Compactness Properties Preserved under 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 surjective continuous mapping.

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


 * Compactness


 * $\sigma$-Compactness


 * Countable Compactness


 * Sequential Compactness


 * Lindelöf Space

Proof of Compactness
Let $T_A$ be compact.

Take an open cover $\mathcal{U}$ of $T_B$.

Since $\phi$ is continuous, $\{\phi^{-1}(U):\ U \in \mathcal{U}\}$ is an open cover of $T_A$.

$T_A$ is compact, so we take a finite subcover $\{\phi^{-1}(U_1),\ldots ,\phi^{-1}(U_n)\}$.

Because $\phi$ is surjective, $\phi(\phi^{-1}(A))=A$ and $\{\phi(\phi^{-1}(U_1)),\ldots ,\phi(\phi^{-1}(U_n))\}=\{U_1,\ldots,U_n\} \subseteq \mathcal U$ is a finite subcover of $T_B$.

Proof of $\sigma$-compactness
Let $T_A$ be $\sigma$-compact.

Then $\displaystyle X_A = \bigcup_{i=1}^\infty S_i$ where $S_i \subseteq X_A$ are compact.

Since $\phi$ is surjective:
 * $\displaystyle \phi(X_A) = X_B = \phi\left({\bigcup_{i=1}^\infty S_i}\right) = \bigcup_{i=1}^\infty \phi(S_i)$ using the proven theorem Image of Union.

We already know that $\phi(S_i)$ is compact for all $i \in \N$, so $X_B$ is the union of a countable number of compact subsets.

This, by definition, implies that $T_B$ is also $\sigma$-compact.