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I've been trying to prove the following exercise from "James Munkers'-a first course in topology":

Let $G$ be a topological group, and let $H\leq G$. Induce the left cosets, $G/H$, with the quotient topology induced by the quotient map $\pi:G\rightarrow G/H$, $x\mapsto xH$. I want to show that $\pi$ is an open map, but my proof seems too simple. The 'proof' is written below:

We know by definition the quotient topology that for all $K\subseteq G$, $K\cdot H$ is open in $G/H$ if and only if $\pi^{-1}[K\cdot H]$ is open in $G$. Since $\pi^{-1}[K\cdot H]=K\cdot H$, we see that if $U\subseteq G$ is open then $\pi[U]=U\cdot H$ is open in $G/H$ because:

$\pi^{-1} [U\cdot H]=U\cdot H=\underset{h\in H}{\bigcup}Uh$ and $Uh$ is open for all $h\in H$. Thus $\pi$ is open.

I would appreciate any corrections to my errors.

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  • $\begingroup$ The key point here is that $\pi^{-1}[K\cdot H]=K$ is not always true. $\endgroup$ – N. Ciccoli Nov 4 '18 at 16:13
  • $\begingroup$ Is it true that $\pi^{-1}[K\cdot H]\supseteq K$? $\endgroup$ – Keen-ameteur Nov 4 '18 at 16:14
  • $\begingroup$ Yes. It is the other inclusion the point. Think about projection $p:\mathbb R^2\to \mathbb R$ on the first component, choose an open subset of the plane and figure out the difference between this two in this case. $\endgroup$ – N. Ciccoli Nov 4 '18 at 16:17
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    $\begingroup$ Yes, this is true. Likewise $\pi^{-1}[K\cdot H]=K\cdot H$ which may be sensibly bigger than $K$. $\endgroup$ – N. Ciccoli Nov 4 '18 at 17:24
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    $\begingroup$ Yes, now it's ok $\endgroup$ – N. Ciccoli Nov 4 '18 at 17:38

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