Lemma 5.8.2. Let $f : X \to Y$ be a continuous map of topological spaces. If $E \subset X$ is an irreducible subset, then $f(E) \subset Y$ is irreducible as well.

**Proof.**
Clearly we may assume $E = X$ (i.e., $X$ irreducible) and $f(E) = Y$ (i.e., $f$ surjective). First, $Y \not= \emptyset $ since $X \not= \emptyset $. Next, assume $Y = Y_1 \cup Y_2$ with $Y_1$, $Y_2$ closed. Then $X = X_1 \cup X_2$ with $X_ i = f^{-1}(Y_ i)$ closed in $X$. By assumption on $X$, we must have $X = X_1$ or $X = X_2$, hence $Y = Y_1$ or $Y = Y_2$ since $f$ is surjective.
$\square$

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