First-Countability is Preserved under Open Continuous Surjection

Theorem
Let $T_A = \left({X_A, \vartheta_A}\right)$ and $T_B = \left({X_B, \vartheta_B}\right)$ be topological spaces.

Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.

If $T_A$ is first-countable, then $T_B$ is also first-countable.

Proof
Let $\phi$ be surjective, continuous and open.

Let $T_A$ be first countable.

Let $b \in X_B$.

Since $\phi$ is surjective there is a point $a \in X_A$ such that $\phi \left({a}\right) = b$.

From the first-countability of $T_A$, there is a neighbourhood base $\mathcal B$, say, of $a$ which is countable.

Let $\mathcal B = \left\{{V_n: n \in \N}\right\}$.

We need to show that $\left\{{\phi \left({V_n}\right): n \in \N}\right\}$ is a neighbourhood base for $b$.

Let $U$ be an open set of $T_B$ that contains $b$.

As $b = \phi \left({a}\right)$ we have that $a \in \phi^{-1} \left({U}\right)$.

From the continuity of $\phi$, we have that $\phi^{-1} \left({U}\right)$ is open.

As $\mathcal B$ is a neighbourhood base, there is an open set $V_n \subseteq \phi^{-1} \left({U}\right)$ such that $a \in V_n$.

$\phi$ is surjective, so from Surjection iff Right Inverse we have that $\phi \left({\phi^{-1} \left({U}\right)}\right) = U$.

So, applying $\phi$ to $V_n$, from Subset of Image we obtain $\phi \left({V_n}\right) \subseteq U$ such that $b \in \phi \left({V_n}\right)$.

This means that $\left\{{\phi \left({V_n}\right): n \in \N}\right\}$ is a neighbourhood base for $b$.

Thus, $T_B$ is first countable.