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 if and only if:
- there exists a open linear transformation $\Lambda : X/N \to Y$ such that:
- $T x = \map \Lambda {\map \pi x}$
- for each $x \in X$.
Proof
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.
$\Box$
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.
$\blacksquare$