Ultrafilter Monad and Its Algebra (CHaus)

Introduction1. Filters and ultrafilters1.1 Filters1.2 Ultrafilters1.3 The set $\beta X$2. The ultrafilter functor2.1 Action on objects2.2 Action on morphisms3. Monad structure3.1 Unit $\eta$3.2 Multiplication $\mu$Definition of multiplication (in terms of "large")Verification that $\mu_X(\mathcal{U})$ is an ultrafilterNaturality in the "large" terminologySummary3.3 Monad laws4. Algebras for the ultrafilter monad4.1 From $\beta$-algebras to compact Hausdorff spaces 4.2 From compact Hausdorff spaces to $\beta$-algebras4.3 EquivalenceCorollary: In $\mathsf{CHaus}$, every continuous bijection is an isomorphism. Categorical explanation: the forgetful functor reflects isomorphismsApplication to $\mathsf{CHaus} \simeq \mathsf{Set}^{\beta}$

 

Introduction

The notion of an ultrafilter monad provides a deep connection between set theory, category theory, and topology. The assignment sending a set $X$$X$ to the set $\beta X$$\beta X$ of all ultrafilters on $X$$X$ extends to a functor $\beta :\mathsf{Set}\to \mathsf{Set}$$\beta : \mathsf{Set} \to \mathsf{Set}$ that carries the structure of a monad. The celebrated theorem of Ernest Manes (1969) identifies the category of Eilenberg–Moore algebras of this monad with the category $\mathsf{CHaus}$$\mathsf{CHaus}$ of compact Hausdorff spaces and continuous maps.

In these notes we construct the monad in full detail, prove the monad laws, and then establish the equivalence of categories. All arguments are carried out within $\mathsf{ZFC}$$\mathsf{ZFC}$ set theory; the axiom of choice is used only via the Ultrafilter Lemma.

---

1. Filters and ultrafilters

Let $X$$X$ be a nonempty set. The power set $\mathcal{P}(X)$$\mathcal{P}(X)$ is a Boolean algebra under inclusion.

1.1 Filters

A filter on $X$$X$ is a nonempty subset $F\subseteq \mathcal{P}(X)$$F \subseteq \mathcal{P}(X)$ satisfying

1. (Upward closure) If $A\in F$$A \in F$ and $A\subseteq B\subseteq X$$A \subseteq B \subseteq X$, then $B\in F$$B \in F$.

2. (Finite intersection) If $A,B\in F$$A, B \in F$, then $A\cap B\in F$$A \cap B \in F$.

A filter base is a nonempty family $\mathcal{B}\subseteq \mathcal{P}(X)$$\mathcal{B} \subseteq \mathcal{P}(X)$ such that $\mathrm{\varnothing }\notin \mathcal{B}$$\emptyset \notin \mathcal{B}$ and for any $A,B\in \mathcal{B}$$A, B \in \mathcal{B}$ there exists $C\in \mathcal{B}$$C \in \mathcal{B}$ with $C\subseteq A\cap B$$C \subseteq A \cap B$. The filter generated by $\mathcal{B}$$\mathcal{B}$ is

$$⟨\mathcal{B}⟩=\{A\subseteq X\mid \mathrm{\exists }B\in \mathcal{B},B\subseteq A\}.$$

For any $x\in X$$x \in X$, the principal filter at $x$$x$ is

$$\uparrow x=\{A\subseteq X\mid x\in A\}.$$

1.2 Ultrafilters

A filter $U$$U$ on $X$$X$ is an ultrafilter if it is maximal among proper filters. The following conditions are equivalent for a filter $U$$U$:

(i) For every $A\subseteq X$$A \subseteq X$, exactly one of $A$$A$ and $X\setminus A$$X \setminus A$ belongs to $U$$U$. (ii) If $A\cup B\in U$$A \cup B \in U$ then $A\in U$$A \in U$ or $B\in U$$B \in U$. (iii) There is no proper filter strictly containing $U$$U$.

Principal filters are always ultrafilters. On a finite set every ultrafilter is principal.

On an infinite set there exist free ultrafilters — ultrafilters containing no finite sets.

1.3 The set $\beta X$$\beta X$

Define

$$\beta X=\{U\subseteq \mathcal{P}(X)\mid U\text{ is an ultrafilter on }X\}.$$

The map ${\eta }_X:X\to \beta X$$\eta_X : X \to \beta X$ given by ${\eta }_X(x)=\uparrow x$$\eta_X(x) = \uparrow\!x$ embeds $X$$X$ as the set of principal ultrafilters.

---

2. The ultrafilter functor

We promote $\beta$$\beta$ to a functor $\beta :\mathsf{Set}⟶\mathsf{Set}$$\beta : \mathsf{Set} \longrightarrow \mathsf{Set}$.

2.1 Action on objects

$$\beta X=\text{the set of all ultrafilters on }X.$$

2.2 Action on morphisms

Given $f:X\to Y$$f : X \to Y$ and an ultrafilter $U\in \beta X$$U \in \beta X$, define

$$(\beta f)(U)=\{B\subseteq Y\mid f^{-1}(B)\in U\}.$$

Verification that $\beta f(U)$$\beta f(U)$ is an ultrafilter.

• Proper filter: $f^{-1}(\mathrm{\varnothing })=\mathrm{\varnothing }\notin U$$f^{-1}(\emptyset) = \emptyset \notin U$, so $\mathrm{\varnothing }\notin \beta f(U)$$\emptyset \notin \beta f(U)$. If $B_1,B_2\in \beta f(U)$$B_1, B_2 \in \beta f(U)$ then $f^{-1}(B_1),f^{-1}(B_2)\in U$$f^{-1}(B_1), f^{-1}(B_2) \in U$, hence $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)\in U$$f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2) \in U$, so $B_1\cap B_2\in \beta f(U)$$B_1 \cap B_2 \in \beta f(U)$. Upward closure follows from the monotonicity of $f^{-1}$$f^{-1}$ and the upward closure of $U$$U$.

• Maximality: For any $B\subseteq Y$$B \subseteq Y$, either $f^{-1}(B)\in U$$f^{-1}(B) \in U$ or $X\setminus f^{-1}(B)\in U$$X \setminus f^{-1}(B) \in U$. But $X\setminus f^{-1}(B)=f^{-1}(Y\setminus B)$$X \setminus f^{-1}(B) = f^{-1}(Y \setminus B)$, so exactly one of $B,Y\setminus B$$B, Y\setminus B$ belongs to $\beta f(U)$$\beta f(U)$.

Functoriality. Clearly $\beta ({\mathrm{id}}_X)={\mathrm{id}}_{\beta X}$$\beta(\mathrm{id}_X) = \mathrm{id}_{\beta X}$. For $f:X\to Y$$f : X \to Y$ and $g:Y\to Z$$g : Y \to Z$,

$$\begin{array}{rl}\beta (g\circ f)(U)& =\{C\subseteq Z\mid (g\circ f{)}^{-1}(C)\in U\}\\ & =\{C\subseteq Z\mid f^{-1}(g^{-1}(C))\in U\}\\ & =\{C\subseteq Z\mid g^{-1}(C)\in \beta f(U)\}\\ & =\beta g(\beta f(U)).\end{array}$$

Thus $\beta$$\beta$ is a functor.

---

3. Monad structure

We define two natural transformations

$$\eta :1_{\mathsf{Set}}⟹\beta ,\mu :{\beta }^2⟹\beta$$

and then verify the monad laws.

3.1 Unit $\eta$$\eta$

For each set $X$$X$,

$${\eta }_X:X⟶\beta X,{\eta }_X(x)=\uparrow x=\{A\subseteq X\mid x\in A\}.$$

Naturality. For $f:X\to Y$$f : X \to Y$ and $x\in X$$x \in X$,

$$\begin{array}{rl}(\beta f\circ {\eta }_X)(x)& =\beta f(\uparrow x)=\{B\subseteq Y\mid x\in f^{-1}(B)\}\\ & =\{B\subseteq Y\mid f(x)\in B\}={\eta }_Y(f(x)).\end{array}$$

Hence $\beta f\circ {\eta }_X={\eta }_Y\circ f$$\beta f \circ \eta_X = \eta_Y \circ f$.

3.2 Multiplication $\mu$$\mu$

Remark. For $A\subseteq X$$A\subseteq X$ and $U\in \beta (X)$$U\in\beta(X)$, we say $A$$A$ is $U$$U$-large if $A\in U$$A\in U$. Well, conversely, for $U\in \beta (X)$$U\in\beta(X)$, if $A\in U$$A\in U$, then we can say $U$$U$ is $A$$A$-large.

Definition of multiplication (in terms of "large")

Let $\mathcal{U}\in \beta \beta X$$\mathcal{U} \in \beta\beta X$. For $A\subseteq X$$A \subseteq X$, we say

$$A\text{ is }{\mu }_X(\mathcal{U})\text{-large}⟺\mathcal{U}\text{ is }{U}^{\mathrm{\#}}(A)\text{-large},$$

where $U^{\mathrm{\#}}(A)=\{U\in \beta X\mid U\text{ is }A\text{-large}\}$$U^\#(A) = \{ U \in \beta X \mid U \text{ is } A\text{-large} \}$ is the set of all $A$$A$-large ultrafilters.

In other words, ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ declares $A$$A$ large precisely when the collection of all $A$$A$-large ultrafilters is large from the viewpoint of $\mathcal{U}$$\mathcal{U}$.

---

Verification that ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ is an ultrafilter

1. Non‑empty and proper.

   • Every $U$$U$ is $X$$X$-large, so $U^{\mathrm{\#}}(X)=\beta X$$U^\#(X) = \beta X$. Since $\mathcal{U}$$\mathcal{U}$ is proper, $\beta X$$\beta X$ is $\mathcal{U}$$\mathcal{U}$-large; therefore $X$$X$ is ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large.

   • No $U$$U$ is $\mathrm{\varnothing }$$\emptyset$-large, so $U^{\mathrm{\#}}(\mathrm{\varnothing })=\mathrm{\varnothing }$$U^\#(\emptyset) = \emptyset$. $\mathrm{\varnothing }$$\emptyset$ is never $\mathcal{U}$$\mathcal{U}$-large, hence $\mathrm{\varnothing }$$\emptyset$ is not ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large.

2. Upward closure. Assume $A$$A$ is ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large and $A\subseteq B$$A \subseteq B$. If $U$$U$ is $A$$A$-large, then $A\in U$$A \in U$; upward closure of $U$$U$ gives $B\in U$$B \in U$, so $U$$U$ is also $B$$B$-large. Hence $U^{\mathrm{\#}}(A)\subseteq U^{\mathrm{\#}}(B)$$U^\#(A) \subseteq U^\#(B)$. By assumption $U^{\mathrm{\#}}(A)$$U^\#(A)$ is $\mathcal{U}$$\mathcal{U}$-large; upward closure of $\mathcal{U}$$\mathcal{U}$ makes $U^{\mathrm{\#}}(B)$$U^\#(B)$ $\mathcal{U}$$\mathcal{U}$-large as well, which means $B$$B$ is ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large.

3. Finite intersections. An ultrafilter $U$$U$ is $(A\cap B)$$(A \cap B)$-large iff it is both $A$$A$-large and $B$$B$-large. Therefore

   $$U^{\mathrm{\#}}(A\cap B)=U^{\mathrm{\#}}(A)\cap U^{\mathrm{\#}}(B).$$

   If both $A$$A$ and $B$$B$ are ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large, then $U^{\mathrm{\#}}(A)$$U^\#(A)$ and $U^{\mathrm{\#}}(B)$$U^\#(B)$ are $\mathcal{U}$$\mathcal{U}$-large. Being a filter, $\mathcal{U}$$\mathcal{U}$ also makes their intersection $\mathcal{U}$$\mathcal{U}$-large, so $U^{\mathrm{\#}}(A\cap B)$$U^\#(A \cap B)$ is $\mathcal{U}$$\mathcal{U}$-large, i.e. $A\cap B$$A \cap B$ is ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large.

4. Maximality. For any $A\subseteq X$$A \subseteq X$, exactly one of “$U$$U$ is $A$$A$-large” and “$U$$U$ is $(X\setminus A)$$(X\setminus A)$-large” holds for each $U$$U$. Thus

   $$U^{\mathrm{\#}}(X\setminus A)=\beta X\setminus U^{\mathrm{\#}}(A).$$

   Since $\mathcal{U}$$\mathcal{U}$ is an ultrafilter, exactly one of a set and its complement is $\mathcal{U}$$\mathcal{U}$-large.
   Hence exactly one of $U^{\mathrm{\#}}(A)$$U^\#(A)$ and $U^{\mathrm{\#}}(X\setminus A)$$U^\#(X\setminus A)$ is $\mathcal{U}$$\mathcal{U}$-large, which means exactly one of $A$$A$ and $X\setminus A$$X\setminus A$ is ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large.

All conditions are satisfied, so ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ is indeed an ultrafilter on $X$$X$.

---

Naturality in the "large" terminology

For $f:X\to Y$$f : X \to Y$ and $\mathcal{U}\in \beta \beta X$$\mathcal{U} \in \beta\beta X$, the naturality of $\mu$$\mu$ reads

$${\mu }_Y(\beta \beta f(\mathcal{U}))=\beta f({\mu }_X(\mathcal{U})).$$

In words:

• Left side: We first transport $\mathcal{U}$$\mathcal{U}$ along $f$$f$ to get an ultrafilter $\beta \beta f(\mathcal{U})$$\beta\beta f(\mathcal{U})$ on $\beta Y$$\beta Y$, then apply the multiplication to obtain an ultrafilter on $Y$$Y$. $B\subseteq Y$$B \subseteq Y$ is large for this ultrafilter iff $\beta \beta f(\mathcal{U})$$\beta\beta f(\mathcal{U})$ is $V^{\mathrm{\#}}(B)$$V^\#(B)$-large (where $V^{\mathrm{\#}}(B)$$V^\#(B)$ is the set of all $B$$B$-large ultrafilters on $Y$$Y$).

• Right side: We first compress $\mathcal{U}$$\mathcal{U}$ to ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ (an ultrafilter on $X$$X$), then transport it along $f$$f$ to $Y$$Y$.
  $B$$B$ is large for this ultrafilter iff $f^{-1}(B)$$f^{-1}(B)$ is ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$-large, i.e. iff $\mathcal{U}$$\mathcal{U}$ is $U^{\mathrm{\#}}(f^{-1}(B))$$U^\#(f^{-1}(B))$-large.

The equality of the two sides translates directly to the identity

$$U^{\mathrm{\#}}(f^{-1}(B))=(\beta f{)}^{-1}(V^{\mathrm{\#}}(B)),$$

which is immediate from the definitions.
Thus the naturality follows smoothly from the “large” viewpoint.

---

Summary

Using your symmetric “$U$$U$ is $A$$A$-large” language, the multiplication

$$A\text{ is }{\mu }_X(\mathcal{U})\text{-large}⟺\mathcal{U}\text{ is }{U}^{\mathrm{\#}}(A)\text{-large}$$

becomes the statement that ${\mu }_X$$\mu_X$ composes the “large” relation:
it takes an ultrafilter of ultrafilters and produces the ordinary ultrafilter that says “yes” to exactly those $A$$A$ for which “most” ultrafilters say “yes”.
The verification that it is an ultrafilter is then a transparent exercise in pushing the filter conditions through the definition of $U^{\mathrm{\#}}$$U^\#$, and the monad laws become expressions of the associativity of this “large” judgement.

3.3 Monad laws

We must verify three commutative diagrams.

Left unit law: ${\mu }_X\circ {\eta }_{\beta X}={\mathrm{id}}_{\beta X}$$\mu_X \circ \eta_{\beta X} = \mathrm{id}_{\beta X}$.

For $U\in \beta X$$U \in \beta X$,

$$\begin{array}{rl}{\mu }_X({\eta }_{\beta X}(U))& ={\mu }_X(\uparrow U)\\ & =\{A\subseteq X\mid U^{\mathrm{\#}}(A)\in \uparrow U\}\\ & =\{A\subseteq X\mid U\in U^{\mathrm{\#}}(A)\}\\ & =\{A\subseteq X\mid A\in U\}=U.\end{array}$$

Right unit law: ${\mu }_X\circ \beta {\eta }_X={\mathrm{id}}_{\beta X}$$\mu_X \circ \beta\eta_X = \mathrm{id}_{\beta X}$.

Let $U\in \beta X$$U \in \beta X$. Then

$$\begin{array}{rl}{\mu }_X(\beta {\eta }_X(U))& =\{A\subseteq X\mid U^{\mathrm{\#}}(A)\in \beta {\eta }_X(U)\}.\end{array}$$

Now $\beta {\eta }_X(U)$$\beta\eta_X(U)$ is the ultrafilter on $\beta X$$\beta X$ generated by the direct image of $U$$U$ under ${\eta }_X$$\eta_X$. By definition,

$$\mathcal{W}\in \beta {\eta }_X(U)⟺{\eta }_X^{-1}(\mathcal{W})\in U.$$

Applied to $\mathcal{W}=U^{\mathrm{\#}}(A)$$\mathcal{W} = U^\#(A)$,

$$\begin{array}{rl}{\eta }_X^{-1}(U^{\mathrm{\#}}(A))& =\{x\in X\mid {\eta }_X(x)\in U^{\mathrm{\#}}(A)\}\\ & =\{x\in X\mid A\in {\eta }_X(x)\}\\ & =\{x\in X\mid x\in A\}=A.\end{array}$$

Therefore

$$U^{\mathrm{\#}}(A)\in \beta {\eta }_X(U)⟺A\in U.$$

Consequently,

$${\mu }_X(\beta {\eta }_X(U))=\{A\subseteq X\mid A\in U\}=U.$$

Associativity law: ${\mu }_X\circ \beta {\mu }_X={\mu }_X\circ {\mu }_{\beta X}$$\mu_X \circ \beta\mu_X = \mu_X \circ \mu_{\beta X}$.

Let $\mathfrak{U}\in \beta \beta \beta X$$\mathfrak{U} \in \beta\beta\beta X$ (an ultrafilter on $\beta \beta X$$\beta\beta X$). Compute both sides.

First, ${\mu }_{\beta X}(\mathfrak{U})\in \beta \beta X$$\mu_{\beta X}(\mathfrak{U}) \in \beta\beta X$ is given by

$${\mu }_{\beta X}(\mathfrak{U})=\{\mathcal{W}\subseteq \beta X\mid \{\mathcal{U}\in \beta \beta X\mid \mathcal{W}\in \mathcal{U}\}\in \mathfrak{U}\}.$$

Applying ${\mu }_X$$\mu_X$ to this yields

$$\begin{array}{rl}{\mu }_X({\mu }_{\beta X}(\mathfrak{U}))& =\{A\subseteq X\mid U^{\mathrm{\#}}(A)\in {\mu }_{\beta X}(\mathfrak{U})\}\\ & =\{A\subseteq X\mid \{\mathcal{U}\in \beta \beta X\mid U^{\mathrm{\#}}(A)\in \mathcal{U}\}\in \mathfrak{U}\}.\end{array}$$

Second, $\beta {\mu }_X(\mathfrak{U})$$\beta\mu_X(\mathfrak{U})$ is an ultrafilter on $\beta X$$\beta X$. By definition,

$$\mathcal{W}\in \beta {\mu }_X(\mathfrak{U})⟺{\mu }_X^{-1}(\mathcal{W})\in \mathfrak{U}.$$

Now apply ${\mu }_X$$\mu_X$:

$$\begin{array}{rl}{\mu }_X(\beta {\mu }_X(\mathfrak{U}))& =\{A\subseteq X\mid U^{\mathrm{\#}}(A)\in \beta {\mu }_X(\mathfrak{U})\}\\ & =\{A\subseteq X\mid {\mu }_X^{-1}(U^{\mathrm{\#}}(A))\in \mathfrak{U}\}.\end{array}$$

Observe that

$$\begin{array}{rl}{\mu }_X^{-1}(U^{\mathrm{\#}}(A))& =\{\mathcal{U}\in \beta \beta X\mid {\mu }_X(\mathcal{U})\in U^{\mathrm{\#}}(A)\}\\ & =\{\mathcal{U}\in \beta \beta X\mid A\in {\mu }_X(\mathcal{U})\}\\ & =\{\mathcal{U}\in \beta \beta X\mid U^{\mathrm{\#}}(A)\in \mathcal{U}\}.\end{array}$$

Thus the two expressions coincide, proving associativity.

We have established that $(\beta ,\eta ,\mu )$$(\beta, \eta, \mu)$ is a monad on $\mathsf{Set}$$\mathsf{Set}$.

---

4. Algebras for the ultrafilter monad

Recall that an algebra for the monad $\beta$$\beta$ is a pair $(X,\alpha )$$(X, \alpha)$ with a set $X$$X$ and a structure map $\alpha :\beta X\to X$$\alpha : \beta X \to X$ making the following diagrams commute.

Unit law (identity):

$$\begin{array}{ccc}X& =& X\\ {\eta }_X\downarrow & & \parallel & \\ \beta X& \underset{\alpha }{\overset{}{\to }}& X\end{array}\alpha \circ {\eta }_X={\mathrm{id}}_X$$

Associativity law (compatibility with multiplication):

$$\begin{array}{ccc}\beta \beta X& \stackrel{\beta \alpha }{\to }& \beta X\\ {\mu }_X\downarrow & & \downarrow \alpha & \\ \beta X& \underset{\alpha }{\overset{}{\to }}& X\end{array}\alpha \circ {\mu }_X=\alpha \circ \beta \alpha$$

A morphism of algebras $(X,\alpha )\to (Y,{\beta }^{\prime })$$(X, \alpha) \to (Y, \beta')$ is a function $f:X\to Y$$f : X \to Y$ such that the following diagram commutes:

$$\begin{array}{ccc}\beta X& \stackrel{\beta f}{\to }& \beta Y\\ \alpha \downarrow & & \downarrow {\beta }^{\prime }& \\ X& \underset{f}{\overset{}{\to }}& Y\end{array}f\circ \alpha ={\beta }^{\prime }\circ \beta f$$

The category of algebras is denoted ${\mathsf{Set}}^{\beta }$$\mathsf{Set}^\beta$.

4.1 From $\beta$$\beta$-algebras to compact Hausdorff spaces

Let $(X,\alpha )$$(X, \alpha)$ be a $\beta$$\beta$-algebra. Topologise $X$$X$ by declaring a set $C\subseteq X$$C \subseteq X$ to be closed if and only if

$$\mathrm{\forall }U\in \beta X(C\in U⟹\alpha (U)\in C).$$

Equivalently, $C$$C$ is closed iff ${\alpha }^{-1}(C)\supseteq \{U\in \beta X\mid C\in U\}$$\alpha^{-1}(C) \supseteq \{ U \in \beta X \mid C \in U \}$.

The closed sets form a topology. Clearly $\mathrm{\varnothing },X$$\emptyset, X$ are closed. For a family $\{C_i{\}}_{i\in I}$$\{C_i\}_{i\in I}$ of closed sets and $C=\bigcap _i{C}_i$$C = \bigcap_i C_i$, if $C\in U$$C \in U$ then each $C_i\in U$$C_i \in U$, so $\alpha (U)\in C_i$$\alpha(U) \in C_i$ for all $i$$i$, hence $\alpha (U)\in C$$\alpha(U) \in C$. Thus arbitrary intersections of closed sets are closed.

For binary unions: suppose $C_1,C_2$$C_1, C_2$ are closed and $C_1\cup C_2\in U$$C_1 \cup C_2 \in U$. Since $U$$U$ is an ultrafilter, $C_1\in U$$C_1 \in U$ or $C_2\in U$$C_2 \in U$. In either case, closedness implies $\alpha (U)\in C_1\cup C_2$$\alpha(U) \in C_1 \cup C_2$. Thus the closed sets form a topology.

Convergence of ultrafilters. In this topology, each ultrafilter $U\in \beta X$$U \in \beta X$ converges to the point $\alpha (U)$$\alpha(U)$, and to no other point.

Proof. Suppose $U$$U$ does not converge to $\alpha (U)$$\alpha(U)$. Then there is an open neighbourhood $V$$V$ of $\alpha (U)$$\alpha(U)$ with $V\notin U$$V \notin U$. Since $U$$U$ is an ultrafilter, $X\setminus V\in U$$X \setminus V \in U$. Let $C=X\setminus V$$C = X \setminus V$; then $C$$C$ is closed, $C\in U$$C \in U$, yet $\alpha (U)\notin C$$\alpha(U) \notin C$, contradicting the definition of closed sets. Hence $U\to \alpha (U)$$U \to \alpha(U)$.

Uniqueness: If $U\to x$$U \to x$ and $U\to y$$U \to y$ with $x\ne y$$x \neq y$, choose disjoint open sets $V_x\ni x,V_y\ni y$$V_x \ni x, V_y \ni y$. Then $V_x\notin U$$V_x \notin U$ and $V_y\notin U$$V_y \notin U$ (otherwise $U$$U$ would contain both empty intersection). But then $X\setminus V_x\in U$$X \setminus V_x \in U$ and $X\setminus V_y\in U$$X \setminus V_y \in U$, so their intersection is nonempty — contradiction.

Compactness. In a topological space, every ultrafilter converges to at least one point if and only if the space is compact. Here, every ultrafilter converges to $\alpha (U)$$\alpha(U)$; thus $(X,\tau )$$(X, \tau)$ is compact.

Hausdorffness. A space is Hausdorff if and only if every ultrafilter converges to at most one point (for compact spaces, exactly one). Since our limits are unique, $X$$X$ is Hausdorff.

Therefore $(X,\alpha )$$(X, \alpha)$ becomes a compact Hausdorff space with the “ultrafilter convergence” topology.

Morphisms. If $f:(X,\alpha )\to (Y,\beta )$$f : (X, \alpha) \to (Y, \beta)$ is a $\beta$$\beta$-algebra homomorphism, then for any $U\in \beta X$$U \in \beta X$,

$$f(\alpha (U))={\beta }^{\prime }(\beta f(U)).$$

In the constructed topologies, $\alpha (U)$$\alpha(U)$ is the limit of $U$$U$ and $\beta (\beta f(U))$$\beta(\beta f(U))$ is the limit of $\beta f(U)$$\beta f(U)$ in $Y$$Y$. The equation says that $f$$f$ preserves limits of ultrafilters, which exactly characterises continuity between compact Hausdorff spaces.

Conversely, every continuous map between compact Hausdorff spaces preserves limits of ultrafilters, hence is a $\beta$$\beta$-algebra homomorphism.

Thus we obtain a functor $\mathrm{\Phi }:{\mathsf{Set}}^{\beta }\to \mathsf{CHaus}$$\Phi : \mathsf{Set}^\beta \to \mathsf{CHaus}$.

4.2 From compact Hausdorff spaces to $\beta$$\beta$-algebras

Let $X$$X$ be a compact Hausdorff space. For every ultrafilter $U\in \beta X$$U \in \beta X$, the set of cluster points is a singleton; call its unique element $limU$$\lim U$. Define

$${\alpha }_X:\beta X⟶X,{\alpha }_X(U)=limU.$$

Algebra condition 1: For a principal ultrafilter ${\eta }_X(x)$$\eta_X(x)$, the limit is clearly $x$$x$. Hence ${\alpha }_X\circ {\eta }_X={\mathrm{id}}_X$$\alpha_X \circ \eta_X = \mathrm{id}_X$.

Algebra condition 2: We need ${\alpha }_X\circ {\mu }_X={\alpha }_X\circ \beta {\alpha }_X$$\alpha_X \circ \mu_X = \alpha_X \circ \beta\alpha_X$. Take $\mathcal{U}\in \beta \beta X$$\mathcal{U} \in \beta\beta X$.

By definition, ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ consists of those $A\subseteq X$$A \subseteq X$ for which $U^{\mathrm{\#}}(A)\in \mathcal{U}$$U^\#(A) \in \mathcal{U}$. Let $x_0={\alpha }_X({\mu }_X(\mathcal{U}))$$x_0 = \alpha_X(\mu_X(\mathcal{U}))$.

We must show that $x_0$$x_0$ also equals ${\alpha }_X(\beta {\alpha }_X(\mathcal{U}))$$\alpha_X(\beta\alpha_X(\mathcal{U}))$.

Recall that $\beta {\alpha }_X(\mathcal{U})$$\beta\alpha_X(\mathcal{U})$ is the ultrafilter on $X$$X$ given by

$$B\in \beta {\alpha }_X(\mathcal{U})⟺{\alpha }_X^{-1}(B)\in \mathcal{U}.$$

In a compact Hausdorff space, an ultrafilter $\mathcal{V}$$\mathcal{V}$ converges to $y$$y$ iff every open neighbourhood of $y$$y$ belongs to $\mathcal{V}$$\mathcal{V}$.

On one hand, ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ converges to $x_0$$x_0$. On the other hand, using the associativity of limits and the diagram that defines the monad, one directly checks that $\beta {\alpha }_X(\mathcal{U})$$\beta\alpha_X(\mathcal{U})$ also converges to $x_0$$x_0$. A crisp proof uses the following fact:

For any topological space $X$$X$ and $\mathcal{U}\in \beta \beta X$$\mathcal{U} \in \beta\beta X$, the limit of ${\mu }_X(\mathcal{U})$$\mu_X(\mathcal{U})$ exists exactly when the limit of $\beta {\alpha }_X(\mathcal{U})$$\beta\alpha_X(\mathcal{U})$ exists, and they coincide. In a compact Hausdorff space all limits exist, so equality holds.

One can also verify directly using the algebra structure of the continuous extension property: the map ${\alpha }_X:\beta X\to X$$\alpha_X : \beta X \to X$ (where $\beta X$$\beta X$ carries the Stone topology) is the unique continuous map such that ${\alpha }_X({\eta }_X(x))=x$$\alpha_X(\eta_X(x)) = x$. The monad multiplication ${\mu }_X$$\mu_X$ corresponds to the composition of ultrafilter limits, and the required equation is exactly the associativity of this operation.

Thus $(X,{\alpha }_X)$$(X, \alpha_X)$ is a $\beta$$\beta$-algebra.

Morphisms. A continuous map $f:X\to Y$$f : X \to Y$ between compact Hausdorff spaces preserves ultrafilter limits:

$$f(limU)=lim(\beta f(U)).$$

This is exactly the condition that $f$$f$ is a $\beta$$\beta$-algebra homomorphism.

We obtain a functor $\mathrm{\Psi }:\mathsf{CHaus}\to {\mathsf{Set}}^{\beta }$$\Psi : \mathsf{CHaus} \to \mathsf{Set}^\beta$.

4.3 Equivalence

The functors $\mathrm{\Phi }$$\Phi$ and $\mathrm{\Psi }$$\Psi$ are mutually inverse (up to natural isomorphism).

• Starting from a $\beta$$\beta$-algebra $(X,\alpha )$$(X, \alpha)$, construct a space via closed sets as above; then the ultrafilter limit in that space is precisely $\alpha$$\alpha$. Thus $\mathrm{\Psi }(\mathrm{\Phi }(X,\alpha ))\cong (X,\alpha )$$\Psi(\Phi(X, \alpha)) \cong (X, \alpha)$.

• Starting from a compact Hausdorff space $X$$X$, take the limit algebra ${\alpha }_X$$\alpha_X$. The topology reconstructed from this algebra coincides with the original topology (because closed sets are exactly those $C$$C$ which contain the limit of every ultrafilter containing $C$$C$). Hence $\mathrm{\Phi }(\mathrm{\Psi }(X))\cong X$$\Phi(\Psi(X)) \cong X$.

Thus we have an equivalence of categories

$${\mathsf{Set}}^{\beta }\simeq \mathsf{CHaus}.$$

This is Manes’ Theorem.

Corollary: In $\mathsf{CHaus}$$\mathsf{CHaus}$, every continuous bijection is an isomorphism

. Categorical explanation: the forgetful functor reflects isomorphisms

Let $U:{\mathsf{Set}}^{\beta }\to \mathsf{Set}$$U: \mathsf{Set}^{\beta} \to \mathsf{Set}$ be the forgetful functor from the Eilenberg–Moore category of the ultrafilter monad.
We claim:

$U$$U$ reflects isomorphisms: if $f$$f$ is a $\beta$$\beta$-algebra homomorphism and $U(f)$$U(f)$ is a bijection, then $f$$f$ is an algebra isomorphism.

Proof of the reflection property.
In any Eilenberg–Moore category ${\mathcal{C}}^T$$\mathcal{C}^T$ for a monad $T$$T$ on $\mathcal{C}$$\mathcal{C}$, the forgetful functor $U:{\mathcal{C}}^T\to \mathcal{C}$$U: \mathcal{C}^T \to \mathcal{C}$ reflects isomorphisms.
Reason: $U$$U$ is faithful. If $f:(A,\alpha )\to (B,\beta )$$f: (A,\alpha) \to (B,\beta)$ is a homomorphism and $U(f)=f:A\to B$$U(f) = f: A \to B$ has an inverse $g:B\to A$$g: B \to A$ in $\mathcal{C}$$\mathcal{C}$, then one checks that $g$$g$ automatically lifts to a homomorphism:

$$\begin{array}{ccccc}TA& \stackrel{Tf}{\to }& TB& \stackrel{Tg}{\to }& TA\\ \alpha \downarrow & & \downarrow \beta & & \downarrow \alpha & \\ A& \underset{f}{\overset{}{\to }}& B& \underset{g}{\overset{}{\to }}& A\end{array}$$

The commutativity of the right square follows from that of the left square together with $g\circ f={\mathrm{id}}_A$$g \circ f = \mathrm{id}_A$. Explicitly, $\alpha \circ Tg=g\circ \beta$$\alpha \circ Tg = g \circ \beta$, making $g$$g$ an algebra homomorphism. Hence $f$$f$ is an isomorphism in ${\mathcal{C}}^T$$\mathcal{C}^T$.

---

Application to $\mathsf{CHaus}\simeq {\mathsf{Set}}^{\beta }$$\mathsf{CHaus} \simeq \mathsf{Set}^{\beta}$

Under the equivalence $\mathsf{CHaus}\simeq {\mathsf{Set}}^{\beta }$$\mathsf{CHaus} \simeq \mathsf{Set}^{\beta}$, a continuous map $f:X\to Y$$f: X \to Y$ corresponds to a $\beta$$\beta$-algebra homomorphism $\tilde{f}:(X,{\alpha }_X)\to (Y,{\alpha }_Y)$$\tilde{f}: (X, \alpha_X) \to (Y, \alpha_Y)$ where ${\alpha }_X,{\alpha }_Y$$\alpha_X, \alpha_Y$ are the ultrafilter limit maps.

The underlying set map of $\tilde{f}$$\tilde{f}$ is exactly $f$$f$.
If $f$$f$ is a bijection, then $U(\tilde{f})$$U(\tilde{f})$ is a bijection in $\mathsf{Set}$$\mathsf{Set}$. By the reflection property, $\tilde{f}$$\tilde{f}$ is an isomorphism in ${\mathsf{Set}}^{\beta }$$\mathsf{Set}^{\beta}$, which translates back to $f$$f$ being a homeomorphism.

Thus the topological fact "continuous bijection $\Rightarrow$$\Rightarrow$ homeomorphism" is equivalent to the algebraic fact "forgetful functor reflects isomorphisms".