Lemma 5.12.7. Let $f : X \to Y$ be a continuous map of topological spaces.
If $X$ is quasi-compact, then $f(X)$ is quasi-compact.
If $f$ is quasi-compact, then $f(X)$ is retrocompact.
Proof. If $f(X) = \bigcup V_ i$ is an open covering, then $X = \bigcup f^{-1}(V_ i)$ is an open covering. Hence if $X$ is quasi-compact then $X = f^{-1}(V_{i_1}) \cup \ldots \cup f^{-1}(V_{i_ n})$ for some $i_1, \ldots , i_ n \in I$ and hence $f(X) = V_{i_1} \cup \ldots \cup V_{i_ n}$. This proves (1). Assume $f$ is quasi-compact, and let $V \subset Y$ be quasi-compact open. Then $f^{-1}(V)$ is quasi-compact, hence by (1) we see that $f(f^{-1}(V)) = f(X) \cap V$ is quasi-compact. Hence $f(X)$ is retrocompact. $\square$
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