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Let E be a topological group and let F be a closed subgroup of E. Define an equivalence relation in E as follows: two elements $a, b$ of E are said to equi- valent iff there is an element $f \in F$ such that a$f$=b. Thus the elements of E are divided into disjoint equivalence classes called the left cosets of F in E. The left coset containing $a \in E$ is obviously the closed set $aF$ of E and we obtain a quotient space B = E/F whose elements are left cosets of F in E and a natural projection $p:E \rightarrow B$. which maps $a \in E$ onto the left coset $aF \in B. B = E/F $is called the quotient space of E by F; it will be called simply a homogeneous space.

Then how to prove that $p$ is an open map? First of all I don't know how what will be the image of an open set under $p$.

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    What is $p^{-1}\bigl(p(U)\bigr)$ for $U \subset E$? – 2017-02-17
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    what's the topology in the quotient space? – 2017-02-17
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    @DanielFischer $p^{-1}(p(U)) = \cup_{g \in F} gU$ – 2017-02-17
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    $$\bigcup_{g\in F} Ug = \bigcup_{u\in U} uF$$ actually. Is that open in $E$? – 2017-02-17
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    Oh yes. thanks for correction. Yeah it is open(being union of open set) – 2017-02-17
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    Wait. I don't think so as F is closed in E, so is uF. And arbitrary union of closed set? – 2017-02-17
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    Yes, $uF$ is closed. But $Ug$ is open and hence $p^{-1}(p(U))$ is open when $U$ is open. – 2017-02-17
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    Ok, but then how $p(U)$ is open ? – 2017-02-18
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    Look at the definition of the quotient topology. $W \subset E/F$ is open if and only if … – 2017-02-18
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    Ah, thanks I got it! $U$ is open in $X$ iff $f^{-1}(U)$ is open where $f :X \rightarrow Y$ is a quotient map and hence $p(U)$ is open above – 2017-02-18

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