Sigma-Compactness is 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 continuous surjection.

If $T_A$ is $\sigma$-compact, then $T_B$ is also $\sigma$-compact.

Proof
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.

From the similar result for compact spaces, we have that $\phi(S_i)$ is compact for all $i \in \N$.

So $X_B$ is the union of a countable number of compact subsets.

Thus, by definition, $T_B$ is also $\sigma$-compact.