Injection is Open Mapping iff Image of Sub-Basis Set is Open
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
- $\forall U \in \SS: f \sqbrk U \in \tau_2$
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
$\blacksquare$