First-Countability is Preserved under Open Continuous Surjection

Theorem

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ 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:

$\map \phi a = b$

From the first-countability of $T_A$, there is a local base $\BB$, say, of $a$ which is countable.

Let $\BB = \set {V_n: n \in \N}$.

We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$.

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

As $b = \map \phi a$ we have that:

$a \in \phi^{-1} \sqbrk U$

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

As $\BB$ is a local base, there is an open set $V_n \subseteq \phi^{-1} \sqbrk U$ such that $a \in V_n$.

$\phi$ is surjective, so from Surjection iff Right Inverse we have that:

$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$

So, applying $\phi$ to $V_n$, from Image of Subset under Mapping is Subset of Image we obtain:

$\phi \sqbrk {V_n} \subseteq U$

such that $b \in \phi \sqbrk {V_n}$.

This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a local base for $b$.

Thus, $T_B$ is first countable.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties