# The general idea of prove openness.

I never really get the idea of proofs involves openness, here's an example:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a local diffeomorphism. Prove that the image of $f$ is an open interval.

So, is the general principle is to show:

• Step n: For an arbitrary $y \in$ $f(U)$, which is the image of an open neighborhood, is in an open neighborhood.

• Step n-1: We find a neighborhood of $y$ open in a open neighborhood, which we chose $\mathbb{R}$ here.

• Step n-2: By the definition of function, $\forall y \in f(U), \exists x \in U: f(x) = y.$ So we choose an open neighborhood of $x, U_x.$

• Step n-3: ..... .

No, you show that $f(U)$ is open for all open subsets $U \subset \mathbb{R}$. Then, since $f$ is continuous, you know that $f(\mathbb{R})$ is a connected open subset of $\mathbb{R}$, hence an open interval.
To show that $f$ is an open mapping, consider an open $U \subset \mathbb{R}$ and any $x \in U$. By assumption, there is an open neighbourhood $V_x \subset U$ of $x$ on which $f$ is a diffeomorphism. That means $f\bigl\lvert_{V_x}$ is open, hence $f(V_x)$ is open. Now, $U = \bigcup\limits_{x \in U} V_x$, and therefore $f(U) = \bigcup\limits_{x \in U} f(V_x)$ is open.
• @user83036 A diffeomorphism is in particular also a local diffeomorphism (and a homeomorphism), so diffeomorphisms are open (and closed) mappings. But here, we don't have a global diffeomorphism given, all that we have is that it is a local diffeomorphism (from that, by connectedness, you can then deduce that it is a global diffeomorphism [to $f(\mathbb{R})$]). – Daniel Fischer Jul 1 '13 at 20:54
• It's not clear to me that $V_x \subseteq U$. So this argument seems to require some adjustment. – Ink Jul 1 '13 at 20:56
• @Ink If $f\bigl\lvert_V$ is a diffeomorphism, so is $f\bigl\lvert_{V \cap U}$. Or restricted to any smaller (open) set. – Daniel Fischer Jul 1 '13 at 20:58