if $L\in\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}\Box{U_{i}}$. Property
2 follows from the characterization of
${(\mathop{\mathcal{P}\ell}_{\mathcal{V}}f)}^{-1}$ given in the first part of
Lemma 9.9.

###### Theorem 9.11

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system of topological spaces, with canonical projective limit
$X,{(p_{i})}_{i\in I}$. If every $X_{i}$ is sober, if $\iota_{X}$ is
surjective, and if ${\mathcal{H}}_{\mathsf{V}}X$ is a projective limit of
${({\mathcal{H}}_{\mathsf{V}}{p_{ij}}\colon{\mathcal{H}}_{\mathsf{V}}{X_{j}}\to{\mathcal{H}}_{\mathsf{V}}{X_{i}})}_{i\sqsubseteq
j\in I}$, then $\mathop{\mathcal{P}\ell}_{\mathcal{V}}X$ is a projective limit
of
${(\mathop{\mathcal{P}\ell}_{\mathcal{V}}{p_{ij}}\colon\mathop{\mathcal{P}\ell}_{\mathcal{V}}{X_{j}}\to\mathop{\mathcal{P}\ell}_{\mathcal{V}}{X_{i}})}_{i\sqsubseteq
j\in I}$.

###### Proof 28

By Proposition 9.10, it suffices to show that the comparison map
$\varphi\colon\mathop{\mathcal{P}\ell}_{\mathcal{V}}X\to Z$ is surjective,
where $Z$ is the canonical projective limit of
${(\mathop{\mathcal{P}\ell}_{\mathcal{V}}{p_{ij}}\colon\mathop{\mathcal{P}\ell}_{\mathcal{V}}{X_{j}}\to\mathop{\mathcal{P}\ell}_{\mathcal{V}}{X_{i}})}_{i\sqsubseteq
j\in I}$. Let ${(L_{i})}_{i\in I}$ be an element of the latter. We form the
quasi-lenses
$(Q_{i},C_{i})\mathrel{\buildrel\text{def}\over{=}}\iota_{X_{i}}(L_{i})$ for
each $i\in I$. For all $i\sqsubseteq j\in I$,
$L_{i}=\mathop{\mathcal{P}\ell}_{\mathcal{V}}{p_{ij}}(L_{j})$, so
$(Q_{i},C_{i})=\mathop{\mathcal{P}\ell}^{\mathrm{q}}_{\mathsf{V}}{p_{ij}}(Q_{j},C_{j})$
by naturality of $\iota$ (Lemma 9.9). Using Theorem 8.8, there is a (unique)
quasi-lens $(Q,C)$ on $X$ such that
$(Q_{i},C_{i})=\mathop{\mathcal{P}\ell}^{\mathrm{q}}_{\mathsf{V}}{p_{i}}(Q,C)$
for every $i\in I$. Since we are assuming that $\iota_{X}$ is surjective,
there is a lens $L$ on $X$ such that $(Q,C)=\iota_{X}(L)$. Then, for every
$i\in I$,
$\iota_{X_{i}}(L_{i})=(Q_{i},C_{i})=\mathop{\mathcal{P}\ell}^{\mathrm{q}}_{\mathsf{V}}{p_{i}}(\iota_{X}(L))=\iota_{X_{i}}(\mathop{\mathcal{P}\ell}_{\mathcal{V}}{p_{i}}(L))$,
by naturality of $\iota$. Since $\iota_{X_{i}}$ is injective (being a
topological embedding, see Lemma 9.2),
$L_{i}=\mathop{\mathcal{P}\ell}_{\mathcal{V}}{p_{i}}(L)$.

One case where $\iota_{X}$ is surjective, or equivalently, a homeomorphism,
where $X$ is as in Theorem 9.11, is when $X$ is weakly Hausdorff, by Lemma
9.2. This happens notably when every $X_{i}$ is locally strongly sober, as we
now argue. The original definition of a locally strongly sober space is a
space in which the collection of limits of every convergent ultrafilter is the
closure of a unique point [13, Definition VI-6.12]. A space is locally
strongly sober if and only if it is sober, coherent, and weakly Hausdorff [22,
Theorem 3.5], and every projective limit of locally strongly sober spaces is
locally strongly sober [23, Theorem 5.1]. We note that every Hausdorff space,
every stably locally compact space is weakly Hausdorff [22, Proposition 2.2],
and since they are all sober and coherent, they are all locally strongly
sober.

Another case where $\iota_{X}$ is surjective is when $X$ is quasi-Polish,
using Proposition 9.5 and Lemma 9.2. Now any projective limit of quasi-Polish
spaces is quasi-Polish [20, Proposition 9.5], so we are in this situation if
every $X_{i}$ is quasi-Polish. We recall that every Polish space is quasi-
Polish.

Hence, combining Theorem 9.11 with Theorem 7.5 (or Corollary 8.11), we obtain
the following.

###### Corollary 9.12

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system of topological spaces, with canonical projective limit
$X,{(p_{i})}_{i\in I}$. If every $X_{i}$ is locally strongly sober, or if
every $X_{i}$ is quasi-Polish, and:

  1. 1.

every $p_{ij}$ is a proper map (e.g., a projection),

  2. 2.

or $I$ has a countable cofinal subset and each $X_{i}$ is locally compact,

then $\mathop{\mathcal{P}\ell}_{\mathcal{V}}X$ is a projective limit of
${(\mathop{\mathcal{P}\ell}_{\mathcal{V}}{p_{ij}}\colon\mathop{\mathcal{P}\ell}_{\mathcal{V}}{X_{j}}\to\mathop{\mathcal{P}\ell}_{\mathcal{V}}{X_{i}})}_{i\sqsubseteq
j\in I}$.

In case 2, we note that the combination of the requirements of $X_{i}$ being
locally compact and locally strongly sober is equivalent to requiring that
$X_{i}$ be stably locally compact [13, Proposition VI-6.15, Corollary
VI-6.16]. Requiring instead that $X_{i}$ be locally compact and quasi-Polish
is equivalent to requiring that $X_{i}$ be locally compact, sober and second-
countable. Indeed, every quasi-Polish space is second-countable by definition,
and conversely every locally compact, sober, second-countable space is quasi-
Polish [5, Theorem 44].

##  10 Subcontinuation functors

All the functors we will consider from now on are subcontinuation functors, in
a sense we define below. (All the functors we have examined until now are
naturally isomorphic to subcontinuation functors, too, but it was easier to
deal with them as we did.) We will see that, whenever $T$ is a subcontinuation
functor, the comparison maps $\varphi\colon TX\to Z$ are topological
embeddings, and even homeomorphisms when $X$ is obtained as a limit of an ep-
system.

For every space $X$, let $\mathcal{L}X$ be the set of lower semicontinuous
maps from $X$ to $\overline{\mathbb{R}}_{+}$, namely the set of continuous
maps from $X$ to $\overline{\mathbb{R}}_{+}$ where the latter is given the
Scott topology of its usual ordering. $\mathcal{L}X$ is ordered pointwise, and
we give it the Scott topology of that ordering.

###### Definition 10.1

A _subcontinuation functor_ $T$ is an endofunctor on $\mathbf{Top}$ such that:

  * •

for every space $X$, $TX$ is a subspace of the space $KX$ of lower
semicontinuous maps from $\mathcal{L}X$ to $\overline{\mathbb{R}}_{+}$, with
the topology generated by subbasic open sets
$[h>r]\mathrel{\buildrel\text{def}\over{=}}\\{F\in TX\mid F(h)>r\\}$,
$h\in\mathcal{L}X$ (the subspace topology induced by the inclusion in the
product $\overline{\mathbb{R}}_{+}^{\mathcal{L}X}$);

  * •

for every morphism $f\colon X\to Y$, $Tf$ maps every $F\in TX$ to the function
$h\in\mathcal{L}Y\mapsto F(h\circ f)$.

$K$ itself, which maps every space $X$ to the space $KX$ of lower
semicontinuous maps from $\mathcal{L}X$ to $\overline{\mathbb{R}}_{+}$, is the
largest subcontinuation functor, which one may call the _continuation_
functor. The name is by analogy with the continuation monad used in the
denotational semantics of programming languages, with answer type
$\overline{\mathbb{R}}_{+}$.

We need the following.

###### Lemma 10.2

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system in $\mathbf{Top}$, with canonical projective limit $X,{(p_{i})}_{i\in
I}$. For every $h\in\mathcal{L}X$,

  1. 1.

there is a largest function $h_{i}\in\mathcal{L}{X_{i}}$ such that $h_{i}\circ
p_{i}\leq h$, for every $i\in I$;

  2. 2.

for all $i\sqsubseteq j\in I$, $h_{i}\circ p_{ij}\leq h_{j}$;

  3. 3.

for all $i\sqsubseteq j\in I$, $h_{i}\circ p_{i}\leq h_{j}\circ p_{j}$;

  4. 4.

for every $r\in\mathbb{R}_{+}$,
$h^{-1}(]r,\infty])=\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}p_{i}^{-1}(h_{i}^{-1}(]r,\infty]))$;

  5. 5.

$\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}(h_{i}\circ p_{i})=h$.

###### Proof 29

1\. Any pointwise supremum of lower semicontinuous maps is lower
semicontinuous, including the empty supremum, which is the constant zero map.
Let
$\mathcal{F}\mathrel{\buildrel\text{def}\over{=}}\\{k\in\mathcal{L}{X_{i}}\mid
k\circ p_{i}\leq h\\}$, and $h_{i}$ be its pointwise supremum. Then
$h_{i}\in\mathcal{L}{X_{i}}$, and it is clear that $h_{i}\circ p_{i}\leq h$,
so $h_{i}$ is the largest element of $\mathcal{F}$.

2\. We have $h_{i}\circ p_{i}=(h_{i}\circ p_{ij})\circ p_{j}\leq h$. By the
maximality of $h_{j}$, $h_{i}\circ p_{ij}\leq h_{j}$.

3\. By item 2, post-composing with $p_{j}$.

4\. Let $V\mathrel{\buildrel\text{def}\over{=}}h_{i}^{-1}(]r,\infty])$.
Letting $\chi_{V}$ be the characteristic function of $V$, $r\chi_{V}$ is a
lower semicontinuous map such that $r\chi_{V}\circ p_{i}\leq h$: indeed, for
every $x\in X$, if $p_{i}(x)\in V$ then $h(x)\geq h_{i}(p_{i}(x))>r$. By the
maximality of $h_{i}$, $r\chi_{V}\leq h_{i}$, which means that every point $x$
of $p_{i}^{-1}(h_{i}^{-1}(]r,\infty]))$ is such that
$r\chi_{V}(p_{i}(x))=r\leq h_{i}(x)$.

4\. Since $h_{i}\circ p_{i}\leq h$ for every $i\in I$,
$h^{-1}(]r,\infty])\supseteq\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}p_{i}^{-1}(h_{i}^{-1}(]r,\infty]))$. For the reverse inclusion, let $x$ be
any point in $h^{-1}(]r,\infty])$. We pick $t\in\mathbb{R}_{+}$ such that
$r<t<h(x)$, and we let
$U\mathrel{\buildrel\text{def}\over{=}}h^{-1}(]t,\infty])$. We recall that
there is a largest open subset $U_{i}$ of $X_{i}$ such that
$p_{i}^{-1}(U_{i})\subseteq U$, for every $i\in I$, and that
$\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}p_{i}^{-1}(U_{i})=U$. Hence
$x\in p_{i}^{-1}(U_{i})$ for some $i\in I$. We note that $t\chi_{U_{i}}\circ
p_{i}\leq h$, since every point mapped by $p_{i}$ into $U_{i}$ is in
$p_{i}^{-1}(U_{i})\subseteq U=h^{-1}(]t,\infty])$. By maximality of $h_{i}$,
$t\chi_{U_{i}}\leq h_{i}$. Then $h_{i}(p_{i}(x))\geq
t\chi_{U_{i}}(p_{i}(x))=t$, since $x\in p_{i}^{-1}(U_{i})$. Since $t>r$, it
follows that $x\in p_{i}^{-1}(h_{i}^{-1}(]r,\infty]))$.

5\. It suffices to observe that, for every $x\in X$, for every $i\in I$, for
every $r\in\mathbb{R}_{+}$, $r<\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}(h_{i}\circ p_{i})(x)$ if and only if
$x\in\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}p_{i}^{-1}(h_{i}^{-1}(]r,\infty]))$, $r<h(x)$ if and only if $x\in
h^{-1}(]r,\infty])$, and to apply item 4.

###### Lemma 10.3

Let $T$ be a subcontinuation functor. Given any projective $T$-situation as
given in Definition 3.1, the comparison map $\varphi$ is a topological
embedding.

###### Proof 30

Let $\varphi\colon TX\to Z$, where $Z,{(q_{i})}_{i\in I}$ is the canonical
projective limit of the projective system $(T{p_{ij}}\colon T{X_{j}}\to
T{X_{i}})_{i\sqsubseteq j\in I}$, and $X,{(p_{i})}_{i\in I}$ is that of
${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$.

For every $h\in\mathcal{L}X$, for every $i\in I$, let
$h_{i}\in\mathcal{L}{X_{i}}$ be the largest such that $h_{i}\circ p_{i}\leq
h$, as given in Lemma 10.2. For every subbasic open subset $[h>r]$ of $TX$,
with $h\in\mathcal{L}X$ and $r\in\mathbb{R}_{+}$, we wish to show that $[h>r]$
is the inverse image under $\varphi$ of some open subset of $Z$. We note that
for every $F\in TX$, $F\in[h>r]$ if and only if
$F(\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}(h_{i}\circ p_{i}))>r$ (by
Lemma 10.2, item 4), if and only if
$\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}F(h_{i}\circ p_{i})>r$ (since
$F$ is Scott-continuous) if and only if $F(h_{i}\circ p_{i})>r$ for some $i\in
I$; so $[h>r]=\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}[h_{i}\circ
p_{i}>r]$. Now we note that $[h_{i}\circ
p_{i}>r]=\varphi^{-1}(q_{i}^{-1}([h_{i}>r]))$. Indeed,
$q_{i}\circ\varphi=T{p_{i}}$, so the elements of
$\varphi^{-1}(q_{i}^{-1}([h_{i}>r]))$ are exactly the elements $F\in TX$ such
that $(q_{i}\circ\varphi)(F)\in[h_{i}>r]$, namely such that
$T{p_{i}}(F)(h_{i})>r$, equivalently such that $F(h_{i}\circ p_{i})>r$; those
are exactly the elements of $[h_{i}\circ p_{i}>r]$. Hence
$[h>r]=\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\varphi^{-1}(q_{i}^{-1}([h_{i}>r]))=\varphi^{-1}(\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}q_{i}^{-1}([h_{i}>r]))$, showing that $\varphi$ is full.

Finally, the preorder of specialization of $TX$ is given by $F\leq F^{\prime}$
if and only if for all $h\in\mathcal{L}X$ and $r\in\mathbb{R}_{+}$,
$F\in[h>r]$ implies $F^{\prime}\in[h>r]$, if and only if $F(h)\leq
F^{\prime}(h)$ for every $h\in\mathcal{L}X$. This is antisymmetric, so $TX$ is
$T_{0}$. It follows that $\varphi$ is a topological embedding.

Directed suprema, in fact arbitrary suprema, of elements of $KX$ are again in
$KX$, where $K$ is the continuation functor, and $X$ is an arbitrary space.
This is because arbitrary suprema of lower semicontinuous maps are lower
semicontinuous. A _subdcpo_ of a dcpo $P$ is a subset $A$ of $P$ such that the
supremum of any directed family $D\subseteq A$, taken in $P$, belongs to $A$.
This entails that $A$ is itself a dcpo, but the property is strictly stronger.
By some abuse of language, we will extend this to subdcpos of $KX$, implicitly
seing the latter as a dcpo.

###### Proposition 10.4

Let $T$ be a subcontinuation functor. Given any projective $T$-situation as
given in Definition 3.1, whose projective system is an ep-system, and if $TX$
is a subdcpo of $KX$, the comparison map $\varphi$ is a homeomorphism.

###### Proof 31

We take the same notations as in Definition 3.1 ($T$-situations). Let $e_{ij}$
be embeddings associated with each of the projections $p_{ij}$. By [20, Lemma
4.1], each $p_{i}$ is a projection, and there are associated embeddings
$e_{i}\colon X_{i}\to X$, such that $e_{j}\circ e_{ij}=e_{i}$ for all
$i\sqsubseteq j\in I$. Moreover, ${(e_{i}(p_{i}(\vec{x})))}_{i\in
I,\sqsubseteq}$ is a monotone net with supremum equal to $\vec{x}$ for every
$\vec{x}\in X$.

Using Lemma 10.3, it remains to show that $\varphi$ is surjective.

Let ${(F_{i})}_{i\in I}$ be any element of $Z$, that is, each $F_{i}$ is in
$TX_{i}$ and $T{p_{ij}}(F_{j})=F_{i}$ for all $i\sqsubseteq j\in I$. The
elements $T{e_{i}}(F_{i})\in TX$ form a monotone net, namely
$T{e_{i}}(F_{i})\leq T{e_{j}}(F_{j})$ for all $i\sqsubseteq j\in I$. Indeed,
this follows from the fact that $e_{i}\circ p_{ij}\leq e_{j}$ (because
$e_{j}\circ e_{ij}=e_{i}$ $e_{ij}\circ p_{ij}\leq\mathrm{id}_{X_{j}}$, and
$e_{i}$ is continuous, hence monotonic). Then, for every $h\in\mathcal{L}X$,
$T{e_{j}}(F_{j})(h)=F_{j}(h\circ e_{j})\geq F_{j}(h\circ e_{i}\circ p_{ij})$
(since composition with the continuous map $h$ is monotonic, and $F_{j}$ is
continuous hence monotonic as well) $=T{p_{ij}}(F_{j})(h\circ
e_{i})=F_{i}(h\circ e_{i})=T{e_{i}}(F_{i})(h)$.

Since $TX$ is a subdcpo of $KX$, the monotone net ${(T{e_{j}}(F_{j}))}_{j\in
I,\sqsubseteq}$ has a pointwise supremum, which is in $TX$. Let us call it
$F$. We show that $\varphi(F)={(F_{i})}_{i\in I}$, or equivalently, that
$T{p_{i}}(F)=F_{i}$ for every $i\in I$. We consider any $h\in\mathcal{L}X$,
and we aim to prove that $T{p_{i}}(F)(h)=F_{i}(h)$, namely that $F(h\circ
p_{i})=F_{i}(h)$, or equivalently, that
$\sup\nolimits^{\scriptstyle\uparrow}_{j\in J}F_{j}(h\circ p_{i}\circ
e_{j})=F_{i}(h)$. By taking $j\mathrel{\buildrel\text{def}\over{=}}i$ and
recalling that $p_{i}\circ e_{i}=\mathrm{id}_{X_{i}}$, we see that the left-
hand side is larger than or equal to the right-hand side. For the other
inequality, we consider any $j\in J$ and we show that $F_{j}(h\circ p_{i}\circ
e_{j})\leq F_{i}(h)$. Let us pick $k\in I$ such that $i,j\sqsubseteq k$. Then
$p_{i}\circ e_{j}\circ p_{jk}\leq p_{ik}$: indeed, $p_{i}\circ
e_{j}=p_{ik}\circ p_{k}\circ e_{k}\circ e_{jk}=p_{ik}\circ e_{jk}$, so
$p_{i}\circ e_{j}\circ p_{jk}\leq p_{ik}\circ e_{jk}\circ p_{jk}\leq p_{ik}$,
using implicitly that continuous maps are monotonic. Therefore $F_{j}(h\circ
p_{i}\circ e_{j})$, which is equal to $T{p_{jk}}(F_{k})(h\circ p_{i}\circ
e_{j})=F_{k}(h\circ p_{i}\circ e_{j}\circ p_{jk})$ is less than or equal to
$F_{k}(h\circ p_{ik})$ (since $F_{j}$ and $h$ are themselves continuous hence
monotonic), and the latter is equal to $T{p_{ik}}(F_{k})(h)=F_{i}(h)$.

##  11 Superlinear previsions and retracts

Previsions form models of mixed non-deterministic and probabilistic choice
[15], and are an elaboration on Walley’s notion of prevision in economics
[51]. We will borrow most of what we need from [19]. A _prevision_ on a space
$X$ is a Scott-continuous map
$F\colon\mathcal{L}X\to\overline{\mathbb{R}}_{+}$ that is _positively
homogeneous_ in the sense that $F(ah)=aF(h)$ for all $a\in\mathbb{R}_{+}$ and
$h\in\mathcal{L}X$. There is a space $\mathbb{P}X$ of previsions on $X$, whose
topology is generated by sets
$[h>r]\mathrel{\buildrel\text{def}\over{=}}\\{F\mid F(h)>r\\}$,
$h\in\mathcal{L}X$, $r\in\mathbb{R}_{+}$.

For example, any continuous valuation $\nu$ on $X$ gives rise to a prevision
$G\colon h\mapsto\int h\,d\nu$. Such a prevision is _linear_ , in the sense
that $G(h+h^{\prime})=G(h)+G(h^{\prime})$ for all
$h,h^{\prime}\in\mathcal{L}X$. Let $\mathbb{P}_{\mathtt{P}}X$ be the subspace
of $\mathbb{P}X$ of linear previsions. Conversely, every linear prevision
$G\in\mathbb{P}_{\mathtt{P}}X$ gives rise to a continuous valuation $U\mapsto
G(\chi_{U})$, where $\chi_{U}$ is the characteristic map of the open set $U$,
and the two constructions are inverse of each other. Additionally, those two
constructions define continuous maps between ${\mathbf{V}}X$ and
$\mathbb{P}_{\mathtt{P}}X$ [48, Satz 4.16]. We will therefore equate
continuous valuations with linear previsions.

A prevision is _sublinear_ (resp., _superlinear_) if and only if
$G(h+h^{\prime})\leq G(h)+G(h^{\prime})$ (resp., $\geq$) for all
$h,h^{\prime}\in\mathcal{L}X$. As in [19], we write
$\mathbb{P}_{{{\mathtt{A}}{\mathtt{P}}}}X$ for the subspace of $\mathbb{P}X$
consisting of sublinear previsions, and
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}X$ for the subspace of $\mathbb{P}X$
consisting of superlinear previsions.

Among the continuous valuations, there are the probability valuations and the
subprobability valuations. Similarly, we say that a prevision $F$ is
_subnormalized_ (resp., _normalized_) iff $F(\mathbf{1}+h)\leq\mathbf{1}+F(h)$
(resp., $=$) for every $h\in\mathcal{L}X$, where $\mathbf{1}$ is the constant
function with value $1$. The homeomorphism between ${\mathbf{V}}X$ and
$\mathbb{P}_{\mathtt{P}}X$ restricts to homeomorphisms between
${\mathbf{V}}_{\leq 1}X$ (resp., ${\mathbf{V}}_{1}X$) and the subspace
$\mathbb{P}_{\mathtt{P}}^{\leq 1}X$ (resp., $\mathbb{P}_{\mathtt{P}}^{1}X$) of
subnormalized (resp., normalized) linear previsions on $X$. We write
$\mathbb{P}_{{{\mathtt{A}}{\mathtt{P}}}}^{\leq 1}X$,
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\leq 1}X$,
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\leq 1}X$,
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{1}X$ for the corresponding spaces of
(sub)normalized, sublinear/superlinear previsions. In general, we write write
$\mathbb{P}_{{{\mathtt{A}}{\mathtt{P}}}}^{\bullet}X$ or
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}X$, where $\bullet$ can be
nothing, “$\leq 1$”, or “$1$”.

All those constructions define endofunctors on $\mathbf{Top}$, whose action
$\mathbb{P}f$ on morphisms $f\colon X\to Y$ is given by
$\mathbb{P}f(F)(h)\mathrel{\buildrel\text{def}\over{=}}F(h\circ f)$. We write
$\mathbb{P}f$ without any $\bullet$ superscript or any subscript
${\mathtt{P}}$, ${{\mathtt{A}}{\mathtt{P}}}$ or ${{\mathtt{D}}{\mathtt{P}}}$
because the action is defined in the same way for all functors. It is easy to
check that $\mathbb{P}f$ is a morphism from
$\mathbb{P}_{\mathtt{P}}^{\bullet}X$ to $\mathbb{P}_{\mathtt{P}}^{\bullet}Y$
for every continuous map $f\colon X\to Y$ and similarly with
${{\mathtt{A}}{\mathtt{P}}}$ or ${{\mathtt{D}}{\mathtt{P}}}$ in place of
${\mathtt{P}}$. Hence all prevision functors are subcontinuation functors in
the sense of Definition 10.1.

Additionally, this construction is compatible with the homeomorphisms
${\mathbf{V}}_{\bullet}X\cong\mathbb{P}_{\mathtt{P}}^{\bullet}X$, meaning that
those homeomorphisms are natural. Explicitly, for every
$F\in\mathbb{P}_{\mathtt{P}}^{\bullet}X$, letting $\nu$ be the associated
continuous valuation defined by
$\nu(U)\mathrel{\buildrel\text{def}\over{=}}F(\chi_{U})$ for every
$U\in{\mathcal{O}}X$, for every continuous map $f\colon X\to Y$, the
continuous valuation $\nu^{\prime}$ associated with $\mathbb{P}f(F)$ is equal
to $f[\nu]$: for every $V\in{\mathcal{O}}Y$,
$\nu^{\prime}(V)=\mathbb{P}f(F)(\chi_{V})=F(\chi_{V}\circ
f)=F(\chi_{f^{-1}(V)})=\nu(f^{-1}(V))=f[\nu](V)$.

Our plan for establishing projective limit preservation theorems for prevision
functors—apart from the case of ep-systems, which will follow from Proposition
10.4—is to rely on the fact that spaces of previsions on $X$ are retracts of
${\mathcal{Q}}_{\mathsf{V}}({\mathbf{V}}_{\bullet}X)$, resp.
${\mathcal{H}}_{\mathsf{V}}({\mathbf{V}}_{\bullet}X)$ under some conditions
[19, Proposition 3.11, Proposition 3.22] and to reuse our limit preservation
theorems for ${\mathcal{Q}}_{\mathsf{V}}$, ${\mathcal{H}}_{\mathsf{V}}$, and
${\mathbf{V}}_{\bullet}$.

A _retraction_ on a category ${\mathbf{C}}$ (of $Y$ _onto_ $X$) is a pair
$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$
such that $r\circ s=\mathrm{id}_{X}$. (Hence an ep-pair is a special case of a
retraction.) We also say that $r$, by itself, is the retraction, with
associated _section_ $s$, and that $X$ is a _retract_ of $Y$.

We call a _natural retraction_
$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$
of a functor $T\colon{\mathbf{C}}\to{\mathbf{D}}$ onto a functor
$S\colon{\mathbf{C}}\to{\mathbf{D}}$ any retraction in the category
${\mathbf{D}}^{\mathbf{C}}$ of functors from ${\mathbf{C}}$ to ${\mathbf{D}}$.
Explicitly, this is a collection of retractions
$\textstyle{SX\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s_{X}}$$\textstyle{TX\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{X}}$,
one for each object $X$ of ${\mathbf{C}}$, which are natural in $X$.

This will be fine for $\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}$, but
we will need the following refinement in the case of
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$. For any subcategory
${\mathbf{K}}$ of ${\mathbf{C}}$, a _${\mathbf{K}}$ -relative_ natural
retraction of $T$ onto $S$ is any collection of
$\textstyle{SX\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s_{X}}$$\textstyle{TX\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{X}}$,
where $X$ ranges over a class of objects of ${\mathbf{C}}$ that contains
$|{\mathbf{K}}|$, and which restricts to a natural retraction on
${\mathbf{K}}$.

Given a diagram $F\colon{\mathbf{I}}\to{\mathbf{C}}$ with a limit
$X,{(p_{i})}_{i\in|{\mathbf{I}}|}$, there is a small category
${\mathbf{I}}_{*}$ obtained by adjoining a fresh object $*$ to ${\mathbf{I}}$,
with unique morphisms from $*$ to all objects of ${\mathbf{I}}_{*}$, and there
is a functor $F_{*}\colon{\mathbf{I}}_{*}\to{\mathbf{C}}$ that extends $F$ and
such that $F_{*}(*)=X$, $F_{*}(*\to X_{i})=p_{i}$ for every
$i\in|{\mathbf{I}}|$. We will call _$F_{*}$ -relative_ natural retraction any
${\mathbf{K}}$-relative natural retraction, where ${\mathbf{K}}$ is a
subcategory of ${\mathbf{C}}$ that contains the image of $F_{*}$. In other
words, an $F_{*}$-relative natural retraction of $T$ onto $S$ is a collection
$\textstyle{SY\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s_{Y}}$$\textstyle{TY\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{Y}}$
of retractions, where $Y$ ranges over at least all the objects $F(i)$,
$i\in|{\mathbf{I}}|$ and $X$, and such that the following naturality
conditions are satisfied: (1) for every morphism $\varphi\colon i\to j$ in
${\mathbf{I}}$, $r_{F(j)}\circ TF(\varphi)=SF(\varphi)\circ r_{F(i)}$ and
$s_{F(j)}\circ SF(\varphi)=TF(\varphi)\circ s_{F(i)}$; (2) for every object
$i$ of ${\mathbf{C}}$, $r_{F(i)}\circ T{p_{i}}=S{p_{i}}\circ r_{X}$ and
$s_{F(i)}\circ S{p_{i}}=T{p_{i}}\circ s_{X}$. (As is customary in category
theory, we write $SF$ for $S\circ F$.)

###### Lemma 11.1

Let $F\colon{\mathbf{I}}\to{\mathbf{C}}$ be a diagram with a limit
$X,{(p_{i})}_{i\in|{\mathbf{I}}|}$, and let $S$ and $T$ be two functors from
${\mathbf{C}}$ to a category ${\mathbf{D}}$. If there is an $F_{*}$-relative
natural retraction
$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{s}$$\textstyle{T\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$,
and if $TX,{(Tp_{i})}_{i\in|{\mathbf{I}}|}$ is a limit of $TF$, then
$SX,{(Sp_{i})}_{i\in|{\mathbf{I}}|}$ is a limit of $SF$.

###### Proof 32

It is clear that $SX,{(S{p_{i}})}_{i\in|{\mathbf{I}}|}$ is a cone of $SF$. In
order to show that it is universal, let $Y,{(q_{i})}_{i\in|{\mathbf{I}}|}$ be
another cone of $SF$. Then $Y,{(s_{F(i)}\circ q_{i})}_{i\in|{\mathbf{I}}|}$ is
a cone of $TF$: for every morphism $\varphi\colon j\to i$ in ${\mathbf{I}}$,
$TF(\varphi)\circ s_{F(j)}\circ q_{j}=s_{F(i)}\circ SF(\varphi)\circ
q_{j}=s_{F(i)}\circ q_{i}$ by $F_{*}$-relative naturality of $s$ and the
definition of a cone of $SF$. By assumption, $TX,{(T{p_{i}})}_{i\in I}$ is a
limit of $TF$, so there is a unique morphism $f\colon Y\to TX$ such that
$T{p_{i}}\circ f=s_{F(i)}\circ q_{i}$ for every $i\in|{\mathbf{I}}|$. Then
$r_{F(i)}\circ T{p_{i}}\circ f=q_{i}$ for every $i\in|{\mathbf{I}}|$, since
$r_{F(i)}$ and $s_{F(i)}$ form a retraction. By $F_{*}$-relative naturality of
$r$, $r_{F(i)}\circ T{p_{i}}\circ f=S{p_{i}}\circ r_{X}\circ f$, so we have
found a morphism $g$ such that $S{p_{i}}\circ g=q_{i}$ for every
$i\in|{\mathbf{I}}|$, namely $r_{X}\circ f$. This is the only one: given any
morphism $g\colon Y\to SX$ such that $S{p_{i}}\circ g=q_{i}$ for every
$i\in|{\mathbf{I}}|$, we must have $s_{F(i)}\circ S{p_{i}}\circ
g=s_{F(i)}\circ q_{i}$ for every $i\in|{\mathbf{I}}|$, namely $T{p_{i}}\circ
s_{X}\circ g=s_{F(i)}\circ q_{i}$ for every $i\in|{\mathbf{I}}|$, by
$F_{*}$-relative naturality of $s$. By the uniqueness of $f$, $f=s_{X}\circ
g$, so $r_{X}\circ f=g$. Hence $g$ is unique.

By Proposition 3.22 of [19], for every topological space $X$, and letting
$\bullet$ be nothing, “$\leq 1$”, or “$1$”, there is a retraction
$r_{{{\mathtt{D}}{\mathtt{P}}}}\colon{\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)\to\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}X$,
defined by
$r_{{{\mathtt{D}}{\mathtt{P}}}}(Q)(h)\mathrel{\buildrel\text{def}\over{=}}\min_{G\in
Q}G(h)$, with associated section $s_{{{\mathtt{D}}{\mathtt{P}}}}$ defined by
$s_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}(F)=\\{G\in\mathbb{P}_{\mathtt{P}}^{\bullet}X\mid
G\geq F\\}$. (The ordering $\leq$ between previsions is the specialization
ordering, which is pointwise, and $\geq$ is the opposite ordering.) We write
them $r_{{{\mathtt{D}}{\mathtt{P}}}\,X}$ and
$s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}$ in order to make the dependency
on $X$ explicit, reserving the notations $r_{{{\mathtt{D}}{\mathtt{P}}}}$ and
$s_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}$ for the families of maps
$r_{{{\mathtt{D}}{\mathtt{P}}}\,X}$, resp.
$s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}$, where $X$ ranges over
topological spaces.

This retraction even cuts down to a homeomorphism between
${\mathcal{Q}}_{\mathsf{V}}^{cvx}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)$ and
$\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$ [19, Theorem 4.15], where
the former denotes the subspace of
${\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)$ consisting of
convex compact saturated subsets of $\mathbb{P}_{\mathtt{P}}^{\bullet}X$. (A
subset $A$ of the latter is _convex_ if and only if for all $G_{1},G_{2}\in
A$, for every $r\in[0,1]$, $rG_{1}+(1-r)G_{2}\in A$.)

###### Lemma 11.2

The transformations $r_{{\mathtt{D}}{\mathtt{P}}}$ and
$s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}$ are natural.

###### Proof 33

Let $f\colon X\to Y$ be any continuous map. For
$r_{{\mathtt{D}}{\mathtt{P}}}$, we need to show that for every
$Q\in{\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)$, for
every $h\in\mathcal{L}Y$,
$r_{{{\mathtt{D}}{\mathtt{P}}}\,Y}({\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f)(Q))(h)=\mathbb{P}f(r_{{{\mathtt{D}}{\mathtt{P}}}\,X}(Q))(h)$.
The left-hand side is equal to
$\min_{G^{\prime}\in{\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f)(Q)}G^{\prime}(h)=\min_{G^{\prime}\in\operatorname{\uparrow\\!}\\{\mathbb{P}f(G)\mid
G\in Q\\}}G^{\prime}(h)=\min_{G^{\prime}\in\\{\mathbb{P}f(G)\mid G\in
Q\\}}G^{\prime}(h)=\min_{G\in Q}\mathbb{P}f(G)(h)$, while the right-hand side
is equal to $r_{{{\mathtt{D}}{\mathtt{P}}}\,X}(Q)(h\circ f)=\min_{G\in
Q}G(h\circ f)$, and those are equal.

For $s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}$, we must show that for every
$F\in\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$,
$s_{{{\mathtt{D}}{\mathtt{P}}}\,Y}^{\bullet}(\mathbb{P}f(F))={\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f)\allowbreak(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F))$.
The left-hand side is in
${\mathcal{Q}}_{\mathsf{V}}^{cvx}(\mathbb{P}_{\mathtt{P}}^{\bullet}Y)$, and we
claim that so is the right-hand side; it suffices to show that it is convex.
We consider any two elements $G^{\prime}_{1}$, $G^{\prime}_{2}$ of
${\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f))(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F))$,
and $r\in[0,1]$. By definition, there are elements $G_{1}$, $G_{2}$ of
$s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F)$ such that
$\mathbb{P}f(G_{1})\leq G^{\prime}_{1}$ and $\mathbb{P}f(G_{2})\leq
G^{\prime}_{2}$. Since $s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F)$ is
convex, $rG_{1}+(1-r)G_{2}\in s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F)$.
It is easy to see that $rG^{\prime}_{1}+(1-r)G^{\prime}_{2}\geq
r\mathbb{P}f(G_{1})+(1-r)\mathbb{P}g(G_{2})=\mathbb{P}f(rG_{1}+(1-r)G_{2})$,
so
$rG^{\prime}_{1}+(1-r)G^{\prime}_{2}\in{\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f)(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F))$.
Since $r_{{{\mathtt{D}}{\mathtt{P}}}\,X}$ and
$r_{{{\mathtt{D}}{\mathtt{P}}}\,Y}$ are homeomorphisms (with domains
${\mathcal{Q}}_{\mathsf{V}}^{cvx}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)$, resp.
${\mathcal{Q}}_{\mathsf{V}}^{cvx}(\mathbb{P}_{\mathtt{P}}^{\bullet}Y)$), in
order to show that
$s_{{{\mathtt{D}}{\mathtt{P}}}\,Y}^{\bullet}(\mathbb{P}f(F))={\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f)(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F))$,
it suffices to show that
$r_{{{\mathtt{D}}{\mathtt{P}}}\,Y}(s_{{{\mathtt{D}}{\mathtt{P}}}\,Y}^{\bullet}(\mathbb{P}f(F)))=r_{{{\mathtt{D}}{\mathtt{P}}}\,Y}({\mathcal{Q}}_{\mathsf{V}}(\mathbb{P}f)(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F)))$.
The left-hand side is equal to $\mathbb{P}f(F)$, and the right-hand side is
equal to
$\mathbb{P}f(r_{{{\mathtt{D}}{\mathtt{P}}}\,X}(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}(F))$
(by naturality of $r_{{\mathtt{D}}{\mathtt{P}}}$), hence to $\mathbb{P}f(F)$.

###### Theorem 11.3

Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. The projective limit of a
projective system ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ of
topological spaces is preserved by
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}$ if and only if it is
preserved by ${\mathbf{V}}_{\bullet}$. In particular, it is preserved under
any of the three sets of conditions of Theorem 4.3.

###### Proof 34

We start with the if direction. Let $X,{(p_{i})}_{i\in I}$ be the canonical
projective limit of ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$.
If ${\mathbf{V}}_{\bullet}X,{({\mathbf{V}}{p_{i}})}_{i\in I}$ is a projective
limit of
${({\mathbf{V}}{p_{ij}}\colon{\mathbf{V}}_{\bullet}{X_{j}}\to{\mathbf{V}}_{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$, then
$\mathbb{P}_{{\mathtt{P}}}^{\bullet}X,{(\mathbb{P}{p_{i}})}_{i\in I}$ is a
projective limit of
${(\mathbb{P}{p_{ij}}\colon\mathbb{P}_{{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{P}}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$. Indeed, we recall that there is a natural homeomorphism between
${\mathbf{V}}_{\bullet}$ and $\mathbb{P}_{\mathtt{P}}^{\bullet}$. The spaces
${\mathbf{V}}_{\bullet}{X_{i}}$ are all sober (see comments around Remark
5.2), hence we can use Theorem 6.4 and conclude that
${\mathcal{Q}}_{\mathsf{V}}{(\mathbb{P}_{{\mathtt{P}}}^{\bullet}X)},{({\mathcal{Q}}_{\mathsf{V}}{(\mathbb{P}{p_{i}})})}_{i\in
I}$ is a projective limit of
$({\mathcal{Q}}_{\mathsf{V}}{(\mathbb{P}{p_{ij}})}\colon{\mathcal{Q}}_{\mathsf{V}}{(\mathbb{P}_{{\mathtt{P}}}^{\bullet}{X_{j}})}\to{\mathcal{Q}}_{\mathsf{V}}{(\mathbb{P}_{{\mathtt{P}}}^{\bullet}{X_{i}})})_{i\sqsubseteq
j\in I}$. We now use Lemma 11.1 with
$S\mathrel{\buildrel\text{def}\over{=}}\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}$
and
$T\mathrel{\buildrel\text{def}\over{=}}{\mathcal{Q}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}$,
and the natural retraction
$(r_{{\mathtt{D}}{\mathtt{P}}},s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet})$—it is
natural by Lemma 11.2.

In the only if direction, if
$\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X,{(\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{i}})}_{i\in
I}$ is a projective limit of the projective system
$(\mathbb{P}{p_{ij}}\colon\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$, then we claim that the comparison map
$\varphi\colon{\mathbf{V}}_{\bullet}X\to Z$ is surjective, where $Z$ is the
canonical projective limit of
${({\mathbf{V}}{p_{ij}}\colon{\mathbf{V}}_{\bullet}{X_{j}}\to{\mathbf{V}}_{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$. This will be enough to show that $\varphi$ is a homeomorphism, using
Proposition 4.2. Because of the natural homeomorphism
${\mathbf{V}}_{\bullet}\cong\mathbb{P}_{\mathtt{P}}^{\bullet}$, we reason with
linear previsions instead of continuous valuations. Let ${(G_{i})}_{i\in I}$
be an element of $Z$. By assumption, there is a (unique) superlinear prevision
$F$ on $X$ such that $\mathbb{P}{p_{i}}(F)=G_{i}$ for every $i\in I$, namely
such that $F(h_{i}\circ p_{i})=G_{i}(h_{i})$ for every
$h_{i}\in\mathcal{L}{X_{i}}$, for every $i\in I$. For every
$h\in\mathcal{L}X$, we build $h_{i}$ as in Lemma 10.2; since $F$ is Scott-
continuous, we obtain that $F(h)=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}F(h_{i}\circ p_{i})=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}G_{i}(h_{i})$. Given any two maps $h,h^{\prime}\in\mathcal{L}X$,
$h+h^{\prime}=\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}(h_{i}\circ
p_{i})+\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}(h^{\prime}_{i}\circ
p_{i})=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}(h_{i}+h^{\prime}_{i})\circ p_{i}$, so a similar argument shows that
$F(h+h^{\prime})=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}F((h_{i}+h^{\prime}_{i})\circ
p_{i})=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}G_{i}(h_{i}+h^{\prime}_{i})$. Since $G_{i}$ is linear, the latter is equal
to $\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}(G_{i}(h_{i})+G_{i}(h^{\prime}_{i}))=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}G_{i}(h_{i})+\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}G_{i}(h^{\prime}_{i})=F(h)+F(h^{\prime})$. Hence $F$ is sublinear. Being in
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}^{\bullet}X$, it is superlinear, hence
linear. Hence $F$ is in $\mathbb{P}_{\mathtt{P}}^{\bullet}X$, and it was built
so that $\mathbb{P}{p_{i}}(F)=G_{i}$ for every $i\in I$, so
$\varphi(F)={(G_{i})}_{i\in I}$.

Superlinear previsions form a model of mixed demonic non-deterministic and
probabilistic choice. Another, earlier model, due to [41, 49, 50, 40], is the
composition ${\mathcal{Q}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}$. We have
already mentioned the fact that
${\mathcal{Q}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}X$ is homeomorphic to
$\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$ for every space $X$ [19,
Theorem 4.15]; naturality was overlooked there, and is dealt with by Lemma
11.2. Together with the natural homeomorphism
${\mathbf{V}}_{\bullet}\cong\mathbb{P}_{\mathtt{P}}^{\bullet}$, this allows us
to obtain the following.

###### Corollary 11.4

Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. The projective limit of a
projective system ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ of
topological spaces is preserved by
${\mathcal{Q}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}$ if and only if it is
preserved by ${\mathbf{V}}_{\bullet}$.

We refer to Theorem 4.3 to what conditions ensure that such limit preservation
results hold.

##  12 Intermission: ${\mathbf{V}}_{\bullet}$ preserves local compactness and
proper maps, and projective limits of $\odot$-consonant spaces

Before we go on with the $\mathbb{P}_{{{\mathtt{A}}{\mathtt{P}}}}$ sublinear
prevision functor, we need to prove a few theorems about the
${\mathbf{V}}_{\bullet}$ functors: that it preserves local compactness, and
that it preserves proper maps. We also need to show that certain spaces known
as $\odot$-consonant (sober) spaces are preserved by projective limits with
proper bonding maps.

###  12.1 On the preservation of local compactness by ${\mathbf{V}}_{\bullet}$

We start with local compactness. It is known that ${\mathbf{V}}_{\leq 1}$
preserves various properties: stable compactness [3, Theorem 39], being a
continuous dcpo [30, Theorem 5.2], being a quasi-continuous dcpo [21, Theorem
5.1], for example. Some of these preservation theorems extend over to
${\mathbf{V}}_{1}$ or to ${\mathbf{V}}$, but a conspicuously absent property
in the list is local compactness. We address this now.

The proof relies on capacities, as studied in [16]. But that paper considers
integrals of lower semicontinuous maps from $X$ to $\mathbb{R}_{+}$ (not
$\overline{\mathbb{R}}_{+}$), hence does not cover $\mathcal{L}X$. Instead, we
will refer to [21, Section 4], where we can find some of the following
information; we will prove the rest. For every monotone map
$\nu\colon{\mathcal{O}}X\to\overline{\mathbb{R}}_{+}$, for every
$h\in\mathcal{L}X$, there is a _Choquet integral_ $\int_{x\in X}h(x)\,d\nu$,
defined as the indefinite Riemann integral
$\int_{0}^{\infty}\nu(h^{-1}(]t,\infty]))\,dt$.

###### Lemma 12.1

The following properties hold.

  1. 1.

The Choquet integral $\int_{x\in X}h(x)\,d\nu$ is linear in $\nu$, monotonic
and even Scott-continuous in $\nu$.

  2. 2.

If $\nu$ is Scott-continuous, then the Choquet integral is Scott-continuous in
$h$.

  3. 3.

If $\nu$ is a continuous valuation, then the Choquet integral is also linear
in $h$.

  4. 4.

For every $U\in{\mathcal{O}}X$, $\int_{x\in X}\chi_{U}(x)\,d\nu=\nu(U)$.

  5. 5.

Given any continuous valuation $\nu^{*}$ on ${\mathcal{Q}}_{\mathsf{V}}X$,
there is a Scott-continuous map
$\nu\colon{\mathcal{O}}X\to\overline{\mathbb{R}}_{+}$ defined by
$\nu(U)\mathrel{\buildrel\text{def}\over{=}}\nu^{*}(\Box U)$ for every
$U\in{\mathcal{O}}X$. Then, for every $h\in\mathcal{L}X$, the map $h^{*}\colon
Q\mapsto\min_{x\in Q}h(x)$ is in $\mathcal{L}{{\mathcal{Q}}_{\mathsf{V}}X}$
and $\int_{x\in
X}h(x)\,d\nu=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}h^{*}(Q)\,d\nu^{*}$.

  6. 6.

For every compact saturated subset $Q$ of $X$, the _unanimity game_
$\mathfrak{u}_{Q}\colon{\mathcal{O}}X\to\overline{\mathbb{R}}_{+}$, which maps
every $U\in{\mathcal{O}}X$ to $1$ if $Q\subseteq U$ and to $0$ otherwise, is a
Scott-continuous map from ${\mathcal{O}}X$ to $\overline{\mathbb{R}}_{+}$.

  7. 7.

Letting
$\nu\mathrel{\buildrel\text{def}\over{=}}\sum_{j=1}^{m}a_{j}\mathfrak{u}_{Q_{j}}$,
where each $Q_{j}$ is compact saturated and $a_{j}\in\mathbb{R}_{+}$, the map
$F\colon\mathcal{L}X\to\overline{\mathbb{R}}_{+}$ defined by:

$\displaystyle F(h)$
$\displaystyle\mathrel{\buildrel\text{def}\over{=}}\int_{x\in
X}h(x)\,d\nu=\sum_{j=1}^{m}a_{j}\min_{x\in Q_{j}}h(x)$

for every $h\in\mathcal{L}X$ is a superlinear prevision.

###### Proof 35

1\. The fact that the Choquet integral is linear in $\nu$, namely that it
commutes with scalar products by non-negative real numbers and with addition
of continuous valuations, follows from the linearity of indefinite Riemann
integration. It is also monotonic in $\nu$. In order to show Scott-continuity,
we consider a directed family ${(\nu_{i})}_{i\in I}$, with (pointwise)
supremum $\nu$, and we observe that $\int_{x\in
X}h(x)\,d\nu=\int_{0}^{\infty}\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\nu_{i}(h^{-1}(]t,\infty]))\,dt$. The key is that the integrand
$t\mapsto\nu_{i}(h^{-1}(]t,\infty]))$ is antitone (all antitone maps are
Riemann-integrable), and that indefinite Riemann integration of antitone maps
$f$ is Scott-continuous in $f$, see [48, Lemma 4.2]. Therefore $\int_{x\in
X}h(x)\,d\nu=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\int_{0}^{\infty}\nu_{i}(h^{-1}(]t,\infty]))\,dt=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\int_{x\in X}h(x)\,d\nu_{i}$.

2\. The proof works as Tix’s original proof of the same result in the special
case where $\nu$ is a continuous valuation [48, Satz 4.4], and also relies on
[48, Lemma 4.2]. Explicitly, let ${(h_{i})}_{i\in I}$ be a directed family in
$\mathcal{L}X$, with (pointwise) supremum $h$. For every $t\in\mathbb{R}_{+}$,
$h^{-1}(]t,\infty])=\\{x\in X\mid\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}h_{i}(x)>t\\}=\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}h_{i}^{-1}(]t,\infty])$. Therefore $\int_{x\in
X}h(x)\,d\nu=\int_{0}^{\infty}\nu(\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}h_{i}^{-1}(]t,\infty]))\,dt=\int_{0}^{\infty}\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\nu(h_{i}^{-1}(]t,\infty]))\,dt=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\int_{0}^{\infty}\nu(h_{i}^{-1}(]t,\infty]))\,dt=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}\int_{x\in X}h_{i}(x)\,d\nu$, using the Scott-continuity of $\nu$ and the
Scott-continuity of indefinite Riemann integration of antitone maps.

3\. This is a result of Tix [48, Satz 4.4].

4\. $\int_{x\in
X}\chi_{U}(x)\,d\nu=\int_{0}^{\infty}\nu(\chi_{U}^{-1}(]t,\infty]))\,dt=\int_{0}^{1}\nu(U)\,dt+\int_{1}^{\infty}0\,dt=\nu(U)$.

5\. This is as with [16, Lemma 7.5]. The fact that $\nu^{*}$ is Scott-
continuous follows from the fact that $\nu$ is, and that the $\Box$ operator
is, too. For the latter, observe that for every directed family
${(U_{i})}_{i\in I}$ of open subsets of $X$, for every
$Q\in{\mathcal{Q}}_{\mathsf{V}}X$,
$Q\in\Box{\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}U_{i}}$ if and only
if $Q\subseteq\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}U_{i}$, which is
equivalent to $Q\subseteq U_{i}$ (namely, $Q\in\Box{U_{i}}$) for some $i\in
I$, because $Q$ is compact.

For every $Q\in{\mathcal{Q}}_{\mathsf{V}}X$, the minimum of $h(x)$ when $x$
ranges over $Q$ is reached, since $Q$ is compact and non-empty. For every
$t\in\mathbb{R}_{+}$, $Q\in{h^{*}}^{-1}(]t,\infty])$ if and only if
$h^{*}(Q)>t$. The latter certainly implies that $h(x)>t$ for every $x\in Q$,
hence that $Q\in\Box{h^{-1}(]t,\infty])}$. Conversely, if
$Q\in\Box{h^{-1}(]t,\infty])}$, then let us pick $x\in Q$ such that $h(x)$ is
the least value reached by $h$ on $Q$; then $h^{*}(Q)=h(x)>t$, so
$Q\in{h^{*}}^{-1}(]t,\infty])$. Hence we have shown that
${h^{*}}^{-1}(]t,\infty])=\Box{h^{-1}(]t,\infty])}$. This implies that $h^{*}$
is in $\mathcal{L}{{\mathcal{Q}}_{\mathsf{V}}X}$, in particular.

Now
$\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}h^{*}(Q)\,d\nu^{*}=\int_{0}^{\infty}\nu^{*}({h^{*}}^{-1}(]t,\infty]))\,dt=\int_{0}^{\infty}\nu^{*}(\Box
h^{-1}(]t,\infty])\,dt=\int_{0}^{\infty}\nu(h^{-1}(]t,\infty])\,dt=\int_{x\in
X}h(x)\,d\nu$.

6\. Monotonicity is clear. For every directed family ${(U_{i})}_{i\in I}$ of
open subsets of $X$,
$\mathfrak{u}_{Q}(\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}U_{i})=1$ if
and only if $Q\in\Box{\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}U_{i}}$,
which is equivalent to the existence of an $i\in I$ such that
$Q\in\Box{U_{i}}$ (equivalently, $\mathfrak{u}_{Q}(U_{i})=1$), as we have seen
at the beginning of the proof of item 5.

7\. That would be a consequence of [16, Propositions 7.2, 7.6], except for the
fact that our functions $h$ take their values in $\overline{\mathbb{R}}_{+}$.
We verify that $\int_{x\in X}h(x)\,d\nu=\sum_{j=1}^{m}a_{j}\int_{x\in
X}h(x)\,d\mathfrak{u}_{Q_{i}}=\sum_{j=1}^{m}a_{j}\min_{x\in Q_{j}}h(x)$: the
first equality is by item 1, and the second one is because $\int_{x\in
X}h(x)\,d\mathfrak{u}_{Q_{i}}$ is equal to
$\int_{0}^{\infty}\mathfrak{u}_{Q_{i}}(h^{-1}(]t,\infty]))\,dt=\int_{0}^{\min_{x\in
Q_{i}}h(x)}1\,dt+\int_{\min_{x\in Q_{i}}h(x)}^{\infty}0\,dt={\min_{x\in
Q_{i}}h(x)}$.

It is easy to see that $F(h)$ is superlinear, because of the laws $\min_{x\in
Q_{i}}ah(x)=a\min_{x\in Q_{i}}h(x)$ (for every $a\in\mathbb{R}_{+}$) and
$\min_{x\in Q_{i}}(h(x)+h^{\prime}(x))\geq\min_{x\in Q_{i}}h(x)+\min_{x\in
Q_{i}}h^{\prime}(x)$. Scott-continuity comes from the fact that
$F(h)=\int_{x\in X}h(x)\,d\nu$, where
$\nu\mathrel{\buildrel\text{def}\over{=}}\sum_{j=1}^{m}a_{j}\mathfrak{u}_{Q_{j}}$,
that $\nu$ is Scott-continuous (by item 6), and by using item 2.

As we will see, ${\mathbf{V}}_{\bullet}$ does not just preserve local
compactness, it maps core-compact spaces to locally compact sober spaces. A
space $X$ is _core-compact_ if and only if ${\mathcal{O}}X$ is a continuous
dcpo; every locally compact space is core-compact [18, Theorem 5.2.9]. The
connection between the two notions can be made more precise as follows. Every
topological space $X$ has a _sobrification_ ${\mathcal{S}}X$ (or $X^{s}$),
which is the free sober space over $X$ [18, Theorem 8.2.44]; then $X$ is core-
compact if and only if ${\mathcal{S}}X$ is locally compact [18, Proposition
8.3.11]. ${\mathcal{S}}X$ can be built as the collection of irreducible closed
subsets, with the topology whose open sets (all of them, not just a base) are
$\diamond U\mathrel{\buildrel\text{def}\over{=}}\\{F\in{\mathcal{S}}X\mid
F\cap U\neq\emptyset\\}$, $U\in{\mathcal{O}}X$. In particular, $\diamond\colon
U\mapsto\diamond U$ is an order-isomorphism between ${\mathcal{O}}X$ and
${\mathcal{O}}{{\mathcal{S}}X}$. This induces a homeomorphism between
${\mathbf{V}}_{\bullet}X$ and ${\mathbf{V}}_{\bullet}{{\mathcal{S}}X}$.

###### Theorem 12.2

For every core-compact space $X$, ${\mathbf{V}}X$ and ${\mathbf{V}}_{\leq 1}X$
are locally compact and sober. If $X$ is also compact, then
${\mathbf{V}}_{1}X$ is locally compact sober.

###### Proof 36

By Remark 5.2 and the text following it, all the spaces
${\mathbf{V}}_{\bullet}X$ are sober.

Replacing $X$ by ${\mathcal{S}}X$ if necessary, we may assume that $X$ is
locally compact and sober. Then the upper Vietoris topology on
${\mathcal{Q}}_{\mathsf{V}}X$ coincides with the Scott topology on
${\mathcal{Q}}X$ (with the reverse inclusion ordering $\supseteq$), by Lemma
8.3.26 of [18], and ${\mathcal{Q}}X$ itself is a continuous dcpo [18,
Proposition 8.3.25]. A fundamental theorem due to Jones [30, Theorem 5.2]
states that for every continuous dcpo $P$, ${\mathbf{V}}_{\leq 1}P$ is a
continuous dcpo under the stochastic ordering, and that a basis is given by
the simple valuations, namely those of the form
$\sum_{i=1}^{n}a_{i}\delta_{x_{i}}$, where each $a_{i}$ is in $\mathbb{R}_{+}$
and $x_{i}\in P$. A similar result holds for ${\mathbf{V}}P$ [13, Theorem
IV-9.16], and for ${\mathbf{V}}_{1}P$ provided that $P$ is also pointed [9,
Corollary 3.3]. We will apply those results to
$P\mathrel{\buildrel\text{def}\over{=}}{\mathcal{Q}}X$, and we notice that if
$X$ is compact, then $P$ is pointed, as $X$ itself will be the least element
of $P$ in that case.

Let $\nu\in{\mathbf{V}}_{\bullet}X$, and let $\mathcal{U}$ be any open
neighborhood of $\nu$. Then $\nu$ is in some finite intersection of subbasic
open sets $\bigcap_{i=1}^{n}[U_{i}>r_{i}]$ that is included in $\mathcal{U}$,
where each $U_{i}$ is open in $X$ and $r_{i}\in\mathbb{R}_{+}$. We consider
$\mu\mathrel{\buildrel\text{def}\over{=}}{\mathbf{V}}{\eta^{\mathcal{Q}}_{X}}(\nu)\in{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}$.
We recall that $\eta^{\mathcal{Q}}_{X}$ is the unit of the
${\mathcal{Q}}_{\mathsf{V}}$ monad, and maps every point $x\in X$ to
$\operatorname{\uparrow\\!}x\in{\mathcal{Q}}_{\mathsf{V}}X$. For every open
subset $U$ of $X$, $\mu(\Box U)=\eta^{\mathcal{Q}}_{X}[\nu](\Box
U)=\nu((\eta^{\mathcal{Q}}_{X})^{-1}(\Box U))=\nu(U)$. It follows that $\mu$
is in $\bigcap_{i=1}^{n}[\Box{U_{i}}>r_{i}]$. The latter is open in the upper
Vietoris topology on
${\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}={\mathbf{V}}_{\bullet}P$,
hence in the Scott topology of the stochastic ordering. Since
${\mathbf{V}}_{\bullet}P$ is a continuous dcpo with a basis of simple
valuations, there is a simple valuation
$\xi^{*}\mathrel{\buildrel\text{def}\over{=}}\sum_{j=1}^{m}a_{j}\delta_{Q_{j}}$
in ${\mathbf{V}}_{\bullet}P$ that is way below $\mu$ and in
$\bigcap_{i=1}^{n}[\Box{U_{i}}>r_{i}]$.

We build a superlinear prevision $F$ on $X$ by letting $F(h)$ be equal to
$\sum_{j=1}^{m}a_{j}\min_{x\in Q_{j}}h(x)$ for every $h\in\mathcal{L}X$. (See
Lemma 12.1, item 7.) Equivalently, $F(h)=\int_{x\in X}h(x)\,d\xi$, where
$\xi\mathrel{\buildrel\text{def}\over{=}}\sum_{j=1}^{m}a_{j}\mathfrak{u}_{Q_{j}}$.
The notations $\xi$, $\xi^{*}$ are justified by the fact that for every
$U\in{\mathcal{O}}X$, $\xi(U)=\xi^{*}(\Box U)$.

Then $s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$ is a compact saturated subset
of $\mathbb{P}_{\mathtt{P}}^{\bullet}X$, as we have seen in Section 11.
Equating $\mathbb{P}_{\mathtt{P}}^{\bullet}X$ with ${\mathbf{V}}_{\bullet}X$,
$s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$ is the subset of those
$\nu^{\prime}\in{\mathbf{V}}_{\bullet}X$ such that for every
$h\in\mathcal{L}X$, $F(h)\leq\int_{x\in X}h(x)\,d\nu^{\prime}$. We claim that
$\nu$ is in the interior of $s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$, and
that $s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$ is included in $\mathcal{U}$;
this will end our proof.

We start by showing that
$s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)\subseteq\mathcal{U}$. Let
$\nu^{\prime}$ be any element of $s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$.
In other words, for every $h\in\mathcal{L}X$, $\int_{x\in
X}h(x)\,d\nu^{\prime}\geq F(h)=\sum_{j=1}^{m}a_{j}\min_{x\in Q_{j}}h(x)$. For
each $i\in\\{1,\cdots,n\\}$, we apply the latter to
$h\mathrel{\buildrel\text{def}\over{=}}\chi_{U_{i}}$. We realize that
$\int_{x\in X}\chi_{U_{i}}(x)\,d\nu^{\prime}=\nu^{\prime}(U_{i})$ (Lemma 12.1,
item 4), and that $\min_{Q_{j}}\chi_{U_{i}}$ is equal to $1$ if
$Q_{j}\subseteq U_{i}$, to $0$ otherwise, so
$F(\chi_{U_{i}})=\sum_{\begin{subarray}{c}1\leq j\leq m\\\ Q_{j}\subseteq
U_{i}\end{subarray}}a_{j}$. Therefore
$\nu^{\prime}(U_{i})\geq\sum_{\begin{subarray}{c}1\leq j\leq m\\\
Q_{j}\subseteq U_{i}\end{subarray}}a_{j}$. We recall that
$\xi^{*}=\sum_{j=1}^{m}a_{j}\delta_{Q_{j}}$ is in
$\bigcap_{i=1}^{n}[\Box{U_{i}}>r_{i}]$, so for every $i\in\\{1,\cdots,n\\}$,
$\xi^{*}(\Box{U_{i}})>r_{i}$, namely $\sum_{\begin{subarray}{c}1\leq j\leq
m\\\ Q_{j}\subseteq U_{i}\end{subarray}}a_{j}>r_{i}$. Hence
$\nu^{\prime}\in\bigcap_{i=1}^{n}[U_{i}>r_{i}]\subseteq\mathcal{U}$.

Next, we verify that $\nu$ is in the interior of
$s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$. We use the fact that $\xi^{*}$ is
way below $\mu$, equivalently that $\mu$ is in the open set $\hbox
to0.0pt{$\uparrow$\hss}\raise 2.15277pt\hbox{$\uparrow$}\xi^{*}$. Since
$\mu={\mathbf{V}}{\eta^{\mathcal{Q}}_{X}}(\nu)$, $\nu$ is in
$({\mathbf{V}}{\eta^{\mathcal{Q}}_{X}})^{-1}(\hbox
to0.0pt{$\uparrow$\hss}\raise 2.15277pt\hbox{$\uparrow$}\xi^{*})$, which is
open since ${\mathbf{V}}{\eta^{\mathcal{Q}}_{X}}$ is continuous. It remains to
show that $({\mathbf{V}}{\eta^{\mathcal{Q}}_{X}})^{-1}(\hbox
to0.0pt{$\uparrow$\hss}\raise 2.15277pt\hbox{$\uparrow$}\xi^{*})$ is included
in $s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$.

For every $\nu^{\prime}\in({\mathbf{V}}{\eta^{\mathcal{Q}}_{X}})^{-1}(\hbox
to0.0pt{$\uparrow$\hss}\raise 2.15277pt\hbox{$\uparrow$}\xi^{*})$, by
definition $\xi^{*}$ is way below, in particular below
$\mu^{\prime}\mathrel{\buildrel\text{def}\over{=}}{\mathbf{V}}{\eta^{\mathcal{Q}}_{X}}(\nu^{\prime})$.
The latter is such that $\mu^{\prime}(\Box
U)=\eta^{\mathcal{Q}}_{X}[\nu^{\prime}](\Box
U)=\nu^{\prime}((\eta^{\mathcal{Q}}_{X})^{-1}(\Box U))=\nu^{\prime}(U)$ for
every $U\in{\mathcal{O}}X$. Hence we may write $\mu^{\prime}$ as
${\nu^{\prime}}^{*}$ and apply Lemma 12.1, item 5, so that $\int_{x\in
X}h(x)\,d\nu^{\prime}=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h(x)\,d{\nu^{\prime}}^{*}$. Since
$\xi^{*}\leq\mu^{\prime}={\nu^{\prime}}^{*}$, the latter is larger than or
equal to $\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h(x)\,d\xi^{*}=\int_{x\in X}h(x)\,d\xi=F(h)$. (We use Lemma 12.1, item 5 on
the pair $\xi$, $\xi^{*}$ for that.) We have shown that $\int_{x\in
X}h(x)\,d\nu^{\prime}\geq F(h)$ for every $h\in\mathcal{L}X$, so
$\nu^{\prime}\in s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}(F)$, as promised.

###  12.2 Proper maps and quasi-adjoints

In order to see that ${\mathbf{V}}$ preserves proper maps, we rely on the
following notion, a very close cousin of the quasi-retractions of [17, Section
4], which were used to characterize proper surjective maps there. We recall
that $\eta^{\mathcal{Q}}_{X}\colon X\to{\mathcal{Q}}_{0\mathsf{V}}X$ is the
unit of the ${\mathcal{Q}}_{0\mathsf{V}}$ monad, and that it maps every $x\in
X$ to $\operatorname{\uparrow\\!}x$. We also recall that the specialization
ordering on spaces of the form ${\mathcal{Q}}_{0\mathsf{V}}X$ is _reverse_
inclusion $\supseteq$.

###### Definition 12.3

A _quasi-adjoint_ to a continuous map $r\colon X\to Y$ is a continuous map
$\varsigma\colon Y\to{\mathcal{Q}}_{0\mathsf{V}}X$ such that:

  1. (a)

$\eta^{\mathcal{Q}}_{Y}\leq{\mathcal{Q}}_{0\mathsf{V}}r\circ\varsigma$, namely
$\operatorname{\uparrow\\!}y\supseteq{\mathcal{Q}}_{0\mathsf{V}}r(\varsigma(y))$
for every $y\in Y$, and

  2. (b)

$\varsigma\circ r\leq\eta^{\mathcal{Q}}_{X}$, namely $x\in\varsigma(r(x))$ for
every $x\in X$.

###### Lemma 12.4

For a continuous map $r\colon X\to Y$, the following are equivalent:

  1. 1.

$r$ is a proper map;

  2. 2.

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

  3. 3.

$r$ has a quasi-adjoint.

The quasi-adjoint $\varsigma$, if it exists, is uniquely determined by
$\varsigma(y)=r^{-1}(\operatorname{\uparrow\\!}y)$ for every $y\in Y$.

###### Proof 37

The equivalence between items 1 and 2 can be found in [13, Lemma VI-6.21].

We make the following observation: $(*)$ for every map $\varsigma\colon
Y\to{\mathcal{Q}}_{0\mathsf{V}}X$ such that
$\varsigma(y)=r^{-1}(\operatorname{\uparrow\\!}y)$ for every $y\in Y$, for
every open subset $U$ of $X$, the complement of $\varsigma^{-1}(\Box U)$ in
$Y$ is equal to $\operatorname{\downarrow\\!}r[F]$, where $F$ is the
complement of $U$. Indeed, for every $y\in Y$, $y\not\in\varsigma^{-1}(\Box
U)$ if and only if $\varsigma(y)=r^{-1}(\operatorname{\uparrow\\!}y)$ is not
included in $U$, if and only if there is an $x\in F$ such that $y\leq r(x)$.

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

It follows that, since $\varsigma(y)\in{\mathcal{Q}}_{0\mathsf{V}}X$ by
assumption, $r^{-1}(\operatorname{\uparrow\\!}y)$ is compact. For every closed
subset $F$ of $X$, we consider its complement $U$. Since $\varsigma$ is
continuous, $\varsigma^{-1}(\Box U)$ is open. But its complement is precisely
$\operatorname{\downarrow\\!}r[F]$, by $(*)$, so
$\operatorname{\downarrow\\!}r[F]$ is closed.

$1\mathrel{\Rightarrow}3$. Let
$\varsigma(y)\mathrel{\buildrel\text{def}\over{=}}r^{-1}(\operatorname{\uparrow\\!}y)$
for every $y\in Y$. This is compact saturated since $r$ is proper, hence
perfect. Hence $\varsigma$ is a map from $Y$ to
${\mathcal{Q}}_{0\mathsf{V}}X$.

We check that $\varsigma$ is continuous. For every open subset $U$ of $X$,
$\varsigma^{-1}(\Box U)$ is the complement of
$\operatorname{\downarrow\\!}r[F]$, where
$F\mathrel{\buildrel\text{def}\over{=}}X\smallsetminus U$, by $(*)$. Since $r$
is proper, $\operatorname{\downarrow\\!}r[F]$ is closed, so
$\varsigma^{-1}(\Box U)$ is open.

Let us check (a). For every $y\in Y$,
$({\mathcal{Q}}_{0\mathsf{V}}r\circ\varsigma)(y)=\operatorname{\uparrow\\!}r[\varsigma(y)]=\operatorname{\uparrow\\!}r[r^{-1}(\operatorname{\uparrow\\!}y)]$.
Every element $y^{\prime}$ of that set is such that $y^{\prime}\geq r(x)$ for
some $x\in X$ such that $r(x)\geq y$, so
$y^{\prime}\in\operatorname{\uparrow\\!}y$.

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

###  12.3 On the preservation of proper maps by ${\mathbf{V}}$

###### Lemma 12.5

For every topological space $X$, there is a continuous map
$\Phi\colon{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}\to\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$
defined by
$\Phi(\mu)(h)\mathrel{\buildrel\text{def}\over{=}}\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h(x)\,d\mu$ for every $h\in\mathcal{L}X$.

###### Proof 38

Given $\mu\in{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}$, we may
define $\nu^{*}\mathrel{\buildrel\text{def}\over{=}}\mu$ and
$\nu(U)\mathrel{\buildrel\text{def}\over{=}}\nu^{*}(\Box U)$ for every
$U\in{\mathcal{O}}X$. Then $\nu$ is Scott-continuous and
$\Phi(\mu)(h)=\int_{x\in X}h(x)\,d\nu$ by Lemma 12.1, item 5, so $\Phi(\mu)$
is Scott-continuous in $h$ by Lemma 12.1, item 2.

We claim that $\Phi(\mu)$ is positively homogeneous. We write
$h^{*}\colon{\mathcal{Q}}_{\mathsf{V}}X\to\overline{\mathbb{R}}_{+}$ for the
map $Q\mapsto\min_{x\in Q}h(x)$. This is in
$\mathcal{L}{{\mathcal{Q}}_{\mathsf{V}}X}$, by Lemma 12.1, item 5.

For every $a\in\mathbb{R}_{+}$, $(ah)^{*}=ah^{*}$. Therefore
$\Phi(\mu)(ah)=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}(ah)^{*}(Q)\,d\mu=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}ah^{*}(Q)\,d\mu=a\Phi(\mu)(h)$,
by linearity of integration (Lemma 12.1, item 3).

We claim that $\Phi(\mu)$ is superlinear. Let $h,h^{\prime}\in\mathcal{L}X$.
Then $\Phi(\mu)(h+h^{\prime})=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}(h(x)+h^{\prime}(x))\,d\mu$. For every $Q\in{\mathcal{Q}}_{\mathsf{V}}X$,
$\min_{x\in Q}(h(x)+h^{\prime}(x))\geq\min_{x\in Q}h(x)+\min_{x\in
Q^{\prime}}h^{\prime}(x)$. Since $\mu$ is a continuous valuation, integration
with respect to $\mu$ is linear (Lemma 12.1, item 3), and monotonic (as a
consequence of Lemma 12.1, item 2) in the integrated function, so
$\Phi(\mu)(h+h^{\prime})\geq\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h(x)\,d\mu+\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h^{\prime}(x)\,d\mu=\Phi(\mu)(h)+\Phi(\mu)(h^{\prime})$.

Hence $\Phi(\mu)$ is a superlinear prevision. We note that $\min_{x\in
Q}(1+h(x))=1+\min_{x\in Q}h(x)$, for every non-empty compact saturated subset
of $X$ and for every $h\in\mathcal{L}X$. If $\mu(X)\leq 1$, then for every
$h\in\mathcal{L}X$,
$\Phi(\mu)(\mathbf{1}+h)=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}(1+h(x))\,d\mu=\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}1\,d\mu+\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h(x)\,d\mu\leq 1+\Phi(\mu)(h)$. Similarly, if $\mu(X)=1$, then
$\Phi(\mu)(\mathbf{1}+h)=1+\Phi(\mu)(h)$. Therefore $\Psi$ is a map from
${\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}$ to
$\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$.

It remains to show that $\Phi$ is continuous. For every $h\in\mathcal{L}X$,
for every $r\in\mathbb{R}_{+}$, $\Phi^{-1}([h>r])=[h^{*}>r]$, where
$h^{*}\colon{\mathcal{Q}}_{\mathsf{V}}X\to\overline{\mathbb{R}}_{+}$ is
defined by $h^{*}(Q)\mathrel{\buildrel\text{def}\over{=}}\min_{x\in Q}h(x)$,
as above. Note that $h^{*}$ is in $\mathcal{L}{{\mathcal{Q}}_{\mathsf{V}}X}$,
by Lemma 12.1, item 5.

###### Corollary 12.6

For every topological space $X$, there is a continuous map from
${\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}$ to
${\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}X}$, which maps every
$\mu\in{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}$ to the collection
of continuous valuations $\nu\in{\mathbf{V}}_{\bullet}X$ such that
$\nu(U)\geq\mu(\Box U)$ for every $U\in{\mathcal{O}}X$.

###### Proof 39

We equate ${\mathbf{V}}_{\bullet}X$ with $\mathbb{P}_{\mathtt{P}}^{\bullet}X$.
Then the map $s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}\circ\Phi$ is
continuous, and maps every
$\mu\in{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}X}$ to the collection
$\\{\nu\in{\mathbf{V}}_{\bullet}X\mid\forall
h\in\mathcal{L}X,\int_{Q\in{\mathcal{Q}}_{\mathsf{V}}X}\min_{x\in
Q}h(x)\,d\mu\leq\int_{x\in X}h(x)\,d\nu\\}$. Let
$\xi\colon{\mathcal{O}}X\to\overline{\mathbb{R}}_{+}$ be defined by $\xi(\Box
U)\mathrel{\buildrel\text{def}\over{=}}\mu(U)$ for every $U\in{\mathcal{O}}X$,
so that we may write $\mu$ as $\xi^{*}$, following the convention of Lemma
12.1, item 5. By this item, for every $\nu\in{\mathbf{V}}_{\bullet}X$,
$\nu\in(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}\circ\Phi)(\mu)$ if and
only if $\int_{x\in X}h(x)\,d\xi\leq\int_{x\in X}h(x)\,d\nu$. By taking
$h\mathrel{\buildrel\text{def}\over{=}}\chi_{U}$ for an arbitrary open subset
$U$ of $X$, the latter implies $\xi(U)\leq\nu(U)$. Conversely, if
$\xi(U)\leq\nu(U)$ for every $U\in{\mathcal{O}}X$, $\int_{x\in
X}h(x)\,d\xi=\int_{0}^{\infty}\xi(h^{-1}(]t,\infty]))\,dt\leq\int_{0}^{\infty}\nu(h^{-1}(]t,\infty]))\,dt=\int_{x\in
X}h(x)\,d\nu$. Hence
$(s_{{{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet}\circ\Phi)(\mu)$ is exactly
$\\{\nu\in{\mathbf{V}}_{\bullet}X\mid\forall
U\in{\mathcal{O}}X,\nu(U)\geq\mu(\Box U)\\}$.

In order to apply the theory of quasi-adjoints, we need to replace
${\mathcal{Q}}_{\mathsf{V}}$ by ${\mathcal{Q}}_{0\mathsf{V}}$ in the result
above. We will do this by using the following trick.

For any topological space $X$, let $X^{\top}$ be the space obtained from $X$
by adding a fresh element $\top$, and whose non-empty open subsets are the
sets $U\cup\\{\top\\}$, $U\in{\mathcal{O}}X$. The specialization preordering
$\leq^{\top}$ of $X^{\top}$ is such that $x\leq^{\top}y$ if and only if
$y=\top$ or $x,y\in X$ and $x\leq y$ in $X$.

As we had said in Section 6, we will reserve the notation $\Box U$ for
$\\{Q\in{\mathcal{Q}}_{\mathsf{V}}X\mid Q\subseteq U\\}$, and use the notation
$\Box_{0}U$ for $\\{Q\in{\mathcal{Q}}_{0\mathsf{V}}X\mid Q\subseteq U\\}$.
Hence $\Box_{0}U=\Box U\cup\\{\emptyset\\}$, for every $U\in{\mathcal{O}}X$.

###### Lemma 12.7

For every topological space $X$, the map $t\colon Q\mapsto Q\cup\\{\top\\}$ is
a homeomorphism of ${\mathcal{Q}}_{0\mathsf{V}}X$ onto
${\mathcal{Q}}_{\mathsf{V}}X$.

###### Proof 40

For every compact saturated subset $Q$ of $X$, $Q\cup\\{\top\\}$ is certainly
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 $U\in{\mathcal{O}}X$, $t^{-1}(\Box_{0}{(U\cup\\{\top\\})})=\Box U$,
so $t$ is full and continuous. Since ${\mathcal{Q}}_{0\mathsf{V}}X$ is
$T_{0}$, $t$ is a topological embedding.

It remains to show that $t$ is surjective. Given any non-empty compact
saturated subset $Q^{\prime}$ of $X^{\top}$, $Q^{\prime}$ must contain some
point, which is below $\top$, so $Q^{\prime}$ must also contain $\top$. But
$\\{\top\\}$ is open in $X^{\top}$, so its complement $X$ is closed in
$X^{\top}$, and therefore
$Q\mathrel{\buildrel\text{def}\over{=}}Q^{\prime}\cap X$ is compact. Since it
is included in the subspace $X$, $Q$ is compact in $X$, too. It is clearly
saturated in $X$, and $Q^{\prime}=Q\cup\\{\top\\}$, so $t$ is surjective.

###### Lemma 12.8

Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. For every topological space
$X$, let $\mathcal{F}_{X}$ be the subset of
${\mathbf{V}}_{\bullet}{(X^{\top})}$ consisting of those elements $\nu$ such
that $\nu(\\{\top\\})=0$. $\mathcal{F}_{X}$ is a closed subspace of
${\mathbf{V}}_{\bullet}{(X^{\top})}$, and there is a continuous map
$c\colon{\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}{(X^{\top})}}\to{\mathcal{Q}}_{0\mathsf{V}}{\mathcal{F}_{X}}$
defined by
$c(\mathcal{Q})\mathrel{\buildrel\text{def}\over{=}}\mathcal{Q}\cap\mathcal{F}_{X}$
for every
$\mathcal{Q}\in{\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}{(X^{\top})}}$.

###### Proof 41

First, the definition of $\mathcal{F}_{X}$ makes sense, and notably the
condition $\nu(\\{\top\\})=0$, because $\\{\top\\}$ is open in $X^{\top}$.
Second, $\mathcal{F}_{X}$ is the complement of $[\\{\top\\}>0]$, hence is
closed in ${\mathbf{V}}_{\bullet}{(X^{\top})}$.

For every
$\mathcal{Q}\in{\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}{(X^{\top})}}$,
$c(\mathcal{Q})$ is compact in ${\mathbf{V}}_{\bullet}{(X^{\top})}$ and
included in $\mathcal{F}_{X}$, hence compact in $\mathcal{F}_{X}$ seen as a
subspace. The specialization ordering of $\mathcal{F}_{X}$ is the restriction
of the stochastic ordering, so $c(\mathcal{Q})$ is saturated in
$\mathcal{F}_{X}$: it suffices to show that for all
$\nu\in\mathcal{Q}\cap\mathcal{F}_{X}$ and $\nu^{\prime}\in\mathcal{F}_{X}$
such that $\nu\leq\nu^{\prime}$, $\nu^{\prime}$ is in $\mathcal{Q}$, which
follows from the fact that $\mathcal{Q}$ is saturated in
${\mathbf{V}}_{\bullet}{(X^{\top})}$. Hence $c$ defines a map from
${\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}{(X^{\top})}}$ to
${\mathcal{Q}}_{0\mathsf{V}}{\mathcal{F}_{X}}$. For every open subset
$\mathcal{U}$ of ${\mathbf{V}}_{\bullet}{(X^{\top})}$,
$c^{-1}(\Box_{0}{(\mathcal{U}\cap\mathcal{F}_{X})})=\\{\mathcal{Q}\in{\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}{(X^{\top})}}\mid\mathcal{Q}\cap\mathcal{F}_{X}\subseteq\mathcal{U}\\}=\Box{([\\{\top\\}>0]\cup\mathcal{U})}$;
indeed, $\mathcal{Q}\cap\mathcal{F}_{X}\subseteq\mathcal{U}$ if and only if
$\mathcal{Q}$ is included in the union of the complement $[\\{\top\\}>0]$ of
$\mathcal{F}_{X}$ with $\mathcal{U}$. Hence $c$ is continuous.

###### Lemma 12.9

Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. For every topological space
$X$, let $\mathcal{F}_{X}$ be as in Lemma 12.8. For every
$\nu\in\mathcal{F}_{X}$, there is a unique $\nu^{-}\in{\mathbf{V}}_{\bullet}X$
such that $i[\nu^{-}]=\nu$, where $i$ is the inclusion map from $X$ into
$X^{\top}$. The map $\\_^{-}\colon\nu\mapsto\nu^{-}$ is continuous from
$\mathcal{F}_{X}$ to ${\mathbf{V}}_{\bullet}X$.

###### Proof 42

Let $\nu\in\mathcal{F}_{X}$. If $\nu^{-}$ exists, then for every
$U\in{\mathcal{O}}X$, we must have
$\nu(U\cup\\{\top\\})=\nu^{-}(i^{-1}(U\cup\\{\top\\})=\nu^{-}(U)$, showing
uniqueness. As far as existence is concerned, we define $\nu^{-}(U)$ as
$\nu(U\cup\\{\top\\})$ for every $U\in{\mathcal{O}}X$. This is a strict map
precisely because $\nu\in\mathcal{F}_{X}$, and it is clear that $\nu^{-}$ is
modular and Scott-continuous. Additionally, $\nu^{-}(X)\leq 1$ if and only if
$\nu(X^{\top})\leq 1$, and similarly with $=$ instead of $\leq$.

This defines a map $\\_^{-}\colon\nu\mapsto\nu^{-}$ from $\mathcal{F}_{X}$ to
${\mathbf{V}}_{\bullet}X$, and it remains to see that it is continuous: the
inverse image of a subbasic open set $[U>r]$, with $U\in{\mathcal{O}}X$ and
$r\in\mathbb{R}_{+}$, is $[U\cup\\{\top\\}>r]$.

We recall that $\Box_{0}U$ denotes $\Box U\cup\\{\emptyset\\}$, and is a
canonical subbasic open subset of ${\mathcal{Q}}_{0\mathsf{V}}X$, where
$U\in{\mathcal{O}}X$.

###### Proposition 12.10

For every topological space $X$, there is a continuous map from
${\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{0\mathsf{V}}X}$ to
${\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}_{\bullet}X}$, which maps every
$\mu\in{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{0\mathsf{V}}X}$ to the collection
of continuous valuations $\nu\in{\mathbf{V}}_{\bullet}X$ such that
$\nu(U)\geq\mu(\Box_{0}U)$ for every $U\in{\mathcal{O}}X$.

###### Proof 43

Let us call $f$ the continuous map from
${\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{\mathsf{V}}{(X^{\top})}}\to{\mathcal{Q}}_{\mathsf{V}}{{\mathbf{V}}_{\bullet}{(X^{\top})}}$
that we obtain from Corollary 12.6 applied to the space $X^{\top}$. We form
the composition:

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
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0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
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164.4128pt\raise 4.50694pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.50694pt\hbox{$\scriptstyle{c}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 183.257pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
183.257pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{{\mathcal{Q}}_{0\mathsf{V}}{\mathcal{F}_{X}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
216.9433pt\raise 7.93056pt\hbox{{}\hbox{\kern 0.0pt\raise
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3.0pt}}}}}}\ignorespaces{\hbox{\kern 239.6415pt\raise 0.0pt\hbox{\hbox{\kern
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where $t$ is from Lemma 12.7, $c$ is from Lemma 12.8, and $\\_^{-}$ is from
Lemma 12.9. This composition is continuous.

Given any $\mu\in{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{0\mathsf{V}}X}$, let
$Q$ be its image by that composition. It remains to show that
$Q=\\{\nu\in{\mathbf{V}}_{\bullet}X\mid\forall
U\in{\mathcal{O}}X,\nu(U)\geq\mu(\Box_{0}U)\\}$. The image of $\mu$ by
${\mathbf{V}}_{\bullet}t$ is the continuous valuation
$\mathcal{U}\mapsto\mu(t^{-1}(\mathcal{U}))=\mu(\\{Q\in{\mathcal{Q}}_{0\mathsf{V}}X\mid
Q\cup\\{\top\\}\in\mathcal{U}\\})$. In particular, for every open subset $U$
of $X$,
${\mathbf{V}}_{\bullet}t(\mu)(\Box{(U\cup\\{\top\\})})=\mu(\\{Q\in{\mathcal{Q}}_{0\mathsf{V}}X\mid
Q\cup\\{\top\\}\subseteq U\cup\\{\top\\}\\})=\mu(\Box_{0}U)$, while
${\mathbf{V}}_{\bullet}t(\mu)(\emptyset)=0$. The image of
${\mathbf{V}}_{\bullet}t(\mu)$ by $f$ is the collection of all
$\nu\in{\mathbf{V}}_{\bullet}{(X^{\top})}$ such that for every
$U\in{\mathcal{O}}X$,
$\nu(U\cup\\{\top\\})\geq{\mathbf{V}}_{\bullet}t(\mu)(\Box{(U\cup\\{\top\\})})$
(and $\nu(\emptyset)\geq{\mathbf{V}}_{\bullet}t(\mu)(\emptyset)$, which is
automatically true); in other words, the collection of all
$\nu\in{\mathbf{V}}_{\bullet}{(X^{\top})}$ such that for every
$U\in{\mathcal{O}}X$, $\nu(U\cup\\{\top\\})\geq\mu(\Box_{0}U)$. This is mapped
by $c$ to the collection $\mathcal{N}$ of all
$\nu\in{\mathbf{V}}_{\bullet}{(X^{\top})}$ satisfying the same condition and
such that $\nu(\\{\top\\})=0$. For any $\nu\in\mathcal{N}$, we have
$\nu^{-}(U)=\nu(U\cup\\{\top\\})\geq\mu(\Box_{0}U)$ for every
$U\in{\mathcal{O}}X$.

The set $Q$ is the upward closure of the collection of continuous valuations
$\nu^{-}$ obtained this way. In particular, for every $\nu^{\prime}\in Q$, for
every $U\in{\mathcal{O}}X$, $\nu^{\prime}(U)$ is larger than or equal to
$\nu^{-}(U)$ for some $\nu\in\mathcal{N}$, and therefore
$\nu^{\prime}(U)\geq\mu(\Box_{0}U)$. Conversely, for every
$\nu^{\prime}\in{\mathbf{V}}_{\bullet}X$ such that
$\nu^{\prime}(U)\geq\mu(\Box_{0}U)$ for every $U\in{\mathcal{O}}X$, let
$\nu\in{\mathbf{V}}_{\bullet}{(X^{\top})}$ be defined by
$\nu(U\cup\\{\top\\})\mathrel{\buildrel\text{def}\over{=}}\nu^{\prime}(U)$ for
every $U\in{\mathcal{O}}X$, and
$\nu(\emptyset)\mathrel{\buildrel\text{def}\over{=}}0$. It is easy to check
that $\nu$ is indeed in ${\mathbf{V}}_{\bullet}{(X^{\top})}$, and that
$\nu(\\{\top\\})=0$, so that $\nu\in\mathcal{F}_{X}$. Additionally,
$\nu^{-}=\nu^{\prime}$. Therefore $\nu^{\prime}$ is in $Q$. Hence $Q$
coincides with the collection
$\\{\nu^{\prime}\in{\mathbf{V}}_{\bullet}X\mid\forall
U\in{\mathcal{O}}X,\nu^{\prime}(U)\geq\mu(\Box_{0}U)\\}$, as promised.

The following somehow generalizes Theorem 6.5 of [17], which states that
${\mathbf{V}}_{1}$ preserves proper surjective maps between stably compact
spaces. We do not deal with surjectivity.

###### Theorem 12.11

Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. For every proper map $r\colon
X\to Y$,
${\mathbf{V}}r\colon{\mathbf{V}}_{\bullet}X\to{\mathbf{V}}_{\bullet}Y$ is
proper.

###### Proof 44

By Lemma 12.4, $r$ has a quasi-adjoint $\varsigma\colon
Y\to{\mathcal{Q}}_{0\mathsf{V}}X$. Let $\varsigma^{\prime}$ be the composition
$\textstyle{{\mathbf{V}}_{\bullet}Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{{\mathbf{V}}\varsigma}$$\textstyle{{\mathbf{V}}_{\bullet}{{\mathcal{Q}}_{0\mathsf{V}}X}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{g}$$\textstyle{{\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}_{\bullet}X}}$,
where $g$ is the continuous map of Proposition 12.10. We check that
$\varsigma^{\prime}$ is a quasi-adjoint to ${\mathbf{V}}r$.

First, we claim that
$\eta^{\mathcal{Q}}_{{\mathbf{V}}_{\bullet}Y}\leq{\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}r}\circ\varsigma^{\prime}$,
namely that for every $\nu\in{\mathbf{V}}_{\bullet}Y$,
$\operatorname{\uparrow\\!}\nu\supseteq{\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}r}(g({\mathbf{V}}\varsigma(\nu)))$.
For every
$\mu^{\prime}\in{\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}r}(g({\mathbf{V}}\varsigma(\nu)))=\operatorname{\uparrow\\!}{\mathbf{V}}r[g(\varsigma[\nu])]$,
there is a $\nu^{\prime}\in g(\varsigma[\nu])$ such that $\mu^{\prime}\geq
r[\nu^{\prime}]$. By definition of $g$, for every $U\in{\mathcal{O}}X$,
$\nu^{\prime}(U)\geq\varsigma[\nu](\Box_{0}U)=\nu(\varsigma^{-1}(\Box_{0}U))$.
Therefore, for every $V\in{\mathcal{O}}Y$, $\mu^{\prime}(V)\geq
r[\nu^{\prime}](V)=\nu^{\prime}(r^{-1}(V))\geq\nu(\varsigma^{-1}(\Box_{0}{r^{-1}(V)}))$.
We now observe that $V\subseteq\varsigma^{-1}(\Box_{0}{r^{-1}(V)})$: for every
$y\in V$, $\operatorname{\uparrow\\!}y$ is included in $V$, and since
$\operatorname{\uparrow\\!}y\supseteq{\mathcal{Q}}_{0\mathsf{V}}r(\varsigma(y))$,
we have $r[\varsigma(y)]\subseteq V$; hence
$\varsigma(y)\in\Box_{0}{r^{-1}(V)}$. Since $\nu$ is monotonic, we conclude
that $\mu^{\prime}(V)\geq\nu(V)$. Since $V$ is arbitrary in ${\mathcal{O}}Y$,
$\mu^{\prime}\geq\nu$. This shows that
$\mu^{\prime}\in\operatorname{\uparrow\\!}\nu$. Since $\mu^{\prime}$ is
arbitrary in
${\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}r}(g({\mathbf{V}}\varsigma(\nu)))$,
we have shown that
$\operatorname{\uparrow\\!}\nu\supseteq{\mathcal{Q}}_{0\mathsf{V}}{{\mathbf{V}}r}(g({\mathbf{V}}\varsigma(\nu)))$.

Second, we claim that $\nu\in\varsigma^{\prime}({\mathbf{V}}r(\nu))$ for every
$\nu\in{\mathbf{V}}_{\bullet}X$. We have
$\varsigma^{\prime}({\mathbf{V}}r(\nu))=g({\mathbf{V}}(\varsigma\circ
r))(\nu))$, and the claim reduces to showing that
$\nu(U)\geq{\mathbf{V}}(\varsigma\circ r)(\nu)(\Box_{0}U)$ for every
$U\in{\mathcal{O}}X$. We compute: ${\mathbf{V}}(\varsigma\circ
r)(\nu)(\Box_{0}U)=(\varsigma\circ r)[\nu](\Box_{0}(U))=\nu((\varsigma\circ
r)^{-1}(\Box_{0}U))$. But $(\varsigma\circ r)^{-1}(\Box_{0}U)\subseteq U$: for
every $x\in(\varsigma\circ r)^{-1}(\Box_{0}U)$, $(\varsigma\circ
r)(x)\subseteq U$, and we conclude since $x\in\varsigma(r(x))$. Since $\nu$ is
monotonic, ${\mathbf{V}}(\varsigma\circ r)(\nu)(\Box_{0}U)\leq\nu(U)$, which
is what we wanted to prove.

We now know that $\varsigma^{\prime}$ is a quasi-adjoint to ${\mathbf{V}}r$,
so ${\mathbf{V}}r$ is proper by Lemma 12.4.

###  12.4 Projective systems consisting of proper maps

###### Proposition 12.12

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system in $\mathbf{Top}$, 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. 1.

every $p_{i}$ is proper; we write $\varsigma_{i}$ for its quasi-adjoint;

  2. 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
$\varsigma_{i}^{-1}(\Box_{0}U)$;

  3. 3.

for every $i\in I$, for every $h\in\mathcal{L}X$, the largest function
$h_{i}\in\mathcal{L}{X_{i}}$ such that $h_{i}\circ p_{i}\leq h$ is
$h^{\dagger}\circ\varsigma_{i}$, where
$h^{\dagger}\in\mathcal{L}{{\mathcal{Q}}_{0\mathsf{V}}X}$ maps every
$Q\in{\mathcal{Q}}_{0\mathsf{V}}X$ to $\min_{x\in Q}h(x)$ if $Q\neq\emptyset$
and $\emptyset$ to $\infty$.

###### Proof 45

We will need to know the following. 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
$\bigcap\nolimits^{\scriptstyle\downarrow}_{j\in J}Q_{j}\subseteq U$ then
$Q_{j}\subseteq U$ for some $j\in J$. It follows that for every filtered
family as above, $\bigcap\nolimits^{\scriptstyle\downarrow}_{j\in J}Q_{j}$ is
compact saturated [18, Proposition 8.3.6]. Every sober space is well-filtered
[18, Proposition 8.3.5].

1\. We fix $i\in I$. Using Lemma 12.4, we will build a quasi-adjoint
$\varsigma_{i}$ to $p_{i}$. We know that, for every $y\in X_{i}$,
$\varsigma_{i}(y)$ must be equal to $p_{i}^{-1}(\operatorname{\uparrow\\!}y)$,
but we will define it differently, so as to make sure that it is compact
saturated, and we will then check that it is equal to
$p_{i}^{-1}(\operatorname{\uparrow\\!}y)$.

Let $y\in X_{i}$. For every $k\in I$ such that $i\sqsubseteq k$,
$p_{ik}^{-1}(\operatorname{\uparrow\\!}y)$ is compact (and saturated) since
$p_{ik}$ is proper. For every $j\sqsubseteq k$, we let
$Q_{jk}\mathrel{\buildrel\text{def}\over{=}}\operatorname{\uparrow\\!}p_{jk}[p_{ik}^{-1}(\operatorname{\uparrow\\!}y)]$.
We claim that for all $j,k,k^{\prime}\in I$ such that $i,j\sqsubseteq
k\sqsubseteq k^{\prime}$, $Q_{jk^{\prime}}\subseteq Q_{jk}$. For every $x\in
Q_{jk^{\prime}}$, there is a point $x^{\prime}\in X_{k^{\prime}}$ such that
$p_{jk^{\prime}}(x^{\prime})\leq x$ and $y\leq p_{ik^{\prime}}(x^{\prime})$.
In other words, $p_{jk}(p_{kk^{\prime}}(x^{\prime}))\leq x$ and $y\leq
p_{ik}(p_{kk^{\prime}}(x^{\prime}))$, showing that
$x\in\operatorname{\uparrow\\!}p_{jk}[p_{ik}^{-1}(\operatorname{\uparrow\\!}y)]=Q_{jk}$.
Hence the family
${(Q_{jk})}_{k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j}$
is filtered. (We write $\operatorname{\uparrow\\!}i$ for the collection of
indices $k\in I$ such that $i\sqsubseteq k$, and similarly for
$\operatorname{\uparrow\\!}j$. Both $\operatorname{\uparrow\\!}i$ and
$\operatorname{\uparrow\\!}j$, as well as their intersection, are cofinal in
$I$, and in particular directed.) Since $X_{j}$ is sober hence well-filtered,
$Q_{j}\mathrel{\buildrel\text{def}\over{=}}\bigcap\nolimits^{\scriptstyle\downarrow}_{k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j}Q_{jk}$
is therefore a compact saturated subset of $X_{j}$.

We verify that for all $j\sqsubseteq j^{\prime}\in I$, $p_{jj^{\prime}}$ maps
$Q_{j^{\prime}}$ to $Q_{j}$. It suffices to verify that it maps
$Q_{j^{\prime}k}$ to $Q_{jk}$ for every
$k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j^{\prime}$. For
every $x\in Q_{j^{\prime}k}$, by definition there is a point $x^{\prime}\in
X_{k}$ such that $p_{j^{\prime}k}(x^{\prime})\leq x$ and $y\leq
p_{ik}(x^{\prime})$. Then $p_{jj^{\prime}}(x)\geq
p_{jj^{\prime}}(p_{j^{\prime}k}(x^{\prime}))=p_{jk}(x^{\prime})$, and $y\leq
p_{ik}(x^{\prime})$, so
$p_{jj^{\prime}}(x)\in\operatorname{\uparrow\\!}p_{jk}[p_{ik}^{-1}(\operatorname{\uparrow\\!}y)]=Q_{jk}$.

Hence ${(p_{jk|Q_{k}}\colon Q_{k}\to Q_{j}))}_{j\sqsubseteq
k\in\operatorname{\uparrow\\!}i}$ is a projective system of compact spaces,
obtained from compact saturated subsets $Q_{i}$ of each $X_{i}$. Each $X_{i}$
is sober, hence also every $Q_{i}$ is sober by Remark 5.2. By Steenrod’s
theorem, its canonical projective limit $Q$ is compact. One can verify that
$Q$ is in fact a compact saturated subset of $X$ [23, Lemma 4.3].

We claim that $Q=p_{i}^{-1}(\operatorname{\uparrow\\!}y)$. By construction,
$Q$ is the collection of tuples
$\vec{x}\mathrel{\buildrel\text{def}\over{=}}{(x_{i})}_{i\in I}$ where each
$x_{j}\in Q_{j}$ and for all $j\sqsubseteq k\in I$, $x_{j}=p_{jk}(x_{j})$. For
each such tuple, $p_{i}(\vec{x})=x_{i}$ is in $Q_{i}$, and $Q_{i}\subseteq
Q_{ii}=\operatorname{\uparrow\\!}p_{ii}[p_{ii}^{-1}(\operatorname{\uparrow\\!}y)]=\operatorname{\uparrow\\!}y$.
Therefore $Q\subseteq p_{i}^{-1}(\operatorname{\uparrow\\!}y)$. Conversely,
let $\vec{x}\mathrel{\buildrel\text{def}\over{=}}{(x_{i})}_{i\in I}$ be any
element of $X$ such that $p_{i}(\vec{x})=x_{i}\in\operatorname{\uparrow\\!}y$.
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
$k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j$,
$x_{j}\in\operatorname{\uparrow\\!}p_{jk}[p_{ik}^{-1}(\operatorname{\uparrow\\!}y)]$.
We simply observe that $x_{j}=p_{jk}(x_{k})$ (hence in particular $x_{j}\geq
p_{jk}(x_{j})$) and $x_{k}\in p_{ik}^{-1}(\operatorname{\uparrow\\!}y)$, since
$p_{ik}(x_{k})=x_{i}\in\operatorname{\uparrow\\!}y$, by assumption.

Using Lemma 12.4, let $\varsigma_{jk}$ be the quasi-adjoint of $p_{jk}$, for
all $j\sqsubseteq k\in I$. We know that
$\varsigma_{jk}(x)=p_{jk}^{-1}(\operatorname{\uparrow\\!}x)$ for every $x\in
X_{j}$. Hence $Q_{jk}$, as defined above, is equal to
${\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))$, and
$Q_{j}=\bigcap\nolimits^{\scriptstyle\downarrow}_{k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j}{\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))$.

For every $y\in X_{i}$, let us define $\varsigma_{i}(y)$ as
$p_{i}^{-1}(\operatorname{\uparrow\\!}y)$, namely as the intersection
$X\cap\prod_{j\in
I}\bigcap\nolimits^{\scriptstyle\downarrow}_{k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j}{\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))$.
We claim that $\varsigma_{i}$ is continuous. It suffices to show that the
inverse image of a basic open set $\Box_{0}{(p_{j}^{-1}(U))}$ ($j\in I$,
$U\in{\mathcal{O}}{X_{j}}$) of $X$ by $\varsigma_{i}$ is open in $X_{i}$. The
elements in that inverse image are the points $y\in X_{i}$ such that
$\bigcap\nolimits^{\scriptstyle\downarrow}_{k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j}{\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))\subseteq
U$. Since $X_{j}$ is sober hence well-filtered, the latter is equivalent to
the existence of
$k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j$ such that
${\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))\subseteq U$. But
${\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))\subseteq U$ is
equivalent to
${\mathcal{Q}}_{\mathsf{V}}{p_{jk}}(\varsigma_{ik}(y))\in\Box_{0}U$, which is
equivalent to
$y\in{({\mathcal{Q}}_{\mathsf{V}}{p_{jk}}\circ\varsigma_{ik})}^{-1}(\Box_{0}U)$.
Therefore
$\varsigma_{i}^{-1}(\Box_{0}{(p_{j}^{-1}(U))})=\bigcup_{k\in\operatorname{\uparrow\\!}i\cap\operatorname{\uparrow\\!}j}{({\mathcal{Q}}_{\mathsf{V}}{p_{jk}}\circ\varsigma_{ik})}^{-1}(\Box_{0}U)$,
which is an open set.

Showing that $\varsigma_{i}$ is quasi-adjoint to $p_{i}$ is now a formality.
For every $y\in X_{i}$,
${\mathcal{Q}}_{0\mathsf{V}}{p_{i}}(\varsigma_{i}(y))=\operatorname{\uparrow\\!}p_{i}[p_{i}^{-1}(\operatorname{\uparrow\\!}y)]\subseteq\operatorname{\uparrow\\!}y$,
and for every $x\in X$,
$\varsigma_{i}(p_{i}(x))=p_{i}^{-1}[\operatorname{\uparrow\\!}p_{i}(x)]$
contains $x$. By Lemma 12.4, it follows that $p_{i}$ is proper.

2\. Since $\varsigma_{i}$ is continuous, $\varsigma_{i}^{-1}(\Box_{0}U)$ is
certainly open. For every $y\in\varsigma_{i}^{-1}(\Box_{0}U)$,
$\varsigma_{i}(y)=p_{i}^{-1}(\operatorname{\uparrow\\!}y)$ is included in $U$.
Since open sets are upwards-closed,
$\varsigma_{i}^{-1}(\Box_{0}U)=\bigcup_{y\in\varsigma_{i}^{-1}(\Box_{0}U)}\operatorname{\uparrow\\!}y$,
so
$p_{i}^{-1}(\varsigma_{i}^{-1}(\Box_{0}U))=\bigcup_{y\in\varsigma_{i}^{-1}(\Box_{0}U)}p_{i}^{-1}(\operatorname{\uparrow\\!}y)\subseteq
U$.

Therefore $\varsigma_{i}^{-1}(\Box_{0}U)\subseteq U_{i}$. In the reverse
direction, for every $y\in U_{i}$,
$\varsigma_{i}(y)=p_{i}^{-1}(\operatorname{\uparrow\\!}y)\subseteq
p_{i}^{-1}(U_{i})\subseteq U$, so $\varsigma_{i}(y)\in\Box_{0}U$.

3\. We have already defined a very similar function we called $h^{*}$ in Lemma
12.1, item 5. But that $h^{*}$ had ${\mathcal{Q}}_{\mathsf{V}}X$ as domain,
while the domain of $h^{\dagger}$ is ${\mathcal{Q}}_{0\mathsf{V}}h$. We had
shown that for every $t\in\mathbb{R}_{+}$,
${h^{*}}^{-1}(]t,\infty])=\Box{h^{-1}(]t,\infty])}$. It immediately follows
that ${h^{\dagger}}^{-1}(]t,\infty])=\Box_{0}{h^{-1}(]t,\infty])}$, a basic
open subset of ${\mathcal{Q}}_{0\mathsf{V}}X$. Hence $h^{\dagger}$ is lower
semicontinuous, namely, in $\mathcal{L}{{\mathcal{Q}}_{0\mathsf{V}}X}$.

Let us write $g$ for $h^{\dagger}\circ\varsigma_{i}$, and $h_{i}$ for the
largest function in $\mathcal{L}{X_{i}}$ such that $h_{i}\circ p_{i}\leq h$,
as described in Lemma 10.2. We have $g\circ
p_{i}=h^{\dagger}\circ\varsigma_{i}\circ p_{i}\leq
h^{\dagger}\circ\eta^{\mathcal{Q}}_{X}$ (by property (b) of quasi-adjoints and
the fact that continuous maps are monotonic), and
$h^{\dagger}\circ\eta^{\mathcal{Q}}_{X}=h$, since for every $x\in X$,
$h^{\dagger}(\eta^{\mathcal{Q}}_{X}(x))=\min_{y\in\operatorname{\uparrow\\!}x}h(y)=h(x)$.
Since $h_{i}$ is the largest element of $\mathcal{L}{X_{i}}$ such that
$h_{i}\circ p_{i}\leq h$, $g\leq h_{i}$. Conversely, the operation
$\\_^{\dagger}$ is monotonic, so
$g=h^{\dagger}\circ\varsigma_{i}\geq(h_{i}\circ
p_{i})^{\dagger}\circ\varsigma_{i}$. For every
$Q\in{\mathcal{Q}}_{0\mathsf{V}}X$, either $Q$ is empty and $(h_{i}\circ
p_{i})^{\dagger}(Q)=0=h_{i}^{\dagger}({\mathcal{Q}}_{\mathsf{V}}{p_{i}}(Q))$,
or $Q$ is non-empty and $(h_{i}\circ p_{i})^{\dagger}(Q)=\min_{x\in
Q}h_{i}(p_{i}(x))=\min_{x\in Q,y\geq
p_{i}(x)}h_{i}(y)=h_{i}^{\dagger}(\operatorname{\uparrow\\!}p_{i}[Q])=h_{i}^{\dagger}({\mathcal{Q}}_{\mathsf{V}}{p_{i}}(Q))$.
Therefore $(h_{i}\circ
p_{i})^{\dagger}=h_{i}^{\dagger}\circ{\mathcal{Q}}_{\mathsf{V}}{p_{i}}$, and
hence $g\geq
h_{i}^{\dagger}\circ{\mathcal{Q}}_{\mathsf{V}}{p_{i}}\circ\varsigma_{i}\geq
h_{i}^{\dagger}\circ\eta^{\mathcal{Q}}_{X}$ (by property (a) of quasi-
adjoints) $=h_{i}$. Hence $g=h_{i}$.

###  12.5 Projective limits of consonant and $\odot$-consonant spaces

In a topological space $X$, for every compact saturated subset $Q$, the
collection $\blacksquare Q$ of all open neighborhoods of $Q$ is a Scott-open
subset of ${\mathcal{O}}X$. Any union of such sets $\blacksquare Q$ 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 $\blacksquare Q\subseteq\mathcal{U}$. As we said in Section
4, the notion arises from [8]; see also [18, Exercise 5.4.12].

###### Proposition 12.13

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ 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
consonant and sober, then so is $X$.

###### Proof 46

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 $p_{i}^{-1}(U_{i})\subseteq U$. Then
${(p_{i}^{-1}(U_{i}))}_{i\in I,\sqsubseteq}$ is a monotone net of open subsets
of $X$, whose union is $U$. Since $\mathcal{U}$ is Scott-open,
$p_{i}^{-1}(U_{i})\in\mathcal{U}$ for some $i\in I$. In other words, $U_{i}$
is in $\mathcal{U}_{i}$.

We use the fact that $X_{i}$ is consonant: there is a compact saturated subset
$Q_{i}$ of $X_{i}$ such that $Q_{i}\subseteq U_{i}$ and every open
neighborhood of $Q_{i}$ is in $\mathcal{U}_{i}$. Using Proposition 12.12, item
1, $Q\mathrel{\buildrel\text{def}\over{=}}p_{i}^{-1}(Q_{i})$ is compact
saturated in $X$.

We note that $Q\subseteq U$. Indeed, for every $x\in Q$, $p_{i}(x)\in
Q_{i}\subseteq U_{i}$, so $x\in p_{i}^{-1}(U_{i})\subseteq U$.

We claim that every open neighborhood $V$ of $Q$ in $X$ lies in $\mathcal{U}$.
Since $Q_{i}$ is upwards-closed it is equal to the union of the sets
$\operatorname{\uparrow\\!}y$ when $y$ ranges over $Q_{i}$. Then
$p_{i}^{-1}(\operatorname{\uparrow\\!}y)=\varsigma_{i}(y)$, where
$\varsigma_{i}$ is the quasi-adjoint of $p_{i}$, so $Q=\bigcup_{y\in
Q_{i}}\varsigma_{i}(y)$. The fact that $Q\subseteq V$ then means that for
every $y\in Q_{i}$, $\varsigma_{i}(y)\in\Box_{0}V$, hence that
$Q_{i}\subseteq\varsigma_{i}^{-1}(\Box_{0}V)$. By definition of $Q_{i}$,
$\varsigma_{i}^{-1}(\Box_{0}V)$ is then in $\mathcal{U}_{i}$. By Proposition
12.12, item 2, $\varsigma_{i}^{-1}(\Box_{0}V)$ is the largest open subset
$V_{i}$ of $X_{i}$ such that $p_{i}^{-1}(V_{i})\subseteq V$. We have seen that
$V_{i}\in\mathcal{U}_{i}$, 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 $\mathcal{U}$.

For every topological space $X$, for every $n\in\mathbb{N}$, let the _copower_
$n\odot X$ be the topological sum (categorical coproduct) of $n$ copies of
$X$. In other words, $n\odot X$ is the collection of pairs $(k,x)$ with $1\leq
k\leq n$ and $x\in X$, with topology generated by the sets $\\{k\\}\times U$,
$U\in{\mathcal{O}}X$. A space $X$ is _$\odot$ -consonant_ if and only if
$n\odot X$ is consonant for every $n\in\mathbb{N}$ [6, Definition 13.1]. There
is an $n\odot\\_$ endofunctor on $\mathbf{Top}$: for every continuous map
$f\colon X\to Y$, $n\odot f$ maps every $(k,x)$ to $(k,f(x))$.

A category ${\mathbf{I}}$ 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 ${\mathbf{C}}$ is a functor from a small
connected category ${\mathbf{I}}$ to ${\mathbf{C}}$. It is clear that every
projective system is a connected diagram. The following says that the copower
functor preserves connected limits in $\mathbf{Top}$.

###### Lemma 12.14

Let $X,{(p_{i})}_{i\in|{\mathbf{I}}|}$ be the canonical limit of a connected
diagram $F\colon{\mathbf{I}}\to\mathbf{Top}$. 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 47

Let $X^{\prime},{(q_{i})}_{i\in|{\mathbf{I}}|}$ be the canonical projective
limit of $(n\odot\\_)\circ F$. The elements of $X^{\prime}$ are the tuples
${((k,x_{i}))}_{i\in I}$ such that $(k,x_{i})=(n\odot p_{ij})(k,x_{j})$ for
all $i\sqsubseteq j\in I$. Note that 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 $X^{\prime}$ is given by the sets $p_{i}^{-1}(\\{k\\}\times
U_{i})$, where $i\in I$, $1\leq k\leq n$, and $U_{i}\in{\mathcal{O}}{X_{i}}$.
The map $f\colon{((k,x_{i}))}_{i\in I}\mapsto(k,{(x_{i})}_{i\in I})$ is
bijective. It is continuous and full because $f^{-1}(\\{k\\}\times
p_{i}^{-1}(U_{i}))=p_{i}^{-1}(\\{k\\}\times U_{i})$. Hence $f$ is a
homeomorphism. Additionally, since $p_{i}$ and $q_{i}$ are both projections
onto coordinate $i$, $(n\odot p_{i})\circ f=q_{i}$.

###### Corollary 12.15

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ 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 48

For every $n\in\mathbb{N}$, $n\odot X,{(n\odot p_{i})}_{i\in I}$ is a
projective limit of $(n\odot p_{ij}\colon n\odot X_{j}\to n\odot
X_{i})_{i\sqsubseteq j\in I}$ by Lemma 12.14. By assumption each space $n\odot
X_{i}$ is consonant. It is sober because any coproduct of sober spaces taken
in $\mathbf{Top}$ is sober [18, Lemma 8.4.2]. By Proposition 12.13, $n\odot X$
must then be consonant. We finally recall that any limit of sober spaces taken
in $\mathbf{Top}$ is sober.

##  13 Sublinear previsions

We proceed in a similar way with the $\mathbb{P}_{{{\mathtt{A}}{\mathtt{P}}}}$
sublinear prevision functor as with the
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}$ superlinear prevision functor. There
is an analogue of the
$(r_{{\mathtt{D}}{\mathtt{P}}},s_{{\mathtt{D}}{\mathtt{P}}}^{\bullet})$
retraction, but it is only ${\mathbf{K}}$-relative natural for some
subcategory ${\mathbf{K}}$, and is not known to be natural. Additionally,
contrarily to ${\mathcal{Q}}_{\mathsf{V}}$, the ${\mathcal{H}}_{\mathsf{V}}$
functor does not preserve all projective limits of sober spaces. The goal of
the intermission of Section 12 is precisely to allow us to overcome those
hurdles.

From [19, Proposition 3.11], there is a map
$r_{{{\mathtt{A}}{\mathtt{P}}}\,X}\colon{\mathcal{H}}_{\mathsf{V}}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)\to\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$
and a map $s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}$ in the other
direction, defined by
$r_{{{\mathtt{A}}{\mathtt{P}}}\,X}(C)(h)\mathrel{\buildrel\text{def}\over{=}}\sup_{G\in
C}G(h)$ for every $h\in\mathcal{L}X$ and
$s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}(F)\mathrel{\buildrel\text{def}\over{=}}\\{G\in\mathbb{P}_{\mathtt{P}}^{\bullet}X\mid
G\leq F\\}$, and they form a retraction under the assumption that
$\mathcal{L}X$ is locally convex. We recall that $\mathcal{L}X$ is equipped
with its Scott topology. That it be _locally convex_ means that for every
$h\in\mathcal{L}X$, every Scott-open neighborhood of $h$ contains a convex
Scott-open neighborhood of $h$. By [6, Lemma 13.6], every $\odot$-consonant
space, hence in particular every LCS-complete space $X$, is such that
$\mathcal{L}X$ is locally convex, in particular every locally compact sober
space, every quasi-Polish space, every Polish space.

Additionally, $r_{{{\mathtt{A}}{\mathtt{P}}}\,X}$ restricts to a
homeomorphism, with inverse $s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}$,
between the subspace
${\mathcal{H}}_{\mathsf{V}}^{cvx}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)\to\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$
of non-empty closed convex subsets of $\mathbb{P}_{\mathtt{P}}^{\bullet}X$ and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$, still under the assumption
that $\mathcal{L}X$ be locally convex [19, Theorem 4.11].

We write $r_{{\mathtt{A}}{\mathtt{P}}}$ for the transformation consisting of
all the maps $r_{{{\mathtt{A}}{\mathtt{P}}}\,X}$, when $X$ varies, and
similarly with $s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$.

###### Lemma 13.1

The transformations $r_{{\mathtt{A}}{\mathtt{P}}}$ and
$s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$ are ${\mathbf{K}}$-relative natural,
where ${\mathbf{K}}$ is the full subcategory of $\mathbf{Top}$ consisting of
spaces $X$ such that $\mathcal{L}X$ is locally convex.

###### Proof 49

We will need to use the following observation: $(*)$ for any lower
semicontinuous map $\psi\colon Z\to\overline{\mathbb{R}}_{+}$, where $Z$ is
any topological space, for every $A\subseteq Z$, $\sup_{z\in
A}\psi(z)=\sup_{z\in cl(A)}\psi(z)$. Indeed, for every $t\in\mathbb{R}$,
$t<\sup_{z\in A}\psi(z)$ if and only if $\psi^{-1}(]t,\infty])$ intersects
$A$, $t<\sup_{z\in A}\psi(z)$ if and only if $\psi^{-1}(]t,\infty])$
intersects $cl(A)$, and those are equivalent conditions since
$\psi^{-1}(]t,\infty])$ is open.

Also: $(**)$ the closure of any convex subset of $\mathcal{L}X$ is convex.
This is a special case of [33, Lemma 4.10 (a)], applied to the semitopological
cone $\mathcal{L}X$; the semitopological part means that scalar multiplication
(by scalars in $\mathbb{R}_{+}$) and addition are separately continuous.

Let $f\colon X\to Y$ be any continuous map, where both $\mathcal{L}X$ and
$\mathcal{L}Y$ are locally convex. Let us start with
$r_{{\mathtt{A}}{\mathtt{P}}}$. We need to show that for every
$C\in{\mathcal{H}}_{\mathsf{V}}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)$, for
every $h\in\mathcal{L}Y$,
$r_{{{\mathtt{A}}{\mathtt{P}}}\,Y}({\mathcal{H}}_{\mathsf{V}}(\mathbb{P}f)(C))(h)=\mathbb{P}f(r_{{{\mathtt{A}}{\mathtt{P}}}\,X}(C))(h)$.
The left-hand side is equal to
$\sup_{G^{\prime}\in{\mathcal{H}}_{\mathsf{V}}(\mathbb{P}f)(Q)}G^{\prime}(h)=\sup_{G^{\prime}\in
cl(\\{\mathbb{P}f(G)\mid G\in
C\\})}G^{\prime}(h)=\sup_{G^{\prime}\in\\{\mathbb{P}f(G)\mid G\in
C\\}}G^{\prime}(h)$ (by $(*)$, since $G^{\prime}\mapsto G^{\prime}(h)$ is
lower semicontinuous, by definition of the weak topology) $=\sup_{G\in
C}\mathbb{P}f(G)(h)=\sup_{G\in C}G(h\circ
f)=r_{{{\mathtt{A}}{\mathtt{P}}}\,X}(C)(h\circ
f)=\mathbb{P}f(r_{{{\mathtt{A}}{\mathtt{P}}}\,X}(C))(h)$.

As far as $s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$ is concerned, we must show
that for every $F\in\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$,
$s_{{{\mathtt{A}}{\mathtt{P}}}\,Y}^{\bullet}(\mathbb{P}f(F))={\mathcal{H}}_{\mathsf{V}}(\mathbb{P}f)\allowbreak(s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}(F))$.
The left-hand side is convex, and we claim that the right-hand side is, too.
Knowing this, we will be able to conclude: since
$r_{{{\mathtt{A}}{\mathtt{P}}}\,Y}$ restricted to
${\mathcal{H}}_{\mathsf{V}}^{cvx}(\mathbb{P}_{\mathtt{P}}^{\bullet}X)$ is a
homeomorphism, it is enough to show that
$r_{{{\mathtt{A}}{\mathtt{P}}}\,Y}(s_{{{\mathtt{A}}{\mathtt{P}}}\,Y}^{\bullet}(\mathbb{P}f(F)))=r_{{{\mathtt{A}}{\mathtt{P}}}\,Y}({\mathcal{H}}_{\mathsf{V}}(\mathbb{P}f)\allowbreak(s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}(F)))$,
and this will follow from the naturality of $r_{{{\mathtt{A}}{\mathtt{P}}}}$.

Hence it remains to show that
${\mathcal{H}}_{\mathsf{V}}(\mathbb{P}f)\allowbreak(s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}(F))$
is convex. This is equal to $cl(A)$, where
$A\mathrel{\buildrel\text{def}\over{=}}\\{\mathbb{P}f(G)\mid G\in
s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}(F)\\}$. Since
$s_{{{\mathtt{A}}{\mathtt{P}}}\,X}^{\bullet}(F)$ is convex and $\mathbb{P}f$
commutes with scalar multiplication and with addition, $A$ is convex. By
$(**)$, $cl(A)$ is convex, too.

We recall that every $\odot$-consonant space, in particular every LCS-complete
space, in particular every locally compact sober space $X$ is such that
$\mathcal{L}X$ is locally convex, whence the following.

###### Corollary 13.2

The transformations $r_{{\mathtt{A}}{\mathtt{P}}}$ and
$s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$ are ${\mathbf{K}}$-relative natural,
where ${\mathbf{K}}$ is the category of $\odot$-consonant spaces, or the
category of LCS-complete spaces, or the category of locally compact sober
spaces.

###### Theorem 13.3

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system of topological spaces, with canonical projective limit
$X,{(p_{i})}_{i\in I}$. Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. If:

  1. 1.

the projective system is an ep-system,

  2. 2.

or $I$ has a countable cofinal subset and each $X_{i}$ is locally compact
sober (and compact, if $\bullet$ is “$1$”),

  3. 3.

or every $X_{i}$ is $\odot$-consonant sober (e.g., LCS-complete) and every
$p_{ij}$ is a proper map,

then
${(\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$ is a projective system of topological spaces, and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X,{(\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{p_{i}})}_{i\in
I}$ is its projective limit, up to homeomorphism.

###### Proof 50

In case 1, we use Proposition 10.4, noticing that
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$ is a subdcpo of $KX$,
namely that pointwise directed suprema of (subnormalized, normalized)
sublinear previsions are again (subnormalized, normalized) sublinear
previsions.

In case 2, we reason as follows. By Theorem 4.3, case 2,
${\mathbf{V}}_{\bullet}X,{({\mathbf{V}}_{\bullet}{p_{i}})}_{i\in I}$ is a
projective limit of
$({\mathbf{V}}_{\bullet}{p_{ij}}\colon{\mathbf{V}}_{\bullet}{X_{j}}\to{\mathbf{V}}_{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$. Using the natural homeomorphism
${\mathbf{V}}_{\bullet}\cong\mathbb{P}_{\mathtt{P}}^{\bullet}$, it follows
that
$\mathbb{P}_{\mathtt{P}}^{\bullet}X,{(\mathbb{P}_{\mathtt{P}}^{\bullet}{p_{i}})}_{i\in
I}$ is a projective limit of
${(\mathbb{P}_{\mathtt{P}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{j}}\to\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$. We now use Theorem 7.5 in order to show that
${\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}X,{({\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{p_{i}})}_{i\in
I}$ is a projective limit of
${({\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{p_{ij}}\colon{\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{j}}\to{\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$. We need to check its assumptions. All the spaces
${\mathbf{V}}_{\bullet}{X_{i}}\cong\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{i}}$
are locally compact and sober by Theorem 12.2, so case 3 of Theorem 7.5
applies. Finally, we conclude by Lemma 11.1. For that, we need to show that
$r_{{\mathtt{A}}{\mathtt{P}}}$ and $s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$
are $F_{*}$-relative natural, where $F$ is the functor defining the diagram of
which we are taking the limit—namely,
$F(i)\mathrel{\buildrel\text{def}\over{=}}X_{i}$, $F(j\to
i)\mathrel{\buildrel\text{def}\over{=}}p_{ij}$. This follows from the fact
that $r_{{\mathtt{A}}{\mathtt{P}}}$ and
$s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$ are natural on the category of
locally compact sober spaces, see Corollary 13.2.

In case 3, finally, we use a similar strategy. Every $\odot$-consonant space
is trivially consonant, so by Theorem 4.3, case 3,
${\mathbf{V}}_{\bullet}X,{({\mathbf{V}}_{\bullet}{p_{i}})}_{i\in I}$ is a
projective limit of
$({\mathbf{V}}_{\bullet}{p_{ij}}\colon{\mathbf{V}}_{\bullet}{X_{j}}\to{\mathbf{V}}_{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$. The latter is a projective system whose bonding maps are proper, by
Theorem 12.11. Also each space ${\mathbf{V}}_{\bullet}{X_{i}}$ is sober, by
Remark 5.2 and the text that follows it. Hence, and reasoning up to the
natural homeomorphism
${\mathbf{V}}_{\bullet}\cong\mathbb{P}_{\mathtt{P}}^{\bullet}$, we can use
Theorem 7.5, case 2, and we obtain that
${\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}X,{({\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{p_{i}})}_{i\in
I}$ is a projective limit of
${({\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{p_{ij}}\colon{\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{j}}\to{\mathcal{H}}_{\mathsf{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$. We conclude by using Lemma 11.1. For that, we need to show that
$r_{{\mathtt{A}}{\mathtt{P}}}$ and $s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$
are $F_{*}$-relative natural, where $F$ is the functor defining the diagram of
which we are taking the limit. This follows from the fact that
$r_{{\mathtt{A}}{\mathtt{P}}}$ and $s_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$
are natural on the category of $\odot$-consonant sober spaces, see Corollary
13.2, and that not only each $X_{i}$, but also $X$ is $\odot$-consonant and
sober, by Corollary 12.15.

Sublinear previsions form a model of mixed angelic non-deterministic and
probabilistic choice. Another, earlier model, due to [41, 49, 50, 40], is the
composition ${\mathcal{H}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}$, where
${\mathcal{H}}_{\mathsf{V}}^{cvx}({\mathbf{V}}_{\bullet}X)$ is the subspace of
${\mathcal{H}}_{\mathsf{V}}({\mathbf{V}}_{\bullet}X)$ consisting of convex
non-empty closed sets. Then
${\mathcal{H}}_{\mathsf{V}}^{cvx}({\mathbf{V}}_{\bullet}X)$ is homeomorphic to
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$ for every space $X$ such
that $\mathcal{L}X$ is convex [19, Theorem 4.15]. As in the demonic case,
naturality was overlooked there, but follows from Lemma 13.1—on the full
subcategory of spaces $X$ such that $\mathcal{L}X$ is convex. Together with
the natural homeomorphism
${\mathbf{V}}_{\bullet}\cong\mathbb{P}_{\mathtt{P}}^{\bullet}$, this allows us
to obtain the following.

###### Corollary 13.4

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system of topological spaces, with canonical projective limit
$X,{(p_{i})}_{i\in I}$. Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. If:

  1. 1.

$I$ has a countable cofinal subset and each $X_{i}$ is locally compact (and
compact, if $\bullet$ is “$1$”),

  2. 2.

or every $X_{i}$ is $\odot$-consonant sober (e.g., LCS-complete) and every
$p_{ij}$ is a proper map,

then
${({\mathcal{H}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}{p_{ij}}\colon{\mathcal{H}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}{X_{j}}\to{\mathcal{H}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$ is a projective system of topological spaces, and
${\mathcal{H}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}X,{({\mathcal{H}}_{\mathsf{V}}^{cvx}{\mathbf{V}}_{\bullet}{p_{i}})}_{i\in
I}$ is its projective limit, up to homeomorphism.

##  14 Forks

We arrive at our final functors, which mix probabilistic and erratic non-
determinism. A _fork_ on a space $X$ is any pair $(F^{-},F^{+})$ of a
superlinear prevision $F^{-}$ on $X$ and of a sublinear prevision $F^{+}$ on
$X$ satisfying _Walley’s condition_ :

$F^{-}(h+h^{\prime})\leq F^{-}(h)+F^{+}(h^{\prime})\leq F^{+}(h+h^{\prime})$

for all $h,h^{\prime}\in\mathcal{L}X$ [15, 36]. A fork is _subnormalized_ ,
resp.  _normalized_ if and only if both $F^{-}$ and $F^{+}$ are.

We write $\mathbb{P}_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}}X$ for the set of
all forks on $X$, and $\mathbb{P}^{\leq
1}_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}}X$,
$\mathbb{P}^{1}_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}}X$ for their subsets
of subnormalized, resp. normalized, forks. The _weak topology_ on each is the
subspace topology induced by the inclusion into the larger space
$\mathbb{P}_{{{\mathtt{D}}{\mathtt{P}}}}X\times\mathbb{P}_{{{\mathtt{A}}{\mathtt{P}}}}X$.
A subbase of the weak topology is composed of two kinds of open subsets:
$[h>r]^{-}$, defined as $\\{(F^{-},F^{+})\mid F^{-}(h)>r\\}$, and $[h>r]^{+}$,
defined as $\\{(F^{-},F^{+})\mid F^{+}(h)>r\\}$, where $h\in\mathcal{L}X$,
$r\in\mathbb{R}^{+}$. The specialization ordering of spaces of forks is the
product ordering $\leq\times\leq$, where $\leq$ denotes the pointwise ordering
on previsions. In particular, all those spaces of forks are $T_{0}$.

It is easy to see that, whether $\bullet$ is nothing, “$\leq 1$”, or “$1$”,
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}$ defines an
endofunctor on $\mathbf{Top}$, whose action on morphisms is given by
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}f\mathrel{\buildrel\text{def}\over{=}}(\mathbb{P}f,\mathbb{P}f)$.

###### Lemma 14.1

Let $\bullet$ be nothing, “$\leq 1$”, or “$1$”, and $T$ be the
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}$ functor. The
comparison map $\varphi\colon TX\to Z$ of any projective $T$-situation is a
topological embedding.

###### Proof 51

Let $Z^{\sharp},{(q_{i}^{\sharp})}_{i\in I}$ be the canonical projective limit
of
$(\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$ and
$\varphi^{\sharp}\colon\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X\to
Z^{\sharp}$ be the comparison map. Similarly with
$Z^{\flat},{(q_{i}^{\flat})}_{i\in I}$ and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$. We also take the notations
($Z$, $\varphi$, $q_{i}$) from Definition 3.1, with
$T\mathrel{\buildrel\text{def}\over{=}}\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}$.

By definition (see Definition 3.1), $\varphi^{\sharp}$ maps every
$F^{-}\in\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$ to
${(\mathbb{P}p_{i}(F^{-}))}_{i\in I}$, $\varphi^{\flat}$ maps every
$F^{+}\in\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$ to
${(\mathbb{P}p_{i}(F^{+}))}_{i\in I}$, and $\varphi$ maps every fork
$(F^{-},F^{+})\in\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$
to ${(\mathbb{P}p_{i}(F^{-}),\mathbb{P}p_{i}(F^{+}))}_{i\in I}$. Also, the
maps $q_{i}^{\sharp}$, $q_{i}^{\flat}$, $q_{i}$ are just projection onto
coordinate $i$, just like $p_{i}$.

As a consequence, for every $i\in I$, for every $h_{i}\in\mathcal{L}{X_{i}}$,
for every $r\in\mathbb{R}_{+}$, for every $\pm\in\\{-,+\\}$,
${\varphi}^{-1}(q_{i}^{-1}([h_{i}>r]^{\pm}))=[h_{i}\circ p_{i}]^{\pm}$.
Indeed, $(F^{-},F^{+})$ is in the left-hand side if and only if
$\mathbb{P}{p_{i}}(F^{\pm})(h_{i})>r$, if and only if $F^{\pm}(h_{i}\circ
p_{i})>r$, if and only if $(F^{-},F^{+})\in[h_{i}\circ p_{i}>r]^{\pm}$.

A subbase of the topology on
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$ is given by the
sets $[h>r]^{\pm}$ where $h\in\mathcal{L}X$, $r\in\mathbb{R}_{+}$, and
$\pm\in\\{+,-\\}$. For every $i\in I$, let $h_{i}$ be the largest map in
$\mathcal{L}{X_{i}}$ such that $h_{i}\circ p_{i}\leq h$, as given in Lemma
10.2. Now $[h>r]^{\pm}$ is the collection of (subnormalized, normalized) forks
$(F^{-},F^{+})$ such that $F^{\pm}(h)>r$, or equivalently such that
$F^{\pm}(h_{i}\circ p_{i})>r$ for some $i\in I$, using item 6 of that lemma
and the Scott-continuity of $F^{\pm}$. In other words,
$[h>r]^{\pm}=\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in I}[h_{i}\circ
p_{i}>r]^{\pm}$, and we have seen that this is equal to
$\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}{\varphi}^{-1}(q_{i}^{-1}([h_{i}>r]^{\pm}))$, hence to
${\varphi}^{-1}(\bigcup\nolimits^{\scriptstyle\uparrow}_{i\in
I}q_{i}^{-1}([h_{i}>r]^{\pm}))$. Therefore $\varphi$ is full.

Since $\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$ is
$T_{0}$, $\varphi$ is a topological embedding.

###### Theorem 14.2

Let $\bullet$ be nothing, “$\leq 1$”, or “$1$”. Let ${(p_{ij}\colon X_{j}\to
X_{i})}_{i\sqsubseteq j\in I}$ be a projective system of topological spaces,
with canonical projective limit $X,{(p_{i})}_{i\in I}$. If
$\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$ is a projective limit of
$(\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$ and if $\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$ is a
projective limit of
${(\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$, then $\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$
is a projective limit of
$(\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$.

###### Proof 52

Let $Z^{\sharp},{(q_{i}^{\sharp})}_{i\in I}$ be the canonical projective limit
of
$(\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{i}})_{i\sqsubseteq
j\in I}$ and
$\varphi^{\sharp}\colon\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X\to
Z^{\sharp}$ be the comparison map. Similarly with
$Z^{\flat},{(q_{i}^{\flat})}_{i\in I}$ and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$, with
$Z^{\natural},{(q_{i}^{\natural})}_{i\in I}$ and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}$. By assumption,
$\varphi^{\sharp}$ and $\varphi^{\flat}$ are homeomorphisms. Relying on Lemma
14.1, it remains to show that $\varphi$ is surjective.

Let ${(F_{i}^{-},F_{i}^{+})}_{i\in I}$ be any element of $Z$. This means that
every $(F_{i}^{-},F_{i}^{+})$ is in
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{i}}$, and that
for all $i\sqsubseteq j\in I$,
$(F_{i}^{-},F_{i}^{+})=\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{ij}}(F_{j}^{-},F_{j}^{+})=(\mathbb{P}{p_{ij}}(F_{j}^{-}),\mathbb{P}{p_{ij}}(F_{j}^{+}))$.
In particular, ${(F_{i}^{-})}_{i\in I}$ is in $Z^{\sharp}$, hence is equal to
$\varphi^{\sharp}(F^{-})$ for some
$F^{-}\in\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$, and
${(F_{i}^{+})}_{i\in I}$ is in $Z^{\flat}$, hence is equal to
$\varphi^{\flat}(F^{+})$ for some
$F^{+}\in\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}X$. Explicitly, this
means that $\mathbb{P}{p_{i}}(F^{+})=F_{i}^{+}$ and
$\mathbb{P}{p_{i}}(F^{-})=F_{i}^{-}$ for every $i\in I$, namely that for every
$h_{i}\in\mathcal{L}{X_{i}}$, $F^{\pm}(h_{i}\circ p_{i})=F_{i}^{\pm}(h_{i})$
(with $\pm$ equal to $-$ or to $+$).

We claim that $(F^{-},F^{+})$ is in
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$. It suffices to
check Walley’s condition. For all $h,h^{\prime}\in\mathcal{L}X$, we write
$h_{i}$ for the largest map in $\mathcal{L}{X_{i}}$ such that $h_{i}\circ
p_{i}\leq h$, for every $i\in I$, and similarly with $h^{\prime}_{i}$. Then:

$\displaystyle F^{-}(h+h^{\prime})$
$\displaystyle=F^{-}(\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}(h_{i}\circ
p_{i})+\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}(h^{\prime}_{i}\circ
p_{i}))$ Lemma 10.2, item 5
$\displaystyle=F^{-}(\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}(h_{i}+h^{\prime}_{i})\circ p_{i})$ $+$ is Scott-continuous
$\displaystyle=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}F^{-}((h_{i}+h^{\prime}_{i})\circ p_{i})$ $F^{-}$ is Scott-continuous
$\displaystyle=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}F_{i}^{-}(h_{i}+h^{\prime}_{i})$
$\displaystyle\leq\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}(F_{i}^{-}(h_{i})+F_{i}^{+}(h^{\prime}_{i}))$ Walley’s condition on
$(F_{i}^{-},F_{i}^{+})$
$\displaystyle=\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}F_{i}^{-}(h_{i})+\sup\nolimits^{\scriptstyle\uparrow}_{i\in
I}F_{i}^{+}(h^{\prime}_{i})$
$\displaystyle=\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}F^{-}(h_{i}\circ
p_{i})+\sup\nolimits^{\scriptstyle\uparrow}_{i\in I}F^{+}(h^{\prime}_{i}\circ
p_{i})$ $\displaystyle=F^{-}(h)+F^{+}(h^{\prime}),$

by using the Scott-continuity of $F^{-}$ and $F^{+}$, and Lemma 10.2, item 5.
The inequality $F^{-}(h)+F^{+}(h^{\prime})\leq F^{+}(h+h^{\prime})$ is proved
similarly.

We have now found an element $(F^{-},F^{+})$ of
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}X$ such that
$\mathbb{P}{p_{i}}(F^{+})=F_{i}^{+}$ and $\mathbb{P}{p_{i}}(F^{-})=F_{i}^{-}$
for every $i\in I$, hence such that
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{i}}(F^{-},F^{+})=(F_{i}^{-},F_{i}^{+})$
for every $i\in I$. Hence $\varphi(F^{-},F^{+})={(F_{i}^{-},F_{i}^{+})}_{i\in
I}$.

We apply Theorem 14.2 and list conditions under which
$\mathbb{P}_{{\mathtt{D}}{\mathtt{P}}}^{\bullet}$ preserves projective limits
(equivalently, ${\mathbf{V}}_{\bullet}$, by Theorem 11.3, hence the conditions
of Theorem 4.3), and under which
$\mathbb{P}_{{\mathtt{A}}{\mathtt{P}}}^{\bullet}$ also preserves projective
limits, and we obtain the following.

###### Corollary 14.3

Let ${(p_{ij}\colon X_{j}\to X_{i})}_{i\sqsubseteq j\in I}$ be a projective
system of topological spaces, with canonical projective limit
$X,{(p_{i})}_{i\in I}$. Let $\bullet$ be nothing, “$\leq 1$” or “$1$”. If:

  1. 1.

the projective system is an ep-system,

  2. 2.

or $I$ has a countable cofinal subset and each $X_{i}$ is locally compact
sober (and compact, if $\bullet$ is “$1$”),

  3. 3.

or every $X_{i}$ is $\odot$-consonant sober (e.g., LCS-complete) and every
$p_{ij}$ is a proper map,

then
${(\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{ij}}\colon\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{j}}\to\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{X_{i}})}_{i\sqsubseteq
j\in I}$ is a projective system of topological spaces, and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X,{(\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}{p_{i}})}_{i\in
I}$ is its projective limit, up to homeomorphism.

##  15 The
$\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}\mathbb{P}_{\mathtt{P}}^{\bullet}$
functor

We will transport this result to the matching model of convex lenses over
spaces of continuous (subprobability, probability) valuations of [50], through
a suitable homeomorphism. However, this will work in a restricted setting.

There is such a homeomorphism between
$\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}\mathbb{P}_{\mathtt{P}}^{\bullet}X$,
the subspace of
$\mathop{\mathcal{P}\ell}_{\mathcal{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}X$ of
convex lenses, and
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}X$, for every
space $X$ such that $\mathcal{L}X$ is locally convex _and_ has a almost open
addition map [19, Theorem 4.17]. The latter property means that for all open
subsets $\mathcal{U}$ and $\mathcal{V}$ of $\mathcal{L}X$,
$\operatorname{\uparrow\\!}(\mathcal{U}+\mathcal{V})=\\{f\in\mathcal{L}X\mid\exists
g\in\mathcal{U},h\in\mathcal{V},f\geq g+h\\}$ is open. When $\bullet$ is
“$1$”, we also need to require $X$ to be compact. The homeomorphism is the
restriction of a retraction
$(r_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}\,X},s_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}\,X}^{\bullet})$,
where
$r_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}\,X}\colon\mathop{\mathcal{P}\ell}_{\mathcal{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}X\to\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}X$
maps every lens $L$ to the fork $(h\mapsto\inf_{G\in L}G(h),h\mapsto\sup_{G\in
L}G(h))$, and
$s_{{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}\,X}\colon\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}X\to\mathop{\mathcal{P}\ell}_{\mathcal{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}X$
maps $(F^{-},F^{+})$ to $\\{G\in\mathbb{P}_{\mathtt{P}}^{\bullet}X\mid
F^{-}\leq G\leq F^{+}\\}$. The fact that it is a retraction is also predicated
on the fact that $\mathcal{L}X$ is locally convex, has an almost open addition
map, and that $X$ is compact if $\bullet$ is “$1$” [19, Proposition 3.32]. We
only consider the homeomorphisms, not the retractions, here.

###### Lemma 15.1

The transformations $r_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}$ and
$s_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}$ between
$\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}\mathbb{P}_{\mathtt{P}}^{\bullet}$
and $\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}$ are
${\mathbf{K}}$-relative natural, where ${\mathbf{K}}$ is the full subcategory
of $\mathbf{Top}$ consisting of spaces $X$ such that $\mathcal{L}X$ is locally
convex, with an almost open addition map, and such that $X$ is compact in case
$\bullet$ is “$1$”.

###### Proof 53

Since these transformations consist of mutually inverse homeomorphisms, it
suffices to show that $r_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}$ is natural.
Lemma 4.6 of [19] states that: $(*)$ for every lens
$L\in\mathop{\mathcal{P}\ell}_{\mathcal{V}}\mathbb{P}_{\mathtt{P}}^{\bullet}X$
(in particular for any convex lens), for every $h\in\mathcal{L}X$,
$\sup_{G\in\operatorname{\uparrow\\!}L}G(h)=\sup_{G\in L}G(h)$ and $\inf_{G\in
cl(L)}G(h)=\inf_{G\in L}G(h)$. The action of the
$\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}$ functor on a morphism $g$ maps
any lens $L$ to $\operatorname{\uparrow\\!}g[L]\cap cl(g[L])$ [50, Proposition
4.33]. Hence, for every continuous map $f\colon X\to Y$, for every
$L\in\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}\mathbb{P}_{\mathtt{P}}^{\bullet}X$,
$r_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}(\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}\mathbb{P}f(L))=(h\mapsto\inf_{G\in\operatorname{\uparrow\\!}(\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L]))}G(h),\allowbreak h\mapsto\sup_{G\in
cl(\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap cl(\mathbb{P}f[L]))}G(h))$.
For every $h\in\mathcal{L}X$,

$\displaystyle\inf_{G\in\operatorname{\uparrow\\!}(\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L]))}G(h)$
$\displaystyle=\inf_{G\in\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L])}G(h)$ by $(*)$
$\displaystyle\geq\inf_{G\in\operatorname{\uparrow\\!}\mathbb{P}f[L]}G(h)$
$\displaystyle\text{since }\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L])\subseteq\operatorname{\uparrow\\!}\mathbb{P}f[L]$
$\displaystyle=\inf_{G\in\mathbb{P}f[L]}G(h)$ by $(*)$
$\displaystyle\geq\inf_{G\in\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L])}G(h)$ $\displaystyle\text{since
}\mathbb{P}f[L]\subseteq\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L]),$

so all those values are equal. In particular,
$\inf_{G\in\operatorname{\uparrow\\!}(\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L]))}G(h)=\inf_{G\in\mathbb{P}f[L]}G(h)$. Similarly,
$\sup_{G\in cl(\operatorname{\uparrow\\!}\mathbb{P}f[L]\cap
cl(\mathbb{P}f[L]))}G(h)=\sup_{G\in\mathbb{P}f[L]}G(h)$. Then
$\inf_{G\in\mathbb{P}f[L]}G(h)=\inf_{G^{\prime}\in L}G^{\prime}(h\circ f)$,
and $\sup_{G\in\mathbb{P}f[L]}G(h)=\sup_{G^{\prime}\in L}G^{\prime}(h\circ
f)$, so
$r_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}(\mathop{\mathcal{P}\ell}_{\mathcal{V}}^{cvx}\mathbb{P}f(L))=(h\mapsto\inf_{G^{\prime}\in
L}G^{\prime}(h\circ f),h\mapsto\sup_{G^{\prime}\in L}G^{\prime}(h\circ f))$.
We compare this to
$\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}f(r_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}(L))=\mathbb{P}_{{\mathtt{A}}{\mathtt{D}}{\mathtt{P}}}^{\bullet}f(h^{\prime}\mapsto\inf_{G^{\prime}\in
L}G^{\prime}(h^{\prime}),h^{\prime}\mapsto\sup_{G^{\prime}\in
L}G^{\prime}(h^{\prime}))$, and we find that those are equal.

Let us write $\mathcal{L}_{\mathrm{co}}X$ for the space $\mathcal{L}X$, but
with the compact-open topology instead of the Scott topology. The compact-open
topology is generated by open subsets
$[Q>r]\mathrel{\buildrel\text{def}\over{=}}\\{h\in\mathcal{L}X\mid\forall x\in
Q,h(x)>r\\}$, where $Q$ ranges over the compact saturated of $X$ and
$r\in\mathbb{R}_{+}$. We can even restrict to basic open subsets $[Q>r]$ where
$Q$ is compact saturated, since $[Q>r]=[\operatorname{\uparrow\\!}Q>r]$.

###### Lemma 15.2

For every weakly Hausdorff, coherent space $X$, addition is almost open on
$\mathcal{L}_{\mathrm{co}}X$, viz., for all open subsets $\mathcal{U}$ and
$\mathcal{V}$ of $\mathcal{L}_{\mathrm{co}}X$,
$\operatorname{\uparrow\\!}(\mathcal{U}+\mathcal{V})$ is open in
$\mathcal{L}_{\mathrm{co}}X$.

###### Proof 54

For every compact saturated subset $Q$ of $X$, let $\langle Q\searrow
r\rangle$ be the function that maps every element of $Q$ to $r$, and all
others to $0$. A _co-step function_ is a pointwise supremum of a finite family
of such functions [10, Section 2]. We also define a relation $\prec$ on
functions from $X$ to $\overline{\mathbb{R}}_{+}$ by $f\prec g$ if and only if
for every $x\in X$, $f(x)\ll g(x)$, where $\ll$ is the way-below relation on
$\overline{\mathbb{R}}_{+}$—namely, $r\ll s$ if and only if $r=0$ or $r<s$.
Finally, we write $\mathop{\Uparrow}f$ for $\\{g\in\mathcal{L}X\mid f\prec
g\\}$.

We claim that the sets $\mathop{\Uparrow}f$ form a base of the compact-open
topology on $\mathcal{L}_{\mathrm{co}}X$, where $f$ ranges over the co-step
functions. Let $f\mathrel{\buildrel\text{def}\over{=}}\sup_{i=1}^{n}\langle
Q_{i}\searrow r_{i}\rangle$, where each $Q_{i}$ is compact saturated and each
$r_{i}\in\mathbb{R}_{+}$. Without loss of generality, we assume that
$r_{i}>0$. Then, for every $g\in\mathcal{L}X$, $f\prec g$ if and only if
$g\in\bigcap_{i=1}^{n}[Q_{i}>r_{i}]$. Indeed, if $f\prec g$, then for every
$i\in\\{1,\cdots,n\\}$, for every $x\in Q_{i}$, $f(x)\geq r_{i}$ and $f(x)\ll
g(x)$, so $g(x)>r_{i}$, using the fact that $r_{i}>0$. Conversely, if
$g\in\bigcap_{i=1}^{n}[Q_{i}>r_{i}]$, then for every $x\in X$, let
$I\mathrel{\buildrel\text{def}\over{=}}\\{i\in\\{1,\cdots,n\\}\mid x\in
Q_{i}\\}$. If $I$ is empty, then $f(x)=0\ll g(x)$. Otherwise, let $i\in I$ be
such that $r_{i}$ is largest. Then $f(x)=r_{i}$, and since
$g\in[Q_{i}>r_{i}]$, we have $f(x)<g(x)$; in any case, $f\prec g$.

Any co-step function $f$ takes only finitely many values, and
$f^{-1}([r,\infty])$ is compact saturated for every
$r\in\mathbb{R}_{+}\smallsetminus\\{0\\}$. Conversely, if $f$ is any function
from $X$ to $\overline{\mathbb{R}}_{+}$ that takes only finitely many values
and is such that $f^{-1}([r,\infty])$ is compact saturated for every
$r\in\mathbb{R}_{+}\smallsetminus\\{0\\}$, then we claim that $f$ is a co-step
function. Indeed, it suffices to list and sort the non-zero values taken by
$f$ as $r_{1}>\cdots>r_{n}>0$, to define
$Q_{i}\mathrel{\buildrel\text{def}\over{=}}f^{-1}([r_{i},\infty])$ for every
$i\in\\{1,\cdots,n\\}$, and to verify that $f=\sup_{i=1}^{n}\langle
Q_{i}\searrow r_{i}\rangle$. In order to see this, we lett
$g\mathrel{\buildrel\text{def}\over{=}}\sup_{i=1}^{n}\langle Q_{i}\searrow
r_{i}\rangle$, and we show that $f^{-1}([r,\infty])=g^{-1}([r,\infty])$ for