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This question has been asked before but in a slightly different patterns of solving_ideas

My point is next:

Let $M$ be a manifold and $p$ a closed point in $M$, such that $p$ is inside an open neighbourhood $U$ homeomorphic to $\mathbb R^n$.

Then: $H_n(M, M - p) = Z, H_i(M, M - p) = 0$ for $i\ne n$.

Proof:

Note that since $p$ is closed, $M - p$ is open. Excision at $p$ yields:

$H_i(M, M - p)$ $\cong$ $H_i(U, U - p)$.

Now it's written: The result holds for $U$, since $U$ is homeomorphic to $\mathbb R^n$, and hence we are done.

But I don't see why result holds for $U$. Everything else is clear for me.

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    You can compute the homology of $(\Bbb R^n, \Bbb R^n-0)$ by the long exact sequence of this pair. Every point $p$ is closed in a manifold, by the way. – 2017-02-12
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    But for that I already need to know homologies of $\mathbb R^n$ and $\mathbb R^n$ \ {0} , right? – 2017-02-12
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    Yes but $\Bbb R^n$ is contractible and $\Bbb R^n-0$ deformation retracts to a sphere,the homology of which you can again compute using induction and Mayer-Vietoris, for example – 2017-02-12
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    Great! Thanks a lot. – 2017-02-12

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