# Quotient map from topological group onto left cosets is open

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.

• The key point here is that $\pi^{-1}[K\cdot H]=K$ is not always true. – N. Ciccoli Nov 4 '18 at 16:13
• Is it true that $\pi^{-1}[K\cdot H]\supseteq K$? – Keen-ameteur Nov 4 '18 at 16:14
• 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. – N. Ciccoli Nov 4 '18 at 16:17
• Yes, this is true. Likewise $\pi^{-1}[K\cdot H]=K\cdot H$ which may be sensibly bigger than $K$. – N. Ciccoli Nov 4 '18 at 17:24
• Yes, now it's ok – N. Ciccoli Nov 4 '18 at 17:38