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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.

‎A‎. ‎Arhangel'skii and M‎. ‎Tkachenko‎, Topological Groups and Related Structures‎, ‎Atlantis Press‎, ‎2008‎.

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Msina Asadzadeh
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    Is it not simply that $\pi^{-1} (\pi U) = U H$? The latter is open, if $U$ is open, agree? – peter a g Nov 09 '15 at 12:59
  • @peterag yeah $\pi^{-1} (\pi U) = U H$, but $UH$ is open in $G$ ? ($G$ is a left topological group. $HU$ is open in $G$.) – Msina Asadzadeh Nov 09 '15 at 13:06
  • OOPS - I hadn't noticed the "LEFT" in the question. Apologies! So what are the definitions: presumably $ h\mapsto g h $ is cts, but I presume that $g\mapsto g^{-1}$ is also cts? – peter a g Nov 09 '15 at 13:08
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    @peterag If inversion were assumed to be continuous, then there would be no meaning to the 'left' in the definition as $hg = (g^{-1}h^{-1})^{-1}$ and then right multiplication would also be continuous. – Matthias Klupsch Nov 09 '15 at 13:14
  • @MatthiasKlupsch - yes, thanks... my fingers typed faster than my small mind could follow, but I did see once I hit 'enter'.... ah well. – peter a g Nov 09 '15 at 13:20
  • @peterag That is, Every left topological group $G$ with continuous inverse is a semitopological group. – Msina Asadzadeh Nov 09 '15 at 15:23

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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.

Alex Ravsky
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