First-Countability is Preserved under Open Continuous Surjection

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Theorem

Let $T_A = \left({S_A, \tau_A}\right)$ and $T_B = \left({S_B, \tau_B}\right)$ 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 $b \in S_B$.

Since $\phi$ is surjective there is a point $a \in S_A$ such that:

$\phi \left({a}\right) = b$

From the first-countability of $T_A$, there is a local 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 local 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 local 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 Image of Subset under Relation is Subset of Image: Corollary 2 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 local base for $b$.

Thus, $T_B$ is first countable.

$\blacksquare$


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