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Let $\Omega \subset \mathbb R^n$ an open and $\varphi:\Omega \to \mathbb R^n$ a local homeomorphism. Prove that $\varphi$ is an open application, that mean that $\varphi(U)$ is open for all $U\subset \Omega $.

Here's my proof. For all $x\in \Omega $, there is $U_x$ open s.t. $\varphi|_{U_x}: U_x\longrightarrow \varphi(U_x)$ is a homeomorphism. Then, $$\Omega =\bigcup_{x\in \Omega }U_x,$$ and thus $$\varphi(\Omega )=\bigcup_{x\in \Omega }\varphi(U_x)$$ which is open. Therefore $\varphi$ is open.

Does it work ?

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  • $\begingroup$ Why $\varphi(U_x)$ is open in $\mathbb R^n$ ? There is a priori no reason... $\endgroup$
    – Surb
    Aug 13, 2018 at 12:29
  • $\begingroup$ @Surb : what ? It's a local homemorphism, i.e. for all open $x\in \Omega $ there is a neighborhood$U_x$ of $x$ s.t. $\varphi:U_x\to \varphi(U_x)$ is a homeomorphism. Therefore $\varphi(U_x)$ is open. $\endgroup$
    – Peter
    Aug 13, 2018 at 12:31
  • $\begingroup$ First, it's not the definition of being a local homeomorphism. Secondly, if this was the definition of being a local homeomorphism, then there is no reason for $\varphi(U_x)$ being open in $\mathbb R^n$. $\endgroup$
    – Surb
    Aug 13, 2018 at 12:33
  • $\begingroup$ en.wikipedia.org/wiki/Local_homeomorphism @Surb is correct. Also, you prove that $\phi(\Omega)$ is open which is not your goal. The idea of the proof is correct, though. $\endgroup$
    – Daniel Robert-Nicoud
    Aug 13, 2018 at 12:34
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    $\begingroup$ Also, $Y$ is always open in $Y$... $\endgroup$
    – Daniel Robert-Nicoud
    Aug 13, 2018 at 12:38

1 Answer 1

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Just to fix thing, I will add an answer. Let $A,B\subset X$. Then $\varphi: A\longrightarrow B$ is a local homeomorphism if for all $x\in A$, there is an open $U$ of $X$ that contain $x$ and an open $W$ in $X$ that contain $\varphi(x)$ s.t. $$\varphi: A\cap U\to B\cap W$$ is a homeomorphism. Now, $\varphi$ will be open mean that if you take an open $O$ of $A\cap U$ (and not of $X$), then $\varphi(O)$ will be open in $B\cap W$ (but not in $X$). In other word, for all open $O'$ of $X$, there is an open $U'$ of $X$ s.t. $$\varphi(O'\cap A\cap U)=W\cap B\cap U'.$$

Now to apply to you problem : $\Omega $ is already open in $\mathbb R^n$. It's a local homeomorphism, therefore, for all $x\in \Omega $ there is $B_x$ open in $\mathbb R^n$ that contain $x$ and $B_{\varphi(x)}$ open of $\mathbb R^n$ that contain $\varphi(x)$ s.t. $$\varphi|_{B_x\cap \Omega }: B_x\cap \Omega \longrightarrow B_{\varphi(x)}\cap \mathbb R^n=B_{\varphi(x)}$$ is a homeomorphism. Therefore, $$\varphi(\Omega) =\varphi\left(\bigcup_{x\in \Omega }B_x\cap \Omega \right)=\bigcup_{x\in \Omega }\varphi(B_x\cap \Omega )=\bigcup_{x\in \Omega }B_{\varphi(x)},$$ which is open in $\mathbb R^n$. Now we are done.

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  • $\begingroup$ Oh yes, I see my mistakes. Very nice, thank you very very much. That helps a lot. I indeed really didn't understand what local homeomorphism was before. Just one thing, is the set $W$ at the beginning of your answer is $\varphi(U)$ or not ? $\endgroup$
    – Peter
    Aug 13, 2018 at 12:56
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    $\begingroup$ You are welcome :-) Since $\varphi|_{A\cap U}$ is injective, you have $\varphi(A\cap U)=\varphi(A)\cap \varphi (U)$. Now, if $\varphi(A)=B$ then you'll have $\varphi(U)=W$, but there is no reason for $\varphi(A)=B$ for reason I said previously. Just in your example, $\varphi(\Omega )\neq \mathbb R^n$ a priori. After, if $\varphi$ is itself an homeomorphism, then of course $\varphi(A)=B$... $\endgroup$
    – Surb
    Aug 13, 2018 at 13:04

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