Continuity Test using Sub-Basis/Proof 2

Theorem
Let $\left({X_1, \tau_1}\right)$ and $\left({X_2, \tau_2}\right)$ be topological spaces.

Let $f: X_1 \to X_2$ be a mapping.

Let $\mathcal S$ be an analytic sub-basis for $\tau_2$.

Suppose that:
 * $\forall S \in \mathcal S: f^{-1} \left({S}\right) \in \tau_1$

where $f^{-1} \left({S}\right)$ denotes the preimage of $S$ under $f$.

Then $f$ is continuous.

Proof
Let $\tau$ be the final topology on $X_2$ with respect to $f$.

By hypothesis, $\mathcal S \subseteq \tau$.

By Synthetic Sub-Basis and Analytic Sub-Basis are Compatible, we have that $\tau_2$ is the topology generated by the synthetic sub-basis $\mathcal S$.

By the definition of the generated topology, we have $\tau_2 \subseteq \tau$.

By the definition of the final topology, it follows that $f$ is continuous.

Also see
As an analytic basis is also an analytic sub-basis, it is seen that Continuity Check using Basis is a special case of this theorem.