Let $X$ be a Banach space, let $Y$ be a closed subspace in $X.$ Let a linear, continuous operator $\pi \colon X \to X/Y$ be defined by $\pi(x) = \overline{x}.$
I'd like to show $\pi$ open.
$\textbf{My attempt:}$
Take an open set $A \subset X$ and take $x \in A$. Therefore exists $\epsilon>0$ such that $B_{(x,\epsilon)} \subset A.$ Now I want to show that $\pi(B_{(x,\epsilon)}) = B_{(\overline{x},\epsilon)}$.
- $\pi(B_{(x,\epsilon)}) \subset B_{(\overline{x},\epsilon)}:$
From continuity, I know the following inequality is true: $||\pi(x)|| \leq||x||$. Now take $y \in B_{(x, \epsilon)}$ and note that $$||\pi(x-y)||=||\pi(x)-\pi(y)|| \leq ||x-y|| < \epsilon.$$ Thefore $\pi(y) \in B_{(\overline{x},\epsilon)}$.
- $B_{(\overline{x},\epsilon)} \subset \pi(B_{(x,\epsilon)}):$
Take $\overline{z} \in B_{(\overline{x},\epsilon)}$. I'm able to show that $\inf_{y \in Y} ||(z-x)+y||< \epsilon$, but can't progress any further.
Any suggestions on how to prove $(\supseteq)$?
Hint: We know that if $S \leq E$ is closed and $E$ is a Banach space, then the quotient $E/S$ is a Banach space as well with $\pi : E \to E/S$ a continuous linear mapping. In general, if you have any relation $\mathcal{R}$ on a set $X$, the mapping $x \in X \mapsto [x] \in X/\mathcal{R}$ is surjective. Now apply a famous theorem on surjective linear continuous mappings between Banach spaces.