Showing that a neighborhood is not homeomorphic to the open subset of $R^n$ 
This is an exercise from Boothby that I am stuck at. It is trivial that the boundary of $H^n$ is a manifold of dimnension $n-1$ because it is homeomorphic to $R^{n-1}$. However I cannot show that no neighborhood in $H^n$ of a point of the boundary can be homeomorphic to an open subset of $R^n$. I think I have to use the invariance of domain, that is, $R^n$ and $R^m$ is homeomorphic if and only if $n=m$. Could anyone please help me how to prove the second part of this exercise? 
This is the definition of $H^n$.

 A: If $p\in\partial H^n$, and $U$ is a neighborhood of $p$, then choose an open convex $V$ containing $p$. Then $V$ and $V\setminus p$ are contractible by convexity to $p+\epsilon v$, where $v$ is any vector pointing into the interior of the half-space, and $\epsilon$ is small enough that $p+\epsilon v\in V$. In fact this gives a homotopy equivalence of pairs $(V,V\setminus \{p\})\simeq (\{p\},\{p\})$ Thus $H^i(V,V\setminus p)=H^i(\{p\},\{p\})=0$ for all $i$. By excision, $H^i(U,U\setminus\{p\})=0$. Thus $U$ has trivial local homology at $p$.
However, if $U$ were homeomorphic to an open subset of $\Bbb{R}^n$, $H^n(U,U\setminus\{p\})=\Bbb{Z}$, since if $B$ is a small ball around the image of $p$ in the image of $U$ in $\Bbb{R}^n$,
which I'll just call $p$ as well, 
we have $H^n(U,U\setminus\{p\})\cong H^n(B,B\setminus \{p\})$, and
from the long exact sequence of the pair, since $B$ is contractible, 
$$H^n(B,B\setminus\{p\})\cong H^{n-1}(B\setminus\{p\})\cong H^{n-1}(S^{n-1}) = \Bbb{Z}.$$
Thus $U$ cannot be homeomorphic to an open subset of $\Bbb{R}^n$.
