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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.

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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 $\mathbb R^{n}$).

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  • $\begingroup$ 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? $\endgroup$
    – Makogan
    Oct 8 '21 at 7:34
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    $\begingroup$ @Makogan Yes, that is correct. $\endgroup$ Oct 8 '21 at 7:34

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