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 proof is correct. However, the notation is a bit confusing. You use $K \cdot H$ to denote two different things: a subset of $G$, as well as a subset of $G/H$.
Below is the same proof with, in my opinion, better notation.
Let $V \subset G$ be open. By the definition of quotient topology, $\pi(V) \subset G/H$ is open if and only if $\pi^{−1}(\pi(V)) \subset G$ is open. It is easy to show that $\pi^{−1}(\pi(V)) = V \cdot H$. Since $V$ is open, $V \cdot h$ is open for every $h \in H$. Since $V \cdot H = \cup_{h\in H} V \cdot h$, $V \cdot H$ is open, and so $\pi(V)$ is open.