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Let $U$ be an open sub-set of $\mathbb{R}^2$, and let $f$ be a continuous function $f:U\rightarrow \mathbb{R}^2$.

I'd like to show that $f$ is an open map, given that for each $u\in U$ exists an open set $V_u$ in $U$ ($V_u\subseteq U$) such that $f\uparrow_{V_u}$ is one-to-one.

I'm not quite sure on where to start here?

And assuming I'll prove this, does this mean that $f$ is homeomorphism? Because if $f$ is open so $f^{-1}$ is continuous, right?

Eric_
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3 Answers3

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Let $ V $ be an open subset of $U$. Write $ V $ as a union of open sets $\{V_i\}_{i\in I}$ on which $f$ is 1:1. Apply the Invariance of Domain theorem to each open set $V_i$. Then recall that the union of open sets is open.

Also note that a map like $f$ isn't a homeomorphism, using Chocosup's example or similar. My favorite similar example is to take the open unit square, stretch it to the rectangle $(0,9)\times(0,1)$, and finally wrap it onto (covering $(0,1)\times(0,1)$ twice) $\{(0,3)\times(0,3)\} - \{[1,2]\times[1,2]\}$.

Edit on decomposing $ V $: For each $ v\in V $ let $ \hat V_v $ be a neighborhood with $ f $ 1:1. Then let $ V_v := \hat V_v \cap V $. Because $\mathbb{R}^2$ is reasonable you could slim down the collection $V_v$ to a countable collection, but it isn't needed here.

Joe Manlove
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  • Thanks! But if I understood it correctly this would show that $f(U)$ is open. If I want to show that $f(A)$ for each open $A\subseteq U$ is also open, can I assume there are ${V_i }_{i\in I} $ such that their union is $A$? I'm just not sure if this is an obvious assumption, or not. – Eric_ Oct 10 '14 at 10:15
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    Whoops, forgot you had used $U $, my bad. Edited to address that and the decomposition. – Joe Manlove Oct 10 '14 at 14:49
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    Assuming that I know something a bit weaker, that for every $u∈U$ apart from a single $u_0$ exists such a $V_u$ so that $f$ is $1:1$ on that $V_u$, can I still prove the same? Does this proof still stands? And if it does, can I assume that apart from $u_0,...,u_n$ and it will still work? (for every natural number of exceptions, not just a single one) – Eric_ Oct 15 '14 at 11:45
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I don't know yet how to prove your thing, but regarding your final question the answer is no. Indeed, you should have to suppose that $f$ is globally one-to-one.

You could consider this example : consider $U = \mathbb{R}^2 \setminus \{0\}$, and $f : (\rho, \theta) \mapsto (\rho, 2\theta)$ in polar coordinates. Then clearly $f$ is not an homeomorphism, but $f$ is open and satisfies your condition.

More generally, any covering map satisfies your properties but is not in general an homeomorphism.

Chocosup
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There is something wrong here!

$f$ open means that: the image by $f$ of any open in $U$ is open.

@Eric: You assume that $f$ is a local bijection: any $x\in U$ admits an open neighborhood $V_x\subset U$ such that $f(V_x)$ is one-to-one. (a constant map is continuous but not a local bijection).

From this assumption you would like to prove that $f$ is open. As suggested by Joe Manlove, form the invariance of domain, you can prove that is a local homeomorphism and open. You can't deduce that $f$ is an homeomorphism unless it's one-to-one.

Another case of interest for you is, when $f$ is a submersion (its differential is everywhere onto), you can prove that $f$ is open.

amine
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