Let $M$ be a closed subspace of the normed vector space $V$, and let $\pi: V \rightarrow V/M$ be the quotient map. Is this necessarily open, or do we require some other hypotheses on $V$ or $M$? (For example, if $V$ is Banach then the Open Mapping Theorem would immediately imply that $\pi$ is open.)
There appears to have been some discussion of the problem here, but that was for general topological spaces. (Also, it seems that the question itself was not answered... it seems that the conclusion is if a continuous surjective map $f: X \rightarrow Y$ is open, then it is a quotient map, which is sort of the converse of what I'm asking.) Anyway, I'm wondering if perhaps the extra constraints on $M$ and $V$, namely that they're vector spaces, is enough to conclude what we want.
Yes, $\pi$ is an open map. Let $U$ be open and $y \in \pi(U)$. Then $y=\pi (x)$ for some $x \in U$. There exists $r>0$ such that $B(x,r) \subset U$. Let us show that $\|w-y\|<r$ implies $w \in \pi (U)$. For any such $w$ we can write $w= \pi (z)$ fro some $z$ so we get $\|\pi(z)-\pi (x)\|<r$. By definition of the norm in the quotient space this implies $\|z-x-m\|<r$ for some $m \in M$. We now get $z-m \in U$ so $\pi(z-m) \in \pi (U)$. But $\pi(m)=0$ so we get $w=\pi (z) \in \pi(U)$ as required.