# Proper maps and quasi-adjoints

Jean Goubault-Larrecq

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Jean Goubault-Larrecq. Proper maps and quasi-adjoints. Topology Proceedings, In press, 68, pp.137-156. hal-05482952

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# PROPER MAPS AND QUASI-ADJOINTS

JEAN GOUBAULT-LARRECQ

**ABSTRACT.** We show that a continuous map is proper if only if it has a quasi-adjoint, namely a left adjoint in the Kleisli category of the Smyth hyperspace monad. As applications, we show that the Smyth and Hoare hyperspace functors, as well as the continuous valuation functor, preserve proper maps. We also show that any projective limit of consonant sober spaces with proper bonding maps is consonant and sober. No separation axiom is assumed.

## 1. INTRODUCTION

The main object of study of this paper is proper maps. We will show that those are characterized as the continuous maps that have what we call a *quasi-adjoint* (Section 3), a natural construction that arises from the consideration of the Kleisli category of the Smyth hyperspace monad $Q_0$ (Section 4). We will apply this in Section 5 and show that the Smyth and Hoare hyperspace functors, as well as the continuous, subprobability and probability valuation functors, preserve proper maps; and in Section 6, where we will show that projective limits of (⊙-)consonant sober spaces with proper bonding maps are (⊙-)consonant and sober.

## 2. PRELIMINARIES

We consider topological spaces that are not necessarily Hausdorff or $T_1$, and we refer to [9]. Every topological space $X$ has a *specialization preordering* $\leq$, defined by $x \leq y$ if and only if every open neighborhood

<footnote n="2010">2010 Mathematics Subject Classification. Primary 54C10.</footnote>

<footnote>Key words and phrases. Proper maps, quasi-adjoints, Smyth hyperspace, Hoare hyperspace, continuous valuations, projective limits.</footnote>
of $x$ contains $y$. We write $y \ge x$ for $x \le y$. The closure of a point $x \in X$
coincides with its downward closure $\downarrow x \stackrel{\text{def}}{=} \{y \in X \mid y \le x\}$. We also
write $\downarrow A$ or $\downarrow_X A$ for $\{y \in X \mid y \le x\}$ for some $x \in A$}, and similarly
$\uparrow A$ or $\uparrow_X A$ for the upward closure $\{y \in X \mid x \le y\}$ for some $x \in A\}$ in
$X$. Every continuous map is monotonic with respect to the underlying
specialization preorderings. A space is $T_0$ if and only if $\le$ is antisymmetric
and $T_1$ if and only if $\le$ is the equality relation.

A subset $A$ of a topological space is *saturated* if and only if it is the
intersection of its open neighborhoods. Equivalently, $A$ is saturated if and
only if $A$ is *upwards-closed*, namely if $A = \uparrow A$.

A subset of a space $X$ is *compact* if and only if every open cover has a finite subcover; we do not require any separation axiom. The compact saturated subsets of a space $X$ form its *Smyth hyperspace* $Q_0X$. We give it the *upper Vietoris topology*, generated from basic open subsets $\Box U \stackrel{\text{def}}{=} \{Q \in Q_0X \mid Q \subseteq U\}$, where $U$ ranges over the open subsets of $X$. We let $QX$ be the subspace of *non-empty* elements of $Q_0X$. The specialization ordering of both $Q_0X$ and $QX$ is reverse inclusion $\supseteq$.

The $\mathcal{Q}$ and $\mathcal{Q}_0$ constructions were studied by a number of people,
starting with Mike Smyth [21], and later by Andrea Schalk [20, Section 7].

There is a monad $(\mathcal{Q}_0, \eta^\mathcal{Q}, \mu^\mathcal{Q})$ on the category **Top** of topological spaces and continuous maps. For every continuous map $f: X \to Y$, $\mathcal{Q}_0f$ maps every $Q \in \mathcal{Q}_0X$ to $\uparrow f[Q]$, where $f[Q]$ denotes the image of $Q$ under $f$. This is continuous because $(\mathcal{Q}_0f)^{-1}(\Box V) = \Box f^{-1}(V)$ for every open subset $V$ of $X$. The unit $\eta^\mathcal{Q}$ is defined by $\eta_X^\mathcal{Q}(x) \stackrel{\text{def}}{=} \uparrow x$ for every $x \in X$, and multiplication $\mu_X^\mathcal{Q}$ maps every $Q \in \mathcal{Q}_0\mathcal{Q}_0X$ to $\bigcup Q$. See Andrea Schalk's PhD thesis for the closely related $\mathcal{Q}$ monad [20, Proposition 7.21].

A continuous map $f: X \to Y$ is *closed* if and only if $\downarrow f[C]$ is closed for every closed subset $C$ of $X$. This definition is from [7, paragraph after Lemma V-5.19]. When $Y$ is $T_1$, we retrieve the usual condition that $f[C]$ be closed. A continuous map $f$ is *perfect* if and only if $f^{-1}(Q)$ is compact saturated for every compact saturated subset $Q$ of $Y$. A *proper* map is a closed, perfect map [7, Definition VI-6.20]. Equivalently, a continuous map $f$ is proper if and only if it is closed (in our sense) and $f^{-1}(\uparrow y)$ is compact for every $y \in Y$ [7, Lemma VI-6.21]. Proper maps are sometimes defined as continuous, closed maps such that the inverse images of one-point sets are compact. On $T_1$ spaces, this is equivalent to our definition.

3. QUASI-ADJOINTS

We recall that the specialization preordering $\leq$ of a Smyth hyperspace $\mathcal{Q}_0Y$ (resp., $\mathcal{Q}Y$) is reverse inclusion. We implicitly extend this ordering pointwise to functions from spaces $X$ to $\mathcal{Q}_0Y$ below.
**Definition 3.1.** A quasi-adjoint to a continuous map $r: X \to Y$ is a continuous map $\varsigma: Y \to Q_0X$ such that:

(a) $\eta_Y^Q \le Q_0r \circ \varsigma$, namely $\uparrow y \supseteq Q_0r(\varsigma(y))$ for every $y \in Y$, and

(b) $\varsigma \circ r \le \eta_X^Q$, namely $x \in \varsigma(r(x))$ for every $x \in X$.

The main result of this paper is the following.

**Theorem 3.2.** For a continuous $r: X \to Y$, the following are equivalent:

(1) $r$ is a proper map;

(2) $\downarrow r[C]$ is closed for every closed subset $C$ of $X$ and $r^{-1}(\uparrow y)$ is compact for every $y \in Y$;

(3) $r$ has a quasi-adjoint $\varsigma$.

*If it exists, $\varsigma$ is uniquely determined by $\varsigma(y) = r^{-1}(\uparrow y)$ for every $y \in Y$.*

*Proof.* The equivalence between items 1 and 2 is from [7, Lemma VI-6.21].

We note that: (*) for every map $\varsigma: Y \to Q_0X$ such that $\varsigma(y) = r^{-1}(\uparrow y)$ for every $y \in Y$, for every open subset $U$ of $X$, the complement of $\varsigma^{-1}(\Box U)$ in $Y$ is equal to $\downarrow r[C]$, where $C$ is the complement of $U$. Indeed, for every $y \in Y$, $y \notin \varsigma^{-1}(\Box U)$ if and only if $\varsigma(y) = r^{-1}(\uparrow y)$ is not included in $U$, if and only if there is an $x \in C$ such that $y \le r(x)$.

(3) $\Rightarrow$ (2). Let $\varsigma$ be a quasi-adjoint of $r$. We claim that $\varsigma(y) = r^{-1}(\uparrow y)$, which will also show the final uniqueness result. For every $x \in \varsigma(y)$, $r(x)$ is in $\uparrow r[\varsigma(y)] \subseteq \uparrow y$ (by (a)), so $y \le r(x)$, namely $x \in r^{-1}(\uparrow y)$. Conversely, for every $x \in r^{-1}(\uparrow y)$, we have $y \le r(x)$. Since $\varsigma$ is continuous hence monotonic, $\varsigma(y) \supseteq \varsigma(r(x))$. By (b), $x \in \varsigma(r(x))$, so $x \in \varsigma(y)$.

Since $\varsigma(y) \in Q_0X$ by assumption, $r^{-1}(\uparrow y)$ is compact. For every closed subset $C$ of $X$, $U \stackrel{\text{def}}{=} X \setminus C$ is open. Since $\varsigma$ is continuous, $\varsigma^{-1}(\Box U)$ is open. But its complement is precisely $\downarrow r[C]$, by (*), so $\downarrow r[C]$ is closed.

(1) $\Rightarrow$ (3). Let $\varsigma(y) \stackrel{\text{def}}{=} r^{-1}(\uparrow y)$ for every $y \in Y$. This is compact saturated since $r$ is proper, hence perfect. Hence $\varsigma$ takes its values in $Q_0X$. For every open subset $U$ of $X$, $\varsigma^{-1}(\Box U)$ is the complement of $\downarrow r[C]$, where $C \stackrel{\text{def}}{=} X \setminus U$, by (*). Since $r$ is proper, $\downarrow r[C]$ is closed, so $\varsigma^{-1}(\Box U)$ is open. Hence $\varsigma$ is continuous.

Let us check (a). For every $y \in Y$, $(Q_0r \circ \varsigma)(y) = \uparrow r[\varsigma(y)] = \uparrow r[r^{-1}(\uparrow y)]$. Every element $y'$ of that set is such that $y' \ge r(x)$ for some $x \in X$ such that $r(x) \ge y$, so $y' \in \uparrow y$.

Let us check (b). For every $x \in X$, we need to check that $x \in \varsigma(r(x)) = r^{-1}(\uparrow r(x))$, or equivalently that $r(x) \ge r(x)$, which is obvious. $\square$

A very close result is due to Martín Escardó [6]. We will comment on this in Remark 4.4 after we have introduced the required categorical apparatus. Another related result is [8, Lemma 4.11], see Remark 4.6.
4. THE KLEISLI CATEGORY OF $Q_0$

Every monad $(T, \eta^T, \mu^T)$ on a category $\mathbf{C}$ has a *Kleisli category* $\mathbf{C}_T$, whose objects are those of $\mathbf{C}$, and whose morphisms from an object $X$ to an object $Y$ are the morphisms from $X$ to $TY$ in $\mathbf{C}$. The identity morphism at $X$ in $\mathbf{C}_T$ is $\eta_X^T$, and the composition of $g: Y \to Z$ and $f: X \to Y$ in $\mathbf{C}_T$ (hence $g: Y \to TZ$ and $f: X \to TY$ in $\mathbf{C}$) is their *Kleisli composition* $g \bullet f \stackrel{\text{def}}{=} \mu_Z^T \circ Tg \circ f$ [19, Chapter VI, Section 5].

Kleisli categories were invented in order to show that every monad arises from an adjunction $L \dashv R$ [17]. The functor $L: \mathbf{C} \to \mathbf{C}_T$ maps every object to itself and every morphism $f: X \to Y$ to $\eta_Y^T \circ f$, as a morphism from $X$ to $Y$ in $\mathbf{C}_T$.

When $T \stackrel{\text{def}}{=} Q_0$ on $\mathbf{C} \stackrel{\text{def}}{=} \mathbf{Top}$, the morphisms $f: X \to Y$ in $\mathbf{Top}_{Q_0}$ (equivalently, $f: X \to Q_0Y$ in $\mathbf{Top}$) can be thought of as certain relations $\{(x, y) \in X \times Y \mid y \in f(x)\}$, and for every morphism $f: X \to Y$, $Lf = \eta_Y^Q \circ f$ is the *epigraph* $\{(x, y) \in X \times Y \mid f(x) \le y\}$ of $f$.

Let $\operatorname{Hom}_{\mathbf{C}}(X, Y)$ be the set of morphisms from $X$ to $Y$ in a category $\mathbf{C}$—the *homset* from $X$ to $Y$. A *preorder-enriched category* is a category in which every homset is equipped with a preordering in such a way that composition is monotonic. A *preordered functor* between preordered categories is a functor that is monotonic on each homset. The category **Top** is preorder-enriched, where for all $f, g \in \operatorname{Hom}_{\mathbf{Top}}(X, Y)$, $f \le g$ if and only if $f$ is pointwise less than or equal to $g$ with respect to the specialization preordering on $Y$. $Q_0$ is a preordered functor: if $f \le g: X \to Y$, then for every $Q \in Q_0X$, $Q_0f(Q) \supseteq Q_0g(Q)$. Indeed, the latter means that $\uparrow g[Q] \subseteq \uparrow f[Q]$, which follows from the fact that every element $g(x)$ with $x \in Q$ is larger than or equal to $f(x)$.

Given a preorder-enriched category $\mathbf{C}$ and a preordered monad $(T, \eta^T, \mu^T)$ on $\mathbf{C}$, meaning a monad whose underlying functor $T$ is preordered, the Kleisli category $\mathbf{C}_T$ is itself preorder-enriched, where $f \le g$ in $\mathbf{C}_T$ if and only if $f \le g$ in $\mathbf{C}$. The fact that Kleisli composition is monotonic follows from the fact that composition in $\mathbf{C}$ is monotonic and $T$ is a preordered functor. In particular, $\mathbf{Top}_{Q_0}$ is preorder-enriched.

In a preorder-enriched category, one can define adjoint morphisms: $g: Y \to X$ is *left adjoint* to $f: X \to Y$, or $f$ is *right adjoint* to $g$, in notation $g \dashv f$, if and only if $\mathrm{id}_Y \le f \circ g$ and $g \circ f \le \mathrm{id}_X$. The following shows how quasi-adjoints arise from this categorical setting.

**Proposition 4.1.** A continuous map $r: X \to Y$ has a quasi-adjoint $\zeta: Y \to Q_0X$ if and only if $Lr = \eta_Y^Q \circ r$ is a right adjoint in $\mathbf{Top}_{Q_0}$, and $\zeta$, seen as a morphism from $Y$ to $X$ in $\mathbf{Top}_{Q_0}$ is its left adjoint.
*Proof.* A left adjoint $\zeta$ to $Lr = \eta_Y^Q \circ r$ is characterized by the fact that the identity on $Y$ in $\mathbf{Top}_{Q_0}$, namely $\eta_Y^Q$, is less than or equal to $Lr \bullet \zeta$, and $\zeta \bullet Lr$ is less than or equal to the identity morphism $\eta_X^Q$ on $X$. But $Lr \bullet \zeta = \mu_Y^Q \circ Q_0 \eta_Y^Q \circ Q_0 r \circ \zeta = Q_0 r \circ \zeta$, by the monad equation $\mu_Y^Q \circ Q_0 \eta_Y^Q = \mathrm{id}_{Q_0 Y}$; and $\zeta \bullet Lr = \mu_X^Q \circ Q_0 \zeta \circ \eta_Y^Q \circ r = \mu_X^Q \circ \eta_{Q_0 X}^Q \circ \zeta \circ r = \zeta \circ r$, by naturality of $\eta^Q$ and the monad equation $\mu_Y^Q \circ \eta_{Q_0 Y}^Q = \mathrm{id}_{Q_0 Y}$. We retrieve the two axioms (a) $\eta_Y^Q \le Q_0 r \circ \zeta$ and (b) $\zeta \circ r \le \eta_X^Q$ of Definition 3.1. $\square$

**Remark 4.2.** Theorem 3.2 suggests defining a $T$-proper morphism in a preorder-enriched category $\mathcal{C}$ with a preordered monad $T$, as a morphism $r: X \to Y$ such that $Lr = \eta_Y^T \circ r$ is a right adjoint in $\mathcal{C}_T$. We have shown that the $Q_0$-proper morphisms on $\mathbf{Top}$ are the proper maps.

**Remark 4.3.** The same argument as in Proposition 4.1, using a general preordered monad $T$ on a general preorder-enriched category $\mathcal{C}$, shows that a morphism $r: X \to Y$ is $T$-proper if and only if $r$ has a $T$-quasi-adjoint $\zeta$—a morphism $\zeta: Y \to TX$ such that $\eta_Y^T \le Tr \circ \zeta$ and $\zeta \circ r \le \eta_X^T$.

**Remark 4.4.** In the setting of Remark 4.2, a $T$-proper morphism is the same thing as a right $T$-arrow, in the words of Escardó [6, Definition 3.1.16]: namely, a morphism $r: X \to Y$ such that $Tr: TX \to TY$ has a left adjoint $\ell: TY \to TX$ in $\mathcal{C}$. Indeed, and relying on Remark 4.3:

* given $r$ and $\ell$ as above, let $\zeta \stackrel{\mathrm{def}}{=} \ell \circ \eta_Y^T$, then:

- since $\mathrm{id}_{TY} \le Tr \circ \ell$, $\eta_Y^T \le Tr \circ \ell \circ \eta_Y^T = Tr \circ \zeta$;

- since $\ell \circ Tr \le \mathrm{id}_{TX}$, $\zeta \circ r = \ell \circ \eta_Y^T \circ r = \ell \circ Tr \circ \eta_X^T$ (by naturality of $\eta^T$) $\le \eta_X^T$;

* conversely, given a $T$-proper morphism $r: X \to Y$, with $T$-quasi-adjoint $\zeta$, let $\ell \stackrel{\mathrm{def}}{=} \mu_X^T \circ T\zeta$, then:

- since $\eta_Y^T \le Tr \circ \zeta$, $\mathrm{id}_{TY} = \mu_Y^T \circ T\eta_Y^T \le \mu_Y^T \circ TTr \circ T\zeta$ (since $T$ is a preordered functor) $= Tr \circ \mu_X^T \circ T\zeta$ (by naturality of $\mu^T$) $= Tr \circ \ell$;

- since $\zeta \circ r \le \eta_X^T$, $\ell \circ Tr = \mu_X^T \circ T\zeta \circ Tr \le \mu_X^T \circ T\eta_X^T$ (since $T$ is a preordered functor) $= \mathrm{id}_{TX}$.

Taking the name from locale theory, Escardó calls finitary any continuous map $f: X \to Y$ between sober spaces such that $\forall_f: \mathcal{O}X \to \mathcal{O}Y$ is Scott-continuous, where for every open subset $U$ of $X$, $\forall_f(U)$ is the largest open subset $V$ of $Y$ such that $f^{-1}(V) \subseteq U$ [6, Definition 2.7.1]. (A Scott-continuous map is one that is monotonic and preserves directed suprema.) By a result of Hofmann and Lawson [13, Proposition 3.3] (or see [6, Proposition 3.2.3]), the finitary maps between sober spaces are exactly the proper maps. Then Escardó shows that the finitary maps between sober spaces are exactly the right $Q_0$-arrows [6, Proposition 3.2.17(1)].
We obtain the same result (without the need for sobriety): the right $Q_0$-arrows are the $Q_0$-proper maps $r$, as we have just argued, namely the continuous maps $r$ such that $Lr$ is a quasi-adjoint, namely the proper maps by Theorem 3.2.

We may vary the monad $T$. For example, we may take $Q$ instead of $Q_0$ for $T$. Let us call $r: X \to Y$ almost surjective if and only if for every $y \in Y$, there is an $x \in X$ such that $y \le r(x)$. If $Y$ is $T_1$, $r$ is almost surjective if and only if it is surjective.

**Proposition 4.5.** A continuous map $r: X \to Y$ has a $Q$-quasi-adjoint $\zeta$ if and only if it is proper and almost surjective; $\zeta$ is uniquely determined by the formula $\zeta(y) \stackrel{\text{def}}{=} r^{-1}(\uparrow y)$ for every $y \in Y$.

*Proof.* Let $r$ have a $Q$-quasi-adjoint $\zeta: Y \to QX$. Up to the inclusion of $QX$ into $Q_0X$, we can consider $\zeta$ as a continuous map from $Y$ to $Q_0X$. For every $y \in Y$, $\uparrow y \supseteq Q_0r(\zeta(y))$, and $Q_0r(\zeta(y)) = \uparrow r[\zeta(y)] = Qr(\zeta(y))$, so $\uparrow y \supseteq Qr(\zeta(y))$. Hence $\eta_Y^Q \le Qr \circ \zeta$. Since we also have $\zeta \circ r \le \eta_X^Q$, $\zeta: Y \to Q_0X$ is quasi-adjoint to $r$. Therefore $r$ is proper, by Theorem 3.2. Additionally, for every $y \in Y$, since $\uparrow y \supseteq Qr(\zeta(y))$ and $\zeta(y)$ is in $QX$, hence non-empty, we can pick $x$ from $\zeta(y)$, and then $r(x) \in Qr(\zeta(y)) \subseteq \uparrow y$, showing that $r$ is almost surjective.

Conversely, if $r$ is proper and almost surjective, then it has a quasi-adjoint $\zeta: Y \to Q_0X$ by Theorem 3.2. For every $y \in Y$, $\zeta(y) = r^{-1}(\uparrow y)$, and the latter is non-empty because $r$ is almost surjective. $\square$

**Remark 4.6.** One can also characterize proper surjective maps: by [8, Lemma 4.11], a continuous map $r: X \to Y$ is proper and surjective if and only if it has a $Q$-quasi-adjoint $\zeta: Y \to QX$ such that $Qr \circ \zeta = \eta_Y^Q$ (not just $\ge$). The continuous maps $r: X \to Y$ such that there is a continuous map $\zeta: Y \to QX$ such that $Qr \circ \zeta = \eta_Y^Q$ were called quasi-retractions there.

## 5. FUNCTORS THAT PRESERVE PROPER MAPS

We will see that several functors from **Top** to **Top** preserve proper maps. The categorical setting for this is the following easy result; $\eta^T F$ is the natural transformation defined as $\eta_{FX}^T$ at each object $X$.

**Theorem 5.1.** Let $\mathbf{C}$ be a preorder-enriched category, $\mathbf{T}$ be a preordered monad on $\mathbf{C}$, and $\mathbf{F}$ be a preordered functor from $\mathbf{C}$ to $\mathbf{C}$. If there is a natural transformation $\lambda: \mathbf{FT} \to \mathbf{TF}$ that is compatible with $\mathbf{T}$ units, namely such that $\lambda \circ F\eta^T = \eta^T F$, then $\mathbf{F}$ preserves $\mathbf{T}$-proper morphisms.
*Proof.* Let $r: X \to Y$ be a $T$-proper morphism in **C**. By Remark 4.3, $r$ has a $T$-quasi-adjoint $\zeta$, namely $\eta_Y^T \leq Tr \circ \zeta$ and $\zeta \circ r \leq \eta_X^T$. We claim that $\lambda_X \circ F\zeta: FY \to TFX$ is a $T$-quasi-adjoint of $Tr$.

We have $\eta_{FY}^T = \lambda_Y \circ F\eta_Y^T$ (by compatibility with $T$ units) $\leq \lambda_Y \circ FTr \circ F\zeta$ (since $\eta_Y^T \leq Tr \circ \zeta$ and $F$ is a preordered functor) $= TFr \circ \lambda_X \circ F\zeta$ ($\lambda$ is natural). Also, $\lambda_X \circ F\zeta \circ Fr = \lambda_X \circ F(\zeta \circ r) \leq \lambda_X \circ F\eta_X^T$ ($\zeta \circ r \leq \eta_X^T$ and $F$ is a preordered functor) $= \eta_{FX}^T$ (compatibility with $T$ units). $\square$

We obtain the following from Theorem 3.2.

**Corollary 5.2.** Let $F$ be a preordered functor from **Top** to **Top**. If there is a natural transformation $\lambda: FQ_0 \to Q_0F$ that is compatible with $Q_0$ units, namely such that $\lambda \circ F\eta^Q = \eta^Q F$, then $F$ preserves proper maps.

### 5.1. The $Q_0$ and $\mathcal{Q}$ functors.

**Proposition 5.3.** $Q_0$ preserves proper maps.

*Proof.* We use Corollary 5.2 on $\lambda_X \stackrel{\text{def}}{=} \eta_{Q_0X}^Q \circ \mu_{Q_0X}^Q : Q_0Q_0X \to Q_0Q_0X$ for every topological space $X$. This is natural in $X$, as a composition of natural transformations, and $\lambda_X \circ Q_0\eta_X^Q = \eta_{Q_0X}^Q$ since $\mu_{Q_0X}^Q \circ Q_0\eta_X^Q = \text{id}_{Q_0X}$, one of the monad laws. $\square$

The same holds for $\mathcal{Q}$, but this requires a bit more effort.

**Lemma 5.4.** For every space $X$, let $i_X^Q: QX \to Q_0X$ be the inclusion map. Let also $c_X^Q: Q_0Q_0X \to Q_0QX$ be defined by $c_X^Q(\mathcal{Q}) \stackrel{\text{def}}{=} \mathcal{Q} \setminus \{\emptyset\}$. Then:

(1) $i_X^Q$ and $c_X^Q$ are well-defined and continuous;

(2) $i_X^Q$ is a natural transformation from $\mathcal{Q}$ to $Q_0$;

(3) $c_X^Q$ is a natural transformation from $Q_0Q_0$ to $Q_0\mathcal{Q}$;

(4) $c_X^Q \circ \eta^Q Q_0 \circ i_X^Q = \eta^Q \mathcal{Q}$.

*Proof.* We will use $\square_0$ for the $\square$ operator on spaces of the form $Q_0Z$, while keeping $\square$ for $QZ$, in the name of disambiguation.

(1) The equation $(i_X^Q)^{-1}(\square_0 U) = \square U$ shows that $i_X^Q$ is continuous.

As far as $c_X^Q$ is concerned, we note that $\{\emptyset\} = \square_0\emptyset$ is open in $Q_0X$, so $QX$ is a closed subspace of $Q_0X$; therefore $c_X^Q(\mathcal{Q}) = \mathcal{Q} \cap QX$ is compact in $QX$. It is upwards-closed in $QX$ because $\mathcal{Q}$ is upwards-closed in $Q_0X$ and $c_X^Q(\mathcal{Q})$ is obtained by removing the top element $\emptyset$ of $Q_0X$. Therefore $c_X^Q(\mathcal{Q}) \in Q_0QX$. The open subsets of $QX$ are the sets of the form $\bigcup_{i \in I} \square_i U_i$ where each $U_i$ is open in $X$, since $\square$ commutes with finite intersections. We can also require every $U_i$ to be non-empty, since $\square\emptyset = \emptyset$. In order to show that $c_X^Q$ is continuous, it then suffices to show that $(c_X^Q)^{-1}(\square_0 \bigcup_{i \in I} \square_i U_i)$ is open. For every $\mathcal{Q} \in Q_0Q_0X$,
$c_X^Q(Q) \in \Box_0 \bigcup_{i \in I} \Box U_i$ if and only if every $Q \in Q \setminus \{\emptyset\}$ is included in some $U_i$, if and only if every $Q \in Q$ is included in some $U_i$ or in $\{\emptyset\} = \Box_0 \emptyset$. Therefore $(c_X^Q)^{-1}(\Box_0 \bigcup_{i \in I} \Box_0 U_i) = \Box_0 (\bigcup_{i \in I} \Box_0 U_i \cup \Box_0 \emptyset)$, which is open.

(2) Let $f: X \to Y$ be a continuous map. Both sides of $i_Y^Q \circ Qf = Q_0f \circ i_X^Q$ map every $Q \in QX$ to $\uparrow f[Q]$.

(3) Let $f: X \to Y$ be an arbitrary continuous map. We show that $c_Y^Q \circ Q_0Q_0f = Q_0Qf \circ c_X^Q$. For every $Q \in Q_0Q_0X$, $Q_0Q_0f(Q) = \uparrow_{Q_0Y}\{\uparrow_Y f[Q] | Q \in Q\}$, so $(c_Y^Q \circ Q_0Q_0f)(Q) = \{Q' \in QY | Q' \neq \emptyset, Q' \subseteq \uparrow_Y f[Q] \text{ for some } Q \in Q\}$; the condition $Q' \neq \emptyset$ comes from the definition of $c_Y^Q$, but is already implied by $Q' \in QY$, so $(c_Y^Q \circ Q_0Q_0f)(Q) = \{Q' \in QY | Q' \subseteq \uparrow_Y f[Q] \text{ for some } Q \in Q\}$. Turning to the other side, $(Q_0Qf \circ c_X^Q)(Q) = \{Q' \in QY | Q' \subseteq \uparrow_Y f[Q] \text{ for some } Q \in Q \text{ with } Q \neq \emptyset\}$. But for every $Q' \in QY$, $Q'$ is non-empty, and $Q' \subseteq \uparrow_Y f[Q]$ implies that $Q$ is non-empty. Therefore $(Q_0Qf \circ c_X^Q)(Q)$ is equal to $\{Q' \in QY | Q' \subseteq \uparrow_Y f[Q] \text{ for some } Q \in Q\}$, too.

(4) For every space $X$, for every $Q \in QX$, $(c_X^Q \circ \eta_{Q_0X}^Q \circ i_X^Q)(Q) = c_X^Q(\{Q' \in Q_0X | Q' \subseteq Q\}) = \{Q' \in Q_0X | Q' \neq \emptyset, Q' \subseteq Q\} = \{Q' \in QX | Q' \subseteq Q\} = \eta_{QX}^Q(Q). \quad \square$

**Proposition 5.5.** *Q preserves proper maps.*

*Proof.* We rely on Lemma 5.4, and we use Corollary 5.2 on $\lambda_X \stackrel{\text{def}}{=} c_X^Q \circ \eta_{Q_0X}^Q \circ \mu_{Q_0X}^Q \circ i_{Q_0X}^Q : QQ_0X \to Q_0QX$ for every topological space $X$. This is natural in $X$, and:

$$ \begin{align*} \lambda_X \circ Q\eta_X^Q &= c_X^Q \circ \eta_{Q_0X}^Q \circ \mu_{Q_0X}^Q \circ i_{Q_0X}^Q \circ Q\eta_X^Q \\ &= c_X^Q \circ \eta_{Q_0X}^Q \circ \mu_{Q_0X}^Q \circ Q_0\eta_X^Q \circ i_X^Q && \text{by naturality of } i^Q \\ &= c_X^Q \circ \eta_{Q_0X}^Q \circ i_X^Q && \text{by the monad equations} \\ &= \eta_{QX}^Q && \text{by Lemma 5.4, item (4).} \end{align*} $$

Hence $F \stackrel{\text{def}}{=} Q$ and $\lambda$ satisfy the requirement of Corollary 5.2. $\square$

We take the opportunity to show a converse to Propositions 5.3 and 5.5.

**Proposition 5.6.** For every continuous map $r: X \to Y$, the following are equivalent: (i) $r$ is proper; (ii) $Q_0r$ is proper; (iii) $Qr$ is proper.

*Proof.* The implication (i) $\Rightarrow$ (ii) is Proposition 5.3.

(ii) $\Rightarrow$ (iii). If $Q_0r$ is proper, then it has a quasi-adjoint $\zeta': Q_0Y \to Q_0Q_0X$. Let $\zeta'': QY \to Q_0QX$ be $c_X^Q \circ \zeta' \circ i_Y^Q$. We have $Q_0Qr \circ \zeta'' = Q_0Qr \circ c_X^Q \circ \zeta' \circ i_Y^Q = c_Y^Q \circ Q_0Q_0r \circ \zeta' \circ i_Y^Q$ since $c$ is natural (see Lemma 5.4, item (3)). This is larger than or equal to $c_Y^Q \circ \eta_{Q_0Y}^Q \circ i_Y^Q$ by axiom (a) of quasi-adjoints, which is equal to $\eta_{QY}^Q$ by Lemma 5.4, item (4).
As for axiom (b), $\zeta'' \circ Qr = c_X^Q \circ \zeta' \circ i_Y^Q \circ Qr = c_X^Q \circ \zeta' \circ Q_0r \circ i_X^Q$
(by naturality of $i^Q$) $\leq c_X^Q \circ \eta_{Q_0X}^Q \circ i_X^Q$ (by axiom (b) of quasi-adjoints)
$= \eta_{QX}^Q$, by Lemma 5.4, item (4).

Therefore $\zeta''$ is a quasi-adjoint of $Qr$.

(iii) $\Rightarrow$ (i). Let $\zeta''$: $QY \to Q_0QX$ be a quasi-adjoint of $Qr$. We define
$\zeta$: $Y \to Q_0X$ as $\mu_X^Q \circ Q_0i_X^Q \circ \zeta'' \circ \eta_Y^Q$. Then:

$$
\begin{align*}
Q_0r \circ \zeta &= Q_0r \circ \mu_X^Q \circ Q_0i_X^Q \circ \zeta'' \circ \eta_Y^Q \\
&= \mu_Y^Q \circ Q_0Q_0r \circ Q_0i_X^Q \circ \zeta'' \circ \eta_Y^Q && \text{by naturality of } \mu^Q \\
&= \mu_Y^Q \circ Q_0i_Y^Q \circ Q_0Qr \circ \zeta'' \circ \eta_Y^Q && \text{by naturality of } i^Q \\
&\geq \mu_Y^Q \circ Q_0i_Y^Q \circ \eta_{QY}^Q \circ \eta_Y^Q && \text{by axiom (a) of quasi-adjoints} \\
&= \mu_Y^Q \circ \eta_{Q_0Y}^Q \circ i_Y^Q \circ \eta_Y^Q && \text{by naturality of } \eta^Q \\
&= i_Y^Q \circ \eta_Y^Q = \eta_Y^Q,
\end{align*}
$$

by the monad equations and the easily verified fact that $i_Y^Q \circ \eta_Y^Q =$
$\eta_Y^Q$: $Y \to Q_0Y$; while:

$$
\begin{align*}
\zeta \circ r &= \mu_X^Q \circ Q_0 i_X^Q \circ \zeta'' \circ \eta_Y^Q \circ r \\
&= \mu_X^Q \circ Q_0 i_X^Q \circ \zeta'' \circ Qr \circ \eta_X^Q && \text{by naturality of } \eta^Q \\
&\le \mu_X^Q \circ Q_0 i_X^Q \circ \eta_{QX}^Q \circ \eta_X^Q && \text{by axiom (b) of quasi-adjoints} \\
&= \mu_X^Q \circ \eta_{Q_0X}^Q \circ i_X^Q \circ \eta_X^Q && \text{by naturality of } \eta^Q \\
&= i_X^Q \circ \eta_X^Q = \eta_X^Q.
\end{align*}
$$

Hence $\zeta$ is a quasi-adjoint of $r$. $\square$

<figure><img src="image_2.png" /></figure>

We end this section with a remark on almost surjective maps.

**Remark 5.7.** Specialized to $T \stackrel{\text{def}}{=} Q$, Theorem 5.1 states that for every
preordered functor $F$ from **Top** to **Top** with a natural transformation
$\lambda: FQ \to QT$ that is compatible with $Q$ units, $F$ preserves $Q$-proper
morphisms, namely proper almost surjective maps (by Proposition 4.5).
In particular, $Q$ itself preserves proper almost surjective maps: the proof
is as with Proposition 5.3, taking $\lambda_X \stackrel{\text{def}}{=} \eta_{QX}^Q \circ \mu_{QX}^Q : QQX \to QQX$.

It is even easier to show that $Q_0$ preserves proper almost surjective
maps: together with Proposition 5.3, it suffices to realize that $Q_0r$ is
almost surjective for every continuous map $r: X \to Y$. Indeed, for every
$Q' \in Q_0Y$, $Q'$ is smaller than or equal to (contains) $Q_0r(Q)$ for some
$Q \in Q_0X$: take $Q \stackrel{\text{def}}{=} \emptyset$, so that $Q_0r(Q) = \emptyset$.
5.2. **The $\mathcal{H}_0$ and $\mathcal{H}$ functors.** For every space $X$, let $\mathcal{H}_0X$ be the set of closed subsets of $X$ and $\mathcal{HX}$ be its subset of non-empty closed subsets. We give them the lower Vietoris topology, whose subbasic open subsets $\diamond U$ are the sets of closed subsets that intersect $U$, for each open subset $U$ of $X$. The specialization ordering is inclusion $\subseteq$.

$\mathcal{H}_0X$ is a standard object in topology, more often equipped with the Vietoris or Fell topologies, though. The $\mathcal{H}$ construction was studied by Schalk [20, Section 6], as well as a localic variant and a variant with the Scott topology. See also [1, Sections 6.2.2, 6.2.3] or [7, Section IV-8]. With the Scott topology, $\mathcal{HX}$ is known as the Hoare powerdomain of $X$.

There is a monad $(\mathcal{H}_0, \eta^\mathcal{H}, \mu^\mathcal{H})$ on **Top**, and a similarly monad $(\mathcal{H}, \eta^\mathcal{H}, \mu^\mathcal{H})$. For every continuous map $f: X \to Y$, $\mathcal{H}_0f$ (resp., $\mathcal{H}f$) maps every $C \in \mathcal{H}_0X$ (resp., $\mathcal{HX}$) to $cl(f[C])$. This is continuous because $(\mathcal{H}_0f)^{-1}(\diamond V) = \diamond f^{-1}(V)$ for every open subset $V$ of $Y$, and similarly for $\mathcal{H}f$. The unit $\eta^\mathcal{H}$ is defined by $\eta_X^\mathcal{H}(x) \stackrel{\text{def}}{=} \downarrow x$ for every $x \in X$. The multiplication $\mu^\mathcal{H}$ is defined by $\mu_X^\mathcal{H}(C) \stackrel{\text{def}}{=} cl(\bigcup C)$.

The functors $\mathcal{H}_0$ and $\mathcal{H}$ are preordered functors: if $f \leq g: X \to Y$, then for every closed subset $C$ of $X$, $cl(f[C]) \subseteq cl(g[C])$. Indeed, it suffices to show that $f[C] \subseteq cl(g[C])$. Every element $y$ of $f[C]$ is equal to $f(x)$ for some $x \in C$, hence is smaller than or equal to $g(x)$. Since closures are closed hence downward-closed, $y = f(x)$ is in $cl(g[C])$.

We recall a construction due to de Brecht and Kawai [4]. For every topological space $X$, there is a continuous function $\sigma_X: \mathcal{H}_0Q_0X \to Q_0\mathcal{H}_0X$ defined by $\sigma_X(C) \stackrel{\text{def}}{=} \{C \in \mathcal{H}_0X \mid \forall Q \in C, C \cap Q \neq \emptyset\}$ for every $C \in \mathcal{H}_0Q_0X$. The sets $\square\diamond U$ form a subbase of the topology on $\mathcal{H}_0Q_0X$ when $U$ ranges over the open subsets of $X$ [4, Lemma 6.1, item (1)], $\sigma_X^{-1}(\square\diamond U) = \diamond\square U$, and $\sigma_X$ is a topological embedding [4, Lemma 6.4].

**Lemma 5.8.** When $X$ varies over the topological spaces, the maps $\sigma_X$ form a natural transformation $\sigma$, which is compatible with $Q_0$ units.

*Proof.* Let $f: X \to Y$ be a continuous map. We wish to show that $\sigma_Y \circ \mathcal{H}_0Q_0f = Q_0\mathcal{H}_0f \circ \sigma_X$. Since $Q_0\mathcal{H}_0Y$ is $T_0$, two elements of this space are equal if and only if they belong to the same subbasic open sets. For every $C \in \mathcal{H}_0Q_0X$, for every open subset $V$ of $Y$, $(\sigma_Y \circ \mathcal{H}_0Q_0f)(C) \in \square\diamond V$ if and only if $\mathcal{H}_0Q_0f(C) \in \diamond\square V$, if and only if $C \in (\mathcal{H}_0Q_0f)^{-1}(\square\diamond V) = \diamond(Q_0f)^{-1}(\square V) = \diamond\square f^{-1}(V)$, while $(Q_0\mathcal{H}_0f \circ \sigma_X)(C) \in \square\diamond V$ if and only if $\sigma_X(C) \in (Q_0\mathcal{H}_0f)^{-1}(\square\diamond V) = \square(\mathcal{H}_0f)^{-1}(\diamond V) = \square\diamond f^{-1}(V)$, if and only if $C \in \diamond\square f^{-1}(V)$.

As far as compatibility with $Q_0$ units is concerned, we must show that $\sigma_X \circ \mathcal{H}_0\eta_X^Q = \eta_{\mathcal{H}_0X}^Q$. Let $C \in \mathcal{H}_0X$. We use the same trick as above. For every open subset $U$ of $X$, $(\sigma_X \circ \mathcal{H}_0\eta_X^Q)(C) \in \square\diamond U$ if and only if
$\mathcal{H}_0\eta_X^Q(C) \in \sigma_X^{-1}(\Box\diamond U) = \diamond\Box U$, if and only if $C \in (\mathcal{H}_0\eta_X^Q)^{-1}(\diamond\Box U) = \diamond(\eta_X^Q)^{-1}(\Box U)$. It is easy to see that $(\eta_X^Q)^{-1}(\Box U) = U$, so $(\sigma_X \circ \mathcal{H}_0\eta_X^Q)(C) \in \Box\diamond U$ if and only if $C \in \diamond U$. Next, $\eta_{\mathcal{H}_0 X}^Q(C) \in \Box\diamond U$ if and only if $C \in (\eta_{\mathcal{H}_0 X}^Q)^{-1}(\Box\diamond U) = \diamond U$. $\square$

Using Corollary 5.2 together with Lemma 5.8, we obtain the following.

**Proposition 5.9.** * $\mathcal{H}_0$ preserves proper maps.*

Let us turn to $\mathcal{H}$. We make an easy observation.

**Lemma 5.10.** Let $r: X \to Y$ be a continuous map, and $\zeta$ map every $y \in Y$ to $r^{-1}(\uparrow y)$. If $\zeta$ takes its values in $Q_0X$ and is continuous, then $\zeta$ is a quasi-adjoint of $r$, and $r$ is proper.

*Proof.* For every $y \in Y$, $Q_0r(\zeta(y)) = \uparrow r[r^{-1}(\uparrow y)]$ is the collection of points larger than or equal to $r(x)$, where $x \in X$ and $r(x) \ge y$; this is included in $\uparrow y$. For every $x \in X$, $\zeta(r(x)) = r^{-1}(\uparrow r(x))$ contains $x$. Hence $\zeta$ satisfies axioms (a) and (b) of Definition 3.1. $\square$

**Proposition 5.11.** For every continuous map $r: X \to Y$, $\mathcal{H}_0r$ is proper if and only if $\mathcal{H}r$ is. In particular, $\mathcal{H}r$ preserves proper maps.

*Proof.* It might seem necessary to write $\diamond_0$ for the $\diamond$ operator on spaces of the form $\mathcal{H}_0Z$, keeping $\diamond$ for spaces $HZ$, as we did with $\square$ in the proof of Lemma 5.4. This is unnecessary because $\diamond_0 U = \diamond U$: the closed sets that intersect $U$ are exactly the non-empty closed sets that intersect $U$.

Let us assume that $\mathcal{H}_0r$ is proper, and let $\zeta': \mathcal{H}_0Y \to Q_0\mathcal{H}_0X$ be a quasi-adjoint of $\mathcal{H}_0r$. By Theorem 3.2, $\zeta'(C) = (\mathcal{H}_0r)^{-1}(\uparrow_{\mathcal{H}_0Y} C)$ for every $C \in \mathcal{H}_0Y$. In particular, if $C \ne \emptyset$, then $\emptyset \not\in \zeta'(C)$, otherwise $\emptyset$ would be in $(\mathcal{H}_0r)^{-1}(\uparrow_{\mathcal{H}_0Y} C)$, meaning that $\mathcal{H}_0r(\emptyset)$, which is empty, would contain $C$. Hence $\zeta'(C)$ is a compact saturated subset of $\mathcal{H}_0X$ that is included in $\mathcal{H}X$; hence it is compact in $\mathcal{H}X$. Since $\mathcal{H}X$ is upwards-closed in $\mathcal{H}_0X$, $\zeta'(C)$ is upwards-closed in $\mathcal{H}X$, hence compact saturated in $\mathcal{H}X$, and therefore belongs to $Q_0\mathcal{H}X$. We therefore define $\zeta''$: $\mathcal{H}X \to Q_0\mathcal{H}X$ as the restriction of $\zeta'$ to $\mathcal{H}X$. For every open subset $U$ of $X$, $\zeta''^{-1}(\Box_0\diamond U) = \zeta'^{-1}(\Box_0\diamond U) \cap \mathcal{H}Y$, so $\zeta''$ is continuous. For every $C \in \mathcal{H}Y$, $\uparrow_{\mathcal{H}Y} C = \uparrow_{\mathcal{H}_0Y} C$, so $(\mathcal{H}r)^{-1}(\uparrow_{\mathcal{H}Y} C) = (\mathcal{H}r)^{-1}(\uparrow_{\mathcal{H}_0Y} C) = (\mathcal{H}_0r)^{-1}(\uparrow_{\mathcal{H}_0Y} C)$ (since $\mathcal{H}_0r$ and $\mathcal{H}r$ are defined by the same formula) $= \zeta'(C) = \zeta''(C)$. By Lemma 5.10, $\mathcal{H}r$ is proper.

For every proper map $r: X \to Y$, $\mathcal{H}_0r$ is proper by Proposition 5.9. We have just seen that this implies that $\mathcal{H}r$ is proper, showing the second part of the Proposition.

We now assume that $\mathcal{H}r$ is proper. Let $\zeta''$: $\mathcal{H}X \to Q_0\mathcal{H}X$ be its quasi-adjoint. By Theorem 3.2, $\zeta''(C) = (\mathcal{H}r)^{-1}(\uparrow_{\mathcal{H}Y} C)$; so, by the same
argument as above, $\zeta''(C) = (\mathcal{H}_0r)^{-1}(\uparrow_{\mathcal{H}_0Y} C)$. Let $\zeta'$ map every $C \in \mathcal{H}_0X$ to $\zeta''(C)$ if $C \neq \emptyset$, and $\emptyset$ to $\mathcal{H}_0X$. Hence $\zeta'(C) = (\mathcal{H}_0r)^{-1}(\uparrow_{\mathcal{H}_0Y} C)$ for every $C \in \mathcal{H}_0X$, including $\emptyset$. We claim that $\zeta'$ takes its values in $\mathcal{Q}_0\mathcal{H}_0X$. For every $C \in \mathcal{H}_0Y$, either $C$ is empty and $\zeta'(C) = \mathcal{H}_0X$, which is the upwards closure of the single point $\emptyset$ in $\mathcal{H}_0$, hence is compact saturated; or $C$ is non-empty, and then $\zeta'(C) = \zeta''(C) \in \mathcal{Q}_0\mathcal{H}X$ is a compact saturated subset of $\mathcal{H}X$; $\mathcal{H}X$ is an upwards-closed subspace of $\mathcal{H}_0X$, so $\zeta'(C)$ is also compact saturated in $\mathcal{Q}_0\mathcal{H}_0X$.

In order to show that $\zeta'$ is continuous, we show that $\zeta'^{-1}(\Box_0\diamond U)$ is open in $\mathcal{H}_0X$ for every open subset $U$ of $X$. For every $C \in \mathcal{H}_0Y$, $C \in \zeta'^{-1}(\Box_0\diamond U)$ if and only if $C = \emptyset$ and $\mathcal{H}_0X \subseteq \diamond U$, or $C \neq \emptyset$ and $\zeta'(C) \subseteq \diamond U$. But $\mathcal{H}_0X \subseteq \diamond U$ is false, since $\emptyset$ is in $\mathcal{H}_0X$ and not in $\diamond U$; and if $C \neq \emptyset$, $\zeta'(C) \subseteq \diamond U$ if and only if $\zeta''(C) \subseteq \diamond U$, since $\zeta'(C) = \zeta''(C)$. Therefore $\zeta'^{-1}(\Box_0\diamond U) = \zeta''^{-1}(\Box_0\diamond U)$, which is open in $\mathcal{H}X$, hence in $\mathcal{H}_0X$, since $\mathcal{H}X = \diamond X$ is open in $\mathcal{H}_0X$. By Lemma 5.10, $\mathcal{H}_0r$ is proper. $\square$

**5.3. The V. functors.** Let us deal with spaces of measures, or rather of continuous valuations, a closely related notion [15, 14]. An introduction to continuous valuations can be found in [7, Section IV-9]; we will give definitions below. In brief, while measures give mass to sets in a $\sigma$-algebras, a continuous valuation gives mass to open sets. Every Borel measure on a second-countable space, and more generally on a hereditarily Lindelöf space, yields a continuous valuation by restricting it to the open sets, as a special case of [2, Theorem 3.1]. Conversely, continuous valuations on a topological space extend to measures on its Borel $\sigma$-algebra under some conditions. One of the first theorems of this kind is due to Jimmie Lawson [18]; see also [16] for more, and [3] for a recent result.

A dcpo (directed complete partial order) is a poset $X$ in which every directed family $D$ has a supremum $\sup D$; $D$ is *directed* if and only if $D \neq \emptyset$ and any two elements of $D$ have an upper bound in $D$. A function $f: X \to Y$ between dcpos is *Scott-continuous* if and only if it is monotonic and preserves suprema of directed families. A subset $U$ of $X$ is *Scott-open* if and only if it is upwards-closed and intersects every directed family $D$ such that $\sup D \in U$.

Let $\mathbb{R}_+$ be the set of extended non-negative real numbers $\mathbb{R}_+ \cup \{\infty\}$, with its usual ordering $\leq$. We give it the Scott topology, whose open sets are the intervals $[t, \infty]$, $t \in \mathbb{R}_+$, plus $\emptyset$ and $\mathbb{R}_+$ itself.

The lattice $\mathcal{O}X$ of open subsets of a topological space $X$ is another example of a dcpo. A *continuous valuation* on a space $X$ is a map $\nu: \mathcal{O}X \to \mathbb{R}_+$ that is *strict* ($\nu(\emptyset) = 0$), *modular* (for all $U, V \in \mathcal{O}X$, $\nu(U) + \nu(V) = \nu(U \cup V) + \nu(U \cap V)$) and *Scott-continuous*. We say that
$\nu$ is a probability valuation if and only if $\nu(X) = 1$, and a subprobability valuation if and only if $\nu(X) \le 1$.

Let $VX$ denote the space of continuous valuations on a space $X$. We give it the weak topology, which is generated by subbasic open sets $[U > r] \stackrel{\text{def}}{=} \{\nu \in VX \mid \nu(U) > r\}$, where $U \in \mathcal{O}X$ and $r \in \mathbb{R}_+$. We define its subspace $V_1X$ of probability valuations and $V_{\le 1}X$ of subprobability valuations similarly. The specialization ordering of each is the stochastic ordering $\le$ given by $\nu \le \nu'$ if and only if $\nu(U) \le \nu'(U)$ for every $U \in \mathcal{O}X$. In general, we will write $V_\bullet X$, where $\bullet$ is nothing, "$\le 1$", or "1".

This defines a functor $V_\bullet$ from **Top** to **Top**. The action of $V_\bullet$ on morphisms $f: X \to Y$ is given by $V_\bullet f(\nu) \stackrel{\text{def}}{=} f[\nu]$ for every $\nu \in V_\bullet X$, where $f[\nu]$ is defined by $f[\nu](V) \stackrel{\text{def}}{=} \nu(f^{-1}(V))$ for every $V \in \mathcal{O}Y$.

The key point here is that there is a natural transformation $\lambda^{\text{DP}}: V_\bullet Q \to QV_\bullet$ where, for every topological space $X$, for every $\xi \in V_\bullet QX$, $\lambda_X^{\text{DP}}(\xi) \stackrel{\text{def}}{=} \{\nu \in V_\bullet X \mid \nu(U) \ge \xi(\Box U) \text{ for every } U \in \mathcal{O}X\}$ [12, Proposition 3.8]. This has a lot more properties than we actually need: $V_\bullet$ is a monad, and $\lambda^{\text{DP}}$ is a weak distributive law of $Q$ over $V_\bullet$. One of the equations that such a weak distributive law satisfies is $\lambda^{\text{DP}} \circ V_\bullet \eta^Q = \eta^Q V_\bullet$, and this is what we have called compatibility with $Q$ units. This is easy to check: for every $\mu \in V_\bullet X$, for every $U \in \mathcal{O}X$, $V_\bullet \eta_X^Q(\mu)(\Box U) = \mu((\eta_X^Q)^{-1}(\Box U) = \mu(U)$, so $\lambda_X^{\text{DP}}(V_\bullet \eta^Q(\mu))(U) = \{\nu \in V_\bullet X \mid \nu(U) \ge \mu(U) \text{ for every } U \in \mathcal{O}X\} = \eta_{V_\bullet X}^Q(\mu)$. By Remark 5.7, we obtain the following; this is related to [8, Theorem 6.5], which states that $V_1$ preserves proper surjective maps between stably compact spaces.

**Proposition 5.12.** $V_\bullet$ preserves proper almost surjective maps.

Let us replace $Q$ by $Q_0$. For every space $X$, let $X^\top$ be obtained by adding a fresh element $\top$ to $X$, and let the open subsets of $X^\top$ be the empty set plus the sets $U \cup \{\top\}$, $U \in \mathcal{O}X$. The specialization preordering of $X^\top$ coincides with that of $X$ on points of $X$, and places $\top$ above all elements of $X$. Let $j_X$ denote the inclusion of $X$ into $X^\top$. This is a topological embedding.

We obtain a functor $\underline{X}^\top$ from **Top** to **Top**, whose action on morphisms $f: X \to Y$ is $f^\top: X^\top \to Y^\top$, mapping every $x \in X$ to $f(x)$ and $\top$ to $\top$.

As earlier, we will write $\Box_0 U$ for $\{Q \in \mathcal{Q}_0 X \mid Q \subseteq U\}$ in order to distinguish it from $\Box U = \{Q \in \mathcal{Q} X \mid Q \subseteq U\}$; we have $\underline{\Box}_0 U = \Box U \cup \{\emptyset\}$.

**Lemma 5.13.** For every topological space $X$, the map $t_X: Q \mapsto Q \cup \{\top\}$ is a homeomorphism of $Q_0 X$ onto $Q(X^\top)$, naturally in $X$. Additionally, $t_X \circ \eta_X^Q = \eta_{X^\top}^Q \circ j_X$.

*Proof.* For every compact saturated subset $Q$ of $X$, $Q \cup \{\top\}$ is saturated and non-empty. Any open cover of $Q \cup \{\top\}$ can be trimmed to one that
does not contain the empty set, hence one of the form $(U_i \cup \{\top\})_{i \in I}$ with each $U_i$ open in $X$; then $(U_i)_{i \in I}$ is an open cover of $Q$, from which we can extract a finite subcover. This shows that $Q \cup \{\top\}$ is compact in $X^\top$.

For every $Q' \in Q(X^\top)$, let $s_X(Q') \stackrel{\text{def}}{=} Q' \cap X$. $X$ is closed in $X^\top$, so $s_X(Q')$ is compact in $X^\top$, hence in $X$. It is clearly upwards-closed in $X$, so $s_X(Q') \in Q_0 X$. We have $t_X(s_X(Q')) = Q'$ because $Q'$ is non-empty and saturated in $X^\top$, hence contains $\top$. For every $Q \in Q_0 X$, $s_X(t_X(Q)) = Q$, too, so $t_X$ is a bijection with inverse $s_X$.

For every $U \in \mathcal{O}X$, $t_X^{-1}(\Box_0(U \cup \{\top\})) = \Box U$, hence also $s_X^{-1}(\Box U) = \Box_0(U \cup \{\top\})$; therefore both $t_X$ and $s_X$ are continuous. For naturality, let $f: X \to Y$ be a continuous map. For every $Q \in Q_0 X$, $Q(f^\top)(t_X(Q)) = \uparrow_{X^\top} f^\top[Q \cup \{\top\}] = \uparrow_X f[Q] \cup \{\top\} = t_Y(Qf(Q))$. Finally, for every $x \in X$, $t_X(\eta_X^Q(x)) = \uparrow_X x \cup \{\top\} = \uparrow_{X^\top} x = \eta_{X^\top}^Q(j_X(x))$. $\square$

**Lemma 5.14.** For every topological space $X$, there is a continuous map $c_X^V: QV_\bullet(X^\top) \to Q_0V_\bullet X$ defined by $c_X^V(Q) \stackrel{\text{def}}{=} \{\nu \in V_\bullet X \mid j_X[\nu] \in Q\}$. It is natural in $X$, and $c_X^V \circ \eta_{V_\bullet(X^\top)}^Q \circ V_\bullet j_X = \eta_{V_\bullet X}^Q$.

*Proof.* Let $\mathcal{F}_X \stackrel{\text{def}}{=} \{\nu \in V_\bullet(X^\top) \mid \nu(\{\top\}) = 0\}$. This definition makes sense, and notably the condition $\nu(\{\top\}) = 0$, because $\{\top\}$ is open in $X^\top$. We give $\mathcal{F}_X$ the subspace topology.

For every $\mu \in \mathcal{F}_X$, let $\mu^-$ map every $U \in \mathcal{O}X$ to $\mu(U \cup \{\top\})$: $\mu^-$ is Scott-continuous, modular, and strict because $\mu(\{\top\}) = 0$. This defines a function $\overline{-}: \mathcal{F}_X \to V_\bullet X$, which is continuous because the inverse image of $[U > r]$ is $[U \cup \{\top\} > r]$. For every $U \in \mathcal{O}X$, $j_X[\mu^-](U \cup \{\top\}) = \mu^-((j_X)^{-1}(U \cup \{\top\})) = \mu^-(U) = \mu(U \cup \{\top\})$, while $j_X[\mu^-](\emptyset) = \mu^-((j_X)^{-1}(\emptyset)) = 0$, so $V_\bullet j_X \circ \overline{-} = \text{id}_{\mathcal{F}_X}$. Conversely, for every $\nu \in V_\bullet X$, $(V_\bullet j_X(\nu))^-$ maps every $U \in \mathcal{O}X$ to $j_X(\nu)[U \cup \{\top\}] = \nu((j_X)^{-1}(U \cup \{\top\})) = \nu(U)$, so $\overline{-} \circ V_\bullet j_X = \text{id}_{V_\bullet X}$. Hence $\mathcal{F}_X$ and $V_\bullet X$ are homeomorphic through $\overline{-}$ and $V_\bullet j_X$, and for every $Q \in QV_\bullet(X^\top)$, $c_X^V(Q)$ is the image of $Q \cap \mathcal{F}_X$ under $\overline{-}$.

$\mathcal{F}_X$ is the complement of $[\{\top\} > 0]$, hence is closed in $V_\bullet(X^\top)$. It follows that for every $Q \in QV_\bullet(X^\top)$, $Q \cap \mathcal{F}_X$ is compact in $V_\bullet(X^\top)$, hence also in $\mathcal{F}_X$. Therefore its image under $\overline{-}$, namely $c_X^V(Q)$, is compact in $V_\bullet X$. For every $\nu \in c_X^V(Q)$, for every $\nu' \in V_\bullet X$ such that $\nu \le \nu'$, it is clear that $j_X[\nu] \le j_X[\nu']$, so $\nu' \in c_X^V(Q)$. Therefore $c_X^V(Q) \in Q_0V_\bullet X$.

For every $U \in \mathcal{O}(V_\bullet X)$, for every $Q \in QV_\bullet(X^\top)$, $Q \in (c_X^V)^{-1}(\Box_0 U)$ if and only if the image of $Q \cap \mathcal{F}_X$ under $\overline{-}$ is included in $U$, if and only if $Q \cap \mathcal{F}_X \subseteq (\overline{-})^{-1}(U)$, equivalently $Q \subseteq (\overline{-})^{-1}(U) \cup \{\top\} > 0$. Hence $(c_X^V)^{-1}(\Box_0 U) = \Box((\overline{-})^{-1}(U) \cup \{\top\} > 0)$: $c_X^V$ is continuous.

Let $f: X \to Y$ be a continuous map. For every $Q \in QV_\bullet(X^\top)$, $c_Y^V(QV_\bullet(f^\top)(Q)) = c_Y^V(\uparrow\{f^\top[\mu] \mid \mu \in Q\})$ is equal to the image of
$\uparrow_{\mathbf{V}_{\bullet}(Y^{\top})} \{f^{\top}[\mu] | \mu \in \mathcal{Q}\} \cap \mathcal{F}_Y$ under $\bar{\cdot}$. Since $\bar{\cdot}$ is a homeomorphism, this is also $\uparrow_{\mathcal{F}_Y} \{(f^{\top}[\mu])^{-} | \mu \in \mathcal{Q}, f^{\top}[\mu] \in \mathcal{F}_Y\}$. Now $f^{\top}[\mu] \in \mathcal{F}_Y$ if and only if $f^{\top}[\mu](\{\top\}) = 0$, if and only if $\mu((f^{\top})^{-1}(\{\top\})) = \mu(\{\top\})$ is equal to 0, and then $(f^{\top}[\mu])^{-}$ maps every $V \in \mathcal{O}Y$ to $\mu((f^{\top})^{-1}(V \cup \{\top\})) = \mu(f^{-1}(V) \cup \{\top\}) = \mu^{-}(f^{-1}(\top)) = f[\mu^{-}]$. Therefore $c_Y^{\mathcal{V}}(\mathcal{Q}\mathbf{V}_{\bullet}(f^{\top})(\mathcal{Q})) = \uparrow_{\mathcal{F}_Y} \{f[\mu^{-}] | \mu \in \mathcal{Q} \cap \mathcal{F}_X\} = \mathcal{Q}_0 \mathbf{V}_{\bullet} f(\{\mu^{-} | \mu \in \mathcal{Q} \cap \mathcal{F}_X\})$. This is equal to $\mathcal{Q}_0 \mathbf{V}_{\bullet} f(c_X^{\mathcal{V}}(\mathcal{Q}))$, showing that $c_X^{\mathcal{V}} \circ \mathcal{Q}\mathbf{V}_{\bullet}(f^{\top}) = \mathcal{Q}_0 \mathbf{V}_{\bullet} f \circ c_X^{\mathcal{V}}$, hence that $c^{\mathcal{V}}$ is a natural transformation.

Finally, for every $\nu \in \mathbf{V}_{\bullet}X$, let $\mu \stackrel{\text{def}}{=} \mathbf{V}_{\bullet}j_X(\nu)$, so that $\nu = \mu^{-}$. Then $c_X^{\mathcal{V}}(\eta_{\mathbf{V}_{\bullet}(X^{\top})}^{\mathcal{Q}}(\mu)) = \{\nu' \in \mathbf{V}_{\bullet}X \mid j_X[\nu'] \ge \mu\}$. Now $j_X[\nu'] \ge \mu$ if and only if for every $U \in \mathcal{O}X$, $j_X[\nu'] (U \cup \{\top\}) \ge \mu(U \cup \{\top\})$ (and $j_X[\nu'](\emptyset) \ge \mu(\emptyset)$, which is trivial), and $j_X[\nu'] (U \cup \{\top\}) = (\mathbf{V}_{\bullet}j_X(\nu'))^{-}(U \cup \{\top\}) = \nu'(U \cup \{\top\});$ so $j_X[\nu'] \ge \mu$ if and only if $\nu' \ge \mu^{-}$. Hence $c_X^{\mathcal{V}}(\eta_{\mathbf{V}_{\bullet}(X^{\top})}^{\mathcal{Q}}(\mu)) = \uparrow_{\mathbf{V}_{\bullet}X} \mu^{-} = \uparrow_{\mathbf{V}_{\bullet}X} \nu$. Therefore $c_X^{\mathcal{V}} \circ \eta_{\mathbf{V}_{\bullet}(X^{\top})}^{\mathcal{Q}} \circ \mathbf{V}_{\bullet}j_X = \eta_{\mathbf{V}_{\bullet}X}^{\mathcal{Q}}$. $\square$

**Theorem 5.15.** $\mathbf{V}_{\bullet}$ preserves proper maps.

*Proof.* We form $\lambda_X \stackrel{\text{def}}{=} c_X^{\mathcal{V}} \circ \lambda_{X^{\top}}^{\text{DP}} \circ \mathbf{V}_{\bullet}t_X$, where $t_X$ is from Lemma 5.13, $c_X^{\mathcal{V}}$ is from Lemma 5.14. Hence $\lambda_X$ is a continuous map from $\mathbf{V}_{\bullet}\mathcal{Q}_0X$ to $\mathcal{Q}_0\mathbf{V}_{\bullet}X$ and is natural in $X$. We have $\lambda_X \circ \mathbf{V}_{\bullet}\eta_X^{\mathcal{Q}} = c_X^{\mathcal{V}} \circ \lambda_{X^{\top}}^{\text{DP}} \circ \mathbf{V}_{\bullet}\eta_{X^{\top}}^{\mathcal{Q}} \circ j_X$ (since $t_X \circ \eta_X^{\mathcal{Q}} = \eta_{X^{\top}}^{\mathcal{Q}} \circ j_X$) $= c_X^{\mathcal{V}} \circ \eta_{\mathbf{V}_{\bullet}(X^{\top})}^{\mathcal{Q}} \circ j_X$ (since $\lambda_X^{\text{DP}}$ is compatible with $\mathcal{Q}$ units) $= \eta_{\mathbf{V}_{\bullet}X}^{\mathcal{Q}}$ (by the last part of Lemma 5.14). Therefore $\lambda$ is compatible with $\mathcal{Q}_0$ units. We conclude by Corollary 5.2. $\square$

**Remark 5.16.** We leave it as an exercise to the reader to show that the map $\lambda_X$ used in the proof of Theorem 5.15 obeys a similar formula as $\lambda_X^{\text{DP}}$, namely $\lambda_X(\xi) = \{\nu \in \mathbf{V}_{\bullet}X \mid \nu(U) \ge \xi(\square_0U) \text{ for every } U \in \mathcal{O}X\}$, for every $\xi \in \mathbf{V}_{\bullet}\mathcal{Q}_0X$.

# 6. PROJECTIVE SYSTEMS CONSISTING OF PROPER MAPS

Our final application has to do with projective limits in **Top**, complementing the results of [11]. A projective system of topological spaces, which we write as $(p_{ij} : X_j \to X_i)_{i \sqsubseteq j \in I}$, is a collection of spaces $X_i$ indexed by a directed preordered set $(I, \sqsubseteq)$, with morphisms $p_{ij} : X_j \to X_i$ (the bonding maps) for all indices $i \sqsubseteq j$, such that $p_{ii} = \text{id}_{X_i}$ and $p_{ij} \circ p_{jk} = p_{ik}$ for all $i \sqsubseteq j \sqsubseteq k$ in $I$. In other words, it is a functor from $(I, \sqsupseteq)$ to **Top**. Every functor from a small category to **Top** has a limit, and limits of projective systems are called projective limits. The canonical projective limit $X$, $(p_i)_{i \in I}$ of $(p_{ij} : X_j \to X_i)_{i \sqsubseteq j \in I}$ is given by $\{\vec{x} \in \prod_{i \in I} X_i \mid \forall i \sqsubseteq j \in I, p_{ij}(x_j) = x_i\}$, with the subspace topology
from the product, and where $p_i$ is projection onto coordinate $i$. $X$ has a
basis of open subsets of the form $p_i^{-1}(U)$ where $i \in I$ and $U \in \mathcal{O}X_i$.

On a historical note, Arthur Harold Stone studied projective limits of hereditary compact spaces (Noetherian spaces) with closed bonding maps [22]. Since every subset of such a space is compact, every closed map between such spaces is proper.

We will also need to know the following. A filtered family of subsets of a space is one that is directed with respect to reverse inclusion. We write $\bigcap_{j \in J} Q_j$ for intersections of filtered families $(Q_j)_{j \in J}$. A well-filtered space $Z$ is a topological space such that for every filtered family $(Q_j)_{j \in J}$ of compact saturated subsets, for every open subset $U$ of $Z$, if $\bigcap_{j \in J} Q_j \subseteq U$ then $Q_j \subseteq U$ for some $j \in J$. It follows that for every filtered family as above, $\bigcap_{j \in J} Q_j$ is compact saturated [9, Proposition 8.3.6]. Every sober space is well-filtered [9, Proposition 8.3.5].

**Proposition 6.1.** Let $(p_{ij}: X_j \to X_i)_{i \subseteq j \in I}$ be a projective system in Top, with canonical projective limit $X$, $(p_i)_{i \in I}$. Let us also assume that each $X_j$ is sober and that each $p_{ij}$ is proper. Then:

(1) every $p_i$ is proper; we write $\zeta_i$ for its quasi-adjoint;

(2) for every $i \in I$, for every $U \in \mathcal{O}X$, the largest open subset $U_i$ of $X_i$ such that $p_i^{-1}(U_i) \subseteq U$ is $\zeta_i^{-1}(\Box U)$.

*Proof*. Before we start, we note that: (*) the sets $\Box p_j^{-1}(U)$ where $j \in I$ and $U \in \mathcal{O}X_j$ form a basis of open subsets of $Q_0 X$. Indeed, let $Q \in \mathcal{Q}_0 X$ and $\mathcal{U}$ be an open neighborhood of $Q$ in $\mathcal{Q}_0 X$. There is an open subset $V$ of $X$ such that $Q \in \Box V \subseteq \mathcal{U}$. We write $V$ as a union $\bigcup_{k \in K} p_{jk}^{-1}(U_k)$, where $j_k \in I$ and $U_k \in \mathcal{O}X_{jk}$. Since $Q$ is compact and included in $\bigcup_{k \in K} \bar{p}_{jk}^{-1}(U_k)$, it is included in a finite sub-union; we may therefore assume $K$ finite. Let $j \in I$ be above every $j_k$ with $k \in K$ (with respect to $\subseteq$), and let $U = \bigcup_{k \in K} p_{jk}^{-1}(U_k)$. For every $\vec{x} \in Q$, $p_{jk}(\vec{x}) \in U_k$ for some $k \in K$; since $p_{jk}(\vec{x}) = p_{jk,j}(p_j(\vec{x}))$, $p_j(\vec{x}) \in \bar{p}_{jk,j}^{-1}(U_k) \subseteq U$. Hence $Q \in \Box p_j^{-1}(U)$. For every $Q' \in \Box p_j^{-1}(U)$, $Q' \subseteq \bar{p}_j^{-1}(U) = \bigcup_{k \in K} \bar{p}_j^{-1}(p_{jk,j}^{-1}(U_k)) = \bigcup_{k \in K} p_{jk}^{-1}(U_k) \subseteq V$, so $\Box p_j^{-1}(U) \subseteq \Box V \subseteq \mathcal{U}$.

(1) We fix $i \in I$. We build a quasi-adjoint $\zeta_i$ to $p_i$.

Let $y \in X_i$. For every $k \in I$ such that $i \subseteq k$, $p_{ik}^{-1}(\uparrow y)$ is compact (and saturated) since $p_{ik}$ is proper. For every $j \subseteq k$, we let $Q_{jk} \stackrel{\text{def}}{=} \uparrow p_{jk}[p_{ik}^{-1}(\uparrow y)]$. We claim that for all $j, k, k' \in I$ such that $i, j \subseteq k \subseteq k'$, $Q_{jk'} \subseteq Q_{jk}$. For every $x \in Q_{jk'}$, there is a point $x' \in X_{k'}$ such that $p_{jk'}(x') \le x$ and $y \le p_{ik'}(x')$. In other words, $p_{jk}(p_{kk'}(x')) \le x$ and $y \le p_{ik}(p_{kk'}(x'))$, showing that $x \in \uparrow p_{jk}[p_{ik}^{-1}(\uparrow y)] = Q_{jk}$. Hence
the family $(Q_{jk})_{k \in \uparrow i \cap \uparrow j}$ is filtered. Since $X_j$ is sober hence well-filtered, $Q_j \stackrel{\text{def}}{=} \bigcap_{k \in \uparrow i \cap \uparrow j} Q_{jk}$ is a compact saturated subset of $X_j$.

We verify that for all $j \subseteq j' \in I$, $p_{jj'}$ maps $Q_{j'}$ to $Q_j$. It suffices to verify that it maps $Q_{j'k}$ to $Q_{jk}$ for every $k \in \uparrow i \cap \uparrow j'$. For every $x \in Q_{j'k}$, by definition there is a point $x' \in X_k$ such that $p_{j'k}(x') \le x$ and $y \le p_{ik}(x')$. Then $p_{jj'}(x) \ge p_{jj'}(p_{j'k}(x')) = p_{jk}(x')$, and $y \le p_{ik}(x')$, so $p_{jj'}(x) \in \uparrow p_{jk}[p_{ik}^{-1}(\uparrow y)] = Q_{jk}$.

Hence $(p_{jk|Q_k} : Q_k \to Q_j)_{j \sqsubseteq k \in \uparrow i}$ is a projective system of compact spaces, obtained from compact saturated subsets $Q_i$ of each $X_i$. Its canonical projective limit $Q$ is then compact saturated in $X$, by [11, Lemma 4.3].

We claim that $Q = p_i^{-1}(\uparrow y)$. By construction, $Q$ is the collection of tuples $\vec{x} \stackrel{\text{def}}{=} (x_i)_{i \in I}$ where each $x_j \in Q_j$ and for all $j \subseteq k \in I$, $x_j = p_{jk}(x_j)$. In other words, $Q = X \cap \prod_{j \in J} Q_j$. For each such tuple, $p_i(\vec{x}) = x_i$ is in $Q_i$, and $Q_i \subseteq Q_{ii} = \uparrow p_{ii}[p_{ii}^{-1}(\uparrow y)] = \uparrow y$. Therefore $Q \subseteq p_i^{-1}(\uparrow y)$. Conversely, let $\vec{x} \stackrel{\text{def}}{=} (x_i)_{i \in I}$ be any element of $X$ such that $p_i(\vec{x}) = x_i \in \uparrow y$. We claim that $\vec{x}$ in $Q$, namely that for every $j \in I$, $x_j \in Q_j$. In turn, we need to show that for every $k \in \uparrow i \cap \uparrow j$, $x_j \in \uparrow p_{jk}[p_{ik}^{-1}(\uparrow y)]$. We simply observe that $x_j = p_{jk}(x_k)$ (hence in particular $x_j \ge p_{jk}(x_j)$) and $x_k \in p_{ik}^{-1}(\uparrow y)$, since $p_{ik}(x_k) = x_i \in \uparrow y$, by assumption.

Using Theorem 3.2, let $\zeta_{jk}$ be the quasi-adjoint of $p_{jk}$, for all $j \subseteq k \in I$. We know that $\zeta_{jk}(x) = p_{jk}^{-1}(\uparrow x)$ for every $x \in X_j$. Hence $Q_{jk}$, as defined above, is equal to $Q_0p_{jk}(\zeta_{ik}(y))$, and $Q_j = \bigcap_{k \in \uparrow i \cap \uparrow j} Q_0p_{jk}(\zeta_{ik}(y))$.

For every $y \in X_i$, let us define $\zeta_i(y)$ as $p_i^{-1}(\uparrow y)$, namely as $X \cap \prod_{j \in J} Q_j = X \cap \prod_{j \in I} \bigcap_{k \in \uparrow i \cap \uparrow j} Q_0p_{jk}(\zeta_{ik}(y))$. We claim that $\zeta_i$ is continuous. It suffices to show that the inverse image of a basic open set of $Q_0X$ by $\zeta_i$ is open in $X_i$. We use the basis given in (*) at the beginning of the proof: $\zeta_i^{-1}(\Box p_j^{-1}(U))$ consists of the points $y \in X_i$ such that $\bigcap_{k \in \uparrow i \cap \uparrow j} Q_0p_{jk}(\zeta_{ik}(y)) \subseteq U$. Since $X_j$ is sober hence well-filtered, the latter is equivalent to the existence of $k \in \uparrow i \cap \uparrow j$ such that $Q_0p_{jk}(\zeta_{ik}(y)) \subseteq U$. But $Q_0p_{jk}(\zeta_{ik}(y)) \subseteq U$ is equivalent to $Q_0p_{jk}(\zeta_{ik}(y)) \in \Box U$, which is equivalent to $y \in (Q_0p_{jk} \circ \zeta_{ik})^{-1}(\Box U)$. Therefore $\zeta_i^{-1}(\Box(p_j^{-1}(U))) = \bigcup_{k \in \uparrow i \cap \uparrow j} (Q_0p_{jk} \circ \zeta_{ik})^{-1}(\Box U)$, which is an open set.

Lemma 5.10 now entails that $\zeta_i$ is quasi-adjoint to $p_i$.

(2) Since $\zeta_i$ is continuous, $\zeta_i^{-1}(\Box U)$ is open. By axiom (b) of quasi-adjoints, $\vec{x} \in \zeta_i(p_i(\vec{x}))$ for every $\vec{x} \in X$. Hence for every $\vec{x} \in p_i^{-1}(\zeta_i^{-1}(\Box U))$, $\vec{x} \in \zeta_i(p_i(\vec{x})) \subseteq U$; namely, $p_i^{-1}(\zeta_i^{-1}(\Box U)) \subseteq U$. In the reverse direction, for every open subset $U_i$ of $X_i$ such that $p_i^{-1}(U_i) \subseteq U$, for every $y \in U_i$, $\zeta_i(y) = p_i^{-1}(\uparrow y) \subseteq p_i^{-1}(U_i) \subseteq U$, so $\zeta_i(y) \in \Box U$. $\square$
In a topological space $X$, the collection $\mathbf{Q}$ of open neighborhoods of a compact saturated subset $Q$ is a Scott-open subset of $\mathcal{O}X$. Any union of such sets $\mathbf{Q}$ is Scott-open, and $X$ is called consonant if and only if the converse holds, namely: for every Scott-open subset $\mathcal{U}$ of $\mathcal{O}X$, for every $U \in \mathcal{U}$, there is a compact saturated subset $Q$ of $X$ such that $Q \subseteq U$ and $\mathbf{Q} \subseteq \mathcal{U}$. The notion arises from [5]; see also [9, Exercise 5.4.12].

**Proposition 6.2.** Let $(p_{ij}: X_j \to X_i)_{i \subseteq j \in I}$ be a projective system of topological spaces, with canonical projective limit $X$, $(p_i)_{i \in I}$. If every $p_{ij}$ is proper and if every $X_i$ is consonant and sober, then so is $X$.

*Proof.* Let $\mathcal{U}$ be a Scott-open subset of $\mathcal{O}X$. For every $i \in I$, let $\mathcal{U}_i$ be the collection of open subsets $U$ of $X_i$ such that $p_i^{-1}(U) \in \mathcal{U}$. Since $p_i^{-1}$ commutes with unions, $\mathcal{U}_i$ is Scott-open in $\mathcal{O}X_i$. Now let $U \in \mathcal{O}X$. For each $i \in I$, let $U_i$ be the largest open subset of $X_i$ such that $p_i^{-1}(U_i) \subseteq U$. The family $(p_i^{-1}(U_i))_{i \in I}$ is directed and its union is $U$ [10, Lemma 3.1]. Since $\mathcal{U}$ is Scott-open, $p_i^{-1}(U_i) \in \mathcal{U}$ for some $i \in I$; namely, $U_i \in \mathcal{U}_i$.

Since $X_i$ is consonant, there is a compact saturated subset $Q_i$ of $X_i$ such that $Q_i \subseteq U_i$ and $\mathbf{Q}Q_i \subseteq \mathcal{U}_i$. Using Proposition 6.1 (1), $Q \stackrel{\text{def}}{=} p_i^{-1}(Q_i)$ is compact saturated in $X$. Let also $\zeta_i$ be the quasi-adjoint of $p_i$.

We have $Q \subseteq U$ because $Q = p_i^{-1}(Q_i) \subseteq p_i^{-1}(U_i) \subseteq U$.

Let $V \in \mathbf{Q}$. Since $Q_i$ is upwards-closed, it is equal to the union of the sets $\uparrow y$ when $y$ ranges over $Q_i$. Then $p_i^{-1}(\uparrow y) = \zeta_i(y)$, so $Q = \bigcup_{y \in Q_i} \zeta_i(y)$. For every $y \in Q_i$, $\zeta_i(y) = p_i^{-1}(\uparrow y) \subseteq p_i^{-1}(Q_i) = Q \subseteq V$, so $Q_i \subseteq \zeta_i^{-1}(\Box V)$. Since $\mathbf{Q}Q_i \subseteq \mathcal{U}_i$, $\zeta_i^{-1}(\Box V)$ is in $\mathcal{U}_i$. By Proposition 6.1 (2), $\zeta_i^{-1}(\Box V)$ is the largest open subset $V_i$ of $X_i$ such that $p_i^{-1}(V_i) \subseteq V$. But $\zeta_i^{-1}(\Box V) \in \mathcal{U}_i$, so by definition of $\mathcal{U}_i$, $p_i^{-1}(V_i)$ is in $\mathcal{U}$. Since $\mathcal{U}$ is upwards-closed, $V$ is in $\mathcal{U}$. Hence $\Box V \subseteq \mathcal{U}$. $\square$

For every topological space $X$, for every $n \in \mathbb{N}$, let the copower $n \odot X$ be the topological sum of $n$ copies of $X$; $n \odot X$ is the collection of pairs $(k, x)$ with $1 \le k \le n$ and $x \in X$, with basic open subsets $\{k\} \times U$, $U \in \mathcal{O}X$. A space $X$ is $\odot$-consonant if and only if $n \odot X$ is consonant for every $n \in \mathbb{N}$ [3, Definition 13.1]. For example, every LCS-complete space is $\odot$-consonant [3, Lemma 13.2]. There is an $n \odot \_ $ endofunctor on **Top**: for every continuous map $f: X \to Y$, $n \odot f$ maps every $(k, x)$ to $(k, f(x))$.

A category **I** is *connected* if and only if it has at least one object, and every two objects are connected by a zig-zag of morphisms. A connected diagram in a category **C** is a functor from a small connected category **I** to **C**. Every projective system is a connected diagram.

**Lemma 6.3.** Let $X$, $(p_i)_{i \in \mathbb{I}}$ be the canonical limit of a connected diagram $F: \mathbb{I} \to \text{Top}$. For every $n \in \mathbb{N}$, $n \cdot X$, $(n \odot p_i)_{i \in I}$ is a projective limit of $(n \odot \_ ) \circ F$.
**Proof.** Let $X'$, $(q_i)_{i \in I}$ be the canonical projective limit of $(n \odot \_) \circ F$. The elements of $X'$ are the tuples $((k, x_i))_{i \in I}$ such that $(k, x_i) = (n \odot p_{ij})(k, x_j)$ for all $i \sqsubseteq j \in I$. The first component $k$ must be the same at all positions $i \in I$, because $\mathbf{I}$ is connected. A base of open subsets of $X'$ is given by the sets $q_i^{-1}(\{k\} \times U_i)$, where $i \in I$, $1 \le k \le n$, and $U_i \in \mathcal{O}X_i$. The map $f: ((k, x_i))_{i \in I} \mapsto (k, (x_i)_{i \in I})$ is a bijection of $X'$ onto $X$. It is continuous because $f^{-1}(\{k\} \times p_i^{-1}(U_i)) = q_i^{-1}(\{k\} \times U_i)$. Its inverse is continuous because the direct image of $\{k\} \times p_i^{-1}(U_i)$ is $q_i^{-1}(\{k\} \times U_i)$. Hence $f$ is a homeomorphism. Additionally, since $p_i$ and $q_i$ are both projections onto coordinate $i$, $(n \odot p_i) \circ f = q_i$. $\square$

**Corollary 6.4.** Let $(p_{ij}: X_j \to X_i)_{i \sqsubseteq j \in I}$ be a projective system of topological spaces, with canonical projective limit $X$, $(p_i)_{i \in I}$. If every $p_{ij}$ is proper and if every $X_i$ is $\odot$-consonant and sober, then so is $X$.

**Proof.** For every $n \in \mathbb{N}$, $n \odot X$, $(n \odot p_i)_{i \in I}$ is a projective limit of $(n \odot p_{ij}: n \odot X_j \to n \odot X_i)_{i \sqsubseteq j \in I}$ by Lemma 6.3. By assumption each space $n \odot X_i$ is consonant. It is sober because any coproduct of sober spaces taken in **Top** is sober [9, Lemma 8.4.2]. By Proposition 6.2, $n \odot X$ must then be consonant. Finally, any limit of sober spaces taken in **Top** is sober [9, Theorem 8.4.13]. $\square$

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UNIVERSITÉ PARIS-SACLAY, CNRS, ENS PARIS-SACLAY, LABORATOIRE MÉTHODES FORMELLES, 91190, GIF-SUR-YVETTE, FRANCE.

E-mail address: jgl@lmf.cnrs.fr