No submersion of a compact manifold in $\mathbb{R}$

Question

There is no submersion of a compact manifold in $\mathbb{R} $? Why?

Answer 1

Use the following facts:

$1).\ $ If $f$ is a submersion, it is an open map because it is locally a projection.

$2).\ $ If $X$ is compact and $Y$ connected, every submersion $f : X \to Y$ is surjective:

$f(X)$ is compact, and $Y$ is Hausdorff, so $f(X)$ is closed. But by $1).\ $, $f(X)$ is open, too, so since $Y$ is connected, $f(X) = Y.$

$3).\ $ Therefore $f(X) = \mathbb R$ which is a contradiction, since $f(X)$ is compact and $\mathbb R$ is not.

Remark: this works for any $\mathbb R^n.$