# Why does a submersion being an open map imply that it's domain and image must be open?

I am trying to understand this problem. All answers I find say more or less the same thing, the canonical projection is an open map, so we are mapping a compact (and hence closed) thing into an open set and hence we have a clopen set, but that must be $$R^n$$ contradiction since $$R^n$$ is not compact.

But all I get from the definition of an open map is, if the domain is open the image is open. I don't see why the canonical projection map MUST map something into an open set.

Like, I can project the closed ball in $$R^3$$ down into the closed disk in $$R^2$$. This is a mapping between 2 closed sets but it's also just removing the last coordinate from elements in the domain, which if I understand correctly is literally what the canonical projection is.

I see no reason or argument that explains why having an open map suddenly guarantees that the image is open.

i.e. if the formal statement is $$f$$ open map $$\iff$$ $$D$$ is open $$\implies$$ $$Im(f)$$ open.

That says nothing about the behaviour of $$f$$ on closed sets. Like I can maybe find a closed set that is a subset of the domain of $$f$$, so the function must be defined in closed sets as well.

In short, I don't see why this argument is true:

Submersions are open maps; but the image of $$M$$ is compact in a Hausdorff space, and hence closed as well. So it's a clopen nonempty set. Since Rn is connected, it's the whole thing. But then $$R^n$$ is the quotient of a compact space, so it's compact, which is not true.

Image of an open map from one topological space to another is open by definition. In your link $$f$$ is an open map on $$M^{n}$$, not on $$\mathbb R^{n}$$. Surely $$M^{n}$$ is open in itself (though it is not open in $$\mathbb R^{n}$$).
• Ohhhh! so since $M^n$ is open, trivially since $M$ is a set and a topology and by definition the universal set of the topology is open, then its image HAS to be open under an open map, correct? Oct 8 '21 at 7:34