#  Bundles of metric structures as left ultrafunctors

Ali Hamad

(June 2024)

###### Abstract

We pursue the study of Ultracategories initiated by Makkai and more recently
Lurie by looking at properties of Ultracategories of complete metric
structures, i.e. coming from continuous model theory, instead of
ultracategories of models of first order theories. Our main result is that for
any continuous theory $\mathbb{T}$, there is an equivalence between the
category of left ultrafunctors from a compact Hausdorff space $X$ to the
category of $\mathbb{T}$-models and a notion of bundle of $\mathbb{T}$-models
over $X$. The notion of bundle of $\mathbb{T}$-models is new but recovers many
classical notions like Bundle of Banach spaces, or (semi)-continuous field of
$\mathrm{C}^{*}$-algebras or Hilbert spaces.

††footnotetext: Affiliation: University of Ottawa, Email<EMAIL_ADDRESS>
###### Contents

  1. Introduction
    1. Outline of results and methodology
  2. 1 Preliminary constructions
    1. 1.1 Some ultracategories constructions
      1. 1.1.1 Ultrasets
      2. 1.1.2 Ultracategories arising from directed colimits
  3. 2 The Ultracategory $\mathsf{k}\text{-}\mathsf{CompMet}$
    1. 2.1 Description of the categorical Fubini transform $\Delta$ in the category $\mathsf{k\text{-}CompMet}$
  4. 3 The bundle (The first functor)
    1. 3.1 Bundles of complete metric spaces
    2. 3.2 The Bundle’s Topology
    3. 3.3 Characterisation of the topology
      1. 3.3.1 Proof of the only if direction
      2. 3.3.2 Proof of the if direction
    4. 3.4 The topology definition satisfies the axioms of 3.4
      1. 3.4.1 Axiom 1
      2. 3.4.2 Axiom 2
      3. 3.4.3 Axiom 3
    5. 3.5 $\mathcal{L}$ is a functor
  5. 4 The inverse functor
    1. 4.1 The inverse functor construction
    2. 4.2 Adjunction
      1. 4.2.1 The counit is an isomorphism
      2. 4.2.2 The unit is an isomorphism
      3. 4.2.3 A nice property of bundles
      4. 4.2.4 Another construction of the left-ultrastructure of $\mathcal{R}(E)$
  6. 5 Generalising to any structure
    1. 5.1 The Ultracategory of Models
      1. 5.1.1 Signature
      2. 5.1.2 Structures and Models
      3. 5.1.3 Ultraproducts and models
    2. 5.2 Bundles of structures
    3. 5.3 Maps of bundles
    4. 5.4 The functor $\mathcal{L}$
      1. 5.4.1 Axiom $3$ (Functions)
      2. 5.4.2 Axiom $4$ (Relations)
      3. 5.4.3 Functoriality of $L$
    5. 5.5 The inverse functor
      1. 5.5.1 Compatibility of function symbols
      2. 5.5.2 Compatibility of relation symbols
    6. 5.6 Adjunction
  7. 6 Models
  8. 7 Functoriality in $\mathsf{CompHaus}$
  9. 8 Examples
    1. 8.1 Banach bundles
      1. 8.1.1 Axiomatisation of Banach spaces
      2. 8.1.2  Definition Banach bundles
      3. 8.1.3 Relating the definition of Banach bundles to our work
    2. 8.2 Bundles of C* algebras
    3. 8.3 Bundles of Hilbert spaces
    4. 8.4 Tracially continuous W* bundles
    5. 8.5 Note regarding the examples
  10. 9 Application: Another proof of Lurie’s result
  11. 10 Bundles of pointed metric spaces

## Introduction

Ultracategories are categories equipped with functors:

$\int_{X}\bullet d\mu:A^{X}\to A,$

for each set $X$ and ultrafilter $\mu$ on $X$, that are meant to axiomatise
the idea of ultraproduct of a family of objects. See Section 1 for the precise
definition. The notion has originally been introduced by Makkai in [20], but a
simplified (non-equivalent) version of the definition was given more recently
by Lurie in [19], which is the notion we use in the present paper.

One very important class of Ultracategories is given by compact Hausdorff
spaces: Manes’ theorem tells us that a compact Hausdorff topology on a set $X$
can be encoded by a function which to every ultrafilter on $X$ associate its
limits, subject to some axioms, and this is exactly what an Ultracategory
structure on a set, seen as a category with no non-identity arrows, is. More
precisely, Lurie shows in [19] that an Ultracategory with no non-identity
morphism is the same as a Compact Hausdorff topological space.

In particular, for each compact Hausdorff space and each ultracategory
$\mathcal{M}$ we can consider _Left Ultrafunctors_ (one of the notions of
morphisms between ultracategories) from $X$ to $\mathcal{M}$, and think of it
as something like a “continuous family of objects of $\mathcal{M}$ indexed by
$X$. For example if $\mathcal{M}=\mathsf{Set}$ is the ultracategory of sets,
then it is shown in [19] that this is the same as a sheaf of sets on $X$.

Both Makkai and Lurie have studied Ultracategories mostly toward application
to the model theory of first order logic. By Lös’ theorem for any first order
theory $\mathbb{T}$, it’s class of models is closed under ultraproduct, and
this endows its category of models with an ultracategory structure. Taking
this ultrastructure into account allows to obtain very strong reconstruction
theorem that allows to recover information about the theory from its category
of models. For example, both Lurie and Makkai prove what they call a
“Conceptual completeness theorem” which says that for a coherent theory
$\mathbb{T}$ (that is a finitary geometric theory), there is an equivalence of
categories between the category of left ultrafunctors from $\mathbb{T}$-Models
to $\mathsf{Set}$ and the classifying topos of the theory $\mathbb{T}$.

However, the theory of Ultracategory has not been studied a lot beyond this.
In this paper we propose to initiate the study of a new class of
Ultracategories: categories of complete metric structures, for example the
ultracategories of Banach spaces, Hilbert spaces, $C^{*}$-Algebras, etc$\dots$

Our main result is, for $X$ a compact Hausdorff space, an equivalence between
the category $\mathrm{Left\ ultrafunctor}(X,\mathcal{M})$, with $\mathcal{M}$
is any category of complete metric structures and a category of “continuous
bundle of objects of $\mathcal{M}$ over $X$”. In many cases, our notions of
bundle recover already studied notion, like the bundles of Banach spaces as
defined in [18, 11] and the continuous fields of $\mathrm{C}^{*}$ algebras
(for various equivalent definitions see [7, 22, 5, 6]). This is of course a
generalization of Lurie’s equivalence between $\mathrm{Left\
ultrafunctor}(X,\mathsf{Set})$ and $\mathrm{Sh}(\mathcal{O}(X))$ and we
recover it as a special case at the very end of the paper.

It turned out that a very good framework to deal with metric structures is
continuous model theory. In our work, we used [12] as a reference for this.
The major advantage of continuous model theory is that it allows us to
consider the notion of ultraproducts and hence makes this category of models
an ultracategory. Although the simplest structure studied by continuous model
theory is a complete metric space bounded by a certain constant, it is capable
of axiomatising many structures like Banach spaces, Hilbert spaces,
$\mathrm{C^{*}}$ algebras, preduals of von Neumann algebras, and Von Neumann
algebras with a faithful normal state (usually called $\sigma$-finite von
Neumann algebras, or $\mathrm{W}^{*}$ probability spaces). For
$\mathrm{C}^{*}$ algebras see [12, page 11 example 2.2.1], for $\sigma$-finite
von Neumann algebras see [4] or a different approach in [15, section 6], for
for tracial von Neumann algebras see [14] and for a general survey on
continuous model theory see [17].

### Outline of results and methodology

Sections 1 and 2 are overview of the literature on Ultracategories and on the
category of complete metric spaces. The goal is only to introduce the
definitions and results we will need throughout the paper.

#### Establishing the equivalence for bounded complete metric spaces

In section 3 and 4 we study the case
$\mathcal{M}=\mathsf{k}\text{-}\mathsf{CompMet}$ the category of complete
metric spaces where the distance function is bounded by a certain $k$ and with
_contractions_ as morphism, that is morphisms satisfying
$d(f(x),f(y))\leqslant d(x,y)$, or $1$-Lipschitz map. In section 3 we defined
what we mean by the category of bundles of complete metric space. The next
step is to define an assignment that gives a bundle of bounded metric space
for each left ultrafunctor from $X$ to $\mathrm{k}\text{-}\mathrm{CompMet}$.
In section 4 we constructed an inverse process to the previous one which leads
us to our first important theorem:

######  Theorem.

4.4 There exists an equivalence of categories between $\mathrm{Left\
ultrafunctors}(X,\mathsf{k\text{-}CompMet})$ and
$\mathrm{Bun}(\mathsf{k\text{-}CompMet},X)$

Here $\mathrm{Bun}(\mathsf{k\text{-}CompMet},X)$ are bundles of bounded
complete metric spaces bounded by $k$ over $X$.

#### Extending the equivalence to structures of continuous model theory

In the context of model theory, one can think of a signature as a theory with
an empty set of axioms, in the sense that $\Sigma$-structures are the model of
the empty theory in the signature $\Sigma$. In section 5 we generalize the
theorem above to the continuous version of such empty theories: We extended
our work from just defining bundles of bounded complete metric spaces to
defining bundles of structure of continuous model theory. And we show that our
previous result can be extended to:

######  Theorem.

5.2 Let $X$ be a compact Hausdorff space the there is an equivalence of
categories between
$\mathrm{Left\ ultrafunctors}(X,\mathsf{CompMet}_{\mathfrak{L}})$ and the
category $\mathrm{Bun}(\mathsf{CompMet}_{\mathfrak{L}},X)$

#### Extending the equivalence to models of continuous model theory

Finally, in section 6 we show that the equivalence above is restricted to one
between “bundle of models of a continuous theory”, i.e. the bundle of
structures in which every fibre is a model of the theory and ultrafunctor
taking value in the category of models. In particular:

######  Theorem.

6.1 Let $X$ be a compact Hausdorff space then there is an equivalence of
categories between
$\mathrm{Left\ ultrafunctors(X,\mathsf{CompMet}_{\mathfrak{L},\mathbb{T}})}$
and the category $\mathrm{Bun}(\mathsf{CompMet}_{\mathfrak{L},\mathbb{T}},X)$

#### Showing that the construction is natural in the Compact Hausdorff space

Let $Y\xrightarrow{f}X$ be a continuous map between compact Hausdorff spaces,
and suppose we have a left ultrafunctor $\mathcal{F}$ from $X$ to
$\mathcal{M}$ where $\mathcal{M}$ is the category of models of some continuous
model theory (could be a simple as complete metric spaces bounded by a certain
$k$ or more complex like Banach spaces), since continuous maps between compact
Hausdorff spaces are a particular example of left ultrafunctors then the
composition $\mathcal{F}\circ f$ gives a left ultrafunctor $Y\to\mathcal{M}$,
and this allows the construction of the category
$\mathrm{CompHaus}_{\mathcal{M}}$ in which the objects are left ultrafunctors
from some compact Hausdorff space $X$ to $\mathcal{M}$ and as morphisms
between $\mathcal{F}:X\rightarrow\mathcal{M}$ and
$\mathcal{G}:Y\rightarrow\mathcal{M}$ consists of a pair $(f,\alpha)$ where
$f$ is a continuous map from $X$ to $Y$ and $\alpha$ is a natural
transformation of let ultrafunctors from $\mathcal{F}$ to $\mathcal{G}\circ
f$, and this category is fibered over $\mathrm{CompHaus}$.

In section 7 we show that the bundle over $Y$ resulting from the composition
$\mathcal{F}\circ f$ is the pullback along $Y\xrightarrow[]{f}X$ in
$\mathsf{Top}$, and we extended the equivalence between
$\mathrm{Bun}(X,\mathcal{M})$ and $\mathrm{left\
ultrafunctors}(X,\mathcal{M})$ to an equivalence between
$\mathrm{CompMet}_{X}$ and $\mathrm{Bun}$ the category of bundles of
$\mathcal{M}$ over any compact Hausdorff space.

#### Examples

Section 8 is dedicated to showing examples where our notion of bundles agrees
with previously existing notions of bundles of metric structures used in
functional analysis. We start with a toy example, bundles of pointed metric
space, a construction that follows nearly the existing ones of Banach spaces
present in [18] and [11], and we showed that this notion corresponds exactly
to our notion of bundles. After that we establish that the two slightly
different notions of bundles already in the literature ([18] and [11]) already
correspond to slightly two different continuous theories of Banach spaces
whose category of models are respectively Banach spaces with linear
contractions and Banach space with linear isometries. After that we show that
bundles of $\mathrm{C}^{*}$ algebras ([24, Appendix C]) is the notion of
bundle that corresponds to the continuous theory of $\mathrm{C}^{*}$-algebras,
we also show that $\mathrm{W}^{*}$-bundles (see [23, section 5] or [1,
subsection 3.1] or [8, definition 2.1]) is the notion of bundles which
corresponds to the continuous model theory of tracial von Neumann algebra[14].

#### Giving an alternative proof of Lurie’s result

As mentioned above, in [19] Lurie shows that for a compact Hausdorff space
over $X$ there is an equivalence of categories between
$\mathrm{Sh}(\mathcal{O}(X))$ and bundles left ultrafunctors from $X$ to
$\mathsf{Set}$. On the other hand it is known that there is an equivalence
between $\mathrm{Sh}(\mathcal{O}(X))$ and the category of étale bundles over
$X$. The category of sets is equivalent to the category of discrete metric
structures which is axiomatisable using continuous model theory. In section 7
we show that bundles of discrete metric spaces are equivalent to étale
bundles, which allows us to write the following chain of equivalences for any
compact Hausdorff space:

$\mathrm{Sh}(\mathcal{O}(X))\simeq\textbf{étale bundles over }\
X\simeq\textbf{bundles of discrete metric spaces over}\ X\simeq\mathrm{Left\
ultrafunctors}(X,\mathsf{Set})$

This allows to reobtain the result shown by Lurie, while giving it an entirely
different proof. Our construction relies on the étale space description of
sheaves while the one given by Lurie uses more or less the functorial
description.

#### A non example

Finally in section 10, we study a new notion of “bundles of pointed complete
spaces over $X$”, and we show that when the $X$ is compact Hausdorff that such
concept is equivalent to left ultrafunctors from $X$ to the category of
pointed complete metric spaces, the reason this does not fit the framework of
bundles of models is that we don’t know if it is possible to have a continuous
first order axiomatisation of complete pointed metric spaces.

##  1 Preliminary constructions

### Definition of an ultracategory

Following [19]

######  Definition 1.1.

An ultrastructure on a category $A$ consists of the following data:

  1. 1.

For every set $X$ a functor from $A^{X}$ to $A$ which we are going to call the
ultraproduct functor and we denote it by

$\int_{X}\bullet\ d\mu$

  2. 2.

Given a set $X$ and family of ultrafilters on $X$ $(\nu_{s})_{s\in S}$ and an
ultrafilter $\mu$ on $S$, we require the existence of a morphism
$\Delta_{\mu,\nu_{\bullet}}:\int_{X}M_{x}d(\int_{S}\nu_{s}d\mu)\rightarrow\int_{S}(\int_{X}M_{x}d\mu)d\mu$,
which is natural in the family $(M_{x})_{x\in X}$, the map
$\Delta_{\mu,\nu_{\bullet}}$ is called the categorical Fubini transform.

  3. 3.

for every principal ultrafilter $\delta_{x_{0}}$ on a set $X$, we require a
natural family of isomorphisms $\epsilon_{X,x_{0}}$ from
$\int_{X}M_{x}d\delta_{x_{0}}$ to $M_{x_{0}}$.

This data is required to satisfy the following axioms :

  1. A

Given a family of ultrafilters $(\nu_{s})_{s\in S}$ on a set $X$, and a family
of objects of $A$, $(M_{x})_{x\in X}$ then the map
$\Delta_{\delta_{s_{0}},\nu_{\bullet}}:\
\int_{X}M_{x}d\int\nu_{s}d\delta_{s_{0}}\rightarrow\int_{S}\int_{X}M_{x}d\nu_{s}d\delta_{s_{0}}$,
is the inverse of the map $\epsilon_{S,s_{0}}$ from
$\int_{S}\int_{X}M_{x}d\nu_{s}d\delta_{s_{0}}$ to $\int_{X}M_{x}d\nu_{s_{0}}$

  2. B

Suppose that we have a monomorphism of sets $f:\ X\rightarrow Y$ then the
categorical Fubini transform from
$\int_{Y}M_{y}df\mu=\int_{Y}M_{y}d\int_{X}\delta_{f(x)}d\mu$ to
$\int_{X}\int_{Y}M_{y}d\delta_{f(x)}d\mu$ is an isomorphism.

  3. C

Suppose that we have a set $R$ and an ultrafilter $\lambda$ on, and suppose we
have $(\mu_{r})_{r\in R}$ a family of ultrafilters on a set $S$, and
$(\nu_{s})_{s\in S}$ is a family of ultrafilters on some set $T$ then the
following diagram commutes:

${\int_{T}M_{t}d\rho}$${\int_{R}(\int_{T}M_{t}d\int_{S}\nu_{s}d\mu_{r})d\lambda}$${\int_{S}\int_{T}M_{t}d\nu_{S}d\int_{R}\mu_{r}d\lambda}$${\int_{R}(\int_{S}(\int_{T}M_{t}d\nu_{s})d\mu_{r}d\lambda}$$\scriptstyle{\Delta_{\lambda,\int_{S}\nu_{s}d\mu_{\bullet}}}$$\scriptstyle{\Delta_{\int_{R}\mu_{r}d\lambda,\nu_{\bullet}}}$$\scriptstyle{\int_{R}\Delta_{\mu_{r},\nu_{\bullet}}d\lambda}$$\scriptstyle{\Delta_{\lambda,\mu_{\bullet}}}$

Here
$\rho=\int_{R}(\int_{S}\nu_{s}d\mu_{r}d\lambda)=\int_{S}\nu_{s}d(\int_{R}\mu_{r}d\lambda)$

In this definition $\int_{X}\nu_{s}d\mu$ is defined by
$B\in\int_{S}\nu_{s}d\mu\iff\\{s\in S:\ B\in\nu_{s}\\}\in\mu$.
Now we define an ultracategory to be a category with an ultrastructure.

### Left ultrafunctors

Suppose that $\sf{M}$ and $\sf{N}$ are two ultracategories, we define a left
ultrafunctors from $\mathsf{M}$ to $\mathsf{N}$ to be a functor equipped with
a left ultrastructure.

######  Definition 1.2.

A left ultrastructure on a functor consists of the following: for every
ultrafilter $\mu$ on a set $X$ and every family of objects $(M_{x})_{x\in X}$
of $\sf{M}$, we have a family of morphisms in $\sf{N}$ which we call all of
them $\sigma_{\mu}$ by abuse of language from $F(\int_{X}M_{x}d\mu)$ to
$\int_{X}F(M_{x})d\mu$.

Such that they satisfy the following axioms :

  1. 0.

The following diagram commutes for every family of morphisms $(\psi_{x})_{x\in
X}$ from $M_{x}$ to $N_{x}$ in $\sf{M}$:

${F(\int_{X}M_{x}d\mu)}$${\int_{X}F(M_{x})d\mu}$${F(\int_{X}N_{x}d\mu)}$${\int_{X}F(N_{x})d\mu}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{F(\int_{X}\psi_{x}d\mu)}$$\scriptstyle{\int_{X}F(\psi_{X}d\mu)}$$\scriptstyle{\sigma_{\mu}}$

  2. 1.

For every principal ultrafilter $\delta_{x_{0}}$ the following diagram
commutes:

${F(\int_{X}M_{x}d\delta_{x_{0}})}$${\int_{X}F(M_{x})d\delta_{{x_{0}}}}$${F(M_{x_{0}})}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{F(\epsilon_{X,x_{0}})}$$\scriptstyle{\epsilon_{X,x_{0}}}$

  3. 2.

For any sets $S$ and $T$ and an ultrafilter $\mu$ on $S$ and a family of
ultrafilters $(\nu_{s})_{s\in S}$ on $T$ indexed by $S$, the following diagram
commutes:

${F(\int_{T}M_{t}d(\int_{S}\nu_{S}d\mu))}$${\int_{T}F(M_{T})d\int_{S}\nu_{s}d\mu}$${F(\int_{S}(\int_{T}M_{t}d\nu_{s})d\mu)}$${\int_{S}F(\int_{T}M_{t}d\nu_{s})d\mu}$${\int_{S}\int_{T}F(M_{t})d\nu_{s}d\mu}$$\scriptstyle{\sigma_{\int_{S}\nu_{s}d\mu}}$$\scriptstyle{F(\Delta_{\mu,\nu_{\bullet}})}$$\scriptstyle{\Delta_{\mu,\nu_{\bullet}}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\int_{S}\sigma_{\nu_{s}}d\mu}$

######  Note.

The dual notion is a right ultrafunctor in which the comparison maps go in the
other direction, we omit writing the axioms which can be found in[19].

######  Definition 1.3.

An ultrafunctor is a left ultrafunctor for which all the comparison maps are
isomorphism.

### Natural transformations of left ultrafunctors

Suppose that $\sf{M}$ and $\sf{N}$ are two ultracategories, and let $F$, $G$
be left ultrafunctors between $\sf{M}$ and $\sf{N}$, a natural transformation
of left ultrafunctors from $F$ to $G$, is a natural transformation $\phi$
satisfying the additional condition: For every family $(M_{i})$ of objects if
$\sf{M}$ and for every ultrafilter $\mu$ on $I$ the following diagram
commutes:

${{F(\int_{I}M_{i}d\mu)}}$${{\int_{I}F(M_{i})d\mu}}$${{G(\int_{I}M_{i}d\mu)}}$${{\int_{I}G(M_{i})d\mu}}$$\scriptstyle{\phi_{\int_{I}M_{i}d\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\sigma^{{}^{\prime}}_{\mu}}$$\scriptstyle{\int_{I}\phi_{M_{i}}d\mu}$

A natural transformation of right ultrafunctors is defined similarly.

###  1.1 Some ultracategories constructions

####  1.1.1 Ultrasets

You may have noticed that at this point we are using the notation
$\int\nu_{s}d\mu$ to denote the ultrafilter defined by $A\in\int\nu_{s}d\mu\
\text{iff}\ \\{s\mid\ A\in\nu_{s}\\}\in\mu$. This notation is not a
coincidence, as this is a special case of ultracategories.

######  Definition 1.4.

An ultraset is a small ultracategory with no non-identity morphisms.

Now the next theorem is due to Lurie[19, theorem 3.1.5]

######  Theorem 1.1.

There is an equivalence of categories between ultrasets(with either left
ultrafunctors or ultrafunctors, they are the same in this case), and the
category of compact Hausdorff spaces with continuous maps.

Suppose that $S$ is a compact Hausdorff space and let $(a_{x})_{x\in X}$ be a
family of points and $\mu$ an ultrafilter on $S$ then this equivalence is
exhibited by defining $\int_{X}a_{x}d\mu$ as the limit of the pushforward of
the ultrafilter $\mu$ by the map $x\mapsto a_{x}$.

Now a particular case of the former is $\beta X$ the set of ultrafilters on
$X$ which is a Stone Space (has a totally separated compact Hausdorff
topology), thus an ultraset, which justifies the notation $\int\nu_{s}d\mu$.

##### Ultrasets corresponding to compact subspaces of $\mathbb{R}$

One particular case of compact Hausdorff space is compact subsets of the real
line $\mathbb{R}$, in this section we are going to give a nice
characterisation of the ultraproduct functor for such sets, which will come in
handy when studying the ultraproduct of metric spaces.

Let $X$ be a set and let $\mu$ be an ultrafilter on $X$ and suppose that
$\phi$ is a function taking values in $M$ where $M$ is a compact subset of
$\mathbb{R}$(we can take $M$ for simplicity to be a closed interval since
every compact can be included in such set and the inclusion is continuous).
Now take the ultraproduct $\int_{X}\phi(x)d\mu$ this is the limit of the
ultrafilter $\phi\mu$ (the pushforward of $\mu$ by $\phi$) which translates to
the fact that $\phi\mu$ contains the neighbourhood filter of
$\int_{X}\phi(x)d\mu$, in other words for arbitrarily small $\epsilon$ the set
$\\{x\in X:\ |\phi(z)-\int_{X}\phi(x)d\mu|<\epsilon\\}\in\mu$ (such
ultraproduct is what is usually referred to as an ultralimit and usually
denote by $\lim_{\mu}\phi(x)$).

Now claim the following:

######  Lemma.

$\int_{X}\phi(x)d\mu=\mathrm{Inf}_{U\in\mu}\mathrm{Sup}_{x\in U}\phi(x)$

###### Proof.

Let us call $m=\int_{X}\phi(x)d\mu$. First let us prove that $m$ is a lower
bound for $\\{\mathrm{Sup}_{x\in U}\phi(x)\mid U\in\mu\\}$, to do this suppose
by contradiction that there exists some $U\in\mu$ such that $m>\sup_{x\in
U}\phi(x)$, let us call $\epsilon=m-\sup_{x\in U}\phi(x)$ then the set
$\\{z\in X\mid|\phi(z)-m|<\epsilon\\}\in\mu$ but $\\{z\in
X\mid|\phi(z)-m|<\epsilon\\}\subseteq\\{z\in X\mid m-\epsilon<\phi(z)\\}$ but
this would imply that $V=\\{z\in X\mid\sup_{x\in U}\phi(x)<\phi(z)\\}\in\mu$,
but $V\bigcap U=\varnothing$ on one hand and on the other $V\bigcap U\in\mu$,
hence a contradiction($\varnothing\in\mu$).
So $m$ is a lower bound for $\\{\mathrm{Sup}_{x\in U}\phi(x),U\in\mu\\}$. To
prove it is the greatest lower bound notice that for any $\epsilon>0$ the set
$V_{\epsilon}=\\{x\in X\mid\phi(x)<m+\epsilon\\}\in\mu$ thus
$\mathrm{Sup}_{x\in V_{\epsilon}}\phi(x)\leq m+\epsilon$ so
$\mathrm{Inf}_{U\in\mu}\mathrm{Sup}_{x\in U}\phi(x)\leq m+\epsilon$ and since
$\epsilon$ was arbitrary then we get that
$\mathrm{Inf}_{U\in\mu}\mathrm{Sup}_{x\in U}\phi(x)\leq m$ and thus since $m$
is a lower bound we get that $m=\mathrm{Inf}_{U\in\mu}\mathrm{Sup}_{x\in
U}\phi(x)$.∎

####  1.1.2 Ultracategories arising from directed colimits

######  Theorem 1.2.

Suppose that we have a category $\mathcal{M}$ that has products and direct
colimits, then in this case it has an ultrastructure given by

$\int_{X}M_{x}d\mu=\varinjlim_{U\in\mu}(\prod_{x\in U}M_{x})$

Here we consider the set of sets of $\mu$ as a direct set by reverse
inclusion.

######  Lemma 1.1.

Suppose that $\mathcal{M}$ is a full subcategory of an ultracategory
$\mathcal{M}^{+}$ which is closed under the ultraproduct functor, then it is
an ultracategory with such restriction of the ultraproduct functor

Theorem 1.2 and lemma 1.1 are just restating proposition 1.3.7 of [19], and a
proof can be found there.

The lemma 1.1 allows us not only to consider categories having directed
colimits and products, but full subcategories of those closed under the
ultraproduct construction given by such directed colimit of products. The main
example of such ultracategories is the ultracategory of models of first order
theory which is a full subcategory of the category of structures of the same
signature (similarity type).

We highlight this construction in the case of set and this extends to all
first order theories. The construction is an application of 1.2:
First we define the ultraproduct of non empty sets by

$\int_{I}M_{i}d\mu=\prod_{I}M_{i}/\sim$

Here $\sim$ identifies tuples that agree on any set of the ultrafilter, and
you can notice that this is just a direct limit of products in $\sf{Set}$. Now
the case where some sets of $(M_{i})$ are empty we have two cases either the
set $\\{i\in I\mid M_{i}=\varnothing\\}\in\mu$ in this case we define
$\int_{I}M_{i}d\mu=\varnothing$, otherwise the set $I^{{}^{\prime}}=\\{i\in
I\mid M_{i}\neq\varnothing\\}\in\mu$ so we define

$\int_{I}M_{i}d\mu=\prod_{I^{{}^{\prime}}}M_{i}/\sim$

In other words we restrict our attention to a set of the ultrafilter for which
the sets are non-empty, so in what follows we are going to denote the elements
of the ultraproduct $\int_{I}M_{i}\mu$ by $(a_{i})_{i\in J}$ where $J\in\mu$.

Now suppose that we have a first order theory with structure
$\langle\mathcal{M}_{1},\dots,\mathcal{M}_{n},\mathcal{R}_{1},\dots,\mathcal{R}_{n},f_{1},\dots,f_{n}\rangle$
and a set of axioms $\mathscr{A}$. The category of structures has
ultraproducts resulting from applying theorem 1.2 and 1.1 which are
constructed as follows: Suppose we have structures $(E^{i})_{i\in I}$, in what
follows we are going to denote by $E^{i}_{\mathcal{M}_{j}}$ the set of sort
$\mathcal{M}_{j}$ corresponding to $E_{i}$.Now let $\mu$ be some ultrafilter
on $I$ and define $\int_{I}E^{i}d\mu$ as follows : For each sort
$\mathcal{M}_{j}$,
$(\int_{I}E^{i}d\mu)_{\mathcal{M}_{j}}=\int_{I}E^{i}_{\mathcal{M}_{j}}d\mu$.

Now for a relation symbol $\mathcal{R}$ with formal domain
$S_{1}\times\dots\times S_{l}$,here each $S_{m}\in\\{M_{j}\\}_{j=1}^{n}$, we
define $\mathcal{R}_{\int_{I}m_{i}d\mu}$ by $((a^{m}_{i})_{i\in I})_{m\in
M}\in\mathcal{R}_{\int_{I}m_{i}d\mu}$ iff $\\{i\in I\
:(a^{m}_{i})\in\mathcal{R}_{E^{i}}\\}\in\mu$.

Now for a function symbol $f$ with formal domain $S_{1}\times\dots S_{l}$ and
formal range $S^{{}^{\prime}}$, we define
$f_{\int_{I}E^{i}d\mu}((a^{m}_{i})_{i\in I})_{m\in M}$ by $(f(a^{m}_{i}))_{\in
I}$.

Then we can regard the category of models in $\sf{Set}$ of $\mathscr{A}$ as a
full subcategory of the category of structures of similarity type
$\langle\mathcal{M}_{1},\dots,\mathcal{M}_{n},\mathcal{R}_{1},\dots,\mathcal{R}_{n},f_{1},\dots,f_{n}\rangle$.
As a result of Łos theorem this subcategory is closed under the categorical
ultraproduct of the category of structures which allows the application of
1.1.

##  2 The Ultracategory $\mathsf{k}\text{-}\mathsf{CompMet}$

Given $k$ a positive real number, we denote by $\mathsf{k\text{-}CompMet}$ the
category of $k$-bounded complete metric spaces, with contractions
($1$-Lipschitz functions) as morphisms. More precisely, the objects are the
complete metric spaces satisfying $d(x,y)\leqslant k$ for all $x,y$, and the
morphisms are functions satisfying $d(f(x),f(y))\leq d(x,y)$ for all $x,y$.

######  Proposition 2.1.

The category $\mathsf{k\text{-}CompMet}$ has all products and all directed
colimits.

This proposition is very classical, we will just give an explicit description
of these products and directed colimits:

The product of a family $(B_{i})_{i\in I}$ of $k$-bounded complete metric
spaces, is computed by taking the products of their underlying sets, and
equipping it with the distance:

$d((b_{i}),(c_{i}))=\mathrm{Sup}_{I}(d(b_{i},c_{i}))$

It should be noted that if we were working with unbounded metric space,
without allowing for the possibility that $d(x,y)$ can be infinite then this
construction would not work and the resulting category would not have all
products. This is the main reason why we work with this specific category
$\mathsf{k\text{-}CompMet}$.

For directed colimits, we consider the category $\mathsf{k\text{-}PsMet}$ of
$k$-bounded pseudo-metric spaces with contractions as morphisms. That is we
remove the requirement that $d(x,y)=0\Rightarrow x=y$, as well as the
completeness requirement.

The category $\mathsf{k\text{-}CompMet}$ is reflexive in
$\mathsf{k\text{-}PsMet}$: to each $k$-bounded pseudo-metric space, one can
associate a metric space by quotienting it by the relation $x\sim y$ if
$d(x,y)=0$, and take the completion of the resulting metric space. Hence
colimits in $\mathsf{k\text{-}CompMet}$ can be obtained by first taking the
colimit in the category $\mathsf{k\text{-}PsMet}$ and then applying this
quotient-completion construction.

Finally, direct colimits in $\mathsf{k\text{-}PsMet}$ are computed as follows:
Let $I$ be a directed set viewed as a category and let $B$ be a functor from
this directed set to the category of $k$ bounded pseudo-metric spaces. Then
one first takes the colimit of the underlying sets:
$\varinjlim_{I}B_{i}\simeq\coprod_{i\in I}B_{i}/\approx$ where $\coprod$
denotes the disjoint union and the equivalence relation is the equivalence
relation generated by: $f\approx g$ iff $f\in B_{i_{1}}$ and $g\in b_{i_{2}}$
and $i_{1}\leq i_{2}$ then $\epsilon_{i_{1},i_{2}}(f)=g$ (here
$\epsilon_{i_{1},i_{2}}$ is the image by the functor $B$ of the morphism
between $i_{1}$ and $i_{2}$ in the directed set viewed as category). And we
equip it with the following pseudo-metric: if $f\in B_{i}$ and $g\in B_{j}$
then $d(f,g)=\mathrm{Inf}_{i,j\leq k}d_{k}(f,g)$. In particular, it should be
noted that $f\in B_{i}$ and $g\in B_{j}$ then, in the colimits in
$\mathsf{k\text{-}CompMet}$, we have that $f=g$ iff $\forall\epsilon>0$, there
exists $k\geqslant i,j$ such that $d_{k}(f,g)<\epsilon$.

We can deduce from this the following by virtue of 1.2:

######  Proposition 2.2.

The category $\mathsf{k\text{-}CompMet}$ has an ultrastructure, where the
ultraproduct functors are given by

$\int_{S}M_{s}d\mu=\varinjlim_{U\in\mu}\left(\prod_{s\in U}M_{s}\right)$

Where $\mu$ is seen as a category with arrow $A\to B$ if $B\subseteq A$.

We can however give a slightly more explicit description of this ultraproduct
construction. We fix $S$ a set and $\mu$ an ultrafilter on $S$ suppose we have
a family of non-empty complete $k$-bounded metric spaces $(M_{s})_{s\in S}$
(that is an object in $\mathsf{k\text{-}CompMet}^{S}$).

We endow the set-theoretic product $\prod_{s\in S}M_{s}$ with the equivalence
relation defined by $(f_{s}\\}\sim(g_{s})$ iff for every $\epsilon>0$ the set
$\\{s\in S:\ d_{s}(f_{s},g_{s})<\epsilon\\}\in\mu$, and the distance given by:

$d((f_{s})_{s\in S},(g_{s})_{s\in
S})=\lim_{\mu}d(f_{s},g_{s})=\operatorname{\mathrm{Inf}}_{M\in\mu}\sup_{s\in
M}d(f_{s},g_{s})$

######  Theorem 2.1.

The distance defined above makes ($\prod_{s\in S}M_{s}/\sim)$ a complete
metric space, which identifies up to canonical isometry with the ultraproduct
$\varinjlim_{U\in\mu}\left(\prod_{s\in U}M_{s}\right)$.

For a proof see [3].

######  Note.

We should be more precise that the construction above would work if the family
of metric spaces $(M_{s})_{s\in S}$ are all non-empty, if some $M_{s}$ are
empty we can do the same trick as in the case of $\mathsf{Set}$ and looking
whether $S^{{}^{\prime}}=\\{s\in S\mid M_{s}=\varnothing\\}$ is in the
ultrafilter or not.

Although we are going to write proofs assuming that no metric space is empty,
this trick can always be used so our proofs also encompass the case where some
metric spaces are allowed to be empty.

###  2.1 Description of the categorical Fubini transform $\Delta$ in the
category $\mathsf{k\text{-}CompMet}$

######  Theorem 2.2.

Let $\mathsf{k\text{-}CompMet}$ denote the category of Complete metric spaces
bounded by a certain $k$ with contractions as morphisms, and let $S,T$ be sets
and let $\nu_{\bullet}=(\nu_{s})_{s\in S}$ be a collection of ultrafilters on
$T$ and let $(M_{t})_{t\in T}$ be a collection of complete metric spaces
indexed by $T$, Then we have:

$\Delta_{\mu,\nu_{\bullet}}((b_{t})_{t\in T})=((b_{t})_{t\in T})_{s\in S}$

###### Proof.

We need to make sure that the map $(b_{t})_{t\in T}\mapsto((b_{t})_{t\in
T})_{s\in S}$ is well defined : Suppose that

$(b_{t})_{t\in T}=(b^{{}^{\prime}}_{t})_{t\in T}$

we know that that for any $\epsilon>0$ the set

$\\{t\in T\
d_{t}(b_{t},b^{{}^{\prime}}_{t})\\}<\epsilon\\}\in\int_{S}\nu_{s}d\mu$

which translates to the fact that for any $\epsilon>0$ the set

$\\{s\in S\ \\{t\in T\
d_{t}(b_{t},b^{{}^{\prime}}_{t})\\}<\epsilon\\}\in\nu_{s}\\}\in\mu$

so we get that for any $\epsilon>0$

$\\{s\in S\ d_{\nu s}((b_{t})_{t\in T}),(b^{{}^{\prime}}_{t})_{t\in
T})<\epsilon\\}\in\mu$

so

$((b^{{}^{\prime}}_{t})_{t\in T})_{s\in S}=((b_{t})_{t\in T})_{s\in S}$

So the map defined the way above is well defined, and we can see that it is a
contraction.

Now the map $\Delta_{\mu,\nu_{\bullet}}$ for the ultrastructure on
$\mathsf{k\text{-}CompMet}$ is the unique map that makes the following diagram
commutes for every set $S_{0}\subseteq S$ satisfying $S_{0}\in\mu$ and every
set $T_{0}\subseteq T$ such that $T_{0}\in\nu_{s}\ \forall s\in S$
[19][proposition 1.2.8]

${{\Pi_{t\in T_{0}}M_{t}}}$${{\prod_{s\in
S_{0}}\int_{S}M_{t}d\nu_{s}}}$${{\int_{T}M_{t}d(\int_{S}\nu_{S}d\mu)}}$${{\int_{S}(\int_{T}M_{t}d\nu_{S})d\mu}}$$\scriptstyle{(q^{T_{0}}_{\nu_{s}})_{s\in
S}}$$\scriptstyle{q^{T_{0}}_{\int_{s}\nu_{s}d\mu}}$$\scriptstyle{\Delta_{\mu,\nu_{\bullet}}}$$\scriptstyle{q^{S_{0}}_{\mu}}$

Now clearly the map $(b_{t})_{t\in T}\mapsto((b_{t})_{t\in T})_{s\in S}$ makes
the diagram above commutes for every set $S_{0}\subseteq S$ satisfying
$S_{0}\in\mu$ and every set $T_{0}\subseteq T$ such that $T_{0}\in\nu_{s}$ for
all $s\in S$.

∎

One particular case which is important to consider is when we have a map of
sets $p$ from $S$ to $T$ and then we consider the family
$\delta_{\bullet}=(\delta_{p(s)})_{s\in S}$ the family of all the principal
ultrafilters of the points in the image of $p$. Then in this case we get the
map

$\Delta_{\mu,\delta_{\bullet}}((b_{t})_{t\in T})=(b_{p(s)})_{s\in S}$

Here $\int_{T}M_{t}d{\delta_{p(s)}}$ was identified with $M_{p(s)}$ (more
precisely, without this identification then the above is a description of the
ultraproduct diagonal map as defined in [19, Notation 1.3.3]).

##  3 The bundle (The first functor)

In this section, we are going to define the category of bundles of complete
metric spaces bounded by some constant $k$ over some compact Hausdorff space
$X$, which we are going to denote by
$\mathsf{Bun}(\mathsf{k\text{-}CompMet},X)$ or alternatively
$\mathsf{Bun}(\mathsf{k\text{-}CompMet})/X$, and construct a functor from the
category
$\text{Left ultrafunctors}(X,\mathsf{k\text{-}CompMet})$, to the category
$\mathsf{Bun}(\mathsf{k\text{-}CompMet},X)$.

###  3.1 Bundles of complete metric spaces

######  Definition 3.1.

A function $f$ from a topological space $E$ to
$\mathbb{R}\bigcup\\{-\infty,\infty\\}$ is said to be upper semi-continuous
(respectively lower semi-continuous) in a point $a$ iff for every $y>f(a)$
(respectively $y<f(a))$ there exists a neighbourhood $V$of $a$ such that for
every $x\in V\ f(x)<y$ (respectively $f(x)>y$).

A function $f$ from a topological space $E$ with values in
$\mathbb{R}\bigcup\\{-\infty,\infty\\}$ is upper semi-continuous (respectively
lower semi-continuous) iff it is upper semi-continuous (respectively lower
semi-continuous) in every point of its domain.

######  Note.

It is easy to see that being upper semi-continuous is equivalent to being
continuous when equipping $[-\infty,+\infty]$ with the topology generated by
$\\{[-\infty,b),\ b\in(-\infty,\infty])\\}$ which is called the left order
topology. The subspace topology of the left order topology of $[0,+\infty]$ is
generated by sets of the form $[0,b)$ with $b\in(0,+\infty]$.

######  Definition 3.2.

let $E$ be a topological space and let $\pi$ be a surjection from $E$ to $X$
such that for each $x\in X$ $\pi^{-1}(x)$ is a metric space with distance
$d_{x}$, let $V$ be an open set then we define

$V_{\epsilon}=\\{f\in E\mid\ \exists g\in V\ \pi(f)=\pi(g)\ \text{and}\
d_{\pi(f)}(f,g)<\epsilon\\}$

######  Definition 3.3.

In the same context as 3.2, let $V,W$ be open sets in $E$ we say that
$V\subseteq_{\epsilon}W$ if there exists $\epsilon>0$ such that $V\subseteq
V_{\epsilon}\subseteq W$

Let $A,B,C$ be topological spaces and let $f,g$ be a continuous map from $A$
to $C$ and $g$ be a continuous map from $B$ to $C$, we define the fibre
product space $A\times_{C}B$ to be the space $\\{\ (x,y)\in A\times B\ \mid
f(x)=g(y)\ \\}$ with the subspace topology of $A\times B$, this space is the
pullback of $f,g$ in the category of topological spaces.

Now We need to give an adequate definition of a continuous family of complete
metric spaces bounded by some constant $k$:

######  Definition 3.4.

A bundle of complete metric space bounded by $k$ is defined to be a triple
$(E,X,\pi)$ with $\pi\ :E\rightarrow X$ a surjection such that for every $x\in
X$ $\pi^{-1}(x)$ is a complete metric space bounded by $k$, if it satisfies
the axioms:

  * •

Axiom(1): The global distance function defined from $E\times_{X}E$ to $[0,k]$
is upper semi-continuous.

  * •

Axiom(2) : $\pi$ is continuous and open.

  * •

Axiom(3) : For every open set $W$ and every $f\in W$ there exist an open
neighbourhood $V$ of $f$ and $\epsilon>0$ such that $V\subseteq_{\epsilon}W$.

######  Definition 3.5.

If $(E,X,\pi)$ and $(E^{{}^{\prime}},X,\pi^{{}^{\prime}})$ are two bundles
with base space $X$, we define a map of bundles $\psi$ to be a continuous map
from $E$ to $E^{{}^{\prime}}$ such that the following diagram commutes:

${E}$${{E^{{}^{\prime}}}}$${X}$$\scriptstyle{\pi}$$\scriptstyle{\psi}$$\scriptstyle{\pi^{{}^{\prime}}}$

and such that for each $x\in X$ the map $\psi|_{\pi^{-1}(x)}$ is a
contraction.

This makes bundles with a fixed base space a category. The case where the base
space is allowed to vary will be treated in section 7.

###  3.2 The Bundle’s Topology

Given a compact Hausdorff space $X$, and a left ultrafunctor $\mathcal{F}$
from $X$ to $\mathsf{k}\text{-}\mathsf{CompMet}$ we want to endow the space
$\coprod_{x\in X}\mathcal{F}(x)$ with a canonical topology making it a bundle
as in our definition 3.4. A common idea usually used in the definition of
bundles, is that the bundle space is some sort of section space (or germs of
section space) to the projection map, the definition that we gave starts from
the realisation that the image by the left-ultrastructure maps of a point in
the base space can be regarded as some sort of “generalised” local section
maps at this point, and hence one can use these to define a topology on the
space $\coprod_{x\in X}\mathcal{F}(x)$ similar in spirit to [9, 13.18]
(constructing a bundle from a family of sections is abundant in functional
analysis literature, the same kind of idea can be seen for example in [8] or
[22, 24]).

Let us call $\mathcal{L}$ the assignment that we are going to define, which
gives a bundle for each left ultrafunctor from a compact Hausdorff set
$X$(ultraset) to $\mathsf{k}\text{-}\mathsf{CompMet}$.

######  Theorem 3.1.

let $X$ be an ultraset, and let $\mathcal{F}$ be a left ultrafunctor from $X$
to $\mathsf{k}\text{-}\mathsf{CompMet}$, let $E=\coprod_{x\in
X}\mathcal{F}(x)$ then there is a unique topology $\tau$ on $E=\coprod_{x\in
X}\mathcal{F}(x)$ such that an ultrafilter $\eta$ converges to a point $f\in
E$ iff:

  1. $C_{1}$

: $\pi\eta$ converges to $\pi f$

  2. $C_{2}$

: for any $\epsilon>0$ if $\sigma_{\pi\eta}(f)=(b_{x})_{x\in X}$ then
$\coprod_{x\in X}B(b_{x},\epsilon)\in\eta$

And this topology is characterized by a set $U$ being open iff for any
ultrafilter $\eta$ converging to a point $f\in U$ then $U\in\eta$.

###### Proof.

######  Lemma.

The condition $C_{2}$ is well defined, that is it does not depend on the
representative of the equivalence class of $(b_{x})_{x\in X}$.

###### Proof.

Suppose $\sigma_{\pi\eta}(f)=(b_{x})_{x\in X}=(b^{{}^{\prime}}_{x})_{x\in X},$
and suppose that for any $\epsilon>0\ \coprod_{x\in
X}B(b_{x},\epsilon)\in\eta$. Let $\epsilon>0$ then :

$S=\\{x\in X:d_{\pi(x)}(b_{x}-b^{{}^{\prime}}_{x})<\epsilon/2\\}\in\pi\eta,$

so

$\coprod_{x\in S}\mathcal{F}(x)\in\eta,$

so

$\coprod_{x\in S}\mathcal{F}(x)\bigcap\coprod_{x\in
X}B(b_{x},\epsilon/2)=\coprod_{x\in S}B(b_{x},\epsilon/2)\in\eta.$

Now let $g\in\coprod_{x\in S}B(b_{x},\epsilon/2)$ then

$d(g,b^{{}^{\prime}}_{x})\leq
d(g,b_{x})+d(b_{x},b^{{}^{\prime}}_{x})<\epsilon/2+\epsilon/2=\epsilon,$

hence

$\coprod_{x\in S}B(b_{x},\epsilon/2)\subseteq\coprod_{x\in
X}B(b^{{}^{\prime}}_{x},\epsilon),$

and this implies that

$\coprod_{x\in}B(b^{{}^{\prime}}_{x},\epsilon)\in\eta$

∎

Now back to the proof of the theorem, the proof relies on [25, Theorem 4.4],
namely every relation satisfying conditions UQ1 and UQ4 of [25] defines a
topology characterized by this relation being the convergence relation on
ultrafilters, we are going to summarise these conditions in the following
theorem :

######  Theorem 3.2.

let $X$ be a set and let $\beta X$ be the set of all ultrafilters on $X$ (the
Stone-Cech compactification of its discrete structure), let $q$ be a relation
on $\beta X\times X$ satisfying the following:

  * •

UQ1: $\forall x\in X$ $\delta_{x}qx$, here $\delta_{x}$ is the principal
ultrafilter at $x$.

  * •

UQ4: If $t:S\rightarrow X$ and $u:S\rightarrow\beta X$ are maps such that
$u(s)\ q\ t(s)$ for every $s\in S$, and if $t\phi\ q\ x$ for an ultrafilter
$\phi$ on $S$, then $\int_{S}u(s)d\phi qx$) .

Here the ultrafilter $\int_{S}u(s)d\phi$ is the ultrafilter defined in 1.

Then there exists a topology on $X$ characterised by being the unique topology
such that the ultrafilter $\phi$ converges to $x\in X$ in the usual sense iff
$\phi qx$.

##### Property UQ1

In what follows let us say that for an ultrafilter $\mu$ on $E$ and a point
$f\in E$, $\mu qf$ if $(\mu,f)$ satisfies conditions $C_{1}$ and $C_{2}$ of
3.1. We need to prove that $\delta_{f}qf$. First it is obvious that
$\pi(\delta_{f})=\delta_{\pi(f)}$ so it converges to $\pi(f)$, so $\delta_{f}$
has property $C_{1}$ of 3.1. Now if $\sigma_{\pi(f)}=(b_{x})_{x\in X}$ then
the equivalence class is completely determined by $b_{\pi(f)}=f$ (using [19,
definition 1.4.1(1)]), so $\delta_{f}$ satisfies property $C_{2}$ of 3.1, so
$\delta_{f}qf$.

#####  Property UQ4

Now let us prove that the convergence relation defined with the two properties
above ($C_{1}$ and $C_{2}$) satisfies the second condition of 3.2: let $S$ be
a set and let $t$ be a map of sets from $T$ to $E$, let $u$ be a map from $S$
to $\beta E$ (where $\beta E$ is the sets of all ultrafilters on E) such that
each $u(s)qt(s)$ (that means satisfies the conditions $C_{1}$ and $C_{2}$),
and suppose $t\mu qf\in E$, we need to show that $\alpha=\int_{S}u(s)d\mu qf$
.

Now showing that $\pi\alpha$ converges to $\pi f$ (in the usual sense of
converging in a topological space) is trivial, to see why we know that we have
a morphism $\Delta_{\mu,\pi u\bullet}$ from $\int_{X}xd\pi\alpha$ to
$\int_{S}(\int_{X}x\ d\pi u(s))d\mu)=\int_{S}\pi(t(s))d\mu$ the latter can be
shown to be equal to $\int_{X}x\ d\pi t\mu=\pi(f)$ (since $t\mu$ satisfies the
2 conditions), since the only morphisms in ultrasets are identities, this
proves that $\pi\alpha$ converges to $\pi(f)$. Thus $\alpha$ satisfies
property $C_{1}$ of 3.1.

So it remains to show that for any $\epsilon>0$ and supposing
$\sigma_{\pi\alpha}(f)=(b_{x})_{x\in X}$ then $\coprod_{x\in
X}B(b_{x},\epsilon)\in\alpha$. Before proceeding let us use the description we
have for the categorical Fubini transforms to find a relation between the 2
maps $\sigma_{\mu}$ and $\sigma_{t\mu}$. By the definition of left
ultrafunctors ([19, definition 1.4.1(2)] if we take the family of ultrafilters
$(\delta_{t(s)})_{s\in S}$, then the following diagram commutes:

${{F(\pi(f))}}$${{\int_{E}F(\pi t(e))dt\mu}}$${{\int_{S}F(\pi
t(s))d\mu}}$$\scriptstyle{\sigma_{t\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\Delta_{\mu,\delta_{t\bullet}}}$

Here we identified $\int_{E}F(\pi t(e))d\delta_{t(s)}$ with $F(\pi(t(s))$
using [19, definition 1.4.1(1)] and also using the fact that
$\int_{S}\delta_{t(s)}d\mu=t\mu$. So for any element $h$ of $F(\pi(f))$ if
$\sigma_{t\mu}(h)=(a_{e})_{e\in E}$ and
$\sigma_{\mu}(h)=(a^{{}^{\prime}}_{s})_{s\in S}$ then
$(a^{{}^{\prime}}_{s})_{s\in S}=(a_{t(s)})_{s\in S}$ (using $\eqref{eq:1})$.
Now we state the following lemma:

######  Lemma 3.1.

Suppose that $\sigma_{\mu}(f)=(q^{{}^{\prime}}_{s})_{s\in S}$ and
$\sigma_{t\mu}(f)=(q_{e})_{e\in E}$ then $(q^{{}^{\prime}}_{s})_{s\in
S}=(q_{t(s)})_{s\in S}=(t(s))_{s\in S}$

###### Proof.

Let $\epsilon>0$, Then since $\mu t$ converges to $f$ then

$\coprod_{x\in X}B(q_{x},\epsilon)\in t\mu,$

and this implies that

$t^{-1}(\coprod_{x\in X}B(q_{x},\epsilon))\in\mu,$

and so

$\\{s\in S\ d_{t(s)}(q_{t(s)},t(s))<\epsilon\\}\in\mu,$

which completes the proof of lemma.∎

Now let $\epsilon>0$, let us start by writing the diagram [19, definition
1.4.1(2)] for the family $(\pi(u(s)))_{s\in S}$ :

${F(\pi(f))}$${\int_{X}F(x)d\pi\alpha}$${\int_{S}F(\pi
t(s))d\mu}$${\int_{S}(\int_{X}F(x)d\pi
u(s))d\mu}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\sigma_{\pi\alpha}}$$\scriptstyle{\Delta_{\mu,u_{\bullet}}}$$\scriptstyle{\int_{S}\sigma_{\pi(u(s))}d\mu}$
since $f\in F(\pi(f))$ and since $\sigma_{\mu}(f)=(t(s))_{s\in S}$ then the
commutativity of the diagram tells us that if
$\sigma_{u(s)}(t(s))=(a^{s}_{x})_{x\in X}$ then

$((a^{s}_{x})_{x\in X})_{s\in S}=((b_{x})_{x\in X})_{s\in S},$

which translates to the fact that for any $\epsilon^{{}^{\prime}}$ we have
that

$\\{s\in S\ \ d_{\pi u(s)}((a^{s}_{x})_{x\in X},(b_{x})_{x\in
X})<\epsilon^{{}^{\prime}}\\}\in\mu,$

in particular

$\\{s\in S\ \ d_{\pi(u(s))}((a^{s}_{x})_{x\in X},(b_{x})_{x\in
X})<\epsilon/2\\}\in\mu.$

We also know that

$\coprod_{x\in X}B(b_{x},\epsilon/2)\in t\mu,$

which implies that

$t^{-1}(\coprod_{x\in X}B(b_{x},\epsilon/2))\in\mu.$

It follows that their intersection

$\\{s\in S\ \ d_{\pi(u(s))}((a^{s}_{x})_{x\in X},(b_{x})_{x\in
X})<\epsilon/2\\}\bigcap t^{-1}(\coprod_{x\in X}B(b_{x},\epsilon/2))\in\mu.$

Now our goal is to show that

$\\{s\in S\ \ d_{\pi(u(s))}((a^{s}_{x})_{x\in X},(b_{x})_{x\in
X})<\epsilon/2\\}\bigcap t^{-1}(\coprod_{x\in
X}B(b_{x},\epsilon/2))\subseteq\\{s\in S\ \coprod_{x\in X}B(x,\epsilon)\in
u(s)\\},$

to do so consider any $s$ in the intersection, we have for such $s$

$t(s)\in B(b_{\pi t(s)},\epsilon/2),$

and on the other hand

$d_{\pi u(s)}((a^{s}_{x})_{x\in X},(b_{x})_{x\in X})<\epsilon/2,$

which implies that the set

$\\{x\in X\ d_{x}(b_{x},a^{s}_{x})<\epsilon/2\\}\in\pi u(s),$

which is equivalent to saying that

$\pi^{-1}((x\in Xd_{x}(b_{x},a^{s}_{x})<\epsilon/2\\})\in u(s).$

Now we already know that

$\coprod_{x\in X}(a^{s}_{x},\epsilon/2)\in u(s),$

thus

$\pi^{-1}((x\in Xd_{x}(b_{x},a^{s}_{x})<\epsilon/2\\})\bigcap\coprod_{x\in
X}(a^{s}_{x},\epsilon/2)\in u(s),$

So it remains to show that this is subset of $\coprod_{x\in
X}B(b_{x},\epsilon),$

to do so let $h\in\pi^{-1}((x\in X\
d_{x}(b_{x},a^{s}_{x})<\epsilon/2\\})\bigcap\coprod_{x\in
X}(a^{s}_{x},\epsilon/2),$ we have that

$d_{\pi(h)}(h,b_{\pi(h)})\leq
d_{\pi(h)}(h,a^{s}_{\pi(h)})+d_{\pi(h)}(a^{s}_{\pi(h)},b_{\pi(h)})<\epsilon/2+\epsilon/2=\epsilon,$

so

$h\in\coprod_{x\in X}B(b_{x},\epsilon),$

and this proves that

$\pi^{-1}((x\in X\ d_{x}(b_{x},a^{s}_{x})<\epsilon/2\\})\bigcap\coprod_{x\in
X}(a^{s}_{x},\epsilon/2)\subseteq\coprod_{x\in X}B(b_{x},\epsilon),$

hence

$\coprod_{x\in X}B(b_{x},\epsilon)\in u(s),$

and thus

$\\{s\in S\ d_{\pi(t(s))}(a^{s}_{x},t_{s})<\epsilon/2\\}\bigcap
t^{-1}(\coprod_{x\in X}B(b_{x},\epsilon/2))\subseteq\\{s\in S\ \coprod_{x\in
X}B(x,w\epsilon)\in u(s)\\},$

so as a result $\coprod_{x\in X}B(b_{x},\epsilon)\in\alpha$, and this is true
for any $\epsilon>0$. So $\alpha$ satisfies property $C_{2}$ of 3.1, so we may
deduce that $\alpha qf$. ∎

######  Note.

By definition the topology characterised by properties $C_{1}$ and $C_{2}$ of
3.1, makes $\pi$ continuous, since the condition $C_{1}$ implies that if
$\eta$ converges to $f$ then $\pi\eta$ converges to $\pi(f)$.

###  3.3 Characterisation of the topology

######  Theorem 3.3.

Let $X$ be a compact Hausdorff space and let $\mathcal{F}:\
X\rightarrow\mathsf{k}\text{-}\mathsf{CompMet}$ be a left ultrafunctor. A set
$C=\coprod_{x\in X}U(x)\subseteq\coprod_{x\in X}\mathcal{F}(x)$ is open in the
topology $\tau$ defined in 3.2, iff it satisfies the following condition:
For every ultrafilter $\mu$ on $X$ converging to a point
$x\in\pi(\coprod_{x\in X}U(x))$ and $\forall g\in U(x)$ if
$\sigma_{\mu}(g)=(b(y))_{y\in X}$ then $\exists W\in\mu$ and $\epsilon>0$ such
that $B_{y}(b_{y},\epsilon)\subseteq U(y)$ for any $y\in W$ (in other words
$\coprod_{x\in W}B(b_{x},\epsilon)\subseteq C$).

######  Note.

Although $\epsilon$ and $W$ depend on the representative of the class of
$\sigma_{\mu}(g)$, their existence does not depend on the representative, so
this condition is well defined.

####  3.3.1 Proof of the only if direction

Let $V=\coprod_{x\in X}U(x)$ be a set satisfying this condition of theorem
3.3. Our goal is to show that $V$ is open by showing that for any ultrafilter
$\mu$ converging to $f\in V$, we have $V\in\mu$. Let $\mu$ be such ultrafilter
converging to some $f\in V$. Now by definition of convergence relation on $E$,
we get that $\pi\mu$ converges to $\pi(f)$. Now we must note that $f\in
U(\pi(f))$ so if $\sigma_{\pi\mu}(f)=(b_{x})_{x\in X}$ then using the
condition of the theorem $\exists W\in\pi\mu\ \text{and}\ \epsilon>0$ such
that for any $x\in W\ B(b_{x},\epsilon)\subseteq U(x)$ so $\coprod_{x\in
W}B(b_{x},\epsilon)\subseteq V$ Now take $\epsilon^{{}^{\prime}}<\epsilon$
then $\coprod_{x\in X}B(b_{x},\epsilon^{{}^{\prime}})\in\mu$ by property
$C_{2}$, but also is $\pi^{-1}(W)$ (since $\pi$ is continuous), so their
intersection $\pi^{-1}(W)\bigcap\coprod_{x\in
X}B(b_{x},\epsilon^{{}^{\prime}})\in\mu$, but now we see that:

$\pi^{-1}(W)\bigcap\coprod_{x\in
X}B(b_{x},\epsilon^{{}^{\prime}})\subseteq\coprod_{x\in
W}B(b_{x},\epsilon)\subseteq V$

thus $V\in\mu$ so $V$ is open.

####  3.3.2 Proof of the if direction

We recall classical results regarding ultrafilters:

######  Lemma 3.2.

let $E$ be set and let $U\subseteq\mathcal{P}(E)$ (the powerset of $E$) be a
set of nonempty subsets of $E$, satisfying the finite intersection property
then there exists an ultrafilter $\nu$ such that $A\in\nu$ for all $A\in U$.

###### Proof.

define the filter $\alpha$ by setting $A\in\alpha\ \text{iff}\ A\supseteq
A^{{}^{\prime}}$ for some $A^{{}^{\prime}}\in U$, now it can be checked that
$\alpha$ is a filter and thus must be contained in some ultrafilter $\nu$. ∎

######  Lemma 3.3.

let $E$ be set and let $U\subseteq\mathcal{P}(E)$ be a set of subsets of $E$,
such that for every finite $U^{{}^{\prime}}\subseteq U$, $\bigcap
U^{{}^{\prime}}\neq\varnothing$ then there exists an ultrafilter $\nu$ such
that $A\in\nu$ for all $A\in U$.

###### Proof.

Apply lemma 3.2 to $\\{\bigcap U^{{}^{\prime}}\mid
U^{{}^{\prime}}\textbf{finite}\ \text{and}\ U^{{}^{\prime}}\subseteq U\\}$ ∎

######  Lemma 3.4.

let $E$ be a set and let $U\subseteq\mathcal{P}(E)$ be a family of subsets,
Let $A\subseteq E$ such that $A$ satisfies the following:

$\forall\mu\in\beta E,(U\subseteq\mu\implies A\in\mu)$

then $\exists\ B_{1},\dots,B_{n}\in U$ such that $B_{1}\bigcap\dots\bigcap
B_{n}\subseteq A$.

###### Proof.

In the case where for some finite family $U^{{}^{\prime}}\subseteq U$,
$\bigcap U^{{}^{\prime}}$ is empty we can say that $\varnothing\subseteq A$.
Otherwise, Assume by contradiction that no such finite family exists, apply
lemma 3.3 to the family $U\bigcup\\{A^{c}\\}$ to get a contradiction, thus
such finite family must exist.∎

Now Let $V=\coprod_{x\in X}U(x)$ be an open set, and let $\eta$ be an
ultrafilter on $X$ converging to a point $x\in\pi(V)$. Now let $f\in U(x)$
suppose that $\sigma_{\eta}(f)=(b_{x})_{x\in X}$, and let $\mu$ be an
ultrafilter on $E$ and suppose that :

(*) $\forall S\in\eta,\ \forall\epsilon>0\coprod_{x\in
S}B(b_{x},\epsilon)\in\mu$

We can show that in this case $\eta=\pi\mu$ and since we assumed the condition
$\eqref{*}$, Then $\mu$ converges to $f$ and thus since $V$ is open, we may
deduce that $V\in\mu$. So we have proved that

$\forall\mu\in\beta E\ ((\forall S\in\eta\ \text{such that}\
\pi^{-1}(S)\in\mu,\forall\epsilon>0\coprod_{x\in S}\
B(b_{x},\epsilon)\subseteq V)\implies V\in\mu)$

Hence we may may deduce by lemma 3.4 that:

$\exists\epsilon>0\ \text{and}\ W\in\eta\ \text{such that}\ \coprod_{x\in
W}B(b_{x},\epsilon)\subseteq V$

So the set $V$ satisfies the condition of theorem Note.

###  3.4 The topology definition satisfies the axioms of 3.4

Now we want to prove that our definition of the topology on $\coprod_{x\in
X}\mathcal{F}(x)$ from theorem 3.1, gives rise to a bundle of complete metric
spaces so will check that our definition satisfies the three axioms of
definition 3.4.

####  3.4.1 Axiom 1

To prove that the distance function is upper semi-continuous. let
$\pi:\coprod_{x\in X}\mathcal{F}(x)\rightarrow X$ be the projection map,
suppose that $\mu$ is an ultrafilter on $E\times_{X}E$ such that $\mu$
converges to $(f,g)$. Now let $r>d(f,g)$. First notice that
$\pi\circ\pi_{1}=\pi\circ\pi_{2}$. Now we have that $\pi_{1}\mu$ converges to
$f$ and $\pi_{2}\mu$ converges to $g$ (simply because projections are
continuous).

Suppose that $\sigma_{\pi\pi_{1}\mu}(f)=(b_{x})_{x\in X}$ and
$\sigma_{\pi\pi_{2}\mu}(g)=(c_{x})_{x\in X}$. Take $\epsilon_{1}$ and
$\epsilon_{2}$ such that they satisfy $r-(\epsilon_{1}+\epsilon_{2})>d(f,g)$,
we know that $\coprod_{x\in X}B(b_{x},\epsilon_{1})\in\pi_{1}\mu$ and that
$\coprod_{x\in X}B(b_{x},\epsilon_{2})\in\pi_{2}\mu$. We also know that
$d_{\pi\pi_{1}\mu}((b_{x})_{x\in X},(c_{x})_{x\in X})\leq d(f,g)$ so that
means that there exists some set $L\in\pi\pi_{1}\mu=\pi\pi_{2}\mu$ such that
$\forall x\in L$ $d(b_{x},c_{x})\leq d(f,g)$. So we deduce that $\coprod_{x\in
L}B(b_{x},\epsilon_{1})\in\pi_{1}\mu$ and that $\coprod_{x\in
L}B(c_{x},\epsilon_{2})\in\pi_{2}\mu$ (by intersecting with $\pi^{-1}(L)$).

Now let $(h,l)\in\pi_{1}^{-1}(\coprod_{x\in
L}B(b_{x},\epsilon_{1}))\bigcap\pi_{2}^{-1}(\coprod_{x\in
L}B(c_{x},\epsilon_{2}))$, then

$d(h,l)<\epsilon_{1}+\epsilon_{2}+d(b_{\pi(h)},c_{\pi(h)})<\epsilon_{1}+\epsilon_{2}+d(f,g)<r$

And thus

$d^{-1}([0,r))\supseteq\pi_{1}^{-1}(\coprod_{x\in
L}B(b_{x},\epsilon_{1}))\bigcap\pi_{2}^{-1}(\coprod_{x\in
L}B(c_{x},\epsilon_{2}))\in\mu$

So $d\mu$ converges to $d(x,y)$ (if we equip $[0,k]$ with the left order
topology of course).

####  3.4.2 Axiom 2

By definition of the topology on $E$, $\pi$ is continuous. Now to prove that
$\pi$ is open, let $V=\coprod_{x\in\pi(V)}U(x)$ be a non-empty open set of
$E$, and let $\eta$ be an ultrafilter on $X$ converging to $x\in\pi(V)$. We
wish to show that $\pi(V)\in\eta$. Since $x\in\pi(V)$ then $U(x)$ is non-empty
then let $a\in U(x)$. If $\sigma_{\eta}(a)=(b_{x})_{x\in X}$ then $\exists
W\in\eta$ and $\epsilon>0$ such that $\coprod_{x\in
W}B(b_{x},\epsilon)\subseteq V$, thus $\pi(\coprod_{x\in
W}B(b_{x},\epsilon))=W\subseteq\pi(V)$. But since $W\in\eta$ thus
$\pi(V)\in\eta$. So $\pi(V)$ is open.

####  3.4.3 Axiom 3

We to prove that our definition of topology associated to left ultrafunctors
satisfies axiom (3) of the definition of bundle topology, which informally
means that if $V\subseteq W$ are open sets in $E=\coprod_{x\in
X}\mathcal{F}(x)$ where $\mathcal{F}$ is a left ultrafunctor from $k$ to the
category $\mathsf{k\text{-}PsMet}$, we can“enlarge" $V$ by some $\epsilon$ and
remain inside $W$ (we have of course given a more formal statement).

Before showing that our construction satisfies axiom $(3)$, we give two
lemmas, which are true for every pair of topological spaces $(E,X)$,
satisfying that there exists a surjection $\pi$ from $E$ to $X$ such that for
every $x\in X$, $\pi^{-1}(x)$ is a metric space and such that the distance
function from $E\times_{X}E$ to $E$ is upper semi-continuous.

######  Definition 3.6.

let $E,X$ be topological spaces and let $\pi$ be a surjection from $E$ to $X$
such that for every $x\in X$ $\pi^{-1}(x)$ is a complete metric space bounded
by some $k$, we call $V\subseteq E$ $\epsilon$-thin iff for every
$f,f^{{}^{\prime}}\in V$, if $\pi(f)=\pi(f^{{}^{\prime}})$ then
$d_{\pi(f)}(f,f^{{}^{\prime}})<\epsilon$.

######  Lemma 3.5.

let $E,X$ be topological spaces and let $\pi$ be a surjection from $E$ to $X$
such that for every $x\in X$ $\pi^{-1}(x)$ is a metric space and such that the
distance function from $E\times_{X}E$ to $E$ is upper semi-continuous then for
any $\epsilon>0$, then $E$ has a basis consisting of $\epsilon$-thin
neighbourhoods.

###### Proof.

The distance $E\times_{X}E$ is upper semi-continuous which implies that for
any $\epsilon>0$ the set $\\{(v,v^{{}^{\prime}})\in E\times_{X}E\
d_{\pi(v)}(v,v^{{}^{\prime}})<\epsilon\ \\}$ is open. This implies that the
sets of form $U_{i}\times_{X}V_{i}$, where $U_{i}$ and $V_{i}$ are open sets
such that: $U_{i}\times_{X}V_{i}\subseteq\\{(v,v^{{}^{\prime}})\in
E\times_{X}E\ d_{\pi(v)}(v,v^{{}^{\prime}})<\epsilon\\}$ form a basis for the
subspace topology of $\\{(v,v^{{}^{\prime}})\in E\times_{X}E\
d_{\pi(v)}(v,v^{{}^{\prime}})<\epsilon\\}$. Now the subspace topology of the
diagonal is generated by the intersection of the diagonal with these basic
open neighbourhoods. So applying the projection map (which is a homeomorphism
between the diagonal of $E\times_{X}E$ and $E$) we may deduce that for any
$\epsilon>0$ $E$ has a topology generated by open sets
$W_{i}=\Delta^{-1}(U_{i}\times_{X}V_{i})$ where $\Delta:E\rightarrow
E\times_{X}E$ is the diagonal map. Now by construction, each set of these
satisfies the following:

$\forall g,g^{{}^{\prime}}\in W_{i}\ \text{if}\ \pi(g)=\pi(g^{{}^{\prime}})\
\text{then}\ d_{\pi(g)}(g,g^{{}^{\prime}})<\epsilon$

∎

######  Lemma 3.6.

Let $E,X$ be topological spaces and let $\pi$ be a surjection from $E$ to $X$
such that for every $x\in X$, $\pi^{-1}(x)$ is a metric space and such that
the distance function from $E\times_{X}E$ to $E$ is upper semi-continuous then
the sets $V_{\epsilon}$ as defined in 3.2 are open in the topology of $E$.

###### Proof.

Take the distance map from $E\times_{X}V$ to $[0,\infty)$, by upper semi-
continuity each $(d|_{E\times_{X}V})^{-1}([0,\epsilon))$ is open in
$E\times_{X}V$, now since $E\times_{X}V$ is open in $E\times_{X}E$, then
$d|_{E\times_{X}V}^{-1}([0,\epsilon))$ is open in the topology of $E\times E$,
so we can apply the first projection (open map) to
$d|_{E\times_{X}V}^{-1}([0,\epsilon))$ to get the open set $V_{\epsilon}$. ∎

Now let $\mathcal{F}$ be a left ultrafunctor from $X$ to the category of
metric spaces bounded by a certain $k$, and let $E=\coprod_{x\in
X}\mathcal{F}(x)$ equipped with the topology defined in 3.2. And let $W$ be an
open set of $E$ and let $f\in W$ and let $\mathcal{N}_{f}$ denote the set of
open neighbourhoods of $f$. We want to show that there exists an open
neighbourhood $V$ of $f$ such that $V\subseteq_{\epsilon}W$ using lemma 3.4.
Take the family of sets
$\\{V_{\epsilon}\mid\epsilon>0,V\in\mathcal{N}_{f}\\}$, let $\mu$ be an
ultrafilter on $E$ and suppose that
$\\{V_{\epsilon}\mid\epsilon>0,V\in\mathcal{N}_{f}\\}\subseteq\mu$, We want to
show that $\mu$ converges to $f$, this will allow us to use lemma 3.4.

First to prove that $\pi\mu$ converges to $\pi f$, notice that for any
neighbourhood $S$ of $\pi(f)$ if we take any $\epsilon>0$
$\pi^{-1}(S)_{\epsilon}=\pi^{-1}(S)$ so $S\in\mu$ thus $\pi\mu$ converges to
$\pi f$. And also suppose that we have chosen a representative $(b_{x})_{x\in
X}$ of the class of $\sigma_{\pi\mu}(f)$ (in other words
$\sigma_{\pi\mu}(f)=(b_{x})_{x\in X}$). Now we regard $b$ as a map from $X$ to
$E$. We will show that $b\pi\mu$ converges to $f$, first notice that $\pi
b\pi\mu$ converges to $\pi(f)$ since $\pi b=\mathrm{id}$. Now we prove that
for any $\epsilon$, $\coprod_{x\in X}B(b_{x},\epsilon)\in b\pi\mu$, this
follows from the fact that $b^{-1}(\coprod_{x\in X}B(b_{x},\epsilon))=X\
\in\pi\mu$ so $b\pi\mu$ converges to $f$.

We have already shown that the distance $E\times_{X}E$ is upper semi-
continuous which implies by lemma 3.5 that for any $r>0$, $E$ has a basis
consisting of $r$-thin neighbourhoods, thus take an open neighbourhood $V$ of
$f$ such that this neighbourhood is $\epsilon/2$-thin and take the set
$V_{\epsilon/2}$.

Since the set $V$ is an open neighbourhood of $f$ and the projection $\pi$ is
open we may deduce that the set $\\{x:b_{x}\in V\\}=\pi(V)\in\pi\mu$ so
$\pi^{-1}((x\in X:b_{x}\in V\\})=\pi^{-1}(\pi(V))\in\mu$. Let us proof that
$\pi^{-1}((x:b_{x}\in V\\})\bigcap V_{\epsilon/2}\subseteq\coprod_{x\in
X}B(b_{x},\epsilon)$. Let $g\in\pi^{-1}((x:b_{x}\in V\\})\bigcap
V_{\epsilon/2}$ now $\exists h\in V$ such that $\pi(g)=\pi(h)$ and such that
$d(g,h)<\epsilon/2$ now $d(g,b_{\pi(g)})\leq
d(g,h)+d(h,b_{\pi(g)})<\epsilon/2+\epsilon/2=\epsilon$. So $\mu$ converges to
$f$, thus $W\in\mu$. So we can apply lemma 3.4.

Thus there exists some neighbourhoods $V^{1},\dots V^{n}$ of $f$ and some
$\epsilon_{1},\dots\epsilon_{n}>0$ such that
$\bigcap_{i=1}^{n}V^{i}\subseteq\bigcap_{i=1}^{n}V_{\epsilon_{i}}^{i}\subseteq
W$ (using lemma 3.4). Now if we call $\epsilon=min_{i=1}^{n}\epsilon_{i}$ and
$V=\bigcap_{i=1}^{n}V^{i}$ (these are not $V$ and $\epsilon$ of the previous
paragraph), then $V\subseteq
V_{\epsilon}\subseteq\bigcap_{i=1}^{n}V_{\epsilon_{i}}^{i}\subseteq W$.

######  Note.

We can also conclude that the sets $V_{\epsilon}$ for $V$ neighbourhood of $f$
and $\epsilon>0$ form a basis for the neighbourhood system at $f$.

###  3.5 $\mathcal{L}$ is a functor

We have described the way $\mathcal{L}$ acts on objects. Now let us describe
how it acts on morphisms:

Let $X$ be a compact Hausdorff space and $\mathcal{F}$ and
$\mathcal{F^{{}^{\prime}}}$ be two left ultrafunctors and let $\nu$ be a
natural transformation of left ultrafunctors between $\mathcal{F}$ and
$\mathcal{F^{{}^{\prime}}}$. Then the induce map of bundles is
$\psi=\mathcal{L}(\nu)$ defined by

$\psi|_{\pi^{-1}(x)}=\nu_{x}$

. Now we need to prove that this induced map $\psi$ is indeed a map of
bundles. Let $E=\mathcal{L}(\mathcal{F})$ and
$E^{{}^{\prime}}=\mathcal{L}(\mathcal{F}^{{}^{\prime}})$, we already have the
commutativity of the diagram:

${E}$${{E^{{}^{\prime}}}}$${X}$$\scriptstyle{\pi^{{}^{\prime}}}$$\scriptstyle{\pi}$$\scriptstyle{\psi}$

by definition, we also already have the fact that $\psi|_{\pi^{-1}(x)}$ is a
contraction by definition.

It remains to show that $\psi$ is continuous, to do so let $\mu$ be an
ultrafilter on $E$ that converges to $f\in E$. Let us show that $\psi\mu$
converges to $\nu_{\pi(f)}(f)$. We start by proving that
$\pi^{{}^{\prime}}\psi\mu$ converges to
$\pi(f)=\pi^{{}^{\prime}}(\nu_{\pi(f)}(f))$ but this follows from the
commutativity of the following diagram:

${E}$${{E^{{}^{\prime}}}}$${X}$$\scriptstyle{\pi^{{}^{\prime}}}$$\scriptstyle{\pi}$$\scriptstyle{\psi}$

Now suppose that
$\sigma^{{}^{\prime}}_{\pi^{{}^{\prime}}\psi\mu}(\nu_{\pi(f)}(f))=\sigma^{{}^{\prime}}_{\pi\mu}(\nu_{\pi(f)}(f))=(b^{{}^{\prime}}_{x})_{x\in
X^{{}^{\prime}}}$ and $\sigma_{\pi\mu}(f)=(b_{x})_{x\in X}$. We know that
since $\nu$ is a natural transformation of left ultrafunctors then the
following diagram commutes:

${{\mathcal{F}(\pi(f))}}$${{\int_{X}\mathcal{F}(x)d\pi\mu}}$${{\mathcal{F^{{}^{\prime}}}(\pi(f))}}$${{\int_{X}\mathcal{F^{{}^{\prime}}}(x)d\pi\mu}}$$\scriptstyle{\sigma_{\pi\mu}}$$\scriptstyle{\nu_{\pi(f)}}$$\scriptstyle{\sigma^{{}^{\prime}}_{\pi\mu}}$$\scriptstyle{\int_{X}\nu_{x}d\mu}$

Which implies that $(b^{{}^{\prime}}_{x})_{x\in X}=(\nu_{x}(b_{x}))_{x\in X}$,
so for simplicity we are going to take $(\nu_{x}(b_{x}))_{x\in X}$ as
representative of the class. Now let $\epsilon>0$ we are going to show that :

$\coprod_{x\in X}B(b_{x},\epsilon)\subseteq\psi^{-1}(\coprod_{x\in
X}B(\nu_{x}(b_{x}),\epsilon)$

Remember that $\psi$ is the map such that $\psi|_{\mathcal{F}(x)}=\nu_{x}$.
Now let $x\in X$, take $g\in\mathcal{F}(x)$ such that $d(g,b_{x})<\epsilon$,
since $\nu_{x}$ is a contraction we get
$d(\nu_{x}(g),\nu_{x}(b_{x}))<\epsilon$ . So

$g\in\psi^{-1}(\coprod_{x\in X}B(\nu_{x}(b_{x}),\epsilon)$

This proves that:

$\coprod_{x\in X}B(b_{x},\epsilon)\subseteq\psi^{-1}(\coprod_{x\in
X}B(\nu_{x}(b_{x}),\epsilon)$

And since $\coprod_{x\in X}B(b_{x},\epsilon)\in\mu$ then :

$\coprod_{x\in X}B(\nu_{x}(b_{x}),\epsilon)\in\psi\mu$

So $\psi\mu$ converges to $\psi(f)$ which completes the Functoriality proof.

##  4 The inverse functor

The first process we defined is a functorial assignment from the category of
left ultrafunctors from a fixed ultraset $X$ to the adequate category
$\mathsf{k}\text{-}\mathsf{CompMet}$.

Now we want to define an inverse process, a functorial assignment
$\mathcal{R}$ that sends a bundle with base space $X$ to a left ultrafunctor,
moreover, we claim that the pair $(\mathcal{L},\mathcal{R})$ is an equivalence
of categories between left ultrafunctors and bundles.

But first let us define $\mathcal{R}$ :

###  4.1 The inverse functor construction

Let $(E,X,\pi)$ be a bundle of complete metric spaces (bounded by some $k$)
and let $X$ be a compact Hausdorff space, Let $\mu$ be an ultrafilter that
converges to $x$. Our goal is to define a left ultrafunctor from $X$ to
$\mathsf{k}\text{-}\mathsf{CompMet}$.

Defining $\mathcal{R}(E)$ as a functor is straightforward: we send each $x$ to
the fibre at $x$, and this is a functor since $X$ has no non identity
morphism. Now we search an adequate left-ultrastructure on this functor, i.e
we wish to construct for every ultrafilter $\mu$ on some set $S$ and every map
of set $f$ from $S$ to $X$ a morphism $\sigma_{\mu}$ from
$F(\int_{S}f(s)d\mu)$ to $\int_{S}F(f(s))d\mu$, which satisfies certain axioms
indicated in [19]. We will see soon that it’s enough to take the case $S=X$
and $f=\mathrm{id}$.

Now we turn to the construction:

######  Definition 4.1.

Let $\mu$ be an ultrafilter on $X$ converging to $x$

For every $W$ neighbourhood of $f\in\pi^{-1}(x)$ define the following:

$A_{W}=\\{\ (b_{y})_{y\in X}\mid\exists U\in\mu,\epsilon>0\ \coprod_{y\in
U}B(b_{y},\epsilon)\subseteq W\ \\}$

The condition “$\exists U\in\mu,\epsilon>0\ \coprod_{y\in
U}B(b_{y},\epsilon)\subseteq W$” is well defined in the sense that it’s
independent of the representative of the class of $(b_{y})_{y\in X}$. Let us
call as usual $\mathcal{N}_{f}$ the set of open neighbourhoods of $f$.

######  Lemma.

The family $(A_{W})_{W\in\mathcal{N}_{f}}$ is a filter basis.

###### Proof.

We need to check that each set of this family is non-empty, and that the
intersection of any two contains a third. Let $W$ be a neighbourhood of $f$,
we want to show that $A_{W}$ is nonempty:
We know that there exist $\epsilon>0$ and an open set $V$ such that
$V\subseteq_{\epsilon}W$ using facts we showed in 3.4.3. Now take any family
$(b_{y})_{y\in\pi(V)}$ such that each $b_{y}\in V$. We already know that
$\pi(V)\in\mu$ since $\mu$ converges to $x$ and the map $\pi$ is open. Now let
us take $\coprod_{x\in\pi(V)}B(b_{x},\epsilon)$ then by our assumption that
$V\subseteq_{\epsilon}W$, we conclude that
$\coprod_{x\in\pi(V)}B(b_{x},\epsilon)\subseteq W$ thus the equivalence class
of the family $(b_{y})_{y\in\pi(V)}$ is in $A_{W}$ thus $A_{W}$ is non empty.
Finally it also clear that $A_{W}\bigcap A_{W^{{}^{\prime}}}=A_{W\bigcap
W^{{}^{\prime}}}$ ∎

######  Theorem 4.1.

the filter $\\{\ B\ \mid B\supseteq A_{W},\ W\in\mathcal{N}_{f}\\}$ is a
Cauchy filter

###### Proof.

Since $E$ is a bundle of complete metric spaces bounded by $k$, then the
hypothesis of lemma 3.5 is satisfied, meaning that there exists an
$\epsilon$-thin neighbourhood of $f$. Now take $L_{\epsilon}$ to be an
$\epsilon$-thin neighbourhood of $f$ then we can directly see that the set
$A_{L_{\epsilon}}$ is of diameter $\leq\epsilon$. ∎

Now since the ultraprdouct $\int_{X}\pi^{-1}(x)$ is complete the constructed
Cauchy filter converges to some element which we are going to denote by
$(b^{f}_{y})_{y\in X}$.

######  Theorem 4.2.

Define $\sigma_{\mu}(f)=(b^{f}_{y})_{y\in X}$ the limit of the Cauchy filter
generated by the family $(A_{W})_{W\in\mathcal{N}(f)}$, then the map
$\sigma_{\mu}$ is a contraction from $\pi^{-1}(x)$ to
$\int_{X}\pi^{-1}(x)d\mu$.

###### Proof.

Suppose that $f$ and $f^{{}^{\prime}}\in\pi^{-1}(x)$ are such that
$d(f,f^{{}^{\prime}})=\epsilon$, since that distance map is upper semi-
continuous there exists a neighbourhood $W_{1}$ of $f$ and a neighbourhood
$W_{2}$ of $f^{{}^{\prime}}$ such that $d(g,g^{{}^{\prime}})<\epsilon$ for
every $(g,g^{{}^{\prime}})\in W_{1}\times_{X}W_{2}$. Now take arbitrary
$\epsilon^{{}^{\prime}}>0$ then $B((b^{f}_{y}),\epsilon^{{}^{\prime}}/2)$
intersects any element of the Cauchy filter in particular $A_{W_{1}}$. Same
thing $B((b^{f^{{}^{\prime}}}_{y}),\epsilon^{{}^{\prime}}/2)$ intersects any
element of the second Cauchy filter converging to
$(b^{f^{{}^{\prime}}}_{y})_{y\in X}$ in particular $A_{W_{2}}$.

######  Lemma 4.1.

Let $f\in E$ and let $\mu$ be an ultrafilter on $X$ converging to $\pi(f)$,
suppose $\sigma_{\mu}(f)=(b_{y}^{f})_{y\in X}$ then for any $\epsilon>0$ and
any $W$ neighbourhood of $f$ if $(g_{y})_{y\in X}\in A_{W}\bigcap
B((b^{f}_{y}),\epsilon)$, there exists some $U\in\mu$ such that
$U\subseteq\pi(W)$ and such that $\forall y\in U$ $g_{y}\in W$ and
$d(g_{y},b^{f}_{y})<\epsilon$.

###### Proof.

Take $(g_{y})_{y\in X}\in A_{W}$, then there exists $U_{1}$ such that
$U_{1}\subseteq\pi(W)$ and $U_{1}\in\mu$ and $\exists\ r>0$ such that
$\coprod_{y\in U_{1}}B(g_{y},r)\subseteq W$. Now since $(g_{y})_{y\in X}\in
B((b^{f}_{y}),\epsilon)$ then there exists $U_{2}\in\mu$ such that
$d(b^{f}_{y},g_{y})<\epsilon$ thus $U=U_{1}\bigcap U_{2}$ will satisfy the
requirements above.

∎

Using the lemma 4.1 if $(g_{y})_{y\in X}\in A_{W_{1}}\bigcap
B((b^{f}_{y}),\epsilon^{{}^{\prime}}/2)$ then there exists some $U\in\mu$ and
$U\subseteq\pi(W_{1})$ such that $\forall y\in U$ $g_{y}\in W_{1}$ and
$d(g_{y},b^{f}_{y})<\epsilon^{{}^{\prime}}/2$. Same thing take $(h_{y})_{y\in
X}\in A_{W_{2}}\bigcap B((b^{f^{{}^{\prime}}}_{y}),\epsilon^{{}^{\prime}}/2)$
such that $\exists V\in\mu$ such that $V\subseteq\pi(W_{2})$ such that
$\forall y\in V$ $h_{y}\in W_{2}$ and
$d(g_{y},b^{f^{{}^{\prime}}}_{y})<\epsilon^{{}^{\prime}}/2$.

Now this would mean that for any $y\in V\bigcap U\in\mu$,
$d(b^{f^{{}^{\prime}}}_{y},b^{f}_{y})<\epsilon+\epsilon^{{}^{\prime}}$ and
since $\epsilon^{{}^{\prime}}$ was arbitrary then when passing to the
ultraproduct $d((b^{f^{{}^{\prime}}}_{y}),(b^{f}_{y})))\leq\epsilon$ so
$\sigma_{\mu}$ is a contraction (and we get for free that it is also
continuous).

∎

######  Note.

We gave the definition for the maps $\sigma_{\mu}$ for ultrafilters on $X$.
Now this can be easily extended to an arbitrary set $S$ as follows : if we
have an ultafilter $\mu$ on a set $S$ a map $M$ of sets from $S$ to $X$ then
if $M\mu$ converges to $x$ and $f$ in $\pi^{-1}(x)$, then if
$\sigma_{M\mu}(f)=(b^{f}_{y})_{y\in X}$ we define
$\sigma_{\mu}(f)=(b^{f}_{M(s)})_{s\in S}$. Notice that this is the only valid
way to define $\sigma_{\mu}$ for arbitrary $S$ to ensure that the following
diagram commutes:

${{F(x)}}$${{\int_{S}F(M_{s})d\mu}}$${{\int_{X}F(y)dM\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\sigma_{M\mu}}$$\scriptstyle{\Delta_{\mu,M}}$

For a proof that this diagram must commute for any left ultrafunctor from an
ultraset to an arbitrary ultracategory, check the end of section 7.

Now let us prove that this gives a Left ultrastructure on the functor
$x\mapsto\pi^{-1}(x)$(which means that we need to check that our definition
satisfies axioms$(0\text{-}1\text{-}2)$ of [19, definition 1]. Axiom $0$ is
automatic since the ultraset does not have any non-identity morphism so it
remains to check axioms $(1\text{-}2)$

##### Axiom 1

Suppose that we have a principal ultrafilter $\delta_{x}$ for some $x\in X$,
let $f$ in $\pi^{-1}(x)$, let us prove that the limit of the Cauchy filter
$\\{A_{W}\\}$ is converging to an element that belongs to the equivalence
class of $f$ which we are going by $(f)$, to do so take any $\epsilon>0$ and
take the open ball $B((f),\epsilon)$ now take any $\epsilon$ thin
neighbourhood $L_{\epsilon}$, now we have that $A_{L_{\epsilon}}\subseteq
B((f),\epsilon)$, thus the Cauchy filter is converging to the $(f)$, and this
also proves the case when we have an ultrafilter on a set $S$ and a map of
sets $M$ from $S$ to $x$ since $M\delta_{s}=\delta_{M(s)}$.

##### Axiom 2

First, let us do the case where we have a collection of ultrafilters on $X$:
Let $(\alpha_{s})_{s\in S}$ be a collection of filters on $X$ each converging
to $x_{s}$ (that means that they define a map $x:s\mapsto x_{s}$, so we will
be writing $x(s)$ instead of $x_{s}$) and let $\mu$ be an ultrafilter on $S$.
We need to check that the following diagram commutes:

${F(z)}$${\int_{X}F(y)d(\int_{S}\alpha_{s}d\mu)}$${\int_{S}F(x_{s})d\mu}$${\int_{S}(\int_{X}F(y)d\alpha_{s})d\mu}$$\scriptstyle{\sigma_{\int_{S}\alpha_{s}d\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\Delta_{\mu,\alpha\bullet}}$$\scriptstyle{\int_{S}\sigma_{\alpha_{s}}d\mu}$

Here $z$ denotes the limit of the ultrafilter $\int_{S}\alpha_{s}d\mu$. Now
suppose that $f\in F(z)$ and let
$\sigma_{\int_{S}\alpha_{S}d\mu}(f)=(a^{f}_{y})_{y\in X}$. By our description
of the categorical Fubini transform we have
$\Delta_{\mu,\alpha\bullet}(\sigma_{\int_{S}\alpha_{S}d\mu}(f))=((a^{f}_{y})_{y\in
X})_{s\in S}$.

Let $\sigma_{x\mu}(f)=(b^{f}_{y})_{y\in X}$ (which implies that
$\sigma_{\mu}(f)=(b^{f}_{x(s)})_{s\in S}$ using Note). Let
$\sigma_{\alpha_{s}}(b^{f}_{x(s)})=(c^{b^{f}_{x(s)}}_{y})_{y\in X}$. Our goal
is to show that
$\Delta_{\mu,\alpha\bullet}(\sigma_{\int_{S}\alpha_{S}d\mu}(f))=\int_{S}\sigma_{\alpha_{s}}(\sigma_{\mu}(f))$
which translates to saying that $((a_{y}^{f})_{y\in X})_{s\in
S}=((c^{b^{f}_{x(s)}}_{y})_{y\in X})_{s\in S}$.

Let $\epsilon>0$, take an $\epsilon/4$-thin open neighbourhood $W_{1}$ of $f$.
Now $A^{x\mu}_{W_{1}}$ must intersect $B((b^{f}_{y})_{y\in X},\epsilon/4)$
since the Cauchy filter converges to $(b^{f}_{y})_{y\in X}$ which implies that
$(b^{f}_{y})_{y\in X}$ is in the closure of every element in the filter. Thus
using 4.1 we can find an element $(h_{y})_{y\in X}$ and a set $U_{2}\in x\mu$
such that $U_{2}\subseteq\pi(W_{1})$ and $d(h_{y},b^{f}_{y})<\epsilon/4$ for
all $y\in U_{2}$ and such that $h_{y}\in W_{1}$ $\forall y\in U_{2}$. And also
$A^{\int_{S}\alpha_{s}d\mu}_{W_{1}}$ must intersect $B((a^{f}_{y})_{y\in
X},\epsilon/4)$ for the same reason. That means we can find an element
$(g_{y})_{y\in X}$ and a set $U_{1}\in\int_{S}\alpha_{s}d\mu$ such that
$U_{1}\subseteq\pi(W_{1})$ and $d(g_{y},a^{f}_{y})<\epsilon/4$ for all $y\in
U_{1}$ and such that $g_{y}\in W_{1}$. Now since
$U_{1}\in\int_{S}\alpha_{s}d\mu$ then the set $H=\\{s\in S\ :\
U_{1}\in\alpha_{s}\\}\in\mu$. And since $U_{2}\in x\mu$ then the set
$J=x^{-1}U_{2}\in\mu$. Now take $s\in J\bigcap H$. Since $s\in J$ then
$x_{s}\in U_{2}$ then $d(h_{x(s)},b^{f}_{x(s)})<\epsilon/4$.

Suppose that $\sigma_{\alpha_{s}}(h_{x(s)})=(k_{y})_{y\in X}$. Now since each
$\sigma_{\alpha_{s}}$ is a contraction then we can deduce that
$d((c^{b^{f}_{x(s)}}_{y})_{y\in X},(k_{y})_{y\in X})<\epsilon/4$, so there
exists some set $U_{3}\in\alpha_{s}$ such that
$d(c^{b^{f}_{x(s)}}_{y},k_{y})<\epsilon/4$ for every $y\in U_{3}$. We know
that, since $W_{1}$ is a neighbourhood of $h_{x(s)}$, the set
$A^{\alpha_{s}}_{W_{1}}$ must intersect any neighbourhood of $(k_{y})_{y\in
X}$, thus we deduce that there exists some $U_{4}\in\alpha_{s}$ and an element
$(l_{y})_{y\in X}$ such that $U_{4}\subseteq\pi(W_{1})$ and
$d(l_{y},k_{y})<\epsilon/4$ for all $y\in U_{4}$ and such that $l_{y}\in
W_{1}$ $\forall y\in U_{4}$ .

Now since $s\in H$ then $U_{1}\in\alpha_{s}$. Take $y\in U_{1}\bigcap
U_{3}\bigcap U_{4}\in\alpha_{s}$. We have

$d(a^{f}_{y},c^{b^{f}_{x(s)}}_{y})\leq\underbrace{d(a^{f}_{y},g_{y})}_{y\in
U_{1}}+\underbrace{d(g_{y},l_{y})}_{W_{1}\ \text{is}\ \epsilon/4\
\text{-thin}}+\underbrace{d(l_{y},k_{y})}_{y\in
U_{4}}+\underbrace{d(k_{y},c^{b^{f}_{x(s)}}_{y})}_{y\in U_{3}}<\epsilon,$

thus

$J\bigcap H\subseteq\\{s\in S\ d_{\alpha(s)}((c^{b^{f}_{x(s)}})_{y\in
X},(a_{y})_{y\in X})<\epsilon,\\}$

and since $J\bigcap H\in\mu$

$\\{s\in S\ d_{\alpha(s)}((c^{b^{f}_{x(s)}})_{y\in X},(a^{f}_{y})_{y\in
X})<\epsilon\\}\in\mu.$

This implies that $(a_{y}^{f})_{y\in X})_{s\in
S}=((c^{b^{f}_{x(s)}}_{y})_{y\in X})_{s\in S}$, and hence
$\Delta_{\mu,\alpha\bullet}(\sigma_{\int_{S}\alpha_{S}d\mu}(f))=\int_{S}\sigma_{\alpha_{s}}(\sigma_{\mu}(f))$
so the diagram commutes.

Now in the more general case when we have a family of ultrafilters
$(\alpha_{s})_{s\in S}$ on some set $T$ and a function $t\mapsto M_{t}$ from
$T$ to $X$. We need to prove that the following diagram commutes:

${F(z)}$${\int_{T}F(M(t))d(\int_{S}\alpha_{s}d\mu)}$${\int_{S}F(x_{s})d\mu}$${\int_{S}(\int_{T}F(M_{t})d\alpha_{s})d\mu}$$\scriptstyle{\sigma_{\int_{S}\alpha_{S}d\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\Delta_{\mu,\alpha\bullet}}$$\scriptstyle{\int_{S}\sigma_{\alpha_{S}}d\mu}$

To do so let $f\in F(z)$ suppose that
$\sigma_{M\int_{S}\alpha_{s}d\mu}=(a^{f}_{y})_{y\in X}$ then
$\sigma_{\int_{S}\alpha_{s}d\mu}=(a_{M(t)}^{f})_{t\in T}$, on the other hand
suppose that $\sigma_{\mu}(f)=(b^{f}_{s})_{s\in S}$. Now for each
$x_{s}=\int_{S}F(M(t))d\alpha_{s}=\int_{X}F(y)dM\alpha_{s}$, suppose
$\sigma_{M\alpha_{s}}(b_{s})=(c_{y}^{b^{f}_{x(s)}})_{y\in X}$, then
$\sigma_{\alpha_{s}}(b^{f}_{s})=(c_{M_{t}}^{b^{f}_{x(s)}})_{t\in T}$ then
$\int_{S}\sigma_{\alpha_{s}d\mu}((b_{s}^{f})_{s\in
S})=((c_{M_{t}}^{b^{f}_{x(s)}})_{t\in T})_{s\in S}$. We already proved that
$((a^{f}_{y})_{y\in X})_{s\in S}=((c_{y}^{b^{f}_{x(s)}})_{y\in X})_{s\in S}$.
We want to prove that $((c_{M_{t}}^{b^{f}_{x(s)}})_{t\in T})_{s\in
S}=((a_{M_{t}}^{f})_{t\in T})_{s\in S}$.

Let $\epsilon>0$, then the first equality means that $\\{s\in S\mid
d((a^{f}_{y})_{y\in X},(c_{y}^{b^{f}_{x(s)}})_{y\in X})<\epsilon\\}\in\mu$.
Now take any $s$ in the set above then since $d((a^{f}_{y})_{y\in
X},(c_{y}^{b^{f}_{x(s)}})_{y\in X})<\epsilon$ then the set $\\{y\mid
d(a^{f}_{y},c_{y}^{b^{f}_{x(s)}})<\epsilon\\}\in M\int_{S}\alpha_{s}d\mu$
which allows us to conclude that $\\{t\in T\mid
d(a^{f}_{M_{t}},c_{M_{t}}^{b^{f}_{x(s)}})<\epsilon\\}\in\int_{S}\alpha_{s}d\mu$
thus $d((a^{f}_{M_{t}}),(c_{M_{t}}^{b^{f}_{x(s)}}))<\epsilon$ so we can deduce
that $\\{s\in S\mid d((a^{f}_{M_{t}})_{t\in
T},(c_{M_{t}}^{b^{f}_{x(s)}}))<\epsilon\\}\in\mu$ which shows that
$((c_{M_{t}}^{b^{f}_{x(s)}})_{t\in T})_{s\in S}=((a_{M_{t}}^{f})_{t\in
T})_{s\in S}$.

###  4.2 Adjunction

We state a basic category theory fact:

######  Lemma 4.2.

Let $\mathcal{L}$ from $C^{{}^{\prime}}$ to $C$ be a functor and let
$\mathcal{R}$ be an assignment on objects from $C$ to $C^{{}^{\prime}}$ such
that $\mathrm{Hom}(\mathcal{L}(X),Y)\simeq\mathrm{Hom}(X,\mathcal{R}(Y))$ for
every object $X\in C^{{}^{\prime}}$ and $Y\in C$ such that this bijection is
natural in $X$. Then $\mathcal{R}$ has a functor structure defined as follows
:

The naturality in $X$ allows us to define a natural transformation $\epsilon$
from $\mathrm{Id}_{C}$ to $\mathcal{L}\mathcal{R}$ (which would be the counit
of adjunction), then if $\sigma\in\mathrm{Hom}(X,X^{{}^{\prime}})$ we define
$\mathcal{R}(\sigma)$ to be the unique map that corresponds to
$\epsilon_{X}\circ\sigma$ by this bijection.

We are going to apply this lemma in our case where $\mathcal{L}$ denotes the
functor from the category of left ultrafunctors between an ultraset and the
ultracategory $\mathsf{k}\text{-}\mathsf{CompMet}$ (with natural
transformations of left ultrafunctors as morphisms as defined in [19]) to
bundles over $X$ as we already defined it, and $\mathcal{R}$ is the reverse
assignment defined above.

######  Theorem 4.3.

Let $\mathcal{F}$ be left ultrafunctor and let $E$ be a bundle, then
$\mathrm{Hom}(\mathcal{L}(\mathcal{F}),E)\simeq\mathrm{Hom}(\mathcal{F},\mathcal{R}(E))$
and this bijection is natural in $\mathcal{F}$.

###### Proof.

Let $\nu$ be a morphism of bundles from $\mathcal{L}(\mathcal{F})$ to $E$ then
define a natural transformation from $\mathcal{F}$ to $\mathcal{R}(E)$ by
$\nu_{x}=\nu|_{\pi^{-1}(x)}$. Naturality is immediate since the category $X$
has no morphisms but identities. Now to check it is really a natural
transformation of left ultrafunctors: Let $\mu$ be an ultrafilter on a set $S$
and $M$ a map of sets from $S$ to $X$ (alternatively a family of points of $X$
indexed by $X$, $(M_{s})_{s\in S}$) such that $M\mu$ converges to $x$ we need
to check that the following diagram commutes :

${{\mathcal{F}(x)}}$${{\int_{S}\mathcal{F}(M_{s})d\mu}}$${{\mathcal{R}{E}(x)}}$${{\int_{X}\mathcal{R}(E)(M_{s})dM\mu}}$$\scriptstyle{\int_{S}\nu_{x}dM\mu}$$\scriptstyle{\sigma_{M\mu}}$$\scriptstyle{\sigma^{{}^{\prime}}_{M\mu}}$$\scriptstyle{\nu_{x}}$

First we observe that it is enough check this diagram in the case where $S=X$,
$M=\mathrm{id}$. Indeed consider the diagram:

${{\int_{S}\mathcal{F}(M_{s})d\mu}}$${{\mathcal{F}(x)}}$${{\int_{X}\mathcal{F}(y)dM\mu}}$${{\int_{S}\mathcal{R}(E)(M_{s})d\mu}}$${{\mathcal{R}{E}(x)}}$${{\int_{X}\mathcal{R}(E)(y)dM\mu}}$$\scriptstyle{\nu_{x}}$$\scriptstyle{\int_{X}\nu_{y}dM\mu}$$\scriptstyle{\sigma_{M\mu}}$$\scriptstyle{\sigma^{{}^{\prime}}_{M\mu}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\sigma^{{}^{\prime}}_{\mu}}$$\scriptstyle{\Delta_{\mu,M}}$$\scriptstyle{\Delta_{\mu,M}}$$\scriptstyle{\int_{S}\nu_{M_{s}}d\mu}$

Our goal is to show that the back square diagram commutes assuming the front
square diagram does (here $x$ is the limit if the ultrafilter $M\mu)$, notice
that the two triangles commute by the property Note. The side square commutes
by naturality of the ultraproduct diagonal map (it is easy to check that the
naturality condition for these maps follows from their definition (composition
of the (natural) categorical Fubini transform and the natural isomorphisms
$\epsilon$).

So we will be restricting our attention to ultrafilters on $X$, and we will be
checking the commutativity of the following diagram (again here $x$ is the
limit of the ultrafilter $\mu$):

${{\mathcal{F}(x)}}$${{\int_{X}\mathcal{F}(y)d\mu}}$${{\mathcal{R}{E}(x)}}$${{\int_{X}\mathcal{R}(E)(y)d\mu}}$$\scriptstyle{{\int_{S}\nu_{y}dM\mu}}$$\scriptstyle{{\sigma_{\mu}}}$$\scriptstyle{{\sigma^{{}^{\prime}}_{\mu}}}$$\scriptstyle{{\nu_{x}}}$

Now take $f\in\mathcal{F}(x)$ and suppose $\nu_{x}(f)=g$ and
$\sigma_{\mu}(f)=(b_{y})_{y\in X}$, Our goal is to show that
$\sigma^{{}^{\prime}}_{\mu}(g)=(\nu_{y}(b_{y}))_{y\in X}$. Suppose that
$\sigma^{{}^{\prime}}_{\mu}(g)=(c^{g}_{y})_{y\in X}$, let $W$ be an
$\epsilon/2$-thin neighbourhood of $g$ then by definition of
$(c^{g}_{y})_{y\in X}$ $A_{W}$ must intersect any neighbourhood of
$(c^{g}_{y})_{y\in X}$ in particular $B((c^{g}_{y})_{y\in X},\epsilon/2)$ thus
there exists $L\in M\mu$ and $(f_{y})_{y\in X}$ such that $\forall y\in L$
each $f_{y}\in W$ and $d(f_{y},c^{g}_{y})<\epsilon/2$. Now since $\nu^{-1}(W)$
is a neighbourhood of $f$ then there exists $H\in M\mu$ and
$\epsilon{{}^{\prime}}>0$ such that $\coprod_{y\in
H}B(b_{y},\epsilon^{{}^{\prime}})\subseteq\nu^{-1}(W)$. Thus for any $y\in H$
we get that $\sigma_{y}(b_{y})\in W$. Now take $y\in H$ (remember that $H\in
M\mu$) then we have $d(\sigma_{y}(b_{y}),c_{y}^{g})\leq
d(\sigma_{y}(b_{y}),f_{y})+d(f_{y},c^{g}_{y})<\epsilon/2+\epsilon/2=\epsilon$.
Thus we get that $(\sigma_{y}(b_{y}))_{y\in X}=(c_{y}^{g})_{y\in X}$, and this
terminates the proof showing the commutativity of the diagram above.

Conversely suppose that we have a natural transformation $\nu$ of left
ultrafunctors from $\mathcal{F}$ to $\mathcal{R}(E)$, we need to show that the
map $\nu$ defined by $\nu|_{\pi^{-1}(x)}=\nu_{x}$ is a continuous map from
$\mathcal{L}(\mathcal{F})$ to $E$ (since the other requirements for being a
map of bundles are automatically satisfied). To do so suppose that $\mu$ is an
ultrafilter on $\mathcal{L}(\mathcal{F})$ that converges to $f$, now to prove
that $\sigma\mu$ converges to $\nu(f)=\nu_{\pi(f)}(f)$ in $E$: We know that
since $\nu$ is a natural transformation of left ultrafunctors then if
$\sigma_{\mu}(f)=(b_{y})_{y\in X}$ then
$\sigma^{{}^{\prime}}_{\mu}(\nu(f))=(\nu_{y}(b_{y}))_{y\in X}$. Also we know
that $\coprod_{y\in X}B(b_{y},\epsilon)\in\mu$ (by definition of the topology
f of $\mathcal{L}(\mathcal{F})$). Now since each $\nu_{y}$ is a contraction
then

$\coprod_{y\in X}B(b_{y},\epsilon)\subseteq\nu^{-1}\coprod_{y\in
X}(\nu_{y}(b_{y}),\epsilon),$

thus

$\coprod_{y\in X}(\nu_{y}(b_{y}),\epsilon)\in\nu\mu.$

Thus we have a map of bundles from $\mathcal{L}(\mathcal{F})$ to $E$. Also, it
is clear that these two processes between
$\mathrm{Hom}(\mathcal{L}(\mathcal{F}),E)$ and
$\mathrm{Hom}(\mathcal{F},\mathcal{R}(E))$ are inverse of each other. Now it
remains to show naturality in $\mathcal{F}$.

######  Notation Convention.

If we have a map $\psi$ in $\mathrm{Hom}(\mathcal{L}(\mathcal{F}),E)$, we will
denote $\widehat{\psi}$ the corresponding map in
$\mathrm{Hom}(\mathcal{F},\mathcal{R}(E))$, conversely if we have a map
$\kappa$ in $\mathrm{Hom}(\mathcal{F},\mathcal{R}(E))$ then we are going to
denote by $\bar{\kappa}$ the corresponding map in
$\mathrm{Hom}(\mathcal{L}(\mathcal{F},E)$.

Now to do the proof let $\nu$ be a natural transformation of left
ultrafunctors from $\mathcal{F^{{}^{\prime}}}$ to $\mathcal{F}$, we need to
show that the following diagram commutes :

${{\mathrm{Hom}(\mathcal{L}(\mathcal{F}),E)}}$${{\mathrm{Hom}(\mathcal{F},\mathcal{R}(E))}}$${{\mathrm{Hom}(\mathcal{L}(\mathcal{F^{{}^{\prime}}}),E)}}$${{\mathrm{Hom}(\mathcal{F^{{}^{\prime}}},\mathcal{R}(E))}}$$\scriptstyle{-\circ\mathcal{L}(\nu)}$$\scriptstyle{-\circ\nu}$

To do so consider a map $\psi$ of bundles from $\mathcal{L}(\mathcal{F})$ to
$E$, we need to show
$\widehat{\psi\circ\mathcal{L}(v)}=\widehat{\psi}\circ\nu$. To do so let $x\in
X$ and let $f\in\mathcal{F^{{}^{\prime}}}(x)$ then

$\widehat{(\psi\circ\mathcal{L}(v))}_{x}(f)=(\psi\circ\mathcal{L}(\nu))(f)=\psi(\nu_{x}(f)),$

on the other hand

$(\widehat{\psi}\circ\nu)_{x}(f)=(\widehat{\psi}_{x}\circ\nu_{x})(f)=\widehat{\psi}_{x}((\nu_{x})(f))=\psi(\nu_{x}(f)),$

so for each $x$

$(\widehat{(\psi\circ\mathcal{L}(v)})_{x}=(\widehat{\psi}\circ\nu)_{x},$

so

$\widehat{(\psi\circ\mathcal{L}(v)}=(\widehat{\psi}\circ\nu),$

so the diagram commutes. ∎

A last thing that we should enlighten is that the functor structure of
$\mathcal{R}$ comes from the adjunction, suppose that we have a map of bundles
$\nu:\ E\rightarrow E^{{}^{\prime}}$, we defined
$R(\nu)=\widehat{(\epsilon_{E}\circ\nu)}$, where $\epsilon_{E}$ is the counit
of adjunction. We are going to give a better description of this map once we
prove that the counit is an isomorphism.

Now we turn into showing our main theorem, in what follows
$\mathrm{Bun}(\mathsf{k\text{-}CompMet},X)$ denotes the category of bundles
with base space $X$.

######  Theorem 4.4.

Let $X$ be compact Hausdorff space then the pair of functors $\mathcal{R}$ and
$\mathcal{L}$ consist and equivalence of categories between $\mathrm{Left\
ultrafunctors}(X,\mathsf{k\text{-}CompMet})$ and
$\mathrm{Bun}(\mathsf{k\text{-}CompMet},X)$

####  4.2.1 The counit is an isomorphism

Let

$\epsilon_{E}:\ \mathcal{L}\mathcal{R}(E)\mapsto E$

be the counit of adjunction.

######  Theorem 4.5.

for every bundle $E$, $\epsilon_{E}$ is a homeomorphism.

###### Proof.

It is clear that $\epsilon_{E}$ is a bijection on sets so it remains to show
that $E$ and $\mathcal{L}\mathcal{R}(E)$ have the same topology.

##### The topology of $E$ is coarser than $\mathcal{L}\mathcal{R}(E)$

First we already get that $\epsilon_{E}$ is continuous from
$\mathcal{L}\mathcal{R}(E)$ to $E$ (by the fact that $\epsilon_{E}$ is a
counit which implies it’s a map of bundles).

##### The topology of $E$ is finer than $\mathcal{L}\mathcal{R}(E)$

For the other direction suppose that $\mu$ is an ultrafilter on $E$ that
converges to $f$, we need to prove that $\mu$ converges to $f$ in the topology
of $\mathcal{L}\mathcal{R}(E)$. Suppose that
$\sigma_{\pi\mu}(f)=(b_{y}^{f})_{y\in\epsilon}$, we need to show that for any
$\epsilon>0$ the set $\coprod_{y\in X}B(b_{y}^{f},\epsilon)\in\mu$, to do this
take an $\epsilon/2$-thin neighbourhood $W$ (in the topology of $E$ of course)
of $f$. Now we know that $A_{W}$ must intersect any neighbourhood of
$(b_{y})_{y\in X}$ in particular $B((b_{y})_{y\in
X},\epsilon^{{}^{\prime}}/2)$, thus there exists $L\in\pi\mu$ and
$(c_{y})_{y\in X}$ such that for each $y\in L$ $c_{y}\in W$ and
$d(c_{y},b_{y}^{f})<\epsilon/2$, so $\pi^{-1}L\in\mu$. On the other hand,
$W\in\mu$ since $\mu$ converges to $f$ in the first topology (topology of
$E$).

Now let us prove that $W\bigcap\pi^{-1}L\subseteq\coprod_{y\in
X}B(b^{f}_{y},\epsilon)$, take $g\in W\bigcap\pi^{-1}L$ then
$d(g,b_{\pi(g)})<d(g,c_{\pi(g)})+d(c_{\pi(g)},b_{\pi(g)})<\epsilon$. Thus
$\coprod_{y\in X}B(b^{f}_{y},\epsilon)\in\mu$. So by the definition of the
topology of $\mathcal{L}\mathcal{R}(E)$ $\mu$ converges to $f$. Thus the two
topologies coincide and $E$ is isomorphic to $\mathcal{L}\mathcal{R}(E)$ as
bundles.

∎

This allows us to describe better how $\mathcal{R}$ acts on morphisms, suppose
that we have a map of bundles $\nu\ E\rightarrow E^{{}^{\prime}}$. Then
$R(\nu)=\widehat{(\epsilon_{E}\circ\nu)}$. More precisely from the fact that
$\epsilon_{E}$ is an isomorphism we get that $R(\nu)_{x}(f)=\nu(f)$ for
$f\in\mathcal{R}(E)(x)$ (which is exactly the thing we expected it to be).

####  4.2.2 The unit is an isomorphism

Now to prove that the unit is an isomorphism consider :

$\eta_{F}:\ \mathcal{F}\mapsto\mathcal{R}\mathcal{L}(\mathcal{F})$

Now the two left ultrafunctors from $X$ to
$\mathsf{k}\text{-}\mathsf{CompMet}$ are the same thing as functors, it
remains to show that they have the same left ultrastructure. but this
immediately follows from $\eta_{F}$ being a natural transformation of left
ultrafunctors which is an isomorphism for every $x\in X$.

####  4.2.3 A nice property of bundles

######  Theorem 4.6.

Let $E$ be a bundle of complete bounded metric spaces, then the induced
topology and the complete metric space topology agree on every fibre.

###### Proof.

Let $W_{x}$ be an open set in the subspace topology of the fibre $E_{x}$ for
some $x\in X$, and let $f\in E_{x}$, there exists an open set $W$ of $E$ such
that $W_{x}=E\bigcap E_{x}$, now we know that there exists a set $V$ such that
$f\in V\subseteq_{\epsilon}W$. In other words $f\in V\subseteq
V_{\epsilon}\subseteq W$. Now by definition of $V_{\epsilon}$
$B(f,\epsilon)\subseteq V_{\epsilon}\bigcap E_{x}$ hence $W_{x}$ is open in
metric topology.

On the other hand, let $(b_{i})$ be a net of elements in $E_{x}$ that
converges to $b$ in the topology of $E$, we need to show that $(b_{i})$
converges to $b$ in metric topology, to do so consider the net $(b_{i},b)\in
E\times_{X}E$, this net converges to $(b,b)$ which satisfy $d(b,b)=0$, by
upper semi-continuity of the distance, for every $\epsilon>0$, there exists a
neighbourhood $W$ in $E\times_{X}E$ and some $i_{0}$ such that every two
points in the same fibre in $W$ have distance $\leq\epsilon$, and such that
for any $i\succ i_{0}$ $(b_{i},b)\in W$, thus $(b_{i})$ converges to $b$ in
metric topology. So both topologies on $E_{x}$ agree. The proof in the last
paragraph is inspired by a similar one in [10]. ∎

####  4.2.4 Another construction of the left-ultrastructure of
$\mathcal{R}(E)$

We give another construction of the left ultrastructure of $\mathcal{R}(E)$
for a bundle $E$, that works only when the bundle $E$ has enough cross
sections.

######  Theorem.

let $\mathcal{M}=\coprod_{x\in X}\mathcal{M}_{x}$ be a bundle of complete
metric spaces bounded by a certain $k$, and let $\mathcal{F}$ be the left
ultrafunctor $x\mapsto\mathcal{M}_{x}$, then for any $x\in X$, if $\mu$ be an
ultrafilter on $X$ converging to $x$, and if $a:U\rightarrow\mathcal{M}$ is a
local continuous section to the projection map $\pi$ (here $U$ open in $X$)
then we claim that $\sigma_{\mu}(f)=(a(x))_{x\in U}$.

######  Note.

It is enough to define a member of the ultraproduct on $U\in\mu$

###### Proof.

Let $W$ be an open neighbourhood of $f$, by continuity of $a$ the ultrafilter
$a\mu$ converges to $f$, that means that for any $\epsilon>0$ $\coprod_{x\in
U}B(a(x),\epsilon)\in\mu$, hence $(a(x))_{x\in U}\in A_{W}$, and the Cauchy
filter converges to $(a(x))_{x\in U}$. ∎

##  5 Generalising to any structure

In this section, we give the construction of bundles of structures of
continuous model theory, a structure of continuous model theory is an
interpretation of sorts, relation, and function symbol not required to satisfy
any axiom. this is a necessary intermediate step before defining bundles of
models of continuous model theory. But we give a necessary introduction to
continuous model theory.

###  5.1 The Ultracategory of Models

We first recall a few concepts from continuous model theory, this exposition
follows mostly [12] and [17].

####  5.1.1 Signature

The signature of continuous model theory consists of the following triplet
$\langle\mathfrak{S},\mathfrak{F},\mathfrak{R}\rangle$ where

  1. 1.

$\mathfrak{S}$ the set of _sorts symbols_ , such that each symbol comes
equipped with a symbol $d_{S}$(should be interpreted as the distance function)
and a constant $k_{S}$ (actual constant not just a symbol) (which should be
interpreted as an upper bound for the distance function).

  2. 2.

$\mathfrak{F}$ the family of _function symbols_ , and for each symbol $f$ we
specify a formal domain $\mathrm{dom}(f)=(S_{1},\dots,S_{n})$ and a formal
range $\mathrm{rng}(f)=S^{{}^{\prime}}$ and a function $\delta_{f}$ which
should be interpreted as the uniform continuity modulus of $f$.

  3. 3.

$\mathfrak{R}$ the set of _relation symbols_ , each equipped with a compact
interval of $\mathbb{R}$(which should be interpreted as the range of these
relations) as well as a uniform continuity modulus $\delta_{\phi}$ for every
$\phi\in\mathfrak{R}$.

######  Note.

we can (and we are going to) treat the distance symbol as a relation symbol.

Now we are in a position to define terms and formulae in continuous model
theory:
As usual the definition is inductive starting by first considering infinitely
many variable symbols for each sort $x_{i}^{S}$ as terms, after that if
$t_{1},\dots,t_{n}$ are terms of sorts $S_{1},\dots,S_{n}$ and $f$ is a
function symbol with range $S^{{}^{\prime}}$ then $f(t_{1},\dots,t_{n})$ is a
term of sort $S^{{}^{\prime}}$. All terms get uniform continuity moduli
inductively. An example of a term is $x^{*}x$ in the language of
$\mathrm{C}^{*}$ algebras (to be more precise to specify the sort in that
language but we will make this more clear in the examples section)
Now for formulae: First we consider atomic formulae: these are defined using
relations symbols,i.e if $t_{1},\dots,t_{n}$ are terms of range
$S_{1},\dots,S_{n}$and $\phi$ is a relation symbol then
$\phi(t_{1},\dots,t_{n})$ is a formula.
Now connectives are just continuous function from $\mathbb{R}^{n}$ to
$\mathbb{R}$ so if $f$ is such a function and $t_{1},..t_{n}$ are connectives
then $f(t_{1},\dots,t_{n})$ is a formula.
Finally we consider quantifiers: if $\phi$ is a formula and
$x^{S}_{i}\in\mathrm{FV}(\phi)$ then $\mathrm{Sup}_{x^{S}_{i}\in S}\phi$ and
$\mathrm{Inf}_{x^{S}_{i}\in S}\phi$ are both formulae. A formula with no free
variable is called a sentence. Again formulae inherit uniform continuity
moduli by their inductive construction. An example of such formulae in the
language of $\mathrm{C}^{*}$ algebras would be: $x^{*}x+2$, $x^{*}yx$,
$\mathrm{Sup}_{x}x^{*}x+y^{*}y$ …

######  Note.

Here free variables of a formula are defined the same way as in the case of
regular model theory.

####  5.1.2 Structures and Models

A $\mathfrak{L}$-structure is triplet
$M=\langle\mathcal{S},\mathcal{F},\mathcal{R}\rangle$ such that for each
symbol in $S$ $\mathfrak{S}$ there is a complete metric in $\mathcal{S}$
bounded by $k_{S}$, and each element of $\mathcal{F}$ is a function $f^{M}$
that corresponds to some element $f\in\mathfrak{F}$ such that if
$\mathrm{dom}(f)$ with formal domain $(S_{1},\dots,S_{n})$ and with formal
range $S^{{}^{\prime}}$ then its interpretation $f^{M}$ has domain
$M^{S_{1}}\times\dots\times M^{S_{N}}$ and range $M^{S^{{}^{\prime}}}$, such
that this function is uniformly continuous with uniform continuity modulus
$\delta_{f}$.

The elements of $\mathcal{R}$ are the interpretations of the relations
symbols, that means that if $\phi\in\mathcal{R}$ with formal domain
$(S_{1},\dots,S_{n})$ and range $B$ a compact interval of $\mathbb{R}$ then
it’s interpretation is a function $\phi^{M}$ with domain
$M^{S_{1}}\times\dots\times M^{S_{N}}$ and with range $B$ such that this
function is uniformly continuous with uniform continuity modulus
$\delta_{\phi}$. In the same manner, we interpret terms and formulae.

Now let $M$ be an $\mathfrak{L}$-structure, and let $\mathbb{T}$ be a set of
sentences in the language $\mathfrak{L}$ then we say that $M$ is a model of
$\mathbb{T}$ if for every $\psi\in\mathbb{T}$ $\psi^{M}=0$, and in this case
we write $M\models\mathbb{T}$. We say that $\mathbb{T}$ is consistent if it
has a model. Notice that if we take $\mathbb{T}=\varnothing$ then its models
in this case are exactly $\mathfrak{L}$-structures.

Let $\mathfrak{L}=\langle\mathfrak{S},\mathfrak{F},\mathfrak{R}\rangle$ be a
Language (or signature, or similarity type), and let $\mathbb{T}$ be family of
sentences in the language $\mathfrak{L}$, we are going to denote by
$\mathrm{CompMet}_{\mathfrak{L}}$ the category of structures of $\mathfrak{L}$
and by $\mathrm{CompMet}_{\mathfrak{L},\mathbb{T}}$ the full subcategory of
models of $\mathbb{T}$. To be more precise we should specify what a morphism
in this category: let $M$ and $N$ be two models then a morphism of models $g$
is a family of morphisms $g^{S}$ for each sort (we will omit the superscript
if the context is clear) iff for every function symbol $f$ with domain
$(S_{1},\dots,S_{n})$ and with range $S^{{}^{\prime}}$, we have that for every
$(a_{1},\dots,a_{n})\in M^{S_{1}}\times\dots\times M^{S_{n}}$ we have
$f^{N}(g^{S_{1}}(a_{1}),\dots,g^{S_{n}}(a_{n}))=g^{S^{{}^{\prime}}}(f^{M}(a_{1},\dots,a_{n}))$.
And for every relation symbol with domain $(S_{1},\dots,S_{n})$ we have the
following
$\phi^{N}(g^{S_{1}}(a_{1}),\dots,g^{S_{n}}(a_{n})\leq\phi^{M}(a_{1},\dots,a_{n})$.

One important particular case of this is when we have only one sort $S$ and
only one relation(the distance relation on this sort), in this case we get a
category equivalent to the category of complete metric spaces bounded by a
certain $k$ with contractions as morphisms, which we denoted by
$\mathsf{k}\text{-}\mathsf{CompMet}$.

####  5.1.3 Ultraproducts and models

In all the previous cases the ultraproduct construction given explicitly in
[12] and [17], makes these categories ultracategories. We think it’s important
to highlight this construction, which is similar to the ultraproduct
construction in usual model theory (after all these are just direct colimits
of products). Of course we assume that the reader is at this point familiar
with the ultraproduct of metric spaces bounded by a certain constant.

Suppose we have a similarity type
$\mathfrak{L}=\langle\mathfrak{S},\mathfrak{F},\mathfrak{R}\rangle$ and a
family of structures $(V^{i})_{i\in I}$ of that similarity type, we define
their ultraproduct as follows for each sort $\mathcal{S}\in\mathfrak{S}$ we
define $(\int_{I}V^{i}d\mu)_{\mathcal{S}}$ by
$(\int_{I}V^{i}d\mu)_{\mathcal{S}}=\int_{I}V^{i}_{\mathcal{S}}d\mu$. Now for
relation symbol we define $\mathscr{R}_{\int_{I}V^{i}d\mu}((a_{i}^{m})_{i\in
I})_{m\in M}=\lim_{\mu}(\mathscr{R}_{V^{i}}(a_{i}^{m}))$. Here $\lim_{\mu}$ is
the ultralimit in $[0,\infty)$( Remember that the family
$(\mathscr{R}_{V^{i}}(a_{i}^{m})))$ is bounded so we can replace $[0,\infty)$
by $[0,k]$ and define this as the limit of the push forward of the ultrafilter
$\mu$ by the map $(a_{i}^{m})_{i\in
I}\mapsto\mathscr{R}_{V^{i}}(a_{i}^{m})_{m\in M}$).Now for function symbols,
things are the same as in usual model theory. That means that for
$f\in\mathfrak{F}$, we define
$f_{\int_{I}(V^{i})d\mu}((a^{i}),\dots(z^{i}))=(f(a^{i},\dots,z^{i}))$. The
fact that models are closed under taking this construction above follows Łos
theorem, an important result in classic model theory is Łos theorem which has
a version in continuous model theory:

######  Theorem 5.1 (Łos).

for any formula $m$ and for any family of structures $\\{M_{x}\\}_{x\in X}$ if
$\mu$ is n ultrafilter on $X$,if we call $M=\int_{X}M_{x}d\mu$ then we have
the following : for any formula $\phi$ and any $\bar{m}=(m_{x})_{x\in X}$ we
have $\phi^{M}(\bar{m})=\int_{X}\phi^{M_{x}}(m_{x})d\mu$.

We see clearly that a similar version of classical Łos theorem is a
consequence of the theorem above if $\phi$ is a sentence and if for every
$x\in X$ $M_{x}\models\phi$ (which is the same thing as saying that
$\phi^{M_{x}}=0$ ) then $\int_{X}M_{x}d\mu\models\phi$.

###  5.2 Bundles of structures

We define what it means to be a bundle of structures:

######  Definition 5.1.

Let $\mathfrak{L}=\langle\mathfrak{S},\mathfrak{F},\mathfrak{R}\rangle$ be a
language, we define a bundle of structures $E$ of that Language with base
space $X$, to be a family of bundles of complete bounded metric spaces
$\\{(E_{S},\pi_{S})\\}_{S\in\mathfrak{S}}$ equipped with with a family of maps
$\\{\pi_{s}\\}_{S\in\mathfrak{S}}$ such that for any $x\in X$,
$(\pi_{S}^{-1}(x))_{S\in\mathfrak{S}}$ is a structure of the language
$\mathfrak{L}$ (so in particular it comes with the function and relation
symbols data) such that the following axioms are satisfied :

For any function symbol $f$ with formal domain $\mathrm{dom}(f)$ and formal
range $\mathrm{rng}(f)$ and for any relation symbol $\phi$ with formal domain
$\mathrm{dom}(\phi)$ we are going to denote by $f^{E}$ and $\phi^{E}$ the
global function and relations respectively (so for any $x\in X$ $f^{E}$
restricts to the interpretation of the function symbol $f$ of
$(\pi_{S}^{-1}(x))_{S\in\mathfrak{S}}$, same thing for relation symbols).

  * •

Axiom(1): Each $\pi_{S}$ is continuous and open.

  * •

Axiom(2): for every $E^{S}$ and for every open set $W\in E_{S}$ and every
$e\in W$ there exist an open set $V$ open neighbourhood of $e$ and
$\epsilon>0$ such that $V\subseteq_{\epsilon}W$.

  * •

Axiom(3): For every function symbol $f$ the map $f^{E}$ is continuous

  * •

Axiom(4): For every relation symbol $\phi$ the map $\phi^{E}$ is upper semi-
continuous.

######  Note.

As stated before we are going to treat the distance symbols as relation
symbols on structures, so in particular, axiom $(4)$ says that the global
distance functions are also upper semi-continuous.

######  Note.

We are going to denote by $E^{S}$ the bundle of structures that corresponds to
a sort $S$ and by $E_{x}$ the fibre over $x$ which is a structure, So
following this convention $E^{S}_{x}$ is the $x$-th fibre of the bundle of
structure corresponding to the sort $S$.

###  5.3 Maps of bundles

Let $E$ and $E^{{}^{\prime}}$ be a two bundles, a morphism $\psi$ in the
category of bundles consists of the following:

For each sort $S$, a map of bundles of bounded metric spaces $\psi^{S}$
between the bundles $E^{S}$ and ${E^{{}^{\prime}}}^{S}$ such that the
following diagram commutes(in $\mathsf{Top}$) :

${E^{S}}$${{E^{{}^{\prime}}}^{S}}$${X}$$\scriptstyle{\psi^{S}}$$\scriptstyle{\pi_{S}}$$\scriptstyle{\pi_{S}}$

and such that for any $x$ $\psi_{x}$ is a map of structures of the language
$\mathfrak{L}$ from $E_{x}$ to $E^{{}^{\prime}}_{x}$.

Now We want to extend the equivalence obtained in section 3(theorem 4.4) to
structures of continuous model theory:
In Other words we want to show the following:

######  Theorem 5.2.

Let $X$ be a compact Hausdorff space then there is an equivalence of
categories between
$\mathrm{Left\ ultrafunctors(X,\mathsf{CompMet}_{\mathfrak{L}})}$ and the
category $\mathrm{Bun}(\mathsf{CompMet}_{\mathfrak{L}},X)$

The rest of section $5$ is devoted to showing the theorem above 5.2.

In order to define this equivalence of categories we are going to expand the
definitions of the functors $\mathcal{L}$ and $\mathcal{R}$ already defined to
the categories above.

###  5.4 The functor $\mathcal{L}$

Let $X$ be a compact Hausdorff space and suppose we have $\mathcal{F}$ a left
ultrafunctor from $X$ to the ultracategory of structures of some language
$\mathfrak{L}$. We know that the functor $\mathcal{F}$ will give rise to a
family of functors $\mathcal{F}^{S}$ for each sort $S$. If we define each
mono-sorted bundle $E_{S}$ to be $\coprod_{X\in X}\mathcal{F}^{S}(x)$ with its
bundle topology given in 3.1, then we have already seen that the first three
axioms are satisfied for this multi-sorted bundle.

It remains to check axioms $3$ and $4$ :

####  5.4.1 Axiom $3$ (Functions)

To prove that axiom $3$ is satisfied by our definition of multi-sorted bundle
suppose that $f$ is a function symbol, and suppose that
$\mathrm{dom}(f)=S_{1}\times\dots\times S_{n}$ and
$\mathrm{rng}(f)=S^{{}^{\prime}}_{1}$. Suppose that $\mu$ is an ultrafilter on
$E^{S_{1}}\times_{X}\dots E^{S_{n}}$(in the case where we have a constant
symbol this space is $X$ the $0$-th product in $\sf{Top}/X$) that converges
converges to $(a^{1},\dots,a^{n})\in M_{y}^{S_{1}}\times\dots\times
M_{y}^{S_{i}}\subseteq E^{S_{1}}\times_{X}\dots\times_{X}E^{S_{n}}$, and
suppose that $f^{M_{y}}(a^{1},\dots,a^{n})=a^{{}^{\prime}}$. Now suppose that
for each $S_{i}$ $\sigma_{\pi_{S_{i}}\mu}^{S_{i}}(a^{i})=(b_{x}^{i})_{x\in
X}$. Now since $\sigma_{\pi_{S_{i}}\mu}$ is a map of $\mathfrak{L}$ structures
we get that
$\sigma_{\pi_{S_{i}}\mu}^{S^{{}^{\prime}}_{1}}(a^{{}^{\prime}})=f^{M}((b_{x}^{1}),\dots,(b_{x}^{n}))$
so we may use $(f^{M_{x}}(b_{x}^{1},\dots,b_{x}^{n}))_{x\in X}$ as
representative of the class of
$\sigma_{\pi_{S_{i}}\mu}^{S_{i}}(a^{{}^{\prime}})$.(this by definition of the
structure of the ultraproduct) (in the case of constant symbol $c$ of sort
$S^{{}^{\prime}}_{1}$ we use $(c_{x})_{x\in X}$ as representative of its
class).

Now let $\epsilon>0$. We know that for any $x$, $f^{M_{x}}$ is uniformly
continuous with uniform continuity modulus independent of $x$, thus we can
deduce that there exist some $\delta$ such that if
$d(m_{x}^{i},b_{x}^{i})<\delta$, we have that
$|f(m_{x}^{1},\dots,m_{x}^{n})-f(b_{x}^{1},\dots,b_{x}^{n})|<\epsilon$. We
want to show that that $f^{M_{y}}\mu$ converges to $a^{{}^{\prime}}$ : We have
that $\coprod_{x\in X}B(b_{x}^{i},\delta)\in\pi_{S_{i}}\mu$. Now take the
following set $\bigcap_{i=1}^{n}\pi_{S_{i}}^{-1}(\coprod_{x\in
X}B(b_{x}^{i},\delta))\in\mu$. If we take
$(l_{1},\dots,l_{n})\in\bigcap_{i=1}^{n}\pi_{S_{i}}^{-1}(\coprod_{x\in
X}B(b_{x}^{i},\delta))$, and suppose that $\pi_{S_{i}}(l_{i})=z$ we have the
following :

$|f^{E}(l_{1},\dots,l_{n})-f^{E}(b_{z}^{1},\dots,b_{z}^{n})|=|f^{M_{z}}(l_{1},\dots,l_{n})-f^{M_{z}}(b_{z}^{1},\dots,b_{z}^{n})|<\epsilon$

then this set satisfies

$\bigcap_{i=1}^{n}\pi_{S_{i}}^{-1}(\coprod_{x\in
X}B(b_{x}^{i},\delta))\subseteq(f^{E})^{-1}(\coprod_{x\in
X}B(f(b_{x}^{1},\dots,b_{x}^{n}),\epsilon))$

thus we get that

$\coprod_{x\in X}B(f(b_{x}^{1},\dots,b_{x}^{n}),\epsilon)\in f^{E}\mu$

thus $f^{E}$ is continuous (In the case we have a constant symbol we have that
$\coprod_{x\in X}B(c_{x},\epsilon)\in c^{E}\mu$ trivially since
$(c^{E})^{-1}\coprod_{x\in X}B(c_{x},\epsilon)=X\in\mu$).

####  5.4.2 Axiom $4$ (Relations)

We are going to denote $\mathcal{F}(x)$ by $M_{x}$. Now we want to prove that
the family of $S$-bundles for $S\in\mathfrak{S}$ satisfies the upper semi-
continuity for each global relation. To do so suppose that $\phi$ is a
relation symbol, and now suppose that $\mu$ is an ultrafilter on $X$ that
converges to $y$, now suppose that $\mathrm{dom}(\phi)=S_{1}\times\dots\times
S_{n}$. From this point forward let us denote by $M$ the ultraproduct
$\int_{X}M_{x}d\mu$.

Let us prove that $\phi^{E}$ is upper semi-continuous: Let $\mu$ be an
ultrafilter on $E^{S_{1}}\times_{X}\dots\times_{X}E^{S_{n}}$ such that $\mu$
converges to $(a^{1},\dots,a^{n})\in M_{y}^{S_{1}}\times\dots\times
M_{y}^{S_{n}}\subseteq E^{S_{1}}\times_{X}\dots\times_{X}E_{S_{n}}$ (for some
$y\in X$) and take $r>0$ such that $\phi^{M_{y}}(a^{1},\dots,a^{i})<r$. Now we
call the quantity $r-\phi^{M_{y}}(a^{1},\dots,a^{n})=\epsilon$. Notice that
for any $i,j$ we get that $\pi_{S_{i}}\mu=\pi_{S_{j}}\mu$ is the same
ultrafilter on $X$, so we’ll call this ultrafilter $\pi_{S_{i}}\mu$ regardless
of which $i$ this ultrafilter come from. Since $\sigma_{\pi_{S_{i}}\mu}$ is a
morphism of $\mathfrak{L}$-structures then if
$\sigma_{\pi_{S_{i}}\mu}^{S_{i}}(a^{i})=(b_{x}^{i})_{x\in X}$ then
$\phi^{M}((b_{x}^{1}),\dots,(b_{x}^{n}))\leq\phi^{M_{y}}(a_{1},\dots,a_{n})$
thus for any $\epsilon^{{}^{\prime}}>0$ there exists $L\in\mu$ such that for
every $x\in L$ we have
$\phi^{M_{x}}(b_{x}^{1},\dots,b_{x}^{n})\leq\phi^{M_{y}}(a^{1},\dots,a^{n})+\epsilon^{{}^{\prime}}$.
So let us pick the $L$ corresponding to $\epsilon^{{}^{\prime}}=\epsilon/2$.

We know that for each $x$ the functions $\phi^{M_{x}}$ are uniformly
continuous with the same uniform continuity modulus(independent of $x$) which
implies that there exists some $\delta$ such that for any $m^{i}_{x}\in
M_{x}^{S_{i}}$ if $d(m^{i}_{x},b_{x}^{i})<\delta$ we get that
$|\phi^{M_{x}}(b_{x}^{1},\dots,b_{x}^{n})-\phi^{M_{x}}(m^{1}_{x},\dots,m^{n}_{x})|<\epsilon/2$.
Let us take the family $\bigcap_{i=1}^{n}\pi_{S_{i}}^{-1}(\coprod_{x\in
L}B(b_{x}^{i},\delta))$. First we know that each $\coprod_{x\in
L}B(b_{x}^{i},\delta)\in\pi_{S_{i}}\mu$ which allows us to deduce that
$\bigcap_{i=1}^{n}(\pi_{S_{i}}^{-1}\coprod_{x\in
L}B(b_{x}^{i},\delta))\in\mu$. Suppose that
$(l^{1},\dots,l^{n})\in\bigcap_{i=1}^{n}\pi_{S_{i}}^{-1}(\coprod_{x\in
L}B(b_{x}^{i},\delta))$, let us call $z=\pi_{S_{i}}(l_{i})$ then we have that:

$\phi^{M_{z}}(l^{1},\dots,l^{n})<\phi^{M_{z}}(b_{z}^{1},\dots,b_{z}^{i})+\epsilon/2\leq\phi^{M_{y}}(a^{1},\dots,a^{n})+\epsilon/2+\epsilon/2=r$

this implies that

$\bigcap_{i=1}^{n}\pi_{S_{i}}^{-1}(\coprod_{x\in
L}B(b_{x}^{i},\delta))\subseteq(\phi^{E})^{-1}([0,r))$

Thus $(\phi^{E})^{-1}([0,r))\in\mu$. Thus we may deduce that $\phi^{E}\mu$
converges to $\phi(a^{1},\dots,a^{n})$ (if we equip $[0,\infty]$ with the left
order topology) thus $\phi^{E}$ is upper semi-continuous.

####  5.4.3 Functoriality of $L$

Since each $\mathcal{L^{S}}$ is a functor by the previous construction we may
deduce that $\mathcal{L}$ defined this way is a functor.

###  5.5 The inverse functor

Now suppose we have a bundle $E$ of structures, we define the inverse functor
by sending a bundle $E$ to the left ultrafunctor $\mathcal{F}(E)$ defined as
follows: for every $x\in X$ we define $\mathcal{F}(x)=E_{x}$ (the fibre at
$x$). Now the left ultrastructure of the functor $\mathcal{R}(E)$ is
constructed from the left-ultrastructure of the restriction of the functor to
each sort as described in 4.1.

Now it remains to check compatibility for both functions and relations symbols
(which means that we are going to show that the $\sigma_{\mu}$ constructed
sort-wise is really a morphism in the category of structures).

####  5.5.1 Compatibility of function symbols

The proof in section 4.1 shows that for each sort $S$ the maps
$\sigma^{S}_{\mu}$ are contractions and thus continuous. Suppose that
$\\{S_{i}\\}_{i=1}^{n}$ is a finite family of sorts. We are going to denote by
$\sigma_{\mu}^{S_{1}\times\dots\times S_{n}}$ the map such that
$\pi_{S_{i}}\circ\sigma_{\mu}^{S_{1}\times\dots\times
S_{n}}=\sigma_{\mu}^{S_{i}}$. Let $\mu$ be an ultrafilter on $X$ that
converges to $y$. As stated before our goal is to show the compatibility of
the morphism $\sigma_{\mu}$. To do so suppose that $f$ is a function symbol,
and suppose that $\mathrm{dom}(f)=S_{1}\times\dots\times S_{n}$ and
$\mathrm{rng}(f)=S^{{}^{\prime}}_{1}$.

Suppose that $(a_{1},\dots,a_{n})\in M_{y}^{S_{1}}\times\dots\times
M_{y}^{S_{n}}$ (in case we have a constant symbol this space is $X$) and
suppose that for each $i$ the already constructed Cauchy filter converges in
$\int_{X}M_{x}^{S_{i}}d\mu$ to $(b^{i}_{x})_{x\in X}$ (this means that
$\sigma_{\mu}^{S_{i}}(a_{i})=(b_{x}^{i})_{x\in X}$).

For simplicity, we are going to call the space $\int_{X}M_{x}^{S_{i}}d\mu=M$.
We know that
$f^{M}((b^{1}_{x}),\dots,(b^{n}_{x}))=(f^{M_{x}}(b^{1}_{x},\dots,b^{n}_{x}))_{x\in
X}$ (by definition), and let us call
$f^{M_{y}}(a_{1},\dots,a_{n})=a^{{}^{\prime}}$ and furthermore we call the
limit of the Cauchy filter corresponding to $a^{{}^{\prime}}$ converges to
$(a_{x})_{x\in X}$ (this means that
$\sigma_{\mu}^{S^{{}^{\prime}}_{1}}(a^{{}^{\prime}})=(a_{x})_{x\in X}$).

Our goal is to show that $(f^{M_{x}}(b^{1}_{x},\dots,b^{n}_{x}))_{x\in
X}=(a_{x})_{x\in X}$ (as equivalence classes of the ultraproduct), to do so
let $\epsilon>0$. Since $f^{M}$ is uniformly continuous with uniform
continuity modulus independent of $x$ then there exists some $\delta$ such
that if for all $i$ if $d((b^{i}_{x}),(h_{i}^{x}))<\delta$ we get that
$d(f(b^{i}_{x}),(h_{i}^{x}))<\epsilon/3$. Now take a neighbourhood $W$ of
$a^{{}^{\prime}}$ that is $\epsilon/3$-thin. Now using Lemma 4.1, we know that
we can find $(g_{x})_{x\in X}$ such that there exists some $U\in\mu$ such that
$U\subseteq\pi(W)$ and such that $\forall x\in x$ $g_{x}\in W$ and
$d(g_{x},a_{x})<\epsilon/3$. Now since $E$ is a bundle, we may deduce that
there exist neighbourhoods $W_{i}$ of each $a_{i}$ such that
$f^{E}(\prod_{i=1}^{n}(W_{i}))\subseteq W$ (using the continuity of $f^{E}$)
(In the case of constant symbol $c$ we deduce the existence of
$W^{{}^{\prime}}$ neighbourhood of $y$, such that for any $x\in
W^{{}^{\prime}}$, $c_{x}\in W$). Now again using the lemma 4.1 we know there
exist $(V_{i})_{i=1}^{n}$ such that each $V_{i}\in\mu$ and such that
$V_{i}\subseteq\pi_{i}(W_{i})$ and $(e^{i}_{x})$ such that for any $x\in
V_{i}$ we have $d(e^{i}_{x},b^{i}_{x})<\delta$ and $e^{i}_{x}\in W_{i}$. Now
take the set $\bigcap_{i=1}^{n}V_{i}\bigcap U\in\mu$, for any $x$ in this set
we have

$d(f^{M_{x}}(e^{1}_{x},\dots,e^{n}_{x}),f(b^{1}_{x},\dots,b^{n}_{x}))<\epsilon/3$

On the other hand we have
$d(g_{x},f^{M_{x}}(e^{1}_{x},\dots,e^{n}_{x}))<\epsilon/3$ this follows from
the fact that $f^{M_{y}}(W_{1}\times_{X}\dots\times_{X}W_{n})\subseteq W$, and
that $W$ is $\epsilon/3$-thin. This implies that for any
$x\in\bigcap_{i=1}^{n}V_{i}\bigcap U$ we have

$d(f^{M_{x}}(b^{1}_{x},\dots,b^{n}_{x}),a_{x})<\epsilon$

thus $(f^{M_{x}}(b^{1}_{x},\dots,b^{n}_{x}))_{x\in X}=(a_{x})_{x\in X}$ as
equivalence classes, thus we get compatibility for each function symbol. In
the case of a constant symbol it suffices to take $W^{{}^{\prime}}\bigcap U$
in the previous argument.

####  5.5.2 Compatibility of relation symbols

Let $\mu$ be an ultrafilter on $X$ that converges to $y$ and let $\phi$ be a
relations symbol such that $\mathrm{dom}(\phi)=S_{1}\times\dots\times S_{n}$.
Suppose that $(a_{1},\dots,a_{n})\in M_{y}^{S_{1}}\times\dots\times
M_{y}^{S_{n}}$ and suppose that for each $i$ we have the already constructed
the Cauchy filter that converges in $\int_{X}M_{x}^{S_{i}}d\mu$ to some
$(b^{i}_{x})_{x\in X}$.(that means that
$\sigma^{S_{i}}_{\mu}(a_{i})=(b_{x}^{i})_{x\in X}$).

Now our objective is to show that
$\phi^{M}((b_{x}^{1}),\dots,(b_{x}^{n}))\leq\phi^{M_{y}}(a_{1},\dots,a_{n})$.
To do so we do an argument by contradiction, suppose it’s not the case then we
have that
$\phi^{M_{y}}(a_{1},\dots,a_{n})<\phi^{M}((b_{x}^{1}),\dots,(b_{x}^{n}))$. Let
us call their difference $\epsilon$. Now by upper semi-continuity there exists
neighbourhoods $W_{i}$ of $a_{i}$ such that
$\phi^{E}(W_{1}\times_{X}\dots\times_{X}W_{n})\subseteq[0,\phi^{M_{y}}(a_{1},\dots,a_{n}))$.
Now using the fact that each $\phi^{M_{x}}$ is continuous with the same
modulus, we get that there exists $\delta$ such that if for every $i$ such
that $|h^{i}_{x}-b^{i}_{x}|<\delta$, we have
$\phi^{M_{x}}(h^{i}_{x})<\epsilon$. Now using the lemma 4.1 we know there
exist a family of sets $\\{V_{i}\\}_{i=1}^{n}$ such that each $V_{i}\in\mu$
and such that $V_{i}\subseteq\pi_{i}(W_{i})$ and $(e^{i}_{x})$ such that for
any $x\in V_{i}$ we have $d(e^{i}_{x},b^{i}_{x})<\delta$ and $e^{i}_{x}\in
W_{i}$.

Now let $\epsilon^{{}^{\prime}}>0$, we know that there exists $U\in\mu$ such
that for any $x\in U$
$\phi^{M}((b^{1}_{x}),\dots,(b^{n}_{x}))<\phi^{M_{x}}(b^{1}_{x},\dots,b^{n}_{x})+\epsilon^{{}^{\prime}}$
(this follows from the fact that that
$\phi^{M}((b^{1}_{x}),\dots,(b^{n}_{x}))=\\\
\int_{X}\phi^{M_{x}}(b_{x}^{1},\dots,b_{x}^{n})d\mu$ and then we apply the
fact that $\int_{X}\phi^{M_{x}}(b_{x}^{1},\dots,b_{x}^{n})d\mu=\\\
\mathrm{Inf}_{U\in\mu}\mathrm{Sup}_{x\in
U}\phi^{M_{x}}(b_{x}^{1},\dots,b_{x}^{n})$). Now take the set
$(\bigcap_{i=1}^{n}V_{i})\bigcap U$, for any $x$ in this set we have
$|\phi(b^{1}_{x},\dots,b^{n}_{x})-\phi(e^{1}_{x},\dots,e^{n}_{x})|<\epsilon$,
but this implies, substituting $\epsilon$ by its value, that
$\phi(e^{1}_{x},\dots,e^{n}_{x})>\phi(a_{1},\dots,a_{n})-\epsilon^{{}^{\prime}}$,
and since $\epsilon^{{}^{\prime}}$ was arbitrary we get that
$\phi(e^{1}_{x},\dots,e^{n}_{x})\geq\phi(a_{1},\dots,a_{n})$ contradiction
with
$\phi^{E}(W_{1}\times_{X}\dots\times_{X}W_{n})\subseteq[0,\phi^{M_{y}}(a_{1},\dots,a_{n}))$.

###  5.6 Adjunction

We have already established that for each sort
$\mathrm{Hom}(\mathcal{L^{S}}(\mathcal{F}^{S},E^{S}))\simeq\mathrm{Hom}(\mathcal{F}^{S},\mathcal{R^{S}}(E^{S}))$,
so the only thing left is to prove that the $\mathrm{Hom}$ functor is
compatible with the structure.

Let us make clear what we exactly mean by the compatibility of the
$\mathrm{Hom}$ functor: We have already established the fact that if we have a
left ultrafunctor $\mathcal{F}$ and a map of bundles $\psi$ from
$\mathcal{L}(\mathcal{F})$ to $E$ then we get a natural transformation of left
ultrafunctors $\widehat{\psi^{S}}$ for each sort. Also we have established
that if we have a natural transformation of left ultrafunctors, we have
already seen that for every sort we get a map of bundles $\bar{\kappa^{S}}$
from $\mathcal{L^{S}}(F^{S})$ to $E^{S}$. And we know that these two processes
are inverses of each other at the level of each sort. So the question is can
we extend this equivalence to the level of the whole structure.

Suppose we have a left ultrafunctor $\mathcal{F}$ and a map of bundles $\psi$
from $\mathcal{L}(\mathcal{F})$ to some bundle $E$ then we get a natural
transformation of left ultrafunctors $\widehat{\psi^{S}}$ for each sort. The
we define for each $x$ the map $\widehat{\psi}_{x}$ by
$\widehat{\psi}_{x}=\psi_{x}$ from
$\mathcal{L}(\mathcal{F})_{x}=\mathcal{F}(x)$ to $E_{x}=\mathcal{R}(E)(x)$.
Now the fact then we have a natural transformation follows from the fact that
for every $x$ the map $\psi_{x}=\widehat{\psi}_{x}$(by definition) is a map of
sorts, and the fact that it’s a natural transformation of left ultrafunctors
follows from the commutativity of this for each ultrafilter on the base space
$X$ converging to arbitrary $y$ :

${\mathcal{F}(y)}$${\int_{X}\mathcal{F}(x)d\mu}$${\mathcal{R}(E)(y)=E_{y}}$${\int_{X}E_{x}d\mu}$$\scriptstyle{\psi_{y}}$$\scriptstyle{\sigma_{\mu}}$$\scriptstyle{\int_{X}\psi_{x}d\mu}$$\scriptstyle{\sigma_{\mu}}$

which means exactly that for every sort the following diagram commutes :

${\mathcal{F}(y)^{S}}$${\int_{X}\mathcal{F}(x)^{S}d\mu}$${\mathcal{R}(E)(y)^{S}=E_{y}^{S}}$${\int_{X}E_{x}^{S}d\mu}$$\scriptstyle{\psi_{y}^{S}}$$\scriptstyle{\sigma_{\mu}^{S}}$$\scriptstyle{\int_{X}\psi_{x}^{S}d\mu}$$\scriptstyle{\sigma_{\mu}^{S}}$

which we already showed. So we get that $\widehat{\psi}$ is well-defined.

Now for the other direction suppose that we have a natural transformation of
left ultrafunctors $\kappa$ from $\mathcal{F}$ to $\mathcal{R}(\mathcal{E})$.
Define $\bar{\kappa}_{x}=\kappa_{x}$, first from the fact that $\kappa$ is a
natural transformation of left ultrafunctors we that for each $x$,
$\bar{\kappa}_{x}=\kappa_{x}$ we get that $\bar{\kappa}_{x}$ is also a map of
structures. The only thing remaining to check that $\bar{\kappa}$ is a map of
bundles is to check that for each sort $\bar{\kappa}^{S}$ is a map of bundles,
but this follows immediately from our work for bounded complete metric spaces.
Finally we know that these two processes are inverses of each other on the
level of each sort thus they are inverses of each other and the two functors
$\mathcal{L}$ and $\mathcal{R}$ are adjoints.

Now the unit and the counit of adjunctions are isomorphisms at the level of
each sort, and hence we get an equivalence of categories between
$\mathrm{Left\ ultrafunctors(X,\mathsf{CompMet}_{\mathfrak{L}})}$ and the
category $\mathrm{Bun}(X,\mathsf{CompMet}_{\mathfrak{L}})$.

##  6 Models

Let $\mathfrak{L}$ be a language Let $\mathbb{T}$ be a theory (set of
sentences) in this language, we define the category of models of $\mathbb{T}$
to be the category whose objects are $\mathfrak{L}$\- structures that are
models of $\mathbb{T}$ (that means for any object $M$ in this category we have
that for any sentence $\phi\in\mathbb{T}$ $\phi^{M}=0$) and having as
morphisms just morphisms of structures.

We can see that the category of models defined this particular way is a full
subcategory of the category of structures, and thus it inherits the
ultrastructures since it’s closed under the ultraproduct functor by Łos
theorem. Let us denote by by $\sf{Met}_{\mathfrak{L}}$ the category of
structures of the language $\mathfrak{L}$, and by
$\sf{Met}_{\mathfrak{L},\mathbb{T}}$ the full subcategory of models of
$\mathbb{T}$. Now we turn to the next important but easy to show lemma:

######  Lemma 6.1.

Let $X$ be a compact Hausdorff space (an ultraset), take the category of left
ultrafunctors from $X$ to $\sf{Met}_{\mathfrak{L}}$, then left ultrafunctors
from $X$ to $\sf{Met}_{\mathfrak{L},\mathbb{T}}$ form a full subcategory of
the previous category.

###### Proof.

This follows immediately from the fact that the condition of being a natural
transformation of left ultrafunctors, doesn’t depend on whether a functor $F$
is taking values in $\sf{Met}_{\mathfrak{L},\mathbb{T}}$ or not. ∎

The important thing regarding this discussion is that we already know that the
concept of full subcategory is carried over by equivalence of categories thus
we get an equivalence between left ultrafunctors from $X$ to models and
between bundles of structure whose every fibre is a model of the theory
$\mathbb{T}$. This inspires our next definition

######  Definition 6.1.

We define a bundle of models of a theory $\mathbb{T}$ in a Language
$\mathfrak{L}$ to be a bundle of structures such that every fibre of the
bundle is a model. As for morphisms of bundles of models, we define them to be
just morphisms of the bundle of structures.

In other words, we can see that the category
$\mathrm{Bun}(X,\sf{Met}_{\mathfrak{L},\mathbb{T}})$ is a full subcategory of
the category of bundles of structures
$\mathrm{Bun}(X,\sf{Met}_{\mathfrak{L},\mathbb{T}})$. This definition allows
us to deduce the following theorem:

######  Theorem 6.1.

Let $X$ be a compact Hausdorff space then the functor $\mathcal{L}$ restricts
to an equivalence of categories between $\mathrm{Left\
ultrafunctors(X,\mathsf{CompMet}_{\mathfrak{L},\mathbb{T}})}$ and the category
$\mathrm{Bun}(X,\mathsf{CompMet}_{\mathfrak{L},\mathbb{T}})$.

##  7 Functoriality in $\mathsf{CompHaus}$

Let $\mathsf{CompHaus}$ denote the category of compact Hausdorff spaces. Let
$\mathcal{M}$ be a category of models of continuous model theory, the category
$\mathsf{CompHaus}_{\mathcal{M}}$ is defined to have as objects: left
ultrafunctors from some compact Hausdorff space $X$ to $\mathcal{M}$ and a
morphism between $\mathcal{F}:X\rightarrow\mathcal{M}$ and
$\mathcal{G}:Y\rightarrow\mathcal{M}$ consists of a pair $(f,\alpha)$ where
$f$ is continuous map from $X$ to $Y$ and $\alpha$ is a natural transformation
of let ultrafunctors from $\mathcal{F}$ to $\mathcal{G}\circ f$ (this
construction is called $\mathrm{Comp}_{\mathcal{M}}$ in [19]). In his paper
Lurie showed this construction to be a stack over $\mathsf{CompHaus}$ with the
latter equipped with the coherent topology. Now we claim the following result
:

######  Theorem 7.1.

The construction $X\mapsto\mathrm{Bun}(\mathcal{M},X)$ depends contravariantly
on $X$ (which means it defines a Grothendieck fibration over $\sf{CompHaus}$).

###### Proof.

First we start by the case where
$\mathcal{M}=\mathsf{k}\text{-}\mathsf{CompMet}$, we did not define what
$X\mapsto\mathrm{Bun}(\mathcal{M},X)$ should do on morphisms so we do that we
define a functor from $\mathrm{Bun}/X$ (which is another way of calling
$\mathrm{Bun}(\mathcal{M},X)$, here $\mathcal{M}$ is fixed to be
$\mathsf{k}\text{-}\mathsf{CompMet}$), to $\mathrm{Bun}/Y$ as follows :
suppose that we have a continuous map $Y\rightarrow X$ and some bundle $E$
over $X$ then we define a bundle $E^{{}^{\prime}}$ over $Y$ as the pullback in
$\mathsf{Top}$ :

${E^{{}^{\prime}}}$${E}$${Y}$${X}$$\scriptstyle{\pi_{2}}$$\scriptstyle{\underline{f}}$$\scriptstyle{\pi_{1}}$$\scriptstyle{f}$

We need of course to verify this is a bundle. Notice that $E^{{}^{\prime}}$ as
a set is equal to $\coprod_{y\in Y}E_{f(y)}$. The fact that the distance
function is upper semi-continuous on
$E^{{}^{\prime}}\times_{Y}E^{{}^{\prime}}$ follows from the following diagram
:

${{E^{{}^{\prime}}\times_{Y}E^{{}^{\prime}}}}$${{E\times_{X}E}}$${{E^{{}^{\prime}}}}$${{E}}$${Y}$${X}$${{[0,k]_{\text{left
order
topology}}}}$$\scriptstyle{\pi_{1}}$$\scriptstyle{\pi_{3}}$$\scriptstyle{f}$$\scriptstyle{\pi_{2}}$$\scriptstyle{\pi_{6}}$$\scriptstyle{\underline{f}}$$\scriptstyle{d}$$\scriptstyle{\langle\underline{f}\circ\pi_{5},\underline{f}\circ\pi_{6}\rangle}$$\scriptstyle{\pi_{4}}$$\scriptstyle{\pi_{5}}$

The distance function on $E^{{}^{\prime}}$ is equal to the composition
$d\circ\langle\underline{f}\circ\pi_{5},\underline{f}\circ\pi_{6}\rangle$,
thus it’s upper semi-continuous (here $[0,k]$ was equipped with the lower
order topology, and $\pi_{1}\dots\pi_{6}$ denotes the pullback maps).

Next we need to show axiom $(2)$ of the definition of bundles is satisfied,
which means that we need to show that $\pi_{1}$ is continuous and open but
this straightforward: $\pi_{1}$ is continuous by definition and open since the
pullback along an open map is an open map.

Finally we need to show axiom $(3)$ of the definition of bundle, suppose that
we have an element $g\in E^{{}^{\prime}}$ contained in some open set $W$ and
suppose without loss of generality that $W$ is basic which means that