Task
Let $M$, $N$ be smooth manifolds of dimension $m$ and $n$ respectively and $f\colon M\longrightarrow N$ a submersion.
Show that $f$ is open.
My Proof
Let $W\subset M$ be open.
Let $p\in W$.
By the theorem of constant rank,
there are a chart $(U,\varphi)$ around $p$
and a chart $(V,\psi)$ around $q := f(p)$
s.t. $f(U)\subset V$, $\varphi(p) = 0$, $\psi(q) = 0$ and
$$
\tilde f(x^1,…,x^m) := \psi\circ f\circ \varphi^{-1}(x^1,…,x^m)
= (x^1,…,x^n).
$$
We can suppose w.l.o.g. that $W\subset U$.
It seems obvious that $\tilde f(W)$ is open,
but I have trouble actually showing this.
Any ideas?
Best Answer
You showed that a submersion $f\colon M \to N$ locally looks like a projection with respect to suitable charts: $$\tilde f = \psi\circ f\circ\varphi^{-1}\colon\mathbb R^m\to\mathbb R^n$$ with $m \ge n$ where $\mathbb{R}^m \cong\mathbb{R}^n \times \mathbb{R}^{m-n}$. It is a known fact in topology that a projection is an open map, see for example the question Projection is an open map.
Now, as both $\psi$ and $\varphi^{-1}$ are diffeomorphisms, the composed map $\tilde f$ is open iff $f$ is open. This concludes the proof.