Weak Local Compactness is Preserved under Open Continuous Surjection

From ProofWiki
Jump to navigation Jump to search

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 weakly locally compact, then $T_B$ is also weakly locally compact.


Proof

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

Let $T_A$ be weakly locally compact.

Take $b \in S_B$.

Let $V$ be a neighbourhood of $b$.

Since $\phi$ is surjective:

$\forall y \in S_B: \exists x \in S_A: x \in \phi^{-1} \left({y}\right)$

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

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

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

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

Thus $T_B$ is weakly locally compact.

$\blacksquare$


Also see


Sources