Factorization of Open Linear Transformation between Topological Vector Spaces

Theorem
Let $K$ be a topological field.

Let $\struct {X, \tau_X}$ and $\struct {Y, \tau_X}$ be topological vector spaces over $K$.

Let $T : X \to Y$ be a linear transformation.

Let $N$ be a vector subspace of $X$ with $N \subseteq \ker T$.

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

Let $\pi : X \to X/N$ be the quotient mapping.

Then $T$ is open :
 * there exists a open linear transformation $\Lambda : X/N \to Y$ such that:
 * $T x = \map \Lambda {\map \pi x}$
 * for each $x \in X$.

Necessary Condition
Suppose that $T$ is open.

From Condition for Mapping from Quotient Vector Space to be Well-Defined, there exists a linear transformation $\Lambda : X/N \to Y$ such that $T x = \map \Lambda {\map \pi x}$ for each $x \in X$.

It remains to show that $\Lambda$ is open.

Let $E \subseteq X/N$.

We want to show that:
 * $\Lambda \sqbrk E = T \sqbrk {\pi^{-1} \sqbrk E}$

Let $y \in \Lambda \sqbrk E$.

So there exists $\map \pi x \in E$ such that $\Lambda {\map \pi x} = y$.

That is, $T x = y$.

Since $\map \pi x \in E$, we have $x \in \pi^{-1} \sqbrk E$.

So we have $y \in T \sqbrk {\pi^{-1} \sqbrk E}$.

Hence we have $\Lambda \sqbrk E \subseteq T \sqbrk {\pi^{-1} \sqbrk E}$.

Now let $z \in T \sqbrk {\pi^{-1} \sqbrk E}$.

Then there exists $x \in \pi^{-1} \sqbrk E$ such that $T x = z$.

Then we have $z = \map \Lambda {\map \pi x}$.

Since $x \in \pi^{-1} \sqbrk E$, we have $\map \pi x \in E$.

So, since $z = \map \Lambda {\map \pi x}$, we have $z \in \Lambda \sqbrk E$.

So we have $T \sqbrk {\pi^{-1} \sqbrk E} \subseteq \Lambda \sqbrk E$.

So we conclude that $\Lambda \sqbrk E = T \sqbrk {\pi^{-1} \sqbrk E}$.

Now let $E$ be an open set in $\struct {X/N, \tau_N}$.

By the definition of the quotient topology, $\pi$ is continuous.

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

Since $T$ is open, we have that $T \sqbrk {\pi^{-1} \sqbrk E}$ is open in $\struct {Y, \tau_Y}$.

That is, $\Lambda \sqbrk E$ is open in $\struct {Y, \tau_Y}$.

Since $E$ was an arbitrary open set in $\struct {X/N, \tau_N}$, we have that $\Lambda$ is open.

Sufficient Condition
Suppose that there exists a open linear transformation $\Lambda : X/N \to Y$ such that:
 * $T x = \map \Lambda {\map \pi x}$

for each $x \in X$.

From Quotient Mapping on Quotient Topological Vector Space is Open Mapping, $\pi$ is open.

Since $\Lambda$ is open, we obtain that $\Lambda \circ \pi = T$ is open by Composition of Open Mappings is Open Mapping.