Factorization of Continuous 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 continuous if and only if:

there exists a continuous 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 continuous.

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 continuous.

Let $V \subseteq Y$.

We want to show that:

$\Lambda^{-1} \sqbrk V = \pi \sqbrk {T^{-1} \sqbrk V}$

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

Then, we have $\map \Lambda {\map \pi x} \in V$.

That is, $T x \in V$.

So, we have $x \in T^{-1} \sqbrk V$.

Then, we have $\map \pi x \in \pi \sqbrk {T^{-1} \sqbrk V}$.

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

Now let $\map \pi x \in \pi \sqbrk {T^{-1} \sqbrk V}$.

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

Then we have, from Quotient Mapping is Linear Transformation, we have $\map \pi {x - z} = {\mathbf 0}_{X/N}$.

So from Kernel of Quotient Mapping, we have $x - z \in N$.

So, we have $x - z \in \ker T$, and so $T x = T z$.

We have $T z \in V$ and $T x = \map \Lambda {\map \pi x}$, so:

$\map \Lambda {\map \pi x} \in V$

So, we have $\map \pi x \in \Lambda^{-1} \sqbrk V$.

So we obtain that $\pi \sqbrk {T^{-1} \sqbrk V} \subseteq \Lambda^{-1} \sqbrk V$.

Hence we have shown:

$\Lambda^{-1} \sqbrk V = \pi \sqbrk {T^{-1} \sqbrk V}$

for all $V \subseteq Y$.

Now let $V$ be open in $\struct {Y, \tau_Y}$.

Since $T$ is continuous, $T^{-1} \sqbrk V$ is open in $\struct {X, \tau_X}$.

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

Hence, $\pi \sqbrk {T^{-1} \sqbrk V}$ is open in $\struct {X/N, \tau_N}$.

We have shown that $\Lambda^{-1} \sqbrk V = \pi \sqbrk {T^{-1} \sqbrk V}$, so:

$\Lambda^{-1} \sqbrk V$ is open in $\struct {X/N, \tau_N}$.

Since $V$ was an arbitrary open set in $\struct {Y, \tau_Y}$, we have that:

$\Lambda : X/N \to Y$ is continuous.

$\Box$

Sufficient Condition

Suppose that there exists a continuous linear transformation $\Lambda : X/N \to Y$ such that:

$T x = \map \Lambda {\map \pi x}$

for each $x \in X$.

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

Since $\Lambda$ is continuous, we have that $\Lambda \circ \pi = T$ is continuous from Composite of Continuous Mappings is Continuous.

$\blacksquare$