}(M, M - x) \cong H_{n}(\mathbb{R^{n}},\mathbb{R^{n}}-x)\cong \mathbb{Z}$$
Each of choice the generator of $H_{n}(M , M - x)$ is called a local orientaion . $M$ has an orientation if every point $x \in X$ has a neighborhood $N$ and a function assigns $x$ to $\mu_{x} \in H_{n}(M,M-x)$ and an element $\mu_{N} \in H_{n}(M , M - N) : i_{*}(\mu_{N})=\mu_{y}$ with homomorphism induced by inclusion :
$$i_{*} : H_{n}(M , M - N) \to H_{n}(M , M - y)$$
For each $y \in N$ . $M$ is said to be orientable if it admids at least one orientation , otherwise , it is called nonorientable
Hint: 1. A manifold $M^n$ is nonorientable if and only if it contains a compact submanifold with boundary $K$ homeomorphic to the total space of the unique nontrivial $B^{n-1}$-bundle over the circle. 2. Use the excision.