Let $G$ is a left topological group and $H$ is a subgroup of $G$. Denote by $G/H$ the set of all left cosets $aH$ of $H$ in $G$ (for each $a\in G$), and endow it with the quotient topology with respect to the canonical mapping $\pi$.
Then the space $G/H$ is called the left coset space of $G$ with respect to $H$.
A left topological group consists of a group $G$ and a topology $\mathfrak{T}$ on the set $G$ such that for all $a\in G$, the left action $\mathfrak{l}_a$ of $a$ on $G$ is a continuous mapping of the space $G$ to itself.
It is not true that a quotient map is necessarily open(Example of quotient mapping that is not open) but in this case, why $\pi$ is open?
EDIT: The following theorem Notes that, $\pi$ is open; But i do not understand why?
Thank you for taking the time.
It seems the following.
Counterexample. Let $G=F(x,y)$ be the free group with two generators $x$ and $y$ and $p$ be a prime number. Let $\mathcal B=\{U_n:n\in\Bbb Z\}$ be the base of the unit of a left topological group $(G,\tau)$, where $U_n=\{x^{mp^n}: m\in\Bbb Z\}$. That is a family $\{gU_n: g\in G, n\in\Bbb N\}$ is a base of the topology $\tau$. Now put $H=\{y^m:m\in\Bbb Z\}$. It is easy to check that a set $U_1H=\pi^{-1}\pi(U_1)$ is not open, so a set $\pi(U_1)$ is not open in the quotient topology too.
The existence of a counterexample to a theorem from Arhangel'skii and Tkachenko’s book seems strange, so I’ll ask my teacher, who is a disciple of Arhangel’skii, about this contradiction.