# Showing linear, continuous operator from Banach space into particular quotient space is open

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.
• I'd have to think about it, that's why I meant this as a hint. Note however that the inclusion you have proved says nothing about openness of $\pi$, in the sense that $\pi$ being open is equivalent to the other inclusion, maybe with a different $\varepsilon$. – Guido A. Apr 1 at 19:29