Weak Local Compactness is Preserved under Open Continuous Surjection

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 continuous mapping which is also an open mapping and a surjection.

If $T_A$ is locally compact, then $T_B$ is also locally compact.

Proof
Let $\phi$ be a mapping which is surjective, continuous and open.

Let $T_A$ be locally compact.

Take $b \in S_B$, and let $V$ be a neighbourhood of $b$.

Since $\phi$ is surjective, $\forall y \in S_B: \exists x \in S_A: x \in \phi^{-1}(y)$.

From the local compactness of $T_A$ and the continuity of $\phi$, there exists a compact neighbourhood $K$ of $x$ such that $\phi(K)\subseteq V$.

Since $K$ is a neighbourhood of $x$, then $x \in K^\circ$ and $y \in \phi(K^\circ)\subseteq \phi(K)$, where $K^\circ$ is the interior of $K$.

$\phi$ is an open mapping and $K^\circ$ is an open set, so $\phi(K^\circ)$ is also open.

Finally we get that $y\in \phi(K)\subseteq V$, where $\phi(K)$ is a compact neighbourhood.

Thus, $T_B$ is locally compact