Show that a submersion is open 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?
 A: 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.
A: Let $W\subset U$ be open.
In particular, $\varphi(W)$ is open and
$$\tilde f(\varphi(W)) = \{(x_1,...,x_n)\mid (x_1,...,x_n,x_{n+1},...,x_{m})\in \varphi(W)\}.$$
Let $x\in\tilde f(\varphi(W))$, i.e. there is $(x_1,...,x_m)\in\varphi(W)$ s.t.
$$x = (x_1,...,x_n)=\tilde f(x_1,...,x_m).$$
Since $\varphi(W)$ is open, there is $\varepsilon>0$ s.t. $$(x_1-\varepsilon,x_1+\varepsilon)\times \dots \times (x_m-\varepsilon,x_m+\varepsilon) \subset \varphi(W)$$
and thus $(x_1-\varepsilon,x_1+\varepsilon)\times\dots\times (x_n-\varepsilon,x_n+\varepsilon) \subset \tilde f(\varphi(W))$.
This proves that $\tilde f(\varphi(W))$ is open.
And since $$f=\psi^{-1}\circ\tilde f\circ \varphi$$ the claim follows.
