There is no submersion of a compact manifold in $\mathbb{R} $? Why?
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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 $\mathbb R^n.$
Matematleta
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Thank you very much! – Ramtin.VA Jun 23 '19 at 02:35
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You are welcome! – Matematleta Jun 23 '19 at 02:36
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sorry one question, why $f(X)$ is a open ? We don’t know X is open. – Ramtin.VA Jun 29 '19 at 19:31
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1We have $f:X\to \mathbb R$, so $X$, being the total space, is both open and closed – Matematleta Jun 29 '19 at 19:34
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I can’t understand sorry, for example X may be a closed manifold, why both closed and open? – Ramtin.VA Jun 29 '19 at 19:38
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1Is $X$ is a topological subspace of another space? If so, what topology is it given? If it is an abstract manifold, then as a total space, it is open and closed in its given topology. – Matematleta Jun 29 '19 at 19:44
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I make sense,thanks – Ramtin.VA Jun 29 '19 at 19:49