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Just to be clear: the objects we're talking about here are smooth ($C^\infty$) manifolds (without a boundary) and submersion is defined as a map between manifolds, which has constant rank that is equal to the dimension of the codomain.

While trying to do an exercise I kept stumbling upon the idea that "submersion is locally a projection and therefore an open map". This idea is not the problem. The "problem" is the theorem that states the following:

Let $M$ and $N$ be smooth manifolds, $dim M=m$, $dim N=n$, and let $f:M\to N$ be a smooth mapping of a constant rank $r$. For every $p\in M$ there is a (smooth) chart $(U, \varphi)$ at $p$ and chart $(V, \psi)$ at $f(p)$, such that $f(U)\subset V$ and such that $f$ has coordinate representation $$\psi \circ f \circ \varphi ^{-1} (x_1,\ldots ,x_r,x_{r+1}, \ldots ,x_m)=(x_1,\ldots ,x_r,0,\ldots ,0)$$

Doesn't this mean that every constant rank mapping, not only submersions, is locally a projection? It seems to me that I've completely misunderstood the idea of "locally being a projection", because that should be something that's very characteristic of submersions. Also, if every constant rank mapping is locally a projection, that would mean there are no constant rank mappings from compact manifolds to euclidean space.

So, my question is: what do people mean when they say "submersion is locally a projection"?

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    $\begingroup$ It's a trivial difference. In one instance, "projection" refers to a map $ (x_1, \ldots, x_m) \mapsto (x_1, \ldots, x_r) $, and in the second instance "projection" refers to a map $ (x_1, \ldots, x_m) \mapsto (x_1, \ldots, x_r, 0, \ldots, 0) $. The difference isn't very interesting. In one case the projection is onto $ \mathbb{R}^r $ and in the second, the projection is onto the subset $ \mathbb{R}^r \subset \mathbb{R}^n $. $\endgroup$ – Jake Mirra Nov 5 '20 at 4:48
  • $\begingroup$ @JakeMirra Oh, I see. But the mapping in the case $r\neq n$ is still open? Because it seems like it is and that would mean there are no constant rank mappings from compact manifolds to euclidean space. Is that even true? $\endgroup$ – blue Nov 5 '20 at 4:59
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    $\begingroup$ No, if $r<n$, the mapping is no longer open. And of course there are constant rank mappings to $\Bbb R^n$ when the rank is less. See @Alekos's example. $\endgroup$ – Ted Shifrin Nov 5 '20 at 6:17
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This local form $(x_1,\ldots, x_r,x_{r+1},\ldots, x_m)\to (x_1,\ldots, x_r,0,\ldots,0)$ is written slightly misleadingly. If we examine a few special cases, we can see that these maps are not all "projections" in the sense which you intend. If $m\ge n=r$, we have that the map is of the form $$ (x_1,\ldots, x_m)\mapsto(x_1,\ldots, x_r)$$ and is indeed locally a bona fide projection. If $m\ge n>r$, then the map looks like $$ (x_1,\ldots, x_m)\mapsto (x_1,\ldots,x_r)\mapsto (x_1,\ldots, x_r,0,\ldots,0)$$ and hence a composition of a projection and then an inclusion. In the case where $r=m\le n$, this map becomes $$ (x_1,\ldots, x_m)\mapsto(x_1,\ldots, x_m,0,\ldots, 0)$$ which is an inclusion. If $r<m\le n$, we get $$ (x_1,\ldots, x_m)\mapsto (x_1,\ldots, x_m,0,\ldots,0)\mapsto (x_1,\ldots, x_r,0,\ldots, 0)$$ which is composition of an inclusion and a projection.

The moral of the story is that there are a few varied behaviors. For an example, take the inclusion of $S^2\hookrightarrow \Bbb{R}^3$. This is a constant rank $2$ map, so the theorem tells us that locally it looks like $(x_1,x_2)\mapsto (x_1,x_2,0)$. I.e. locally it is a standard inclusion. It is an example of an inclusion of a compact manifold into Euclidean space, and does not contradict the theorem.

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