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Every quasi-compact Hausdorff space has a canonical extremally disconnected cover

Lemma 5.26.9. Let $X$ be a quasi-compact Hausdorff space. There exists a continuous surjection $X' \to X$ with $X'$ quasi-compact, Hausdorff, and extremally disconnected. If we require that every proper closed subset of $X'$ does not map onto $X$, then $X'$ is unique up to isomorphism.

Proof. Let $Y = X$ but endowed with the discrete topology. Let $X' = \beta (Y)$. The continuous map $Y \to X$ factors as $Y \to X' \to X$. This proves the first statement of the lemma by Example 5.26.8.

By Lemma 5.26.5 we can find a quasi-compact subset $E \subset X'$ surjecting onto $X$ such that no proper closed subset of $E$ surjects onto $X$. Because $X'$ is extremally disconnected there exists a continuous map $f : X' \to E$ over $X$ (Proposition 5.26.6). Composing $f$ with the map $E \to X'$ gives a continuous selfmap $f|_ E : E \to E$. Observe that $f|_ E$ has to be surjective as otherwise the image would be a proper closed subset surjecting onto $X$. Hence $f|_ E$ has to be $\text{id}_ E$ as otherwise Lemma 5.26.7 shows that $E$ isn't minimal. Thus the $\text{id}_ E$ factors through the extremally disconnected space $X'$. A formal, categorical argument (using the characterization of Proposition 5.26.6) shows that $E$ is extremally disconnected.

To prove uniqueness, suppose we have a second $X'' \to X$ minimal cover. By the lifting property proven in Proposition 5.26.6 we can find a continuous map $g : X' \to X''$ over $X$. Observe that $g$ is a closed map (Lemma 5.17.7). Hence $g(X') \subset X''$ is a closed subset surjecting onto $X$ and we conclude $g(X') = X''$ by minimality of $X''$. On the other hand, if $E \subset X'$ is a proper closed subset, then $g(E) \not= X''$ as $E$ does not map onto $X$ by minimality of $X'$. By Lemma 5.26.4 we see that $g$ is an isomorphism. $\square$

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