I am trying to understand submersions and I ran into this explanation:
A submersion is an open map. This is because locally there exists
co-ordinates in which the submersion can be written as the projection
onto the first few co-ordinates, which is an open map.
Why is this open? All I know about submersion is that by the local submersion theorem, around any point $p$ you can locally turn your submersion into the canonical projection onto the lesser dimensional space. i.e. you get $(x^1, …, x^m) \rightarrow(x^1, … , x^n)$ with $n < m$.
I don't see why this is open. For example take a closed ball in $\mathbb{R}^3$ and project it down onto its shadow in $\mathbb{R}^2$, well that is certainly a canonical projection since all we did is take out the last coordinate as required, but the image is the closed disk, so we are just mapping a closed set onto a closed set.
What am I misunderstanding? Is the theorem actually stating that the coordinates in the submersion span the whole of $\mathbb{R}^m$ and $\mathbb{R}^n$?
Best Answer
If you want to check whether a map is open, why are you starting with a closed set? Of course you won't get anywhere. Assume that $m>n$ and consider the projection $\pi\colon \Bbb R^m \to \Bbb R^n$. Let $\Omega\subseteq\Bbb R^m$ be open and let's show that $\pi[\Omega]\subseteq \Bbb R^n$ is open. Write $\Bbb R^m = \Bbb R^n \times \Bbb R^{m-n}$, so that $\pi$ reads $\pi(x,y) = x$. Let $x\in \pi[\Omega]$. Then there is $y\in \Bbb R^{m-n}$ such that $(x,y)\in \Omega$. Now since $\Omega$ is open there is $\epsilon > 0$ such that $B((x,y), \epsilon) \subseteq \Omega$. Then we claim that $B(x,\epsilon) \subseteq \pi[\Omega]$. Let $x' \in B(x,\epsilon)$, so that $\|x-x'\|<\epsilon$. Then $\|(x,y) - (x',y)\| = \|x-x'\| < \epsilon$, which says that $(x',y) \in B((x,y),\epsilon)$, so that $(x',y) \in \Omega$. This means that $x'\in \pi[\Omega]$, as desired.
(Here, we use the norm on $\Bbb R^m = \Bbb R^n \times \Bbb R^{m-n}$ given by $\|(x,y)\| = \|x\| + \|y\|$. Recall that all norms in finite dimensional vector spaces are equivalent.)