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


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

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

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:

By assumption:
 * $\forall S \in \FF : f \sqbrk S \in \tau_2$

By :
 * $f \sqbrk B \in \tau_2$

Hence we have shown that:

Let $W \in \tau_1$.

By definition of sub-basis:
 * $\exists \AA \subseteq \BB : W = \bigcup \AA$

We have:

From $(1)$ above:
 * $\forall B \in \AA : f \sqbrk B \in \tau_2$

By :
 * $f \sqbrk W \in \tau_2$

Hence we have shown that:
 * $\forall W \in \tau_1 : f \sqbrk W \in \tau_2$

Hence by definition, $f$ is an open mapping