For me this appears in the context of Hatcher exercise 6.a chapter 3.3. But the question stands on its own.
Let $M$ be an $n$ dimensional closed manifold, and $p\in M$ an arbitrary point, my goal is to describe as much as possible the homology of $M\backslash\{p\}$.
So far here are my thoughts, up to homotopy this is the same as the homology of $A=M\backslash D^n$ where $D^n$ is an open disc. And so we have $M=A\cup \bar{D^n}$ with $\bar{D^n}\cap A=S^{n-1}$.
So we can use Mayer Vietoris to study $H_i(A)$. For me the cases $1<i<n-1$ all follow without any extra effort, and the case $i=1$ also follows without too much trouble. It is the two remaining cases which give me trouble $$...0\to H_n(A)\to H_n(M)\to H_{n-1}(S^{n-1})\to H_{n-1}(A)\to H_{n-1}(M)\to 0$$
If $M$ is non orientable we immediately solve $H_n(A)=0$, and have $0\to\mathbb{Z}\to H_{n-1}(A)\to \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}^r\to 0$. I don't think there is an argument to solve this extension problem in general, but perhaps in this specific case where the maps come from a topological situation it is possible?
In the orientable case we get $H_n(A)<H_n(M)$ and so is isomorphic to $\mathbb{Z}$ or $\{0\}$. Furthermore in the former case because the next group in the exact sequence is also $\mathbb{Z}$, which implies the injection $0\to H_n(A)\to H_n(M)$ is an isomorphism, thus for the last undetermined homology group we get the short exact sequence
$$0\to \mathbb{Z}\to H_{n-1}(A)\to \mathbb{Z}^r\to 0$$
This sequence splits, so $H_{n-1}(A)=H_{n-1}(M)\oplus \mathbb{Z}$.
In the case where $H_n(A)=0$ we get the exact sequence $$0\to H_n(M)\to H_{n-1}(S^{n-1})\to H_{n-1}(A)\to H_n{n-1}(M)\to 0$$ Which becomes $0\to \mathbb{Z} \xrightarrow{\cdot d} \mathbb{Z}\to H_{n-1}(A)\to \mathbb{Z}^r\to 0$. We can solve this extension problem and get $H_{n-1}(A)=\mathbb{Z}/d\mathbb{Z}\oplus H_{n-1}(M)$, but I don't think $d$ can be determined in full generality.
First of all did I make any "silly mistakes" making this whole rambling useless. Second of all does anybody have any idea to compute $H_{n-1}(A)$ in the non-orientable case?
Any help is appreciated, thank you very much.