2
$\begingroup$

enter image description here

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 $H^n$.

enter image description here

$\endgroup$
  • $\begingroup$ Unless I've misread the question, this is the so called "invariance of boundary". Proving this is decidedly non-trivial, and isn't done until chapter 17 of Lee's textbook on smooth manifolds, or the second section of Hatcher's book on Algebraic Topology. $\endgroup$ – Duncan Ramage Jan 5 '18 at 5:40
  • $\begingroup$ The question assumes invariance of domain. Isn't it possible to just use the invariance theorem to prove it? $\endgroup$ – Keith Jan 5 '18 at 5:55
  • $\begingroup$ Invariance of domain and invariance of boundary are different theorems. Unfortunately, I think jgon's solution is the simplest out there. $\endgroup$ – Duncan Ramage Jan 5 '18 at 6:01
3
$\begingroup$

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 $\Bbb{R}^n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.