Lemma 5.26.4. Let $f : X \to Y$ be a continuous map of Hausdorff quasi-compact topological spaces. If $Y$ is extremally disconnected, $f$ is surjective, and $f(Z) \not= Y$ for every proper closed subset $Z$ of $X$, then $f$ is a homeomorphism.

Proof. By Lemma 5.17.8 it suffices to show that $f$ is injective. Suppose that $x, x' \in X$ are distinct points with $y = f(x) = f(x')$. Choose disjoint open neighbourhoods $U, U' \subset X$ of $x, x'$. Observe that $f$ is closed (Lemma 5.17.7) hence $T = f(X \setminus U)$ and $T' = f(X \setminus U')$ are closed in $Y$. Since $X$ is the union of $X \setminus U$ and $X \setminus U'$ we see that $Y = T \cup T'$. By Lemma 5.26.2 we see that $y$ is contained in the closure of $Y \setminus T$ and the closure of $Y \setminus T'$. On the other hand, by Lemma 5.26.3, this intersection is empty. In this way we obtain the desired contradiction. $\square$

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