Quotient Mapping on Quotient Topological Vector Space is Open Mapping

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

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

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

Let $\struct {X/N, \tau_N}$ be the quotient topological vector space of $X$ modulo $N$.

Let $\pi : \struct {X, \tau} \to \struct {X/N, \tau_N}$ be the quotient mapping.

Then $\pi$ is an open mapping.

Proof
Let $V \in \tau$.

We prove that:
 * $\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$

Let $x \in \pi^{-1} \sqbrk {\pi \sqbrk V}$.

Then we have $\map \pi x \in \pi \sqbrk V$.

So there exists $v \in V$ such that $\map \pi x = \map \pi v$.

Hence, since $\pi$ is a linear transformation from Quotient Mapping is Linear Transformation, we have:
 * $\map \pi {x - v} = {\mathbf 0}_{X/N}$

From Kernel of Quotient Mapping, we then have:
 * $x - v \in N$

so that:
 * $x \in v + N$

So we have:
 * $x \in V + N$

and hence:
 * $\pi^{-1} \sqbrk {\pi \sqbrk V} \subseteq N + V$

Conversely, if $x + y \in N + V$ for $x \in N$ and $y \in V$, we have:
 * $\map \pi {x + y} = \map \pi x + \map \pi y = \map \pi y$

from Quotient Mapping is Linear Transformation and Kernel of Quotient Mapping.

Hence, we have $\map \pi {x + y} \in \pi \sqbrk V$, giving $x + y \in \pi^{-1} \sqbrk {\pi \sqbrk V}$.

So we obtain:
 * $\pi^{-1} \sqbrk {\pi \sqbrk V} \subseteq N + V$

and hence:
 * $N + V = \pi^{-1} \sqbrk {\pi \sqbrk V}$

From Sum of Set and Open Set in Topological Vector Space is Open, $N + V \in \tau$.

So $\pi^{-1} \sqbrk {\pi \sqbrk V}$ is open in $\struct {X, \tau}$.

From the definition of the quotient topology, it follows that $\pi \sqbrk V$ is open in $\struct {X/N, \tau_N}$.