Injection is Open Mapping iff Image of Sub-Basis Set is Open

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Theorem

Let $\struct {X_1, \tau_1}$ and $\struct {X_2, \tau_2}$ be topological spaces.


Let $\SS \subseteq \powerset {X_1}$ be a sub-basis of $\tau_1$.


Let $f: X_1 \to X_2$ be an injection.


Then:

$f$ is an open mapping

if and only if:

$\forall U \in \SS: f \sqbrk U \in \tau_2$


Proof

Necessary Condition

Let $f$ be an open mapping.

By definition of open mapping:

$\forall U \in \tau_1 : f \sqbrk U \in \tau_2$

By definition of sub-basis:

$\SS \subseteq \tau_1$

Hence:

$\forall U \in \SS: f \sqbrk U \in \tau_2$

$\Box$


Sufficient Condition

Let $f$ satisfy:

$\forall U \in \SS: f \sqbrk U \in \tau_2$


Let $\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \FF \text{ is finite} }$


Let $B \in \BB$.

By definition of $\BB$:

$\ds \exists \FF \subseteq \SS, \FF \text{ is finite} : B = \bigcap \FF$

From Image of Intersection under Injection:

\(\ds f \sqbrk B\) \(=\) \(\ds f \sqbrk {\bigcap \set {S : S \in \FF} }\)
\(\ds \) \(=\) \(\ds \bigcap \set {f \sqbrk S : S \in \FF}\)

By assumption:

$\forall S \in \FF : f \sqbrk S \in \tau_2$

By Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets:

$f \sqbrk B \in \tau_2$

Hence we have shown that:

\(\text {(1)}: \quad\) \(\ds \forall B \in \BB: \, \) \(\ds f \sqbrk B\) \(\in\) \(\ds \tau_2\)


Let $W \in \tau_1$.

By definition of sub-basis:

$\exists \AA \subseteq \BB : W = \bigcup \AA$

We have:

\(\ds f \sqbrk W\) \(=\) \(\ds f \sqbrk {\bigcup \set{B : B \in \AA} }\)
\(\ds \) \(=\) \(\ds \bigcup \set{f \sqbrk B : B \in \AA}\) From Image of Union under Mapping

From $(1)$ above:

$\forall B \in \AA : f \sqbrk B \in \tau_2$

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:

$f \sqbrk W \in \tau_2$

Hence we have shown that:

$\forall W \in \tau_1 : f \sqbrk W \in \tau_2$


Hence by definition, $f$ is an open mapping

$\blacksquare$