Show that a submersion is open

Question

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?

Answer 1

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.

Answer 2

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.