Openness of canonical projection of quotient topological group

Question

I'm studying the topological groups, and have difficulties with understanding why for a canonical projection onto quotient $\rho$ for a subgroup $H$ a preimage of image of an open set $U$ $\rho^{-1}(\rho(U))$ is $UH$. All the following reflections seem to be pretty clear

Answer 1

This has nothing to do with topology. Given a group $G$ and a subgroup $H$ of $G$, one defines $G/H$ to be the set of left cosets of $H$ in $G$. A left coset of $H$ in $G$ is a subset of $G$ having the form $$gH = \{gh \mid h \in H\} \text{ with } g \in G .$$ Thus we have $$G/H = \{gH \mid g \in G \}.$$ Changing perspective, we can define an equivalence relation $∼$ on $G$ by $x∼y$ if $x=yh$ for some $h \in H$. The equivalence classes of this equivalence relation are exactly the left cosets of $H$ in $G$, and each element $g \in G$ is contained in the left coset $gH$. Thus the left cosets of $H$ form a partition of $G$.

Let $\rho : G \to G/H$ denote the quotient map $\rho(g) = gH$. Then by definition for all $g \in G$ $$\rho^{-1}(\rho(g)) = gH .$$ This is true for any equivalence relation $\sim$ on an arbitrary set $X$: The quotient map $\rho : X \to X/{\sim}$ given by $\rho(x) = [x]_\sim =$ equivalence class of $x$ with respect to $\sim$ has the property that $\rho^{-1}(\rho(x)) =[x]_\sim$ for all $x \in X$.

We conclude that for all $M \subset G$ $$\rho^{-1}(\rho(M)) = \rho^{-1}(\rho(\bigcup_{m \in M}\{m\})) =\rho^{-1}(\bigcup_{m \in M}\rho(\{m\})) = \bigcup_{m \in M}\rho^{-1}(\rho(\{m\})) = \bigcup_{m \in M}\rho^{-1}(\rho(m)) \\= \bigcup_{m \in M} mH = MH .$$