Open Set in Quotient Topological Vector Space

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \tau}$ be a topological vector space over $\GF$.

Let $N$ be a linear subspace of $X$.

Let $X/N$ be the quotient vector space of $X$ modulo $N$.

Let $\tau_N$ be the quotient topology on $X/N$.

Let $U \subseteq X/N$.

Then we have $U \in \tau_N$ if and only if there exists $V \in \tau$ such that $U = \pi \sqbrk V$.

Proof

Necessary Condition

Suppose that $U \in \tau_N$.

From the definition of the quotient topology, $\pi^{-1} \sqbrk U \in \tau$.

Let $V = \pi^{-1} \sqbrk U$.

Since $\pi$ is surjective, we have $\pi \sqbrk {\pi^{-1} \sqbrk U} = U$ and so $U = \pi \sqbrk V$.

Hence we have shown that there exists $V \in \tau$ such that $U = \pi \sqbrk V$.

$\Box$

Sufficient Condition

Suppose that there exists $V \in \tau$ such that $U = \pi \sqbrk V$.

From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, we have that $\pi$ is an open mapping.

So $U \in \tau_N$ as the image of an open set under $\pi$.

$\blacksquare$