TITLE: Quotient of a Banach space $X$ gets quotient topology under standard norm induced from $X$.
QUESTION [5 upvotes]: Let $X$ be a Banach space and $Y$ be a closed subspace of $X$. Then there is a canonical norm on $X/Y$. I want to show that this norm induces quotient topology on $X/Y$. Any hint/solution? I was trying to show that $\pi :X\to X/Y$ is open, so we need to show that $\exists$ $\epsilon >0$ s.t. $B_{X/Y}(0,\epsilon) \subset \pi(B_{X}(0,1)).$ Can someone please help?

REPLY [6 votes]: First, if $x\in X$, then
$$
\|\pi(x)\|_{X/Y}=\|x+Y\|_{X/Y}=\inf_{y\in Y}\|x+y\|_X\le \|x\|_X,
$$
and hence $\pi$ is bounded.
Next, as 
$$
\pi\big(B_X(0,1)\big)=\left\{x+Y:x\in B_X(0,1)\right\},
$$
let $\hat z\in B_{X/Y}(0,1)$. Then $\hat z=z+Y$, with 
$$
\|\hat z\|=\|\pi(z)\|_{X/Y}=\|z+Y\|_{X/Y}=\inf_{y\in Y}\|z+y\|_X<1,
$$
and hence there exists a $y\in Y$, such that $\|z+y\|_X<1$, i.e., $z+y\in B_X(0,1)$, and
hence
$$
\hat z=\pi(y+z)\in \pi\big(B_X(0,1)\big).
$$
Thus
$$
B_{X/Y}(0,1)\subset \pi\big(B_X(0,1)\big),
$$
and hence $\pi$ is open.