Quotient Mapping on Quotient Topological Vector Space is Open Mapping

Theorem

Let $K$ be a topological field.

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

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 $V \in \tau$.

From Quotient Mapping is Linear Transformation, $\pi$ is a linear transformation.

From Preimage of Image of Linear Transformation, we have:

$\pi^{-1} \sqbrk {\pi \sqbrk V} = \ker \pi + V$

From Kernel of Quotient Mapping, we have $\ker \pi = N$ and so:

$\pi^{-1} \sqbrk {\pi \sqbrk V} = N + V$

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

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

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

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.41$: Theorem