3

[I put a PS at the bottom of the text, as I think problem's solved]

There is a simple idea that a restriction of a homeomorphism to a subset of its domain yields a new homeomorphism onto its image. I am having problems with certain cases, where I think that this applies, but my texts conclude that the result is just continuous and bijective. When the domain is in $R^n$, domain invariance is then used.

One example is thishomeomorphic manifolds proof by John. He restricts 3 homeomorphisms, concludes that the restriction are homeomorphisms and that is it. I solved this question the same way, the maps are homeomorphic, why wouldn't the restriction of the middle homeomorphism $h$ not be ? But this proof does not seem to use domain invariance, so I wonder if it's correct.

As usual, there probably is a simple insight here that I am lacking, but I am not getting myself on the right track, so any hint appreciated.

What I have seen looks like john's answer, but was really: Mand N are homeomorphic manifolds with dimensions m and n. Obtain a contradiction if m > n.

$h:M\rightarrow N$ is a homeomorphism. for $h(x) \in N$ there is an open V in N, an open $O_v$ in $R^n$, and a homeomorphism $\psi_N :V\rightarrow O_v$.

$h^{-1}(V)$ is open in M, so there is an open $U\subset h^{-1}(V)$ around x, an open $0_u$ in $R^m$ and a homeomorphism $\psi_M : U \rightarrow O_u$.

Again, $h(U)\subset V $ is open, and the restriction $\psi_N|_{h(U)}$ is a homeomorphism onto its image $O_v'$.

My text says here that $\psi_M\ o\ h^{-1} \ o\ \psi_N|_{h(U)}^{-1}$ (edited: I should use $ \psi_N|_{h(U)} \ o\ h\ o\ \psi_M^{-1}$) is continuous and injective and uses domain invariance to conclude it is open. I fail to see why the restriction of h to U would not be a homeomorphism, and the composition is not just a composition of three homeomorphisms, which is open, or what else is the reason that domain invariance is used. The domain invariance proposition which is used is: any continuous injective function from an open $O\subset R^n$ into $R^n$ is open.

PS: following freakish' hint, I think I found the answer that all the restrictions are indeed homeomorphisms, but it is not useful to know because I need an embedding into a higher dimension, and that is where I need Brouwer's domain invariance

HongGame
  • 33
  • Without knowing *exactly* what your course text says, it's not really worth anyone's efforts to make a guess; you are as likely as not to respond "that's not what the text says" (e.g. the answer of @Esgariath). – Lee Mosher Aug 21 '22 at 14:12
  • hmm, you are completely correct. The question comes down to the question why the linked solution by john is not correct. I ll see if I can rephrase things – HongGame Aug 21 '22 at 14:25
  • 1
    I think you misunderstand the proof. Or maybe the author badly explains it. The invariance of domain is used to prove that $m$ cannot be different from $n$, not that the composition is a homeomorphism. Which isn't even relevant, it is enough if it is continuous and injective. – freakish Aug 21 '22 at 15:06
  • Thanks to your hint I looked a bit further an d I think I found it: The composition is open, but what I missed is the embedding. The function I SHOULD use is the inverse of what I put above, it should be $\psi_N o h o \psi_M^{-1}$ but after this, thats what I missed, there is an embedding $R^n -> R^m$. What I did was prove it is open in Rn, but I need domain invariance to prove it is open in $R^m$, and that is what you mean with it is not useful to know that the composition (before embedding) is open – HongGame Aug 21 '22 at 16:05

1 Answers1

0

Suppose we have $h:A\to X$ homeomorphism, $B\subseteq A, Y=h(B)$. Let $f = h_{\restriction B}:B \to Y$. You claim that $f$ is continous and bijective. Now consider $h^{-1}$. It is a homeomorphism as well, so $g = (h^{-1})_{\restriction Y}$ is continous bijection as well. I think you can already see that $g = f^{-1}$, so $f$ is open.

Esgeriath
  • 2,481
  • Thanx. But my claim would be that the restriction of f is open. So all 3 functions in john's answer would be homeomorphisms, but the proof in my course text concludes that the composition is just continuous and bijective, and I do not see why. I changed my post a little in the hope it is clearer. btw, basically John seems to prove the invariance of dimension without using domain invariance – HongGame Aug 21 '22 at 14:05