## 5.26 Extremally disconnected spaces

The material in this section is taken from [Gleason] (with a slight modification as in ). In Gleason's paper it is shown that in the category of quasi-compact Hausdorff spaces, the “projective objects” are exactly the extremally disconnected spaces.

Definition 5.26.1. A topological space $X$ is called extremally disconnected if the closure of every open subset of $X$ is open.

If $X$ is Hausdorff and extremally disconnected, then $X$ is totally disconnected (this isn't true in general). If $X$ is quasi-compact, Hausdorff, and extremally disconnected, then $X$ is profinite by Lemma 5.22.2, but the converse does not hold in general. For example the $p$-adic integers $\mathbf{Z}_ p = \mathop{\mathrm{lim}}\nolimits \mathbf{Z}/p^ n\mathbf{Z}$ is a profinite space which is not extremally disconnected. Namely, if $U \subset \mathbf{Z}_ p$ is the set of nonzero elements whose valuation is even, then $U$ is open but its closure is $U \cup \{ 0\}$ which is not open.

Lemma 5.26.2. Let $f : X \to Y$ be a continuous map of topological spaces. Assume $f$ is surjective and $f(E) \not= Y$ for all proper closed subsets $E \subset X$. Then for $U \subset X$ open the subset $f(U)$ is contained in the closure of $Y \setminus f(X \setminus U)$.

Proof. Pick $y \in f(U)$ and let $V \subset Y$ be any open neighbourhood of $y$. We will show that $V$ intersects $Y \setminus f(X \setminus U)$. Note that $W = U \cap f^{-1}(V)$ is a nonempty open subset of $X$, hence $f(X \setminus W) \not= Y$. Take $y' \in Y$, $y' \not\in f(X \setminus W)$. It is elementary to show that $y' \in V$ and $y' \in Y \setminus f(X \setminus U)$. $\square$

Lemma 5.26.3. Let $X$ be an extremally disconnected space. If $U, V \subset X$ are disjoint open subsets, then $\overline{U}$ and $\overline{V}$ are disjoint too.

Proof. By assumption $\overline{U}$ is open, hence $V \cap \overline{U}$ is open and disjoint from $U$, hence empty because $\overline{U}$ is the intersection of all the closed subsets of $X$ containing $U$. This means the open $\overline{V} \cap \overline{U}$ avoids $V$ hence is empty by the same argument. $\square$

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. $\square$

Lemma 5.26.5. Let $f : X \to Y$ be a continuous surjective map of Hausdorff quasi-compact topological spaces. There exists a quasi-compact subset $E \subset X$ such that $f(E) = Y$ but $f(E') \not= Y$ for all proper closed subsets $E' \subset E$.

Proof. We will use without further mention that the quasi-compact subsets of $X$ are exactly the closed subsets (Lemma 5.12.5). Consider the collection $\mathcal{E}$ of all quasi-compact subsets $E \subset X$ with $f(E) = Y$ ordered by inclusion. We will use Zorn's lemma to show that $\mathcal{E}$ has a minimal element. To do this it suffices to show that given a totally ordered family $E_\lambda$ of elements of $\mathcal{E}$ the intersection $\bigcap E_\lambda$ is an element of $\mathcal{E}$. It is quasi-compact as it is closed. For every $y \in Y$ the sets $E_\lambda \cap f^{-1}(\{ y\} )$ are nonempty and closed, hence the intersection $\bigcap E_\lambda \cap f^{-1}(\{ y\} ) = \bigcap (E_\lambda \cap f^{-1}(\{ y\} ))$ is nonempty by Lemma 5.12.6. This finishes the proof. $\square$

Proposition 5.26.6. Let $X$ be a Hausdorff, quasi-compact topological space. The following are equivalent

1. $X$ is extremally disconnected,

2. for any surjective continuous map $f : Y \to X$ with $Y$ Hausdorff quasi-compact there exists a continuous section, and

3. for any solid commutative diagram

$\xymatrix{ & Y \ar[d] \\ X \ar@{..>}[ru] \ar[r] & Z }$

of continuous maps of quasi-compact Hausdorff spaces with $Y \to Z$ surjective, there is a dotted arrow in the category of topological spaces making the diagram commute.

Proof. It is clear that (3) implies (2). On the other hand, if (2) holds and $X \to Z$ and $Y \to Z$ are as in (3), then (2) assures there is a section to the projection $X \times _ Z Y \to X$ which implies a suitable dotted arrow exists (details omitted). Thus (3) is equivalent to (2).

Assume $X$ is extremally disconnected and let $f : Y \to X$ be as in (2). By Lemma 5.26.5 there exists a quasi-compact subset $E \subset Y$ such that $f(E) = X$ but $f(E') \not= X$ for all proper closed subsets $E' \subset E$. By Lemma 5.26.4 we find that $f|_ E : E \to X$ is a homeomorphism, the inverse of which gives the desired section.

Assume (2). Let $U \subset X$ be open with complement $Z$. Consider the continuous surjection $f : \overline{U} \amalg Z \to X$. Let $\sigma$ be a section. Then $\overline{U} = \sigma ^{-1}(\overline{U})$ is open. Thus $X$ is extremally disconnected. $\square$

Lemma 5.26.7. Let $f : X \to X$ be a surjective continuous selfmap of a Hausdorff topological space. If $f$ is not $\text{id}_ X$, then there exists a proper closed subset $E \subset X$ such that $X = E \cup f(E)$.

Proof. Pick $p \in X$ with $f(p) \not= p$. Choose disjoint open neighbourhoods $p \in U$, $f(p) \in V$ and set $E = X \setminus U \cap f^{-1}(V)$. Then $p \not\in E$ hence $E$ is a proper closed subset. If $x \in X$, then either $x \in E$, or if not, then $x \in U \cap f^{-1}(V)$ and writing $x = f(y)$ (possible as $f$ is surjective) we find $y \in V \subset E$ and $x \in f(E)$. $\square$

Example 5.26.8. We can use Proposition 5.26.6 to see that the Stone-Čech compactification $\beta (X)$ of a discrete space $X$ is extremally disconnected. Namely, let $f : Y \to \beta (X)$ be a continuous surjection where $Y$ is quasi-compact and Hausdorff. Then we can lift the map $X \to \beta (X)$ to a continuous (!) map $X \to Y$ as $X$ is discrete. By the universal property of the Stone-Čech compactification we see that we obtain a factorization $X \to \beta (X) \to Y$. Since $\beta (X) \to Y \to \beta (X)$ equals the identity on the dense subset $X$ we conclude that we get a section. In particular, we conclude that the Stone-Čech compactification of a discrete space is totally disconnected, whence profinite (see discussion following Definition 5.26.1 and Lemma 5.22.2).

Using the supply of extremally disconnected spaces given by Example 5.26.8 we can prove that every quasi-compact Hausdorff space has a “projective cover” in the category of quasi-compact Hausdorff spaces.

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. $\square$

Remark 5.26.10. Let $X$ be a quasi-compact Hausdorff space. Let $\kappa$ be an infinite cardinal bigger or equal than the cardinality of $X$. Then the cardinality of the minimal quasi-compact, Hausdorff, extremally disconnected cover $X' \to X$ (Lemma 5.26.9) is at most $2^{2^\kappa }$. Namely, choose a subset $S \subset X'$ mapping bijectively to $X$. By minimality of $X'$ the set $S$ is dense in $X'$. Thus $|X'| \leq 2^{2^\kappa }$ by Lemma 5.25.1.

Comment #3047 by Jakub Byszewski on

In Lemma 5.25.7 there is a missing assumption that $f$ is surjective. Without this assumption the result is not true with obvious counterexamples (for example $X$ has no isolated points and $f$ is a constant map).

This does not affect the way this lemma is used in the proof of Lemma 5.26.9 as the map $f|_{E}$ considered there is surjective.

Comment #3152 by on

Dear Jakub Byszewski, thanks very much for catching this error. I fixed the error, gave the proof of the lemma, and explained why the lemma applies as you indicated. See this change and this change.

Comment #4331 by suggestion_bot on

I think it would be good to insert the following basic lemma somewhere close to the beginning of this section.

Lemma. Any open subset of an extremally disconnected space is extremally disconnected.

Comment #4463 by on

@Mark Bowron: thanks for your contribution and fixed here.

@suggestion_bot: seems just a bit too obvious doesn't it? If we use it somewhere in a meaningful way, then perhaps it makes sense.

Comment #6063 by Alex on

It looks like there is a typo in the last paragraph of Proposition 5.26.6. I think it should read: "Let $\sigma$ be a section. Then $\overline{U} = \sigma^{-1}(U)$ is open ..."

Comment #6064 by Laurent Moret-Bailly on

Just after Definition 08YI, there is a more direct way to see that $\mathbb{Z}_p$ is not extremally disconnected: the set $U$ of nonzero elements with even valuation is open, but its closure $U\cup \{ 0 \}$ isn't.

(By the way, the braces around $0$ don't show in the preview).

Comment #6195 by on

@#6063. No I think it is fine as is, but it is sort of a bit tricky. Namely, the inverse image $\sigma^{-1}(\overline{U})$ is both open and closed and contained in $\overline{U}$ and contains $U$. Hence it must equal $\overline{U}$ and also be open.

@#6064. Thanks very much! I have added this in this commit.

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