Countability Axioms 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 continuous mapping which is also an open mapping.

If $T_A$ has one of the following properties, then $T_B$ has the same property:


 * First-Countability


 * Second-Countability

Proof of First-Countability
Assume that $\phi$ is surjective, continuous and open. Also assume that $T_A$ is first countable.

Take $b\in X_B$, since $\phi$ is surjective there is a point $a\in X_A$ such that $\phi(a)=b$.

From the first-countability of $T_A$, there is a countable neighbourhood base of $a$, let's call it $\left\{V_n\ :\ n\in\mathbb{N}\right\}$.

$\phi$ is open, so $\left\{\phi(V_n)\ :\ n\in\mathbb{N}\right\}$ is a family of neighbourhoods of $b$.


 * Let $U$ be an open set that contains $b$, from the continuity of $\phi$, $\phi^{-1}(U)$ is open and $a\in\phi^{-1}(U)$. Using the neighbourhood base, there is an open set $V_n\subseteq \phi^{-1}(U)$.


 * Applying $\phi$ and using that $\phi$ is surjective we obtain $\phi(V_n)\subseteq U$. This means that $\left\{\phi(V_n)\ :\ n\in\mathbb{N}\right\}$ is a neighbourhood base for $b$.

Thus, $T_B$ is first countable.

Proof of Second-Countability
Assume that $\phi$ is surjective, continuous and open. Also assume that $T_A$ is second countable.

From the second-countability of $T_A$, there is a countable base, let's call it $\left\{V_n\ :\ n\in\mathbb{N}\right\}$.

$\phi$ is open, so $\left\{\phi(V_n)\ :\ n\in\mathbb{N}\right\}$ is a family of open sets.


 * Let $U$ be an open set of $T_B$, from the continuity of $\phi$, $\phi^{-1}(U)$ is open. Using the base, there is an open set $V_n\subseteq \phi^{-1}(U)$.


 * Applying $\phi$ and using that $\phi$ is surjective we obtain $\phi(V_n)\subseteq U$. This means that $\left\{\phi(V_n)\ :\ n\in\mathbb{N}\right\}$ is a base.

Thus, $T_B$ is second-countable.