# On positive face structures
and positive-to-one computads

Marek Zawadowski
Instytut Matematyki, Uniwersytet Warszawski
ul. S.Banacha 2,
00-913 Warszawa, Poland
<EMAIL_ADDRESS>
###### Abstract

We introduce the notion of a positive face structure. The positive face
structures to positive-to-one computads are like simple graphs, c.f. [MZ], to
free ${\omega}$-categories over ${\omega}$-graphs. In particular, they allow
to give an explicit combinatorial description of positive-to-one computads.
Using this description we show, among other things, that positive-to-one
computads form a presheaf category with the exponent category being the
category of principal positive face structures. We also present the Harnik
argument in this context showing that the ${\omega}$-categories are monadic
over positive-to-one computads with the ’free functor’ being the inclusion
${\bf Comp}^{+/1}\rightarrow{{\omega}Cat}$.

###### Contents

  1. 1 Introduction
  2. 2 Positive hypergraphs
  3. 3 Positive face structures
  4. 4 Atlas for $\gamma$ and $\delta$
  5. 5 Combinatorial properties of positive face structures
  6. 6 The ${\omega}$-categories generated by the positive face structures
  7. 7 Normal positive face structures
  8. 8 Decomposition of positive face structures
  9. 9 $S^{*}$ is a positive-to-one computad
  10. 10 The inner-outer factorization in ${{\cal C}types}^{+/1}_{\omega}$
  11. 11 The terminal positive-to-one computad
  12. 12 A description of the positive-to-one computads
  13. 13 Positive-to-one computads form a presheaf category
  14. 14 The principal pushouts
  15. 15 Yet another full nerve of the ${\omega}$-categories
  16. 16 A monadic adjunction
  17. 17 Appendix

##  1 Introduction

In this paper we present a combinatorial description of the category of the
positive-to-one computads ${\bf Comp}^{+/1}$. We show, that this category is a
presheaf category and we describe its exponent category in a very simple
combinatorial way as the category of positive principal face structures ${\bf
pFs}^{+/1}$, see section 3. However the proof of that requires some extended
studies of the category of all positive face structures. Intuitively, the
(isomorphism classes of) positive face structures correspond to the types of
arbitrary cells in positive-to-one computads. The notion of a positive face
structure in the main notion introduced in this paper. We describe in a
combinatorial way, the embedding functor ${\bf e}:{\bf
Comp}^{+/1}\rightarrow{{\omega}Cat}$ of the category of positive-to-one
computads into the category into the ${\omega}$-categories as the left Kan
extension along a suitable functor ${\bf j}$, and its right adjoint as the
restriction along ${\bf j}$. We end by adopting an argument due to V.Harnik to
show that the right adjoint to ${\bf e}$ is monadic. This approach does not
cover the problem of the cells with empty domains which is important for both
Makkai’s multitopic categories and Baez-Dolan opetopic categories. However it
keeps something from the simplicity of the Joyal’s $\theta$-categories, i.e.
the category of positive face structures or rather the category of positive
computypes ${{\cal C}types}^{+/1}_{\omega}$, the full image of the former in
${{\omega}Cat}$ is not much more complicated than the category of simple
${\omega}$-categories, the dual of the category of disks, c.f. [J], [MZ],
[Be]. In this sense this paper may be considered as a step towards a
comparison of these two approaches.

### Positive face structures

Positive face structures represent all possible shapes of cells in positive-
to-one computads. A positive face structure $S$ of dimension $2$ can be
pictured as a figure

$s_{6}\vphantom{s_{4}}$$\vphantom{s_{6}}s_{4}$$x_{6}$$s_{5}$$x_{8}$$x_{7}$$\Downarrow
a_{2}$$\Downarrow
a_{1}$$s_{7}\vphantom{\phantom{s_{7}}}$$x_{4}$$x_{9}$$x_{5}$$s_{2}\vphantom{s_{0}}$$\vphantom{s_{2}}s_{0}$$x_{0}$$s_{1}$$x_{2}$$x_{1}$$\Downarrow
a_{0}$$S:$$s_{3}\vphantom{\phantom{s_{2}}}$$x_{3}$

and a positive face structure $T$ of dimension $3$ can be pictured as a figure

$t_{3}\vphantom{t_{1}}$$\vphantom{t_{3}}t_{1}$$y_{3}$$t_{2}$$y_{5}$$y_{4}$$\Downarrow\\!b_{3}$$t_{4}\vphantom{}$$y_{6}$$\vphantom{}t_{0}$$y_{1}$$y_{0}$$y_{2}$$\Downarrow\\!b_{2}$$\Downarrow\\!b_{1}$$\Longrightarrow$$\beta$$t_{3}\vphantom{t_{1}}$$\vphantom{t_{3}}t_{1}$$t_{2}$$y_{5}$$y_{4}$$\Downarrow\\!b_{0}$$t_{4}\vphantom{}$$y_{6}$$\vphantom{}t_{0}$$y_{1}$$y_{0}$

They have faces of various dimensions that fit together so that it make sense
to compose them in a unique way. By $S_{n}$ we denote faces of dimension $n$
in $S$. Each faces $a$ has a face $\gamma(a)$ as its codomain and a non-empty
set of faces $\delta(a)$ as its domain. In $S$ above we have for $a_{1}$

$\gamma(a_{1})=x_{4}\;\;\;{\rm
and}\;\;\;\delta(a_{1})=\\{x_{5},x_{6},x_{9}\\}$

and in $T$ we have for $\beta$

$\gamma(\beta)=b_{0}\;\;\;{\rm
and}\;\;\;\delta(\beta)=\\{b_{1},b_{2},b_{3}\\}$

This is all the data we need. Moreover, these (necessarily finite) data
satisfy four conditions (see Section 3 for details). Below we explain them in
an intuitive way.

Globularity. This is the main condition. It relate the sets that are obtained
by duble application of $\gamma$ and $\delta$. They are

$\gamma\gamma(a)=\gamma\delta(a)-\delta\delta(a),\hskip
42.67912pt\delta\gamma(a)=\delta\delta(a)-\gamma\delta(a).$

Let us look how it works for two faces $a_{1}$ and $\beta$. In case of the
face $a_{1}$ we have

$\gamma\delta(a_{1})=\\{s_{3},s_{4},s_{6}\\},\hskip
42.67912pt\delta\delta(a_{1})=\\{s_{4},s_{6},s_{7}\\}$
$\gamma\gamma(a_{1})=s_{3},\hskip 42.67912pt\delta\gamma(a_{1})=\\{s_{7}\\}$

So we have indeed

$\delta\delta(a_{1})-\gamma\delta(a_{1})=\\{s_{4},s_{6},s_{7}\\}-\\{s_{3},s_{4},s_{6}\\}=\\{s_{7}\\}=\delta\gamma(a_{1})$
$\gamma\delta(a_{1})-\delta\delta(a_{1})=\\{s_{3},s_{4},s_{6}\\}-\\{s_{4},s_{6},s_{7}\\}=\\{s_{3}\\}=\\{\gamma\gamma(a_{1})\\}$

Similarly for the face $\beta$ we have

$\gamma\gamma(\beta)=y_{0},\hskip
42.67912pt\delta\gamma(\delta)=\\{y_{1},y_{4},y_{5},y_{6}\\}$
$\gamma\delta(\beta)=\\{y_{0},y_{2},y_{3}\\},\hskip
42.67912pt\delta\delta(\beta)=\\{y_{1},y_{2},y_{3},y_{4},y_{5},y_{6}\\}$

and hence

$\gamma\delta(\beta)-\delta\delta(\beta)=\\{y_{0},y_{2},y_{3}\\}-\\{y_{1},y_{2},y_{3},y_{4},y_{5},y_{6}\\}=\\{y_{0}\\}=\\{\gamma\gamma(\beta)\\}$
$\delta\delta(\beta)-\gamma\delta(\beta)=\\{y_{1},y_{2},y_{3},y_{4},y_{5},y_{6}\\}-\\{y_{0},y_{2},y_{3}\\}=\\{y_{1},y_{4},y_{5},y_{6}\\}=\delta\gamma(\beta)$

As we see in both cases $a_{1}$ and $\beta$ the first actual formula is a bit
more baroque (due to the curely brackets around $\gamma\gamma(a_{1})$,
$\gamma\gamma(\beta)$) that in the globularity condition stated above. However
in the following we omit curely bracket on purpose to have a simpler notation
hoping that it will contribute to simplicity without messing things up.

Using $\delta$’s and $\gamma$’s we can define two binary relations $<^{+}$ and
$<^{-}$ on faces of the same dimension which are transitive closures of the
relations $\lhd^{+}$ and $\lhd^{-}$, respectively. $a\lhd^{+}b$ holds iff
there is a face $\alpha$ such that $a\in\delta(\alpha)$ and
$\gamma(\alpha)=b$, and $a\lhd^{-}b$ holds iff $\gamma(a)\in\delta(b)$. We
call $<^{+}$ the upper order and $<^{-}$ the lower order. The following three
conditions refer to these relations.

Strictness. In each dimension, the relation $<^{+}$ is a strict order. The
relation $<^{+}$ on $0$-dimensional faces is required to be a linear order.

Disjointness. This condition says that no two faces can be comparable with
respect to both orders $<^{+}$ and $<^{-}$.

Pencil linearity. This final condition says that the sets of cells with common
codomain ($\gamma$-pencil) and the sets of cells that have the same
distinguished cell in the domain ($\delta$-pencil) are linearly ordered by
$<^{+}$.

The morphism of positive face structures are functions that preserves
dimensions and operations $\gamma$ and $\delta$. The size of a positive face
structure $S$ is defined as an infinite sequence of natural numbers
$size(S)=\\{size(S)_{k}\\}_{k\in{\omega}}=\\{S_{k}-\delta(S_{k+1})\\}_{k\in{\omega}}$
(almost all equal $0$). We order the sequences lexicographically with higher
dimensions being more important. The induction on the size of face structures
is a convenient way of reasoning about positive face structures. Dimension of
a face structure $S$ is the index of the largest non-zero number in the
sequence $size(S)$. If for $k\leq dim(S)$ ($k<dim(S)$), $size(S)_{k}=1$ then
$S$ is principal (normal). The normal positive face structures plays role of
the pasting diagram in [HMP] and the principal positive face structures plays
role of the (positive) multitopes. Note that, contrary to [HMP], we do not
consider either the empty-domain multitopes or the pasting diagrams. The
precise connection between these two approaches will be described elsewhere.
On positive face structures we define operations of the domain, codomain, and
special pushouts which plays the role of composition. With these operations
(isomorphisms classes of) the positive face structures form the terminal
positive-to-one computad, and at the same time the monoidal globular category
in the sense of Batanin.

### Categories and functors

We shall define the following categories

${\bf Comp}^{+/1}\vphantom{{{\omega}Cat}}$$\vphantom{{\bf
Comp}^{+/1}}{{\omega}Cat}$${\bf e}$${\bf Fs}^{+/1}\vphantom{{{\cal
C}types}^{+/1}_{\omega}}$$\vphantom{{\bf Fs}^{+/1}}{{\cal
C}types}^{+/1}_{\omega}$${\bf j}$$(-)^{*}$${\bf pFs}^{+/1}\vphantom{{\cal
S}}$$\vphantom{{\bf pFs}^{+/1}}{\cal S}$${\bf i}$${\bf k}$

where ${\bf pFs}^{+/1}$ is the category of principal positive face structures,
${\bf Fs}^{+/1}$ is the category of positive face structures, ${\cal S}$ is
the category of simple categories c.f. [MZ], $(-)^{*}$ is the embedding
functor of positive face structures into positive-to-one computads, ${\bf e}$
is the inclusion functor, ${{\cal C}types}^{+/1}_{\omega}$ is the full image
of the composition functor $(-)^{*};{\bf e}$, with the non-full embedding
${\bf j}$.

Having these functor we can form the following diagram

$Set^{({\bf pFs}^{+/1})^{op}}\vphantom{sPb({\cal
S}^{op},Set)}$$\vphantom{Set^{({\bf pFs}^{+/1})^{op}}}sPb({\cal
S}^{op},Set)$$sPb(({\bf Fs}^{+/1})^{op},Set)\vphantom{sPb(({{\cal
C}types}^{+/1}_{\omega})^{op},Set)}$$\vphantom{sPb(({\bf
Fs}^{+/1})^{op},Set)}sPb(({{\cal C}types}^{+/1}_{\omega})^{op},Set)$$Lan_{\bf
j}$${\bf i}^{*}$$Ran_{\bf k}$${\bf k}^{*}$$Ran_{\bf i}$${\bf
Comp}^{+/1}\vphantom{{{\omega}Cat}}$$\vphantom{{\bf
Comp}^{+/1}}{{\omega}Cat}$${\bf
e}$$\widehat{(-)}$$\widetilde{(-)}$$\widehat{(-)}$$\widetilde{(-)}$${\bf
j}^{*}$

in which all the vertical arrows comes in pairs and they are adjoint
equivalences of categories. The unexplained categories in the diagram above
are: $sPb(({\bf Fs}^{+/1})^{op},Set)$ \- the category of the special pullback
preserving morphisms from $({\bf Fs}^{+/1})^{op}$ to $Set$ and natural
transformations, $sPb({\cal S}^{op},Set)$ \- the category of the special
pullback preserving morphisms from ${\cal S}^{op}$ to $Set$ and natural
transformations, $sPb(({{\cal C}types}^{+/1}_{\omega})^{op},Set)$ \- the
category of the special pullbacks preserving morphisms from $({{\cal
C}types}^{+/1}_{\omega})^{op}$ to $Set$ and natural transformations.

The functors $\widehat{(-)}:{\bf Comp}^{+/1}\rightarrow sPb(({\bf
Fs}^{+/1})^{op},Set)$ and $\widehat{(-)}:{{\omega}Cat}\rightarrow sPb(({{\cal
C}types}^{+/1}_{\omega})^{op},Set)$ are defined similarly, due to the
embeddings $(-)^{*}$ in the previous diagram. For a computad $Q$ and an
${\omega}$-category $C$, $\widehat{Q}$ and $\widehat{C}$ are presheaves so
that for a positive face structure $S$ we have

$\widehat{Q}={\bf Comp}^{+/1}(S^{*},Q)\hskip
42.67912pt\widehat{C}={{\omega}Cat}(S^{*},C)$

The adjoint functors $\widetilde{(-)}$ that produce ${\omega}$-categories are
slightly more complicated. They are defined in Sections 13 and 15. The other
functors are standard. The functors ${\bf i}^{*}$, ${\bf j}^{*}$, ${\bf
k}^{*}$ are inverse image functors. $Ran_{\bf i}$ and $Ran_{\bf k}$ are the
right Kan extensions along ${\bf i}$, ${\bf k}$, respectively and $Lan_{\bf
j}$ is the left Kan extension along ${\bf j}$.

Since we have ${\bf e};\widehat{(-)}=\widehat{(-)};Lan_{\bf j}$, and
$\widehat{(-)}$’s are equivalences of categories, the functor $Lan_{\bf j}$ is
like ${\bf e}$ but moved into a more manageable context. In fact we have a
very neat description of this functor.

### The content

Since the paper is quite long I describe below the content of each section to
help the reading. Sections 2 and 3 introduce the notion of a positive hyper-
graph and positive face structure. Section 4 is concerned with establishing
what kind of inclusions hold between iterated applications of $\gamma$’s and
$\delta$’s. Section 5 contains many technical statements concerning positive
face structure. All of them are there because they are needed afterwards.
Section 6 describes the embedding $(-)^{*}:{\bf
Fs}^{+/1}\rightarrow{{\omega}Cat}$ i.e. it’s main goal is to define an
${\omega}$-category $S^{*}$ for any positive face structure $S$. Section 7
describes rather technical but useful properties of normal positive face
structures. In section 8 we study a way we can decompose positive face
structures if they are at all decomposable. Any positive face structure is
either principal or decomposable. This provides a way of proving the
properties of positive face structures by induction on the size. Using this in
section 9 we show that the ${\omega}$-category $S^{*}$ and in fact the whole
functor $(-)^{*}$ end up in ${\bf Comp}^{+/1}$. The next two short sections
10, 11 describe just what is in their titles: factorization in ${{\cal
C}types}^{+/1}_{\omega}$, and the terminal positive-to-one computad in terms
of positive face structures. Section 12 gives an explicite description of all
the cells in a given positive-to-one computad with the help of positive face
structures. In other words, it describe in concrete terms the functor
$\overline{(-)}:{\bf Comma}^{+/1}_{n}\longrightarrow{{\bf Comp}^{+/1}_{n}}$.
Section 13 establishes the equivalence of categories between ${\bf
Comp}^{+/1}$ and the category of presheaves over ${\bf pFs}^{+/1}$. In Section
14 the principal pullbacks are introduced and the V.Harnik’s argument in the
present context is presented. Section 15 describes a full nerve functor

$\widehat{(-)}:{{\omega}Cat}\longrightarrow Set^{({{\cal
C}types}^{+/1}_{\omega})^{op}}$

and identifies its essential image as the special pullbacks preserving
functors. Section 16 describes the inclusion functor as the left Kan extension

$Lan_{\bf j}:sPb(({\bf Fs}^{+/1})^{op},Set)\longrightarrow sPb(({{\cal
C}types}^{+/1}_{\omega})^{op},Set)$

with the formulas involving just coproduct (and no other colimits). This gives
as a corollary the fact that ${\bf e}:{\bf
Comp}^{+/1}\rightarrow{{\omega}Cat}$ preserves connected limits. Then it is
shown that the right adjoint to $Lan_{\bf j}$

${\bf j}^{*}:sPb(({{\cal C}types}^{+/1}_{\omega})^{op},Set)\longrightarrow
sPb(({\bf Fs}^{+/1})^{op},Set)$

(and hence the right adjoint to ${\bf e}:{\bf
Comp}^{+/1}\rightarrow{{\omega}Cat}$) is monadic. In Appendix we recall the
definition of the category of positive-to-one computads.

### Acknowledgments

I am very grateful to Victor Harnik and Mihaly Makkai for the conversations
concerning matters of this paper.

The diagrams for this paper were prepared with a help of catmac1 of Michael
Barr.

### Notation

In the paper we will use both directions of compositions of morphisms in
categories. But each time we will write which way we compose the morphism. So,
for the composition of a pair of morphism $x\stackrel{{\scriptstyle
f}}{{\longrightarrow}}y\stackrel{{\scriptstyle g}}{{\longrightarrow}}z$ we can
write either $f;g$ or $g\circ f$. ${\omega}$ is the set of natural numbers.

##  2 Positive hypergraphs

A positive hypergraph $S$ is a family $\\{S_{k}\\}_{k\in{\omega}}$ of finite
sets of faces, a family of functions $\\{\gamma_{k}:S_{k+1}\rightarrow
S_{k}\\}_{k\in{\omega}}$, and a family of total relations
$\\{\delta_{k}:S_{k+1}\rightarrow S_{k}\\}_{0\leq k<n}$. Moreover
$\delta_{0}:S_{1}\rightarrow S_{0}$ is a function and only finitely many among
sets $\\{S_{k}\\}_{k\in{\omega}}$ are non-empty. As it is always clear from
the context we shall never use the indices of the functions $\gamma$ and
$\delta$.

A morphism of positive hypergraphs $f:S\longrightarrow T$ is a family of
functions $f_{k}:S_{k}\longrightarrow T_{k}$, for $k\in{\omega}$, such that
the diagrams

$S_{k}\vphantom{T_{k}}$$\vphantom{S_{k}}T_{k}$$f_{k}$$S_{k+1}\vphantom{T_{k+1}}$$\vphantom{S_{k+1}}T_{k+1}$$f_{k+1}$$\gamma$$\gamma$$S_{k}\vphantom{T_{k}}$$\vphantom{S_{k}}T_{k}$$f_{k}$$S_{k+1}\vphantom{T_{k+1}}$$\vphantom{S_{k+1}}T_{k+1}$$f_{k+1}$$\delta$$\delta$

commute, for $k\in{\omega}$. The commutation of the left hand square is the
commutation of the diagram of sets an functions but in case of the right hand
square we mean more than commutation of a diagram of relations, i.e. we demand
that for any $a\in S_{\geq 1}$, $f_{a}:\delta(a)\longrightarrow\delta(f(a))$
be a bijection, where $f_{a}$ is the restriction of $f$ to $\delta(a)$. The
category of positive hypergraphs is denoted by ${\bf Hg}^{+1}$.

When convenient and does not lead to confusions, for $a\in S_{k}$ we sometime
treat $\gamma(a)$ as an element of $S_{k-1}$ and sometimes as a subset
$\\{\gamma(a)\\}$ of $S_{k-1}$.

Before we go on, we need some notation. Let $S$ be a positive hypergraph.

  1. 1.

The dimension of $S$ is $max\\{k\in{\omega}:S_{k}\neq\emptyset\\}$, and it is
denoted by $dim(S)$.

  2. 2.

The sets of faces of different dimensions are assumed to be disjoint (i.e.
$S_{k}\cap S_{l}=\emptyset$, for $k\neq l$). $S$ is also used to mean the set
of all faces of $S$, i.e. $\bigcup_{k=0}^{n}S_{k}$; the notation $A\subseteq
S$ mean that $A$ is a set of some faces of $S$; $A_{k}=A\cap S_{k}$, for
$k\in{\omega}$.

  3. 3.

If $a\in S_{k}$ then the face $a$ has dimension $k$ and we write $dim(a)=k$.

  4. 4.

For $a\in S_{\geq 1}$ the set $\theta(a)=\delta(a)\cup\gamma(a)$ is the set of
codimension $1$ faces in $a$.

  5. 5.

$S_{\geq k}=\bigcup_{i\geq k}S_{i}$, $\;\;\;\;\;S_{\leq k}=\bigcup_{i\leq
k}S_{i}$. The set $S_{\leq k}=\bigcup_{i\leq k}S_{i}$ is closed under $\delta$
and $\gamma$ so it is a sub-hypergraph of $S$, called $k$-truncation of $S$.

  6. 6.

The image of $A\subseteq S$ under $\delta$ and $\gamma$ will be denoted by

$\delta(A)=\bigcup_{a\in A}\delta(a),\hskip
14.22636pt\gamma(A)=\\{\gamma(a)\,:\,a\in A\\},$

respectively. In particular
$\delta\delta(a)=\bigcup_{x\in\delta(a)}\delta(x)$,
$\gamma\delta(a)=\\{\gamma(x):x\in\delta(a)\\}$.

  7. 7.

$\iota(a)=\delta\delta(a)\cap\gamma\delta(a)$ is the set of internal faces of
the face $a\in S_{\geq 2}$.

  8. 8.

On each set $S_{k}$ we introduce two binary relations $<^{S_{k},-}$ and
$<^{S_{k},+}$, called lower and upper order, respectively. We usually omit $k$
and even $S$ in the superscript.

    1. (a)

$<^{S_{0},-}$ is the empty relation. For $k>0$, $<^{S_{k},-}$ is the
transitive closure of the relation $\lhd^{S_{k},-}$ on $S_{k}$, such that
$a\lhd^{S_{k},-}b$ iff $\gamma(a)\in\delta(b)$. We write $a\perp^{-}b$ iff
either $a<^{-}b$ or $b<^{-}a$, and we write $a\leq^{-}b$ iff either $a=b$ or
$a<^{-}b$.

    2. (b)

$<^{S_{k},+}$ is the transitive closure of the relation $\lhd^{S_{k},+}$ on
$S_{k}$, such that $a\lhd^{S_{k},+}b$ iff there is $\alpha\in S_{k+1}$, such
that $a\in\delta(\alpha)$ and $\gamma(\alpha)=b$. We write $a\perp^{+}b$ iff
either $a<^{+}b$ or $b<^{+}a$, and we write $a\leq^{+}b$ iff either $a=b$ or
$a<^{+}b$.

    3. (c)

$a\not\perp b$ if both conditions $a\not\perp^{+}b$ and $a\not\perp^{-}b$
hold.

  9. 9.

Let $a,b\in S_{k}$. A lower path $a_{0},\ldots,a_{m}$ from $a$ to $b$ in $S$
is a sequence of faces $a_{0},\ldots,a_{m}\in S_{k}$ such that $a=a_{0}$,
$b=a_{m}$ and for $\gamma(a_{i-1})\in\delta(a_{i})$, $i=1,\ldots,m$.

  10. 10.

Let $x,y\in S_{k}$. An upper path $x,a_{0},\ldots,a_{m},y$ from $x$ to $y$ in
$S$ is a sequence of faces $a_{0},\ldots,a_{m}\in S_{k+1}$ such that
$x\in\delta(a_{0})$, $y=\gamma(a_{m})$ and $\gamma(a_{i-1})\in\delta(a_{i})$,
for $i=1,\ldots,m$.

  11. 11.

The iterations of $\gamma$ and $\delta$ will be denoted in two different ways.
By $\gamma^{k}$ and $\delta^{k}$ we mean $k$ applications of $\gamma$ and
$\delta$, respectively. By $\gamma^{(k)}$ and $\delta^{(k)}$ we mean the
application as many times $\gamma$ and $\delta$, respectively, to get faces of
dimension $k$. For example in $a\in S_{5}$ then
$\delta^{3}(a)=\delta\delta\delta(a)\subseteq S_{2}$ and
$\delta^{(3)}(a)=\delta\delta(a)\subseteq S_{3}$.

  12. 12.

For $l\leq k$, $a,b\in S_{k}$ we define $a<_{l}b$ iff
$\gamma^{(l)}(a)<^{-}\gamma^{(l)}(b)$.

  13. 13.

A face $a$ is unary iff $\delta(a)$ is a singleton.

We have

###### Lemma 2.1

If $S$ is an hypergraph and $k\in{\omega}$, then $<^{S_{k+1},-}$ is a strict
partial order iff $<^{S_{k},+}$ is a strict partial order.

##  3 Positive face structures

To simplify the notation, we treat both $\delta$ and $\gamma$ as functions
acting on faces as well as on sets of faces, which means that sometimes we
confuse elements with singletons. Clearly, both $\delta$ and $\gamma$ when
considered as functions on sets are monotone.

A positive hypergraph $S$ is a positive face structure if it is non-empty,
i.e. $S_{0}\neq\emptyset$ and

  1. 1.

Globularity: for $a\in S_{\geq 2}$:

$\gamma\gamma(a)=\gamma\delta(a)-\delta\delta(a),\hskip
42.67912pt\delta\gamma(a)=\delta\delta(a)-\gamma\delta(a);$

  2. 2.

Strictness: for $k\in{\omega}$, the relation $<^{S_{k},+}$ is a strict order;
$<^{S_{0},+}$ is linear;

  3. 3.

Disjointness: for $k>0$,

$\perp^{S_{k},-}\cap\perp^{S_{k},+}=\emptyset$

  4. 4.

Pencil linearity: for any $k>0$ and $x\in S_{k-1}$, the sets

$\\{a\in S_{k}\;|\;x=\gamma(a)\\}\;\;\;\;{\rm and}\;\;\;\;\\{a\in
S_{k}\;|\;x\in\delta(a)\\}$

are linearly ordered by $<^{S_{k},+}$.

$S$ is a weak positive face structure if $S$ is a globular, strict, disjoint,
positive hypergraph (i.e. pencil linearity is not required to hold).

The category of (weak) positive face structures is the full subcategory of
${\bf Hg}^{+1}$ whose objects are the (weak) positive face structures and is
denoted by denoted by ${\bf Fs}^{+/1}$ (${\bf wFs}^{+/1}$).

Remarks.

  1. 1.

The reason why we call the first condition ’globularity’ is that it will imply
the usual globularity condition in the ${\omega}$-categories generated by
positive face structures.

  2. 2.

For each $k\in{\omega}$, the $k$-truncation of a weak positive face structure
$S$ is again a weak positive face structure $S_{\leq k}$. In particular, any
$k$-truncation of a positive face structure $S$ is a weak positive face
structure $S_{\leq k}$, but it does not necessarily satisfy local linearity
condition.

  3. 3.

Note that if we were to assume that each positive face structure has a single
cell of dimension $-1$ then linearity of $<^{S_{0},+}$ would become a special
case of pencil linearity.

  4. 4.

The fact that, for $x\in S_{k-1}$, the set $\\{a\in S_{k}\;|\;x=\gamma(a)\\}$
is linearly ordered is sometimes referred to as $\gamma$-linearity of
$<^{S_{k},+}$, and the fact that the set $\\{a\in
S_{k}\;|\;\;x\in\delta(a)\\}$ is linearly ordered is sometimes referred to as
$\delta$-linearity of $<^{S_{k},+}$.

  5. 5.

If $S$ has dimension $n$, as hypergraph, then we say that $S$ is an $n$-face
structure.

  6. 6.

The size of positive face structure $S$ is the sequence natural numbers
$size(S)=\\{|S_{n}-\delta(S_{n+1})|\\}_{n\in{\omega}}$, with almost all being
equal $0$. We have an order $<$ on such sequences, so that
$\\{x_{n}\\}_{n\in{\omega}}<\\{y_{n}\\}_{n\in{\omega}}$ iff there is
$k\in{\omega}$ such that $x_{k}<y_{k}$ and for all $l>k$, $x_{l}=y_{l}$. This
order is well founded and many facts about positive face structures will be
proven by induction on the size.

  7. 7.

Let $S$ be a positive face structure. $S$ is $k$-principal iff
$size(S)_{l}=1$, for $l\leq k$. $S$ is principal iff $S$ is
$dim(S)$-principal. $S$ is normal iff $S$ is $(dim(S)-1)$-principal. $S$ is
$k$-normal iff $S$ is normal and $dim(S)=k$. By ${\bf pFs}^{+/1}$ (${\bf
nFs}^{+/1}$) we denote full subcategories of ${\bf Fs}^{+/1}$ whose objects
are principal (normal) positive face structures.

##  4 Atlas for $\gamma$ and $\delta$

We have an easy

###### Lemma 4.1

Let $S$ be a positive face structure, $a\in S_{n}$, $n>1$. Then

  1. 1.

the sets $\delta\gamma(a)$, $\iota(a)$, and $\gamma\gamma(a)$ are disjoint;

$\delta\gamma$$\iota$$\gamma\gamma$

  2. 2.

$\delta\delta(a)=\delta\gamma(a)\cup\iota(a)$;

  3. 3.

$\gamma\delta(a)=\gamma\gamma(a)\cup\iota(a)$.

Proof. These are immediate consequences of globularity. $~{}~{}\Box$

Moreover

###### Lemma 4.2

Let $S$ be a positive face structure, $a\in S_{n}$, $n>2$. Then we have

  1. 1.

$\delta\gamma\gamma(a)\subseteq\delta\gamma\delta(a)\subseteq\delta\gamma\gamma(a)\cup\iota\gamma(a)=\delta\delta\gamma(a)=\delta\delta\delta(a)$;

  2. 2.

$\gamma\gamma\gamma(a)\subseteq\gamma\gamma\delta(a)\subseteq\gamma\gamma\gamma(a)\cup\iota\gamma(a)=\gamma\delta\gamma(a)=\gamma\delta\delta(a)$.

Proof. From globularity we have
$\gamma\gamma(\alpha)\subseteq\gamma\delta(\alpha)$. Thus by monotonicity of
$\delta$ and $\gamma$ we get

$\gamma\gamma\gamma(\alpha)\subseteq\gamma\gamma\delta(\alpha)\hskip
14.22636pt{\rm and}\hskip
14.22636pt\delta\gamma\gamma(\alpha)\subseteq\delta\gamma\delta(\alpha)\hskip
14.22636pt{\rm and}\hskip
14.22636pt\gamma\gamma\delta(\alpha)\subseteq\gamma\delta\delta(\alpha).$

Similarly, as we have from globularity:
$\delta\gamma(\alpha)\subseteq\delta\delta(\alpha)$ it follows by monotonicity
of $\delta$ and $\gamma$:

$\gamma\delta\gamma(\alpha)\subseteq\gamma\delta\delta(\alpha)\hskip
14.22636pt{\rm and}\hskip
14.22636pt\delta\delta\gamma(\alpha)\subseteq\delta\delta\delta(\alpha)\hskip
14.22636pt{\rm and}\hskip
14.22636pt\delta\gamma\delta(\alpha)\subseteq\delta\delta\delta(\alpha).$

The equalities

$\delta\gamma\gamma(a)\cup\iota\gamma(a)=\delta\delta\gamma(a)\hskip
14.22636pt{\rm and}\hskip
14.22636pt\gamma\gamma\gamma(a)\cup\iota\gamma(a)=\gamma\delta\gamma(a)$

follow easily from Lemma 4.1.

Thus it remains to show that:

  1. 1.

$\delta\delta\gamma(a)\supseteq\delta\delta\delta(a)$,

  2. 2.

$\gamma\delta\gamma(a)\supseteq\gamma\delta\delta(a)$.

Both inclusions can be proven similarly. We shall show the first only.

Suppose contrary, that there is
$u\in\delta\delta\delta(a)-\delta\delta\gamma(a)$. Let $x\in\delta(a)$ be
$<^{-}$-minimal element in $\delta(a)$ such that there is $s\in\delta(x)$ with
$u\in\delta(s)$. If $s\in\delta\gamma(a)$ then $u\in\delta\delta\gamma(a)$
contrary to the supposition. Thus $s\not\in\delta\gamma(a)$. Since
$\delta\gamma(a)=\delta\delta(a)-\gamma\delta(a)$ it follows that
$s\in\gamma\delta(a)$. Hence there is $x^{\prime}\in\delta(a)$ with
$\gamma(x^{\prime})=s$. In particular $x^{\prime}<^{-}x$. Moreover

$u\in\delta(s)=\delta\gamma(x^{\prime})\subseteq\delta\delta(x^{\prime}).$

Then there is $s^{\prime}\in\delta(x^{\prime})$ so that
$u\in\delta(s^{\prime})$. This contradicts the $<^{-}$-minimality of $x$.
$~{}~{}\Box$

From Lemma 4.2 we get

###### Corollary 4.3

Let $S$ be a positive face structure, $a\in S_{n}$, $n>2$, $k<n$. Then, with
$\xi^{l}$ and $\xi^{\prime l}$ being two fixed strings of $\gamma$’s and
$\delta$’s of length $l$, we have

  1. 1.

$\gamma^{k}(a)\subseteq\gamma\xi^{k-1}(a)$;

  2. 2.

$\delta\xi^{k-1}(a)\subseteq\delta^{k}(a)$;

  3. 3.

$\delta^{k}(a)\cap\gamma^{k}(a)=\emptyset$;

  4. 4.

$\xi^{k}(a)\subseteq\gamma^{k}(a)\cup\delta^{k}(a)$;

  5. 5.

$\delta^{2}\xi^{k-2}(a)=\delta^{2}\xi^{\prime k-2}(a)$, (e.g.
$\delta^{k}(a)=\delta^{2}\gamma^{k-2}(a)$);

  6. 6.

$\gamma\delta\xi^{k-2}(a)=\gamma\delta\xi^{\prime k-2}(a)$, (e.g.
$\gamma\delta^{k-1}(a)=\gamma\delta\gamma^{k-2}(a)$);

  7. 7.

$\xi^{k-2}\delta\gamma(a)=\xi^{k-2}\delta^{2}(a)$, for $k>2$;

  8. 8.

$\delta^{k}(a)=\delta\gamma^{k-1}(a)\cup\iota\gamma^{k-2}(a)$, for $k>1$.

##  5 Combinatorial properties of positive face structures

### Local properties

###### Proposition 5.1

Let $S$ be a positive face structure, $k>0$ and $\alpha\in S_{k}$,
$a_{1},a_{2}\in\delta(\alpha)$, $a_{1}\neq a_{2}$. Then we have

  1. 1.

$a_{1}\not\perp^{+}a_{2}$;

  2. 2.

$\delta(a_{1})\cap\delta(a_{2})=\emptyset$ and
$\gamma(a_{1})\neq\gamma(a_{2})$.

Proof. Ad 1. Suppose contrary that there are $a_{1},a_{2}\in\delta(\alpha)$
such that $a_{1}<^{+}a_{2}$. So we have an upper path

$a_{1},\beta_{1},\ldots,\beta_{r},a_{2}$

and hence a lower path

$\beta_{1},\ldots,\beta_{r},\alpha.$

In particular $\beta_{1}<^{-}\alpha$. As
$a_{1}\in\delta(\beta_{1})\cap\delta(\alpha)$ by $\delta$-linearity we have
$\beta_{1}\perp^{+}\alpha$. But then
$(\alpha,\beta_{1})\in\perp^{+}\cap\perp^{-}\neq\emptyset$ i.e. $S$ does not
satisfy the disjointness. This shows 1.

Ad 2. This is an immediate consequence of 1. If $a_{1},a_{2}\in\delta(\alpha)$
and either $\gamma(a_{1})=\gamma(a_{2})$ or
$\delta(a_{1})\cap\delta(a_{2})\neq\emptyset$ then by pencil linearity we get
that $a_{1}\perp^{+}a_{2}$, contradicting 1. $~{}~{}\Box$

After proving the above proposition we can introduce more notation. Let $S$ be
a positive face structure, $n\in{\omega}$.

  1. 1.

For a face $\alpha\in S_{n+2}$, we shall denote by
$\rho(\alpha)\in\delta(\alpha)$ be the only face in $\delta(\alpha)$, such
that $\gamma(\rho(\alpha))=\gamma\gamma(\alpha)$.

  2. 2.

$X\subseteq S_{n+1}$, $a,b\in S_{n}$ and $a,\alpha_{1},\ldots,\alpha_{k},b$ be
an upper path in $S$. We say that it is a path in $X$ (or $X$-path) if
$\\{\alpha_{k},\ldots,\alpha_{k}\\}\subseteq X$.

###### Lemma 5.2

Let $S$ be a positive face structure, $n\in{\omega}$, $\alpha\in S_{n+2}$,
$a,b\in S_{n+1}$, $y\in\delta\delta(\alpha)$. Then

  1. 1.

there is a unique upper $\delta(\alpha)$-path from $y$ to
$\gamma\gamma(\alpha)$;

  2. 2.

there is a unique $x\in\delta\gamma(\alpha)$ and an upper
$\delta(\alpha)$-path from $x$ to $y$ such that $\gamma(x)=\gamma(y)$;

  3. 3.

if $t\in\delta(y)$ there is an $x\in\delta\gamma(\alpha)$ and an upper
$\delta(\alpha)$-path from $x$ to $y$ and $t\in\delta(x)$;

  4. 4.

If $a<^{+}b$ then $\gamma(a)\leq^{+}\gamma(b)$.

Proof. Ad 1. Uniqueness follows from Lemma 5.1. To show the existence, let us
suppose contrary that there is no $\delta(\alpha)$-path from $y$ to
$\gamma\gamma(\alpha)$. We shall construct an infinite upper
$\delta(\alpha)$-path from $y$

$y,a_{1},a_{2},\ldots$

As $y\in\delta\delta(\alpha)$ there is $a_{1}\in\delta(\alpha)$ such that
$y\in\delta(a_{1})$. So now suppose that we have already constructed
$a_{1},\ldots,a_{k}$. By assumption $\gamma(a_{k})\neq\gamma\gamma(\alpha)$,
so $\gamma(a_{k})\in\delta\delta(\alpha)$. Hence there is
$a_{k+1}\in\delta(\alpha)$ such that $\gamma(a_{k})\in\delta(a_{k+1})$. This
ends the construction of the path.

As in positive face structure there are no infinite paths, this is a
contradiction and if fact there is a $\delta(\alpha)$-path from $y$ to
$\gamma\gamma(\alpha)$.

Ad 2. Suppose not, that there is no $x\in\delta\gamma(\alpha)$ as claimed. We
shall construct an infinite descending lower $\delta(\alpha)$-path

$\ldots\lhd^{-}a_{1}\lhd^{-}a_{0}$

such that $\gamma(a_{0})=y$, $\gamma\gamma(a_{n})=\gamma(y)=t$, for
$n\in{\omega}$.

By assumption
$y\not\in\delta\gamma(\alpha)=\delta\delta(\alpha)-\gamma\delta(\alpha)$. So
$y\in\gamma\delta(\alpha)$. Hence there is $a_{0}\in\delta(\alpha)$ such, that
$\gamma(a_{0})=y$. Now, suppose that the lower $\delta(\alpha)$-path

$a_{k}\lhd^{-}a_{k-1}\lhd^{-}\ldots\lhd^{-}a_{0}$

has been already constructed. By globularity, we can pick $z\in\delta(a_{k})$,
such that $\gamma(z)=t$. By assumption
$z\not\in\delta\gamma(\alpha)=\delta\delta(\alpha)-\gamma\delta(\alpha)$. So
$z\in\gamma\delta(\alpha)$. Hence there is $a_{k+1}\in\delta(\alpha)$ such,
that $\gamma(a_{k+1})=z\in\delta(a_{k})$. Clearly, $\gamma\gamma(a_{k+1})=t$.
This ends the construction of the path. But by strictness such a path has to
be finite, so there is $x$ as needed.

Ad 3. This case is similar. We put it for completeness.

Suppose not, that there is no $x\in\delta\gamma(\alpha)$ as above. We shall
construct an infinite descending lower $\delta(\alpha)$-path

$\ldots\lhd^{-}a_{1}\lhd^{-}a_{0}$

such that $\gamma(a_{0})=y$, $t\in\delta\gamma(a_{n})$, for $n\in{\omega}$.

By assumption
$y\not\in\delta\gamma(\alpha)=\delta\delta(\alpha)-\gamma\delta(\alpha)$. So
$y\in\gamma\delta(\alpha)$. Hence there is $a_{0}\in\delta(\alpha)$ such, that
$\gamma(a_{0})=y$. Now, suppose that the lower $\delta(\alpha)$-path

$a_{k}\lhd^{-}a_{k-1}\lhd^{-}\ldots\lhd^{-}a_{0}$

has been already constructed. By globularity, we can pick $z\in\delta(a_{k})$,
such that $t\in\delta(z)$. By assumption
$z\not\in\delta\gamma(\alpha)=\delta\delta(\alpha)-\gamma\delta(\alpha)$. So
$z\in\gamma\delta(\alpha)$. Hence there is $a_{k+1}\in\delta(\alpha)$ such,
that $\gamma(a_{k+1})=z\in\delta(a_{k})$. Clearly,
$t\in\delta\gamma(a_{k+1})$. This ends the construction of the path. But by
strictness such a path has to be finite, so there is $x$ as needed.

Ad 4. The essential case is when $a\lhd^{+}b$. This follows from 1. Then use
the induction. $~{}~{}\Box$

###### Lemma 5.3

Let $S$ be a positive face structure, $n>1$, $\alpha\in S_{n+1}$, and $a,b\in
S_{n}$ such that $a<^{+}b$. Then

  1. 1.

$\iota\delta(\alpha)=\iota\gamma(\alpha)$ ;

  2. 2.

$\iota(a)\subseteq\iota(b)$;

  3. 3.

$\iota(a)\cup\gamma\gamma(a)\subseteq\iota(b)\cup\gamma\gamma(b)$;

  4. 4.

$\delta\delta(a)\subseteq\delta\delta(b)$;

  5. 5.

$\theta\theta(a)\subseteq\theta\theta(b)$.

Proof. Ad 1. $\iota\delta(\alpha)\subseteq\iota\gamma(\alpha)$:

Fix $a\in\delta(\alpha)$ and $t\in\iota(a)$. Thus there are $x,y\in\delta(a)$
such, that $\gamma(x)=t\in\delta(y)$. By Lemma 5.2 2,3 there are
$x^{\prime},y^{\prime}\in\delta\gamma(\alpha)$ such, that
$x^{\prime}\leq^{+}x$, $y^{\prime}\leq^{+}y$ and
$\gamma(x^{\prime})=t\in\delta(y^{\prime})$. Thus $t\in\iota\gamma(\alpha)$
and the first inclusion is proved.

$\iota\delta(\alpha)\supseteq\iota\gamma(\alpha)$:

Fix $t\in\iota\gamma(\alpha)$. In particular, there are
$x,y\in\delta\gamma(\alpha)$, so that $\gamma(x)=t\in\delta(y)$. Suppose that
$t\not\in\iota\delta(\alpha)$. We shall build an infinite
$\delta(\alpha)$-path

$a_{1}\lhd^{-}a_{2}\ldots$

such that $\gamma\gamma(a_{i})=t$ for $i\in{\omega}$.

Since $\delta\gamma(\alpha)\subseteq\delta\delta(\alpha)$, there is
$a_{1}\in\delta(\alpha)$ such, that $x\in\delta(a_{1})$. Since
$t\not\in\iota\delta(\alpha)$, it follows that $\gamma\gamma(a_{1})=t$.
Suppose now that we have already constructed the path

$a_{1}\lhd^{-}a_{2}\ldots a_{k}$

as above. We have
$\gamma\gamma(a_{k})=t\lhd^{+}\gamma(y)\leq^{+}\gamma\gamma\gamma(\alpha)$. So
$\gamma(a_{k})\neq\gamma\gamma(\alpha)$ and
$\gamma(a_{k})\in\delta\delta(\alpha)$. Then there is
$a_{k+1}\in\delta(\alpha)$ such, that $\gamma(a_{k})\in\delta(a_{k+1})$.
Again, as $t\not\in\iota\delta(\alpha)$, it follows that
$\gamma\gamma(a_{k+1})=t$. This ends the construction of the path. Since, by
strictness, such a path cannot exists we get the other inclusion.

Ad 2. Since the inclusion is transitive, it is enough to consider the case
$a\lhd^{+}b$, i.e. there is an $\alpha\in S_{n+1}$ such, that
$a\in\delta(\alpha)$ and $b=\gamma(\alpha)$. Then by 1. we have

$\iota(a)\subseteq\iota\delta(\alpha)=\iota\gamma(\alpha)=\iota(b)$

Ad 3. As above it is enough to consider the case $a\lhd^{+}b$, i.e. that there
is $\alpha\in S_{n+1}$ such that $a\in\delta(\alpha)$ and $\gamma(\alpha)=b$.
By 2. we need to show that
$\gamma\gamma(a)\subseteq\iota(b)\cup\gamma\gamma(b)$. We have

$\gamma\gamma(a)\in\gamma\gamma\delta(\alpha)\subseteq\iota\gamma(\alpha)\cup\gamma\gamma\gamma(\alpha)=\iota(b)\cup\gamma\gamma(b).$

Ad 4. Again it is enough to consider the case $a\lhd^{+}b$, i.e. that there is
$\alpha\in S_{n+1}$ such that $a\in\delta(\alpha)$ and $\gamma(\alpha)=b$. We
have

$\delta\delta(a)\subseteq\delta\delta\delta(\alpha)=\delta\delta\gamma(\alpha)=\delta\delta(b).$

Ad 5. This follows from 3. and 4. $~{}~{}\Box$

### Global properties

###### Lemma 5.4

Let $S$ be a positive face structure, $n\in{\omega}$, $a,b\in S_{n}$,
$a<^{+}b$. Then, there is an upper $S_{n+1}-\gamma(S_{n+2})$-path from $a$ to
$b$.

Proof. Let $a,\alpha_{1},\ldots,\alpha_{k},b$ be an upper path is $S$. By
Lemma 5.2 we can replace each face $\alpha_{i}$ in this path which is not in
$S-\gamma(S)$ by a sequence of faces which are $<^{+}$-smaller. Just take
$\Gamma\in S_{n+2}$, such that $\gamma(\Gamma)=\alpha_{i}$ and take instead of
$\alpha_{i}$ a path in $\delta(\Gamma)$ from $\gamma(\alpha_{i-1})$ (if $i=0$
then from $a$) to $\gamma(\alpha_{i})$. Repeated application of this procedure
will eventually yield the required path. $~{}~{}\Box$

###### Lemma 5.5

Let $S$ be a positive face structure, $n>0$, $a\in S_{n}$, $\alpha\in
S_{n+1}$, and either $\gamma(a)\in\iota(\alpha)$ or
$\delta(a)\cap\iota(\alpha)\neq\emptyset$. Then $a<^{+}\gamma(\alpha)$.
Moreover, if $\alpha\in S-\gamma(S)$ then there is a unique
$a^{\prime}\in\delta(\alpha)$ such that $a\leq^{+}a^{\prime}$.

Proof. If $a\in\delta(\alpha)$ there is nothing to prove. So we assume that
$a\not\in\delta(\alpha)$.

We begin with the second part of Lemma, i.e. we assume that $\alpha\in
S_{n+1}-\gamma(S_{n+2})$.

Let $\gamma(a)\in\iota(\alpha)$. Thus there are $b,c\in\delta(\alpha)$ such
that $\gamma(a)=\gamma(b)\in\delta(c)$. In particular $a<^{-}c$. By
$\gamma$-linearity either $b<^{+}a$ or $a<^{+}b$. Suppose that $b<^{+}a$. Then
we have an $(S-\gamma(S))$-upper path $b,\beta_{0},\ldots,\beta_{r},a$. As
$b\in\alpha\cap\beta_{0}$ and $\alpha,\beta_{0}\in S-\gamma(S)$, we have
$\alpha=\beta_{0}$. But then $c\in\delta(\alpha)=\delta(\beta_{0})$ and hence
$c<^{+}\gamma(\beta_{0})\leq^{+}a$. So we get $a<^{-}c$ and $c<^{+}a$
contradicting the disjointness of $\perp^{+}$ and $\perp^{-}$. Thus we can put
$a^{\prime}=b$ and we have $a<^{+}a^{\prime}$. The uniqueness of $a^{\prime}$
follows from the fact that $\gamma(a)=\gamma(a^{\prime})$.

The case $\delta(a)\cap\iota(\alpha)\neq\emptyset$ is similar and we put it
for completeness. Thus there are $b,c\in\delta(\alpha)$ such that
$\gamma(b)\in\delta(a)\cap\delta(c)$. In particular $b<^{-}a$. By
$\delta$-linearity either $c<^{+}a$ or $a<^{+}c$. Suppose that $c<^{+}a$. Then
we have an $(S-\gamma(S))$-upper path $c,\beta_{0},\ldots,\beta_{r},a$. As
$c\in\alpha\cap\beta_{0}$ and $\alpha,\beta_{0}\in S-\gamma(S)$, we have
$\alpha=\beta_{0}$. But then $b\in\delta(\alpha)=\delta(\beta_{0})$ and hence
$b<^{+}\gamma(\beta_{0})\leq^{+}a$. So we get $b<^{-}a$ and $b<^{+}a$
contradicting the disjointness of $\perp^{+}$ and $\perp^{-}$. Thus we can put
$a^{\prime}=c$ and we have $a<^{+}a^{\prime}$. The uniqueness of $a^{\prime}$
follows from the fact that $\gamma(b)\in\delta(a^{\prime})$ and
$a^{\prime}\in\delta(\alpha)$ and Lemma 5.1.

The first part of the Lemma follows from the above, Lemma 5.2.4 and the
following Claim.

Claim. If $\alpha\in S_{n+1}$ and $x\in\iota(\alpha)$ then there is an
$\alpha^{\prime}\in S_{n+1}$ such that $\alpha^{\prime}\leq^{+}\alpha$,
$x\in\iota(\alpha^{\prime})$ and $\alpha^{\prime}\not\in\gamma(S_{n+2})$.

Proof of the Claim. Suppose that Claim is not true. To get a contradiction, we
shall build an infinite descending $\gamma(S_{n+2})$-path

$\ldots\lhd^{+}\alpha_{1}\lhd^{+}\alpha_{0}=\alpha$

such that $x\in\iota(\alpha_{i})$, for $i\in{\omega}$.

We put $\alpha_{0}=\alpha$. Suppose that we have already constructed
$\alpha_{0},\ldots,\alpha_{k}\in\gamma(S_{n+2})$. Hence there is $\beta\in
S_{n+2}$ such, that $\gamma(\beta)=\alpha_{k}$. Since
$\iota\delta(\beta)=\iota\gamma(\beta)=\iota(\alpha_{k})$, there is
$\alpha_{k+1}\in\delta(\beta)$ such, that $x\in\iota(\alpha_{k+1})$. This ends
the construction of the infinite path and the proof of the Claim and the
Lemma. $~{}~{}\Box$

###### Corollary 5.6

Let $S$ be a positive face structure. If $a\in S-\delta(S)$ then $\gamma(a)\in
S-\iota(S)$ and $\delta(a)\subseteq S-\iota(S)$.

Proof. Let $a\in S_{n}$ and $\alpha\in S_{n+2}$. If either
$\gamma(a)\in\iota(\alpha)$ or $\delta(a)\cap\iota(\alpha)\neq\emptyset$ then
by Lemma 5.5 $a<^{+}\gamma(\alpha)$. Thus $a\in\delta(S)$. $~{}~{}\Box$

A lower path $b_{0},\ldots,b_{m}$ is a maximal path if
$\delta(b_{0})\subseteq\delta(S)-\gamma(S)$ and
$\gamma(b_{m})\in\gamma(S)-\delta(S)$, i.e. if it can’t be extended either
way.

######  Lemma 5.7 (Path Lemma)

Let $k\geq 0$, $B=(a_{0},\ldots,a_{k})$ be a maximal $S_{n}$-path in a
positive face structure $S$, $b\in S_{n}$, $0\leq s\leq k$, $a_{s}<^{+}b$.
Then there are $0\leq l\leq s\leq p\leq k$ such, that

  1. 1.

$a_{i}<^{+}b$ for $i=l,\ldots,p$;

  2. 2.

$\gamma(a_{p})=\gamma(b)$;

  3. 3.

either $l=0$ and $\delta(a_{0})\subseteq\delta(b)$ or $l>0$ and
$\gamma(a_{l-1})\in\delta(b)$;

  4. 4.

$\gamma(a_{i})\in\iota(S)$, for $l\leq i<p$.

Proof. Let $0\leq l\leq p\leq k$ be such that $a_{i}<^{+}b$ for $l\leq i\leq
p$ and either $l=0$ or $a_{l-1}\not<^{+}b$ and either $p=k$ or
$a_{k+1}\not<^{+}b$. We shall show that $l$ and $p$ have the properties stated
in the Lemma. From the very definition the property $1$ holds.

We shall show 2. Take an upper $(S-\gamma(S))$-path from $a_{p}$ to $b$:
$a_{p},\beta_{0},\ldots,\beta_{r},b$. If
$\gamma(a_{p})=\gamma\gamma(\beta_{i})$, for $i=0,\ldots,r$ then
$\gamma(a_{p})=\gamma\gamma(\beta_{r})=\gamma(b)$ and we are done. So suppose
contrary and let

$i_{0}=min\\{i:\gamma(a_{p})\neq\gamma\gamma(\beta_{i})\\}$

Then there are $a,c\in\delta(\beta_{i_{0}})$ such that
$\gamma(a_{p})=\gamma(a)\in\delta(c)$ (NB. $a=a_{p}$ if $i_{0}=0$ and
$a=\gamma(\beta_{i_{0}-1})$ otherwise). In particular
$\gamma(a_{p})\in\iota(\beta_{i_{0}})$. As $\gamma(a_{p})\in\delta(S)$, we
have $p<k$. Thus $\gamma(a_{p})\in\delta(a_{p+1})\cap\iota(\beta_{i_{0}})$,
and by Lemma 5.5 $a_{p+1}<^{+}c<^{+}b$. But this contradicts the choice of
$p$. So the property 2. holds.

Now we shall show 3. Take an upper $(S-\gamma(S))$-path from $a_{l}$ to $b$:
$a_{l},\beta_{0},\ldots,\beta_{r},b$. We have two cases: $l=0$ and $l>0$.

If $l=0$ then there is no face $a\in S$ such that $\gamma(a)\in\delta(a_{l})$.
As $\delta(a_{l})\subseteq\delta\delta(\beta_{0})$ we must have
$\delta(a_{l})\in\delta\gamma(\beta_{i})$, for $i=0,\ldots,r$. Hence
$\delta(a_{l})\subseteq\delta\gamma(\beta_{r})=\delta(b)$ and 3. holds in this
case.

Now suppose that $l>0$. If $\gamma(a_{l-1})\in\delta\gamma(\beta_{i})$, for
$i=0,\ldots,r$, then $\gamma(a_{l-1})\in\delta\gamma(\beta_{r})=\delta(b)$ and
3. holds again. So suppose contrary , and let

$i_{1}=min\\{i:\gamma(a_{l-1})\not\in\delta\gamma(\beta_{i})\\}$

Then there are $a,c\in\delta(\beta_{i_{1}})$ such that
$\gamma(a_{l-1})=\gamma(a)\in\delta(c)$ (NB: $c=a_{l}$ if $i_{1}=0$ and
$c=\gamma(\beta_{i_{1}-1})$ otherwise). In particular
$\gamma(a_{l-1})\in\iota(\beta_{i_{1}})$, and by Lemma 5.5 we have
$a_{l-1}<^{+}a<^{+}b$ contrary to the choice of $l$. Thus 3. holds in this
case as well.

Finally, we shall show 4. Let $l\leq j\leq p$ and
$a_{j},\beta_{0},\ldots,\beta_{r},b$ be an upper $(S-\delta(S))$-path from
$a_{j}$ to $b$. As $a_{j}<^{-}a_{p}$ and $a_{p}<^{+}b$ we have
$\gamma(a_{j})\neq\gamma(b)$. So we can put

$i_{2}=min\\{i:\gamma(a_{j})\neq\gamma\gamma(\beta_{i})\\}$

But then
$\gamma(a_{j})\in\gamma\delta(\beta_{i_{2}})-\gamma\gamma(\beta_{i_{2}})=\iota(\beta_{i_{2}})$.
Therefore $\gamma(a_{j})\in\iota(S)$ and 4. holds. $~{}~{}\Box$

###### Lemma 5.8

Let $S$ be a positive face structure, $n\in{\omega}$, $x,y\in S_{n}$,
$x<^{+}y$. Then if $x,y\not\in\iota(S_{n+2})$, then there is an upper path
from $x$ to $y$ contained in $S_{n+1}-\delta(S_{n+2})$.

Proof. Assume $x,y\in(S-\iota(S))$ and $x<^{+}y$. Let

$x,b_{0},\ldots,b_{k},y$

be an upper path from $x$ to $y$ with the longest possible initial segment
$b_{0},\ldots,b_{l}$ in $S-\delta(S)$.

We need to show that $k=l$. So suppose contrary, that $l<k$. By Corollary 5.6
$\gamma(b_{l})not\in\iota(S)$. Let $\alpha$ be a face such that
$b_{l+1}\in\iota(\alpha)$. In particular $b_{l+1}<^{+}\gamma(\alpha)$. As
$\gamma(b_{l})\not\in\iota(S)$ we have
$\gamma(b_{l})\in\delta\gamma(\alpha)\;(=\delta\delta(\alpha)-\iota(\alpha))$.
So $\gamma(b_{l})\in\delta(b_{l+1}\cap\delta\gamma(\alpha)$.

Let $a$ be the $<^{+}$-largest element of the set $\\{b\in
S:\gamma(b_{l})\in\delta(b)\\}$. Then $b_{l+1}\leq a$. By a similar argument
as above for $b_{l+1}$ we get that $a\not\in\delta(S)$. By Lemma 5.7 there is
$p$ such that $l+1\leq p\leq k$ such that $\gamma(b_{p})=\gamma(a)$ (Note that
the facts that $y\not\in\iota(S)$ and Lemma 5.7.4 are needed here.). Thus we
have an upper path from $x$ to $y$,
$x,b_{0},\ldots,b_{l},a,b_{p+1},\ldots,b_{k},y$ with a longer initial segment
in $S-\delta(S)$. But this means in fact that in fact $l=k$, as required.
$~{}~{}\Box$

### Order

###### Lemma 5.9

Let $S$ be a positive face structure, $n\in{\omega}$, $a,b\in S_{n}$. Then we
have

  1. 1.

If $a<^{+}b$ then for any $x\in\delta(a)$ there is $y\in\delta(b)$ such that
$y\leq^{+}x$;

  2. 2.

If $a<^{+}b$ and $\gamma(a)=\gamma(b)$ then for any $y\in\delta(b)$ there is
$x\in\delta(a)$ such that $y\leq^{+}x$;

  3. 3.

If $\gamma(a)=\gamma(b)$ then either $a=b$ or $a\perp^{+}b$;

  4. 4.

If $\gamma(a)<^{+}\gamma(b)$ then either $a<^{+}b$ or $a<^{-}b$;

  5. 5.

If $a<^{+}b$ then $\gamma(a)\leq^{+}\gamma(b)$;

  6. 6.

If $a<^{-}b$ then $\gamma(a)<^{+}\gamma(b)$;

  7. 7.

If $\gamma(a)\perp^{-}\gamma(b)$ then $a\not\perp^{-}b$ and $a\not\perp^{+}b$.

Proof. Ad 1. Let $a<^{+}b$ and $x\in\delta(a)$. We have two cases: either
$x\in\gamma(S)$ or $x\not\in\gamma(S)$.

In the first case there is $a^{\prime}\in$, such that $\gamma(a^{\prime})=x$.
Let $a_{0},\ldots,a_{k}$ be a maximal path containing $a^{\prime},a$, say
$a_{s-1}=a^{\prime}$ and $a_{s}=a$, where $0<s\leq k$. As $a_{s}<^{+}b$, by
Lemma 5.7 there is $l\leq s$ and $y\in\delta(a_{l})\cap\delta(b)$. Clearly,
$y\leq^{+}x$.

In the second case consider an upper path from $a$ to $b$:
$a,\beta_{0},\ldots,\beta_{r},b$. We have
$x\in\delta(a)\subseteq\delta\delta(\beta_{0})$. As $x\not\in\gamma(S)$ so
$x\not\in\gamma\delta(\beta_{0})$, and hence
$x\in\delta\delta(\beta_{0})-\gamma\delta(\beta_{0})=\delta\gamma(\beta_{0})$.
Thus we can define

$r^{\prime}=max\\{i:x\in\delta\gamma(\beta_{i})\\}$

If we had $r^{\prime}<r$ then again we would have
$x\in\delta\delta(\beta_{r^{\prime}+1})-\gamma\delta(\beta_{r^{\prime}+1})=\delta\gamma(\beta_{r^{\prime}+1})$,
contrary to the choice of $r^{\prime}$. So $r^{\prime}=r$ and
$x\in\delta\gamma(\beta_{r})=\delta(b)$. Thus we can put $y=x$.

Ad 2. Fix $a<^{+}b$ such that $\gamma(a)=\gamma(b)$ and $y\in\delta(b)$. We
need to find $x\in\delta(a)$ with $y\leq^{=}$. Take a maximal
$(S-\gamma(S))$-path $a_{0},\ldots,a_{k}$ passing through $y$, i.e. there is
$0\leq j\leq k$ such that $y\in\delta(a_{j})$ and if $y\in\gamma(S)$ then
moreover $j>0$ and $\gamma(a_{j-1})=y$. Since $a_{j}\not\in\gamma(S)$ by
$\delta$-linearity $a_{j}<^{+}b$. Thus by Lemma 5.7 there is $j\leq p\leq k$
such that $\gamma(a_{p})=\gamma(b)=\gamma(a)$. Since $a_{p}\not\in\gamma(S)$
by $\gamma$-linearity we have $a_{p}\leq^{+}a$. If $a_{p}=a$ then we can take
as the face $x$ either $y$ if $p=0$ or $\gamma(a_{p-1})$ if $p>0$. So assume
now that $a_{p}<^{+}a$. Again by Lemma 5.7 there is $0\leq l\leq p$ such that
either $l=0$ and $\delta(a_{0})\subseteq\delta(a)$ or $l>0$ and
$\gamma(a_{l-1})\in\delta(a)$. As $a_{l}$ is the first face in the path
$a_{0},\ldots,a_{k}$ such that $a_{l}<^{+}a$ and $a_{j}$ is the first face in
the path $a_{0},\ldots,a_{k}$ such that $a_{l}<^{+}b$ and moreover $a<^{+}b$
it follows that $j\leq l$. Thus in this case we can take as the face $x$
either $y$ if $l=0$ or $\gamma(a_{l-1})$ if $l>0$.

Ad 3. This is an immediate consequence of $\gamma$-linearity.

Ad 4. Suppose $\gamma(a)<^{+}\gamma(b)$. So there is an upper path

$\gamma(a),c_{1},\ldots,c_{k},\gamma(b)$

with $k>0$. We put $c_{0}=a$. We have $\gamma(c_{k})=\gamma(b)$ so by
$\gamma$-linearity $c_{k}\perp^{+}b$. So we have two cases: either
$b<^{+}c_{k}$ or $c_{k}<^{+}b$.

If $b<^{+}c_{k}$ then by Lemma 5.7 for any maximal path that contains $b$ and
the face $c_{k}$ we get that $c_{k-1}<^{-}b$. Thus we have $a<^{-}b$.

If $c_{k}<^{+}b$ then by Lemma 5.7 for any maximal path that extends
$c_{0},c_{1},\ldots,c_{k}$ and face $b$ we get that either there is $0\leq
i<k$ such that $\gamma(c_{i})\in\delta(b)$ and then $a<^{-}b$ or else
$a=c_{0}<^{+}b$.

Ad 5. This is repeated from Lemma 5.2.

Ad 6. Suppose $a<^{-}b$. Then there is a lower path

$a=a_{0},a_{1},\ldots,a_{k}=b$

with $k>0$. Then we have an upper path

$\gamma(a)=\gamma(a_{0}),a_{1},\ldots,a_{k},\gamma(a_{k})=\gamma(b).$

Hence $\gamma(a)<^{+}\gamma(b)$.

Ad 7. Easily follows from 5 and 6. $~{}~{}\Box$

###### Proposition 5.10

Let $S$ be a positive face structure, $a,b\in S_{n}$, $a\neq b$. Let
$\\{a_{i}\\}_{0\leq i\leq n}$, $\\{b_{i}\\}_{0\leq i\leq n}$ be two sequences
of codomains of $a$ and $b$, respectively, so that

$a_{i}=\gamma^{(i)}(a)\hskip 56.9055ptb_{i}=\gamma^{(i)}(b)$

(i.e. $dim(a_{i})=i$), for $i=0,\ldots,n$. Then there are two numbers $0\leq
l\leq k\leq n$ such that either

  1. 1.

$a_{i}=b_{i}$ for $i<l$,

  2. 2.

$a_{i}<^{+}b_{i}$ for $l\leq i\leq k$,

  3. 3.

$a_{i}<^{-}b_{i}$ for $k+1=i\leq n$,

  4. 4.

$a_{i}\not\perp b_{i}$ for $k+2\leq i\leq n$,

or the roles of $a$ and $b$ are interchanged.

Proof. The above conditions we can present more visually as:

$a_{0}=b_{0},\ldots,a_{l-1}=b_{l-1},\hskip 14.22636pta_{l}<^{+}b_{l},\ldots
a_{k}<^{+}b_{k},$ $a_{k+1}<^{-}b_{k+1},\hskip 14.22636pta_{k+2}\not\perp
b_{k+2},\ldots,a_{n}\not\perp b_{n}.$

These conditions we will verify from the bottom up. Note that by strictness
$<^{S_{0},+}$ is a linear order. So either $a_{0}=b_{0}$ or
$a_{0}\perp^{+}b_{0}$. In the later case $l=0$. As $a\neq b$ then there is
$i\leq n$ such that $a_{i}\neq b_{i}$. Let $l$ be minimal such, i.e.
$l=min\\{i:a_{i}\neq b_{i}\\}$. By Lemma 5.9 3., $a_{l}\perp^{+}b_{l}$. So
assume that $a_{l}<^{+}b_{l}$. We put $k=max\\{i\leq n:a_{i}<^{+}b_{i}\\}$. If
$k=n$ we are done. If $k<n$ then by Lemma 5.9 4., we have
$a_{k+1}<^{-}b_{k+1}$. Then if $k+1<n$, by Lemma 5.9 5\. 6. 7.,
$a_{i}\not\perp b_{i}$ for $k+2\leq i\leq n$. This ends the proof.
$~{}~{}\Box$

From the above Proposition we get immediately

###### Corollary 5.11

Let $S$ be a positive face structure, $a,b\in S_{n}$, $a\neq b$. Then either
$a\perp^{+}b$ or there is a unique $0\leq l\leq k$ such that
$a\perp^{-}_{l}b$, but not both.

The above Corollary allows us to define an order $<^{S}$ (also denoted $<$) on
all cells of $S$ as follows. For $a,b\in S_{n}$,

$a<^{S}b\;\;\;{\rm iff}\;\;\;a<^{+}b\;\;{\rm
or}\;\;\exists_{l}\;\;a<^{-}_{l}b.$

###### Corollary 5.12

For any positive face structure $S$, and $n\in{\omega}$, the relation $<^{S}$
restricted to $S_{n}$ is a linear order.

Proof. We need to verify that $<^{S}$ is transitive.

Let $a,b,c\in S_{n}$. There are some cases to consider.

If $a<^{+}b<^{+}c$ then clearly $a<^{+}c$.

If $a<^{+}b<^{-}_{l}c$ then we have
$\gamma^{(l)}(a)<^{+}\gamma^{(l)}(b)<^{-}\gamma^{(l)}(c)$. By Lemma 5.2 4.,
and transitivity of $<^{-}$ we have $\gamma^{(l)}(a)<^{-}\gamma^{(l)}(c)$.
Hence $a<^{-}_{l}c$.

If $a<^{-}_{l}b<^{+}c$ then, by use Lemma 5.7, either the maximal lower path
from that contains both $\gamma^{(l)}(a)$ and $\gamma^{(l)}(b)$ passes between
$\gamma^{(l)}(a)$ and $\gamma^{(l)}(b)$ through a face in
$\delta\gamma^{(l+1)}(c)$ and $a<^{-}_{l}c$ or it does not passes through a
face $\delta\gamma^{(l+1)}(c)$ and then $\gamma^{(l)}(a)<^{+}\gamma^{(l)}(c)$.
Thus by Proposition 5.10 either there is $l\leq l^{\prime}\leq n$ such that
$\gamma^{(l^{\prime})}(a)<^{+}\gamma^{(l^{\prime})}(c)$ and hence
$a<^{-}_{l^{\prime}}c$ or $a<^{+}c$.

Finally suppose that $a<^{-}_{k}b<^{-}_{l}c$.

If $k=l$ then clearly $a<_{l}c$.

If $k>l$ then $\gamma^{(l)}(a)<^{+}\gamma^{(l)}(b)<^{-}\gamma^{(l)}(c)$ and,
by the previous argument, $\gamma^{(l)}(a)<^{-}\gamma^{(l)}(c)$, i.e.
$a<^{-}_{l}c$.

Finally, assume that $k<l$, i.e.
$\gamma^{(k)}(a)<^{-}\gamma^{(k)}(b)<^{+}\gamma^{(k)}(c)$. Then, by Path
Lemma, either $\gamma^{(k)}(a)<^{-}\gamma^{(k)}(c)$ or
$\gamma^{(k)}(a)<^{+}\gamma^{(k)}(c)$. In the latter case, by Proposition
5.10, either $a<^{+}b$ or there is $k^{\prime}$, such that $k<k^{\prime}\leq
n$ and $\gamma^{(k^{\prime})}(a)<^{+}\gamma^{(k^{\prime})}(c)$. In any case we
have $a<c$, as required. $~{}~{}\Box$

From the proof of the above corollary we get

###### Lemma 5.13

Let $S$ be a positive face structure, $a\in S_{n}$, $a$. Then the set

$\\{b\in S_{n}\,:\,a\leq^{+}b\\}$

is linearly ordered by $\leq^{+}$.

Proof. Suppose $a\leq^{+}b,b^{\prime}$. If we where to have
$b<^{-}_{l}b^{\prime}$ for some $l\leq n$ then, by Corollary 5.12 we would
have $a<^{-}_{l}b^{\prime}$ which is a contradiction. $~{}~{}\Box$

###### Corollary 5.14

Any morphism of positive face structures is one-to-one. Moreover any
automorphism of positive face structures is an identity.

Proof. By Corollary 5.12, the (strict, linear in each dimension) order $<^{S}$
is defined internally using relations $<^{-}$ and $<^{+}$ that are preserved
by any morphism. Hence $<^{S}$ must be preserved by any morphism, as well.
From this observation the Corollary follows. $~{}~{}\Box$

###### Lemma 5.15

Let $S$ be a positive face structure, $a,b\in S_{n}$, $a$. Then

  1. 1.

if $\iota(a)\cap\iota(b)\neq\emptyset$ then $a\perp^{+}b$;

  2. 2.

if $\emptyset\neq\iota(a)\subseteq\iota(b)\neq\iota(a)$ then $a<^{+}b$;

  3. 3.

if $a\perp^{-}b$ then $\iota(a)\cap\iota(b)=\emptyset$.

Proof. 2\. is an easy consequence of 1. and Lemma 5.3. 3. is an easy
consequence of 1. and Disjointness. We shall show 1.

Assume that $u\in\iota(a)\cap\iota(b)$. Thus there are $x,y\in\delta(a)$ and
$x^{\prime},y^{\prime}\in\delta(b)$ such that
$\gamma(x)=\gamma(x^{\prime})=u\in\delta(y)\cap\delta(y^{\prime})$. If
$x=x^{\prime}$ then by Local linearity $a\perp^{+}b$, as required. So assume
that $x\neq x^{\prime}$. Again by Local linearity $x\perp^{+}x^{\prime}$, say
$x^{\prime}\perp^{+}x$. Thus there is a $T-\gamma(T)$-path
$x^{\prime},a_{1},\ldots,a_{k},x$. As, for $i=1,\ldots,k$,
$\gamma\gamma(a_{i})=u$ and $\gamma\gamma(b)\not\in\iota(b)\ni u$, we have
that $\gamma(a_{i})\neq\gamma(b)$ and $a_{i}\neq b$. Once again by Local
linearity $a_{0}\perp^{+}b$ and by Path Lemma $a_{i}<b$, for $i=1,\ldots,k$
with $\gamma(a_{k})\neq\gamma(b)$. As $\gamma(a_{k})=x\in\delta(a)$, again by
Path Lemma $a<^{+}b$, as well. $~{}~{}\Box$

###### Proposition 5.16

Let $S$ be a positive face structure, $a,b\in S_{k}$, $\alpha\in S_{k+1}$, so
that $\alpha$ is a $<^{+}$-minimal element in $S_{k+1}$, and
$a\in\delta(\alpha)$, $b=\gamma(\alpha)$. Then $b$ is the $<^{+}$-successor of
$a$.

Proof. Assume that $\alpha$ is a $<^{+}$-minimal element in $S_{k+1}$. Suppose
that there is $c\in S_{k}$ such that $a<^{+}c<^{+}b$. Thus we have an upper
path

$a,\beta_{1},\ldots,\beta_{i},c,\beta_{i+1},\ldots,\beta_{l},b.$

Hence $\beta_{1}<^{-}\beta_{l}$. Moreover
$a\in\delta(\beta_{1})\cap\delta(\alpha)$ and
$\gamma(\beta_{l})=b=\gamma(\alpha)$. Thus both $\beta_{1}$ and $\beta_{l}$
are $<^{+}$-comparable with $\alpha$. Since $\alpha$ is $<^{+}$-minimal we
have $\alpha<^{+}\beta_{1},\beta_{l}$. By Lemma 5.13,
$\beta_{1}\perp^{+}\beta_{l}$. But then we have
$(\beta_{1},\beta_{l})\in\perp^{+}\cap\perp^{-}\neq\emptyset$, contradicting
strictness. $~{}~{}\Box$

###### Proposition 5.17

Let $T$ be a positive face structure and $X\subseteq T$ a subhypergraph of
$T$. Then $X$ is a positive face structure iff the relation $<^{X,+}$ is the
restriction of $<^{T,+}$ to $X$.

Proof. Assume that $X$ is a subhypergraph of a positive face structure $T$.
Then $X$ satisfies axioms of globularity, disjointness, and strictness of the
relations $<^{X_{k},+}$ for $k>0$.

Clearly, if $<^{X_{k},+}\;=\;<^{T_{k},+}\cap(X_{k})^{2}$ then the relation
$<^{X_{0},+}$ is linear, the relations $<^{X_{k},+}$, for $k>0$, satisfy
pencil linearity, i.e. $X$ is a positive face structure.

Now we assume that the subhypergraph $X$ of positive face structure $T$ is a
positive face structure. We shall show that for $k\in{\omega}$, $a,b\in X_{k}$
we have $a<^{X_{k},+}b$ iff $a<^{T_{k},+}b$. Since $X$ is a subhypergraph
$a<^{X_{k},+}b$ implies $a<^{T_{k},+}b$. Thus it is enough to show that if
$a<^{T_{k},+}b$ then $a\perp^{X_{k},+}b$. We shall prove this by induction on
$k$. For $k=0$ it is obvious, since $<^{X_{0},+}$ is linear. So assume that
for faces $x,y\in X_{l}$, with $l<k$ we already know that $x<^{X_{l},+}y$ iff
$x<^{T_{l},+}y$. Fix $a,b\in T_{k}$ such that $a<^{T_{k},+}b$. Then by Lemma
5.9.2 $\gamma(a)\leq^{T_{k-1},+}\gamma(b)$ and hence by inductive hypothesis
$\gamma(a)\leq^{X_{k-1},+}\gamma(b)$. Thus we have an upper $X$-path
$a=a_{r},\gamma(a),a_{r-1}\ldots,a_{1},\gamma(b)$, with $r\geq 1$. As
$a<^{T_{k},+}b$, by Path Lemma $a_{i}<^{T_{k},+}b$ for $i=1,\ldots,r$. Again
by induction on $r$ we shall show that $a_{i}<^{X_{k},+}b$, for
$i=1,\ldots,r$. As $\gamma(a_{1})=\gamma(b)$ by pencil linearity we have
$a_{1}\perp^{X_{k},+}b$. So $a_{1}<^{X_{k},+}b$. Suppose that
$a_{i}<^{X_{k},+}b$, for $i\leq l<r$. Let
$a_{l},\alpha_{s},\ldots,\alpha_{1},b$ be an upper $T$-path. As
$a_{l+1}<^{T_{k},+}b$, we cannot have $\gamma(a_{l+1})\in\delta(b)$.
Therefore, for some $j\geq 1$, $\gamma(a_{l+1})\in\iota(\alpha_{j})$. So by
Lemma 5.5, $a_{l+1}<^{X_{k},+}\gamma(\alpha_{j})\leq^{X_{k},+}b$, as required.
$~{}~{}\Box$

###### Lemma 5.18

Let $T$ be a positive face structure, $a,b,\alpha\in T$. If
$a\in\delta(\alpha)$ and $a<^{+}b<^{+}\gamma(\alpha)$ then $b\in\iota(T)$.

Proof. Assume that $a,b,\alpha\in T$ are as in the assumption of the Lemma.
Thus we have an upper path $a,\alpha_{0},\ldots,\alpha_{r},b$. As
$a\in\delta(\alpha)\cap\delta(\alpha_{0})$, by pencil linearity we have
$\alpha\perp^{+}\alpha_{0}$. If $\alpha<^{+}\alpha_{0}<^{-}\alpha_{r}$ then
$\gamma(\alpha)\leq^{+}\gamma(\alpha_{r})=b$ contradicting our assumption.
Thus $\alpha_{0}<^{+}\alpha$. Then by Path Lemma, since
$b=\gamma(\alpha_{r})<^{+}\gamma(\alpha)$, we have $\alpha_{r}<^{+}\alpha$ and
$b\in\iota(T)$, as required. $~{}~{}\Box$

### Some equations

###### Proposition 5.19

Let $S$ be a positive face structure $0<k\in{\omega}$. Then

  1. 1.

$\iota(S_{k+1})=\iota(S_{k+1}-\delta(S_{k+2}))$;

  2. 2.

$\delta(S_{k})=\delta(S_{k}-\gamma({S_{k+1}}))$;

  3. 3.

$\gamma(S_{k})=\gamma(S_{k}-\gamma({S_{k+1}}))$;

  4. 4.

$\delta(S_{k})=\delta(S_{k}-\iota({S_{k+2}}))$;

  5. 5.

$\delta(S_{k})=\delta(S_{k}-\delta({S_{k+1}}))\cup\iota(S_{k+1})$.

Proof. In all the above equations the inclusion $\supseteq$ is obvious. So in
each case we need to check the inclusion $\subseteq$ only.

Ad 1. Let $s\in\iota(S_{k+1})$, i.e. there is $a\in S_{k+1}$ such that
$s\in\iota(a)$. By strictness, there is $b\in S_{k+1}$ such that $a<^{+}b$ and
$b\not\in\delta(S_{k+1})$. By Lemma 5.3, we have

$s\in\iota(a)\subseteq\iota(b)\subseteq\iota(S_{k+1}-\delta(S_{k+2}))$

as required.

Ad 2. Let $x\in\delta(S_{k})$. Let $a\in S_{k}$ be the $<^{+}$-minimal element
in $S_{k}$ such that $x\in\delta(a)$. We shall show that $a\in
S_{k}-\gamma(S_{k+1})$. Suppose contrary that there is an $\alpha\in S_{k+1}$
such that $a=\gamma(\alpha)$. The by globularity

$x\in\delta(a)=\delta\gamma(\alpha)=\delta\delta(\alpha)-\gamma\delta(\alpha).$

So there is $b\in\delta(\alpha)$ such that $x\in\delta(b)$. As $b<^{+}a$ this
contradicts the minimality of $a$.

Ad 3. This is similar to the previous one but simpler.

Ad 4. Since $\iota({S_{k+2}})\subseteq\gamma({S_{k+1}})$ 4\. follows from 2.

Ad 5. Let $x\in\delta(S_{k})$. Let $a\in S_{k}$ be the $<^{+}$-maximal element
in $S_{k}$ such that $x\in\delta(a)$. If $a\not\in\delta(S_{k+1})$ then
$x\in\delta(S_{k}-\delta({S_{k+1}}))$ as required. So assume that
$a\in\delta(S_{k+1})$, i.e. there is $\alpha\in S_{k+1}$ such that
$a\in\delta(\alpha)$. Thus $x\in\delta\delta(\alpha)$. As
$a<^{+}\gamma(\alpha)$, by choice of $a$ we have
$x\not\in\delta\gamma(\alpha)\;(=\delta\delta(\alpha)-\gamma\delta(\alpha))$.
So $x\in\gamma\delta(\alpha)$ and hence
$x\in\iota(\alpha)\subseteq\iota(S_{k+1})$, as required. $~{}~{}\Box$

##  6 The ${\omega}$-categories generated by the positive face structures

Let $T^{*}$ ($T^{*}_{n}$) be the set of all face substructures of the face
structure $T$ (of dimension at most $n$). We introduce operations

${\bf d}^{(n)},{\bf c}^{(n)}:T^{*}\longrightarrow T^{*}_{n}$

of the $n$-th domain and the $n$-th codomain. For $S$ in $(T^{*})_{\geq n}$
the faces $n$-th domain ${\bf d}^{(n)}S$ are:

  1. 1.

$({\bf d}^{(n)}S)_{k}=\emptyset,$ for $k>n$,

  2. 2.

$({\bf d}^{(n)}S)_{n}=S_{n}-\gamma(S_{n+1})$,

  3. 3.

$({\bf d}^{(n)}S)_{k}=S_{k}$, for $k<n$.

and faces $n$-th codomain ${\bf c}^{(n)}S$ are:

  1. 1.

$({\bf c}^{(n)}S)_{k}=\emptyset,$ for $k>n$,

  2. 2.

$({\bf c}^{(n)}S)_{n}=S_{n}-\delta(S_{n+1})$,

  3. 3.

$({\bf c}^{(n)}S)_{n-1}=S_{n-1}-\iota(S_{n+1})$,

  4. 4.

$({\bf c}^{(n)}S)_{k}=S_{k}$, for $k<n-1$.

If $k\in{\omega}$, $dim(S)=k+1$, we write ${\bf d}S$ for ${\bf d}^{(k)}(S)$,
and ${\bf c}S$ for ${\bf c}^{(k)}(S)$.

We have

###### Lemma 6.1

Let $S$ be a positive face structure. Then

  1. 1.

if $dim(S)\geq k$ then ${\bf d}^{(k)}(S)$, ${\bf c}^{(k)}(S)$ are positive
face structures of dimension $k$;

  2. 2.

if $dim(S)>k$ then ${\bf d}{\bf d}^{(k+1)}(S)={\bf d}^{(k)}(S)$ , ${\bf d}{\bf
d}^{(k+1)}(S)={\bf d}^{(k)}(S)$;

  3. 3.

if $dim(S)>1$ then ${\bf d}{\bf d}S={\bf d}{\bf c}S$, ${\bf c}{\bf d}S={\bf
c}{\bf c}S$.

  4. 4.

For any $k\in{\omega}$ and $\alpha\in S_{k}$, the least sub-hypergraph of $S$
containing the face $\alpha$ is again a positive face structure of dimension
$k$; it is denoted by $[\alpha]$. Moreover, if $k>0$, then

${\bf c}[\alpha]=[\gamma(\alpha)],\hskip 28.45274pt{\bf
d}[\alpha]=[\delta(\alpha)]$

where $[\delta(\alpha)]$ is the least sub-hypergraph of $S$ containing the set
of face $\delta(\alpha)$.

Proof. Ad 1. Clearly, ${\bf d}^{(k)}S$ is a sub-hypergraph $S$ and ${\bf
c}^{(k)}S$ is a sub-hypergraph $S$ by Corollary 5.6. Any sub-hypergraph $T$ of
a positive face structure $S$ satisfies the conditions of globularity,
strictness (possibly without $<^{T_{0},+}$ being linear), and disjointness.

By Lemma 5.4, for $a,b\in{\bf d}^{(k)}S_{l}$ we have $a<^{S_{l},+}b$ iff
$a<^{{\bf d}^{(k)}S_{l},+}b$. Moreover, by Lemma 5.8, for $a,b\in{{\bf
c}^{(k)}S}_{l}$ we have $a<^{S_{l},+}b$ iff $a<^{{{\bf c}^{(k)}S}_{l},+}b$.
Hence by Lemma 5.17 both ${\bf d}^{(k)}S$ and ${\bf c}^{(k)}S$ are positive
face structures.

Ad 2. Fix a positive face structure $S$ and $k\in{\omega}$ such that
$dim(S)>k$. Then the faces of ${\bf c}^{(k+1)}(S)$, ${\bf c}{\bf
c}^{(k+1)}(S)$, and ${\bf c}^{(k)}(S)$ are as in the table

$\begin{array}[]{|c|c|c|c|}dim&{\bf c}^{(k+1)}(S)&{\bf c}{\bf
c}^{(k+1)}(S)&{\bf c}^{(k)}(S)\\\ \hline\cr
k+1&S_{k+1}-\delta(S_{k+2})&\emptyset&\emptyset\\\ \hline\cr
k&S_{k}-\iota(S_{k+2})&(S_{k}-\iota(S_{k+2}))-\delta(S_{k+1}-\delta(S_{k+2}))&S_{k}-\delta(S_{k+1})\\\
\hline\cr
k-1&S_{k-1}&S_{k-1}-\iota(S_{k+1}-\delta(S_{k+2}))&S_{k-1}-\iota(S_{k+1})\\\
\hline\cr l&S_{l}&S_{l}&S_{l}\end{array}$

where $l<k-1$. Moreover the faces of ${\bf d}^{(k+1)}(S)$, ${\bf d}{\bf
d}^{(k+1)}(S)$, and ${\bf d}^{(k)}(S)$ are as in the table

$\begin{array}[]{|c|c|c|c|}dim&{\bf d}^{(k+1)}(S)&{\bf d}{\bf
d}^{(k+1)}(S)&{\bf d}^{(k)}(S)\\\ \hline\cr
k+1&S_{k+1}-\gamma(S_{k+2})&\emptyset&\emptyset\\\ \hline\cr
k&S_{k}&S_{k}-\gamma(S_{k+1}-\gamma(S_{k+2}))&S_{k}-\gamma(S_{k+1})\\\
\hline\cr l&S_{l}&S_{l}&S_{l}\end{array}$

where $l<k$. Thus the equalities in question all follow from Lemma 5.19.

Ad 3. Let $dim(S)=n>1$. Note that both $({\bf d}{\bf d}S)_{n-2}$ and $({\bf
d}{\bf c}S)_{n-2}$ are the sets of all $<^{+}$-minimal elements in $S_{n-2}$,
i.e. they are equal and the equation ${\bf d}{\bf d}S={\bf d}{\bf c}S$ holds.

To see that ${\bf c}{\bf d}S={\bf c}{\bf c}S$ holds, note first that both
$({\bf c}{\bf d}S)_{n-2}$ and $({\bf c}{\bf c}S)_{n-2}$ are the sets of all
$<^{+}$-maximal elements in $S_{n-2}$. Moreover

$({\bf c}{\bf d}S)_{n-3}=S_{n-3}-\iota(S_{n-1}-\gamma(S_{n})),$ $({\bf c}{\bf
c}S)_{n-3}=S_{n-3}-\iota(S_{n-1}-\delta(S_{n})).$

Now the equality ${\bf c}{\bf d}S={\bf c}{\bf c}S$ follows from the following
equalities

$\iota(S_{n-1}-\gamma(S_{n}))=\iota(S_{n-1})=\iota(S_{n-1}-\delta(S_{n})).$

Both equalities follow from Lemma 5.3. We shall show the first equality only.

Suppose contrary, that there is $x\in\iota(S_{n-1})$ such that
$x\not\in\iota(S_{n-1}-\gamma(S_{n}))$. Let $a\in S_{n-1}$ be a
$<^{+}$-minimal face such that $x\in\iota(a)$. Since
$x\not\in\iota(S_{n-1}-\gamma(S_{n}))$, there is $\alpha\in S_{n}$ such that
$a=\gamma(\alpha)$. By Lemma 5.3 we have

$\iota\delta(\alpha)=\iota\gamma(\alpha)=\iota(a).$

Therefore, there is $a^{\prime}\in\delta(\alpha)$ such that
$x\in\iota(a^{\prime})$. Clearly $a^{\prime}\lhd^{+}a$, and hence $a$ is not
$<^{+}$-minimal contrary to the supposition. This ends the proof of the first
equality above.

Ad 4. Fix $\alpha\in S_{k}$. We need to show that $[\alpha]$ is a positive
face structure. The globularity, strictness (except for linearity of
$<^{[\alpha]_{0},+}$), and disjointness are clear.

The linearity of $<^{[\alpha]_{0},+}$. If $k\leq 2$ it is obvious. Put
$a=\gamma^{(k+2)}(\alpha)$. Using Corollary 4.3, we have

$[\alpha]_{0}=\delta^{(k)}(\alpha)\cup\gamma^{(k)}(\alpha)=$
$=\delta\delta(\gamma^{(n-2)}(\alpha))\cup\gamma\gamma(\gamma^{(n-2)}(\alpha))=\delta\delta(a)\cup\gamma\gamma(a)$

Thus it is enough to assume that $k=2$. But in this case, as we mentioned, the
linearity of $<^{[\alpha]_{0},+}$ is obvious.

The $\gamma$-linearity of $[\alpha]$. The proof proceeds by induction on
$k=dim(\alpha)$. For $k\leq 2$, the $\gamma$-linearity is obvious. So assume
that $k>2$ and that for $l<k$ and $a\in S_{l}$.

First we shall show that ${\bf c}([\alpha])=[\gamma(\alpha)]$. We have

${\bf
c}([\alpha])_{k-1}=(\gamma(\alpha)\cup\delta(\alpha))-\delta(\alpha)=\gamma(\alpha)=[\gamma(\alpha)]_{k-1}$
${\bf
c}([\alpha])_{k-2}=(\gamma\gamma(\alpha)\cup\delta\delta(\alpha))-\iota(\alpha)=\delta\gamma(\alpha)\cup\gamma\gamma(\alpha)=[\gamma(\alpha)]_{k-2}$

and for $l<k-2$

${\bf
c}([\alpha])_{l}=\gamma^{(l)}(\alpha)\cup\delta\gamma^{(l)}(\alpha))=\gamma^{(l)}(\alpha)\cup\delta^{(l)}\gamma(\alpha)=[\gamma(\alpha)]_{l}$

Note that the definition of ${\bf c}(H)$ make sense for any positive
hypergraph $H$ and in the above argument we haven’t use the fact (which we
don’t know yet) that $[\alpha]$ is a positive face structure.

Thus, for $l<k-2$, $[\alpha]_{l}=[\gamma(\alpha)]_{l}$. By induction,
$[\gamma(\alpha)]$ is a positive face structure, and hence $[\alpha]_{l}$ is
$\gamma$-linear $l<k-2$. Clearly $[\alpha]_{l}$ is$\gamma$-linear, for
$l=k-1,k$. Thus it remains to show the $\gamma$-linearity of $k-2$-cells in
$[\alpha]$.

Fix $t\in[\alpha]_{k-3}$, and let

$\Gamma_{t}=\\{x\in[\alpha]_{k-2}:\gamma(x)=t\\}.$

We need to show that $\Gamma_{t}$ is linearly ordered by $<^{+}$. We can
assume that
$t\in\gamma([\alpha]_{n-2})=\gamma\delta\delta(\alpha)=\gamma\delta\gamma(\alpha)$
(otherwise $\Gamma_{t}=\emptyset$ is clearly linearly ordered by $<^{+}$). By
Proposition 5.1 there is a unique $x_{t}\in\delta\gamma(\alpha)$ such that
$\gamma(x_{t})=t$. From Lemma 5.2.2 we get easily the following Claim.

Claim 1. For every $x\in\Gamma_{t}$ there is a unique upper
$\delta(\alpha)$-path from $x_{t}$ to $x$.

Now fix $x,x^{\prime}\in\Gamma_{t}$. By the Claim 1, we have the unique upper
$\delta(\alpha)$-path

$x_{t},a_{0},\ldots,a_{l},x,\;\;\;\;x_{t},a^{\prime}_{0},\ldots,a^{\prime}_{l},x^{\prime}.$

Suppose $l\leq l^{\prime}$. By Proposition 5.1, for $i\leq l$,
$a_{i}=a^{\prime}_{i}$. Hence either $l=l^{\prime}$ and $x=x^{\prime}$ or
$l<l^{\prime}$ and

$x,a_{l+1},\ldots,a_{l^{\prime}},x^{\prime}$

is a $\delta(\alpha)$-upper path. Hence either $x=x^{\prime}$ or
$x\perp^{+}x^{\prime}$ and $[\alpha]_{k-2}$ satisfy the $\gamma$-linearity, as
required.

The proof is of the $\delta$-linearity of $[\alpha]$ is very similar to the
one above. For the same reasons the only non-trivial thing to check is the
condition for $(k-2)$-faces. We pick $t\in\delta\delta(\alpha)$ and consider
the set

$\Delta_{t}=\\{x\in[\alpha]_{k-2}:t\in\delta(x)\\}.$

Then we have a unique $y_{t}\in\delta\gamma(\alpha)$ such that
$t\in\delta(y_{t})$. From Lemma 5.2.3 we get the following Claim.

Claim 2. For every $y\in\Delta_{t}$ there is a unique upper
$\delta(\alpha)$-path from $y_{t}$ to $y$.

The $\delta$-linearity of the $(k-2)$-faces in $[\alpha]$ can be proven from
Claim 2 similarly as the $\gamma$-linearity from Claim 1.

It remains to verify the equalities

${\bf c}[\alpha]=[\gamma(\alpha)],\hskip 28.45274pt{\bf
d}[\alpha]=[\delta(\alpha)].$

The first one we already checked on the way. To see that the second equality
also hold we calculate

${\bf
d}[\alpha]_{k-1}=(\gamma(\alpha)\cup\delta(\alpha))-\gamma(\alpha)=\delta(\alpha)=[\delta(\alpha)]_{k-1}$
${\bf
d}[\alpha]_{k-2}=(\gamma\gamma(\alpha)\cup\delta\delta(\alpha))=\gamma\delta(\alpha)\cup\delta\delta(\alpha)=[\delta(\alpha)]_{k-2}$

and for $l<k-2$

${\bf
d}[\alpha]_{l}=\gamma^{(l)}(\alpha)\cup\delta^{(l)}(\alpha))=\gamma^{(l)}\delta(\alpha)\cup\delta^{(l)}(\alpha)=[\delta(\alpha)]_{l}$

So the second equality holds as well.$~{}~{}\Box$

###### Lemma 6.2

Let $S$ and $T$ be positive face structures such that ${\bf
c}^{(k)}S\subseteq{\bf d}^{(k)}T$. Then the pushout $S+_{k}T$ in ${\bf
Fs}^{+/1}$ of $S$ and $T$ over ${{\bf c}^{(k)}S}$ exists. Moreover, if ${\bf
c}^{(k)}S=S\cap T$ then the diagram of inclusions in ${\bf Fs}^{+/1}$

${\bf c}^{(k)}S\vphantom{T}$$\vphantom{{\bf c}^{(k)}S}T$$S\vphantom{S\cup
T}$$\vphantom{S}S\cup T$

is the pushout.

Proof. Assume that ${\bf c}^{(k)}S=S\cap T\subseteq{\bf d}^{(k)}T$. Let $S\cup
T$ be the obvious sum of $S$ and $T$ as positive hypergraphs. The fact that
$S\cup T$ is a pushout in ${\bf Hg}^{+1}$ is obvious. Thus the only thing we
need to verify that $S\cup T$ is a positive face structure.

First we write in details the condition ${\bf c}^{(k)}S=S\cap T\subseteq{\bf
d}^{(k)}T$:

  1. 1.

$S_{l}\cap T_{l}=\emptyset$, for $l>k$,

  2. 2.

$S_{k}-\delta(S_{k+1})\subseteq T_{k}-\gamma(T_{k+1})$,

  3. 3.

$S_{k-1}-\iota(S_{k+1})\subseteq T_{k-1}$,

  4. 4.

$S_{l}\subseteq T_{l}$, for $l<k-1$.

Now we describe the orders $<^{+}$ in $S\cup T$:

$<^{(S\cup
T)_{l},+}=\left\\{\begin{array}[]{ll}<^{S_{l},+}+<^{T_{l},+}&\mbox{for
$l>k$,}\\\ <^{S_{l},+}+_{(S_{k}-\delta(S_{k+1}))}<^{T_{l},+}&\mbox{for
$l=k$,}\\\ <^{S_{l},+}+_{(S_{k-1}-\iota(S_{k+1}))}<^{T_{l},+}&\mbox{for
$l=k-1$,}\\\ <^{T_{l},+}&\mbox{for $l<=k-1$.}\end{array}\right.$

We shall comment on these formulas. For $l>k$ the formulas say that the order
$<^{+}$ in $(S\cup T)_{l}$ is the disjoint sum of the orders in $S_{l}$ and
$T_{l}$. This is obvious.

For $l<k-1$ the order $<^{+}$ in $(S\cap T)_{l}$ is just the order
$<^{T_{l},+}$. The only case that requires an explanation is $l=k-2$. So
suppose that $a,b\in T_{k-2}$ and $a<^{(S\cup T)_{k-2},+}b$. So we have an
upper path

$a,\alpha_{1},\ldots,\alpha_{m},b$

such that $\alpha_{i}\in(S\cup T)_{k-1}=\iota(S_{k+1})\cup T_{k-1}$. By Lemma
5.4, we can assume that if $\alpha_{i}\in S_{k-1}$ then
$\alpha_{i}\not\in\gamma(S_{k})$. But then $\alpha_{i}\not\in\iota(S_{k+1})$.
So in fact $\alpha_{i}\in T_{k-1}$, as required.

The most involved are the formulas for $<^{(S\cap T)_{l},+}$ for $l=k$ and
$l=k-1$. In both cases the comparison in $S\cup T$ involves orders both from
$S$ and $T$. In the former case we have that, for $a,b\in(S\cup T)_{k}$, we
have

$a<^{(S\cup T)_{k},+}b\mbox{ iff }\hskip 199.16928pt$
$\left\\{\begin{array}[]{ll}\mbox{either }a,b\in T_{k}&\mbox{ and
}a<^{T_{k},+}b,\\\ \mbox{or }a,b\in S_{k}&\mbox{ and }a<^{S_{k},+}b,\\\
\mbox{or }a\in\delta(S_{k+1}),\;b\in T_{k}&\mbox{ and }\exists_{a^{\prime}\in
S_{k}-\delta(S_{k+1})}a<^{S_{k},+}a^{\prime}\mbox{ and
}a^{\prime}\leq^{T_{k},+}b.\end{array}\right.$

The orders $<^{S_{k},+}$ and $\leq^{T_{k},+}$ are glued together along the set
$S_{k}-\delta(S_{k+1})$ which is the set of $<^{S_{k},+}$-maximal elements in
$S_{k}$ and at the same time it is contained in the set of
$<^{T_{k},+}$-minimal elements $T_{k}-\gamma(T_{k+1})$. This is obvious when
we realize that $\delta(S_{k+1})\cap\gamma(T_{k+1})=\emptyset$.

In the later case we have for $x,y\in(S\cup T)_{k-1}$ we have

$x<^{(S\cup T)_{k-1},+}y\mbox{ iff }\hskip 199.16928pt$
$\left\\{\begin{array}[]{ll}\mbox{either }x,y\in S_{k-1}&\mbox{ and
}x<^{S_{k-1},+}y,\\\ \mbox{or }x,y\in T_{k-1}&\mbox{ and }x<^{T_{k-1},+}y,\\\
\mbox{or }x\in\iota(S_{k+1}),\;y\in T_{k}&\mbox{ and }\exists_{x^{\prime}\in
S_{k-1}-\iota(S_{k+1})}x<^{S_{k},+}x^{\prime}\mbox{ and
}x^{\prime}\leq^{T_{k},+}y,\\\ \mbox{or }x\in
T_{k},\;y\in\iota(S_{k+1})&\mbox{ and }\exists_{x^{\prime}\in
S_{k-1}-\iota(S_{k+1})}x<^{T_{k},+}x^{\prime}\mbox{ and
}x^{\prime}\leq^{S_{k},+}y.\end{array}\right.$

The order $<^{S_{k-1},+}$ is ’plugged into’ the order $\leq^{T_{k-1},+}$ along
the set $S_{k}-\iota(S_{k+1})$.

To show that these formulas hold true we argue by cases. Assume that
$x,y\in(S\cup T)_{k-1}$ and that $x<^{(S\cup T)_{k-1},+}y$ i.e. there is an
upper path

$x,a_{1},\ldots,a_{m},y$

with $a_{i}\in(S\cup T)_{k},$ for $i=1,\ldots,m$.

First suppose that $x,y\in S_{k-1}$ and that the set
$\\{a_{i}\\}_{i}\not\subseteq S_{k}$. Let
$a_{i_{0}},a_{i_{0}+1},\ldots,a_{i_{1}}$ be a maximal subsequence of
consecutive elements of the path $a_{1},\ldots,a_{m}$ such that
$\\{a_{i}\\}_{i_{0}\leq i\leq i_{1}}\subseteq T_{k}$. Thus it is an upper path
in $T_{k}$ from $\bar{x}$ to $\bar{y}=\gamma(a_{i_{1}})$, where

$\bar{x}=\left\\{\begin{array}[]{ll}x&\mbox{if }i_{0}=1,\\\
\gamma(a_{i_{0}-1})&\mbox{otherwise.}\end{array}\right.$

Note that from maximality of the path $a_{i_{0}},\ldots,a_{i_{1}}$ follows
that both $\bar{x},\bar{y}\in S_{k-1}-\iota(S_{k+1})$. As we have
$\bar{x}<^{T_{k-1},+}\bar{y}$ from Corollary 5.11 we have
$\bar{x}\not\perp^{T_{l},-}\bar{y}$, for all $l<k-1$. Clearly
$\perp^{S_{l},-}\subseteq\perp^{T_{l},-}$. Thus
$\bar{x}\not\perp^{S_{l},-}\bar{y}$, for all $l<k-1$, as well. But then again
by Corollary 5.11 we have that $\bar{x}\perp^{S_{k-1},+}\bar{y}$. If we were
to have $\bar{y}<^{S_{k-1},+}\bar{x}$ then, as $\bar{x},\bar{y}\in
S_{k-1}-\iota(S_{k+1})$, we would have $\bar{y}<^{T_{k-1},+}\bar{x}$. But this
would contradict the strictness of $<^{T_{k-1},+}$. So we must have
$\bar{x}<^{S_{k-1},+}\bar{y}$. In this way we can replace the upper path
$a_{1},\ldots,a_{m}$ in $(S\cup T)_{k}$ from $x$ to $y$ by an upper path from
$x$ to $y$ in $S_{k}$.

Next, suppose that $x,y\in T_{k-1}$ and that the set
$\\{a_{i}\\}_{i}\not\subseteq T_{k}$. Let
$a_{i_{0}},a_{i_{0}+1},\ldots,a_{i_{1}}$ be a maximal subsequence of
consecutive elements of the path $a_{1},\ldots,a_{m}$ such that
$\\{a_{i}\\}_{i_{0}\leq i\leq i_{1}}\subseteq S_{k}$. Thus it is an upper path
in $S_{k}$ from $\bar{x}$ to $\bar{y}=\gamma(a_{i_{1}})$, where

$\bar{x}=\left\\{\begin{array}[]{ll}x&\mbox{if }i_{0}=1,\\\
\gamma(a_{i_{0}-1})&\mbox{otherwise.}\end{array}\right.$

Note that from maximality of the sequence $a_{i_{0}},\ldots,a_{i_{1}}$ follows
that both $\bar{x},\bar{y}\in S_{k-1}-\iota(S_{k+1})\subseteq T_{k-1}$. Thus
by Lemma 5.8 there is an upper path from $\bar{x}$ to $\bar{y}$ in
$S_{k-1}-\delta(S_{k})\subseteq T_{k-1}$. In this way we can replace the upper
path $a_{1},\ldots,a_{m}$ in $(S\cup T)_{k}$ from $x$ to $y$ by an upper path
from $x$ to $y$ in $T_{k}$.

Thus we have justified the first two cases of the above formula. The following
two cases are easy consequences these two. This end the description of the
orders in $S\cup T$.

From these descriptions follows immediately that $<^{(S\cup T),+}$ is strict
for all $l$. It remains to show the pencil linearity. Both $\gamma$\- and
$\delta$-linearity of $l$-cells, for $l<k-1$ or $l>k$, are obvious.

To see the $\gamma$-linearity of $k$-cells assume $a\in S_{k}$ and $b\in
T_{k}$, such that $\gamma(a)=\gamma(b)$. Let $\bar{a}\in S_{k}$ be the
$<^{S_{k},+}$-maximal $k$-cells, such that $\gamma(a)=\gamma(\bar{a})$. Then
$\bar{a}\in{\bf c}^{(k)}(S)_{k}\subseteq{\bf d}^{(k)}(T)_{k}$. So $\bar{a}\in
T_{k}$ is $<^{T_{k},+}$-minimal $k$-cells, such that
$\gamma(\bar{a})=\gamma(b)$. Thus

$a\leq^{S_{k},+}\bar{a}\leq^{T_{k},+}b.$

Thus the $\gamma$-linearity of $k$-cells holds. The proof of
$\delta$-linearity of $k$-cells is similar.

Finally, we need to establish the $\gamma$\- and $\delta$\- linearity of
$(k-1)$-cells in $S\cup T$.

In order to prove the $\gamma$-linearity, let $x\in\iota(S_{k+1})$ and $y\in
T_{k-1}$ such that $\gamma(x)=\gamma(y)$. We need to show that $x\perp^{(S\cup
T)_{k-1},+}y$.

Let $\alpha_{0}\in S_{k+1}$ such that $x\in\iota(\alpha_{0})$,
$a\in\delta(\alpha_{0})$ such that $x=\gamma(a)$ and let
$\alpha_{0},\ldots,\alpha_{l}$ be a lower path in $S_{k+1}$ such that
$\gamma(\alpha_{l})\in T_{k}$. Since $x\in\iota(\alpha_{0})$, then
$x\in\gamma\delta(\alpha_{0})$ and, by Lemma 4.2

$\gamma(x)\in\gamma\gamma\delta(\alpha_{0})\subseteq\iota\gamma(\alpha_{0}).$

As $\gamma(\alpha_{0})\leq^{+}\gamma(\alpha_{l})$, by Lemma 5.3, we have
$\gamma(x)\in\iota\gamma(\alpha_{l})\cup\gamma\gamma\gamma(\alpha_{l})$. Thus
we have two cases:

  1. 1.

$\gamma(x)\in\iota\gamma(\alpha_{l})$,

  2. 2.

$\gamma(x)=\gamma\gamma\gamma(\alpha_{l})$.

Case 1: $\gamma(x)\in\iota\gamma(\alpha_{l})$.

By Lemma 5.2.2, there is a unique $z\in\delta\gamma(\alpha_{l})$ such that
$\gamma(z)=\gamma(x)$ and $z<^{+}x$. As $\gamma(\alpha_{l})\in T_{k}$, so
$z\in T_{k-1}$. If $y<^{T_{k-1},+}z$ then indeed $y<^{(S\cup T)_{k-1},+}z$, as
required. By $\gamma$-linearity in $T_{k-1}$, it is enough to show that it is
impossible to have $z<^{T_{k-1},+}y$.

Suppose contrary, that there is an upper path $z,b_{0},\ldots,b_{r},y$ in $T$.
Since $\gamma(\alpha_{l})$ is $<^{+}$-minimal in $T$ (as $\alpha_{l}\in S$)
and $z\in\delta\gamma(\alpha_{l})\cap\delta(b_{0})$, by $\delta$-linearity in
$T_{k}$ we have $\gamma(\alpha_{l})<^{+}b_{0}$. By Lemma 5.3, we have

$\gamma(x)\in\iota\gamma(\alpha_{l})\subseteq\iota(b_{0})\subseteq\iota(b_{r})$

But $\gamma(b_{r})=y$ so $\gamma\gamma(b_{r})=\gamma(y)=\gamma(x)$. In
particular $\gamma(x)\not\in\iota(b_{r})$ and we get a contradiction.

Case 2: $\gamma(x)=\gamma\gamma\gamma(\alpha_{l})$.

By Lemma 5.2.2 there is $z\in\delta\gamma(\alpha_{l})$ such that
$\gamma(x)=\gamma(z)(=\gamma\gamma\gamma(\alpha_{l}))$, so that we have

$z<^{S_{k-1},+}x<^{S_{k-1},+}\gamma\gamma(\alpha_{l}).$

As $\gamma(\alpha_{l})\in T_{k}$ and it is $<^{+}$-minimal in $T_{k}$, by
Proposition 5.16, there is no face $y\in T_{k-1}$ so that

$z<^{T_{k-1},+}y<^{T_{k-1},+}\gamma\gamma(\alpha_{l}).$

So if $y\in T_{k-1}$ and $\gamma(y)=\gamma(x)$ then either

$y\leq^{T_{k-1},+}z<^{S_{k-1},+}x\;\;\;\mbox{ or
}\;\;\;x<^{S_{k-1},+}\gamma\gamma(\alpha_{l})\leq^{S_{k-1},+}y.$

In either case $x\perp^{{(S\cup T)_{k-1},+}}y$, as required. This ends the
proof of $\gamma$-linearity of $(k-1)$-faces in $(S\cup T)$.

Finally, we prove the $\delta$-linearity of $(k-1)$-faces in $S\cup T$. Let
$x\in\iota(S_{k+1})$ and $y\in T_{k-1}$, $t\in T_{k-2}$ such that
$t\in\delta(x)\cap\delta(y)$. We need to show that $x\perp^{(S\cup
T)_{k-1},+}y$.

Let $\alpha\in S_{k+1}$ such that $x\in\iota(\alpha_{0})$,
$a\in\delta(\alpha_{0})$ such that $x=\gamma(a)$, and let
$\alpha_{0},\ldots,\alpha_{l}$ be a lower path in $S_{k+1}$ such that
$\gamma(\alpha_{l})\in T_{k}$. As $x\in\iota(\alpha_{0})$, using Lemma 4.2 we
have

$t\in\delta(x)\subseteq\delta\gamma\delta(\alpha_{0})\subseteq\delta\gamma\gamma(\alpha_{0})\cup\iota\gamma(\alpha_{0}).$

As $\gamma(\alpha_{0})<^{+}\gamma(\alpha_{l})$, by Lemma 5.3, we have two
cases:

  1. 1.

$t\in\iota\gamma(\alpha_{l})$,

  2. 2.

$t\in\delta\gamma\gamma(\alpha_{l})$.

Case 1: $t\in\iota\gamma(\alpha_{l})$.

By Lemma 5.2.3, there is a unique $z\in\delta\gamma(\alpha_{l})$ such that
$t\in\delta(z)$ and $z<^{+}x$. As $\gamma(\alpha_{l})\in T_{k}$, so $z\in
T_{k-1}$. If $y<^{T_{k-1},+}z$ then indeed $y<^{(S\cup T)_{k-1},+}z$, as
required. By $\delta$-linearity in $T_{k}$, it is enough to show that it is
impossible to have $z<^{T_{k-1},+}y$.

Suppose contrary, that there is an upper path in $T$

$z,b_{0},\ldots,b_{r},y.$

Since $\gamma(\alpha_{l})$ is $<^{+}$-minimal in $T_{k}$ and
$z\in\delta\gamma(\alpha_{l})\cap\delta(b_{0})$, by $\delta$-linearity of
$k$-faces in $T$ we have $\gamma(\alpha_{l})<^{+}b_{0}$. By Lemma 5.3, we have

$t\in\iota\gamma(\alpha_{l})\subseteq\iota(b_{0})\subseteq\ldots\subseteq\iota(b_{r}).$

But $\gamma(b_{r})=y$, so $t\in\delta(y)\subseteq\delta\gamma(b_{r})$. In
particular $t\not\in\iota(b_{r})$ and we get a contradiction.

Case 2: $t\in\delta\gamma\gamma(\alpha_{l})$.

By Lemma 5.2.3 there is $z\in\delta\gamma\gamma(\alpha_{l})$ such that
$t\in\delta(z)$ and we have

$z<^{S_{k-1},+}x<^{S_{k-1},+}\gamma\gamma(\alpha_{l}).$

As $\gamma(\alpha_{l})\in T_{k}$, and it is $<^{+}$-minimal face in $T_{k}$,
by Lemma 5.16, there is no face $y\in T_{k-1}$ such that

$z<^{T_{k-1},+}y<^{T_{k-1},+}\gamma\gamma(\alpha_{l}).$

So if $y\in T_{k-1}$ and $t\in\delta(y)$ then either

$y\leq^{T_{k-1},+}z<^{S_{k-1},+}x\;\;\;\mbox{ or
}\;\;\;x<^{S_{k-1},+}\gamma\gamma(\alpha_{l})\leq^{S_{k-1},+}y.$

In either case $x\perp^{{(S\cup T)_{k-1},+}}y$, as required. This ends the
proof of $\delta$-linearity of $(k-1)$-faces in $(S\cup T)$ and the whole
proof that $S\cup T$ is a positive face structure. $~{}~{}\Box$

Let $S$ and $T$ be positive face structures such that ${\bf c}^{(k)}S={\bf
d}^{(k)}T$. Then the pushout

${\bf c}^{(k)}S\vphantom{T}$$\vphantom{{\bf
c}^{(k)}S}T$$S\vphantom{S+_{k}T}$$\vphantom{S}S+_{k}T$

is called special pushouts

Now we shall describe an ${\omega}$-category $S^{*}$ generated by the face
structure $S$.

The set $S^{*}_{n}$ of $n$-cell of $S^{*}$ is the set of all positive face
sub-structures of $S$ of dimension at most $n$, for $n\in{\omega}$. The domain
and codomain operations in $S^{*}$ are restricted operations

${\bf d}^{(k)},{\bf c}^{(k)}:S^{*}_{n}\longrightarrow S^{*}_{k}$

of the $k$-th domain and the $k$-th codomain, for $k\leq n$. The identity
operation

${\bf i}^{(n)}:S^{*}_{k}\longrightarrow S^{*}_{n}$

is an inclusion and the composition map

${\bf m}_{n,k,n}:S^{*}_{n}\times_{S^{*}_{k}}S^{*}_{n}\longrightarrow
S^{*}_{n}$

is the sum, i.e. if $X$, $Y$ are sub-face structures of $S$ of dimension at
most $n$ such that ${\bf c}^{(k)}X={\bf d}^{(k)}Y$ then

${\bf m}_{n,k,n}(X,Y)=X+_{k}Y=X\cup Y.$

###### Corollary 6.3

Let $S$ be a weak positive face structure. Then $S^{*}$ is an
${\omega}$-category. In fact, we have a functor

$(-)^{*}:{\bf Fs}^{+/1}\longrightarrow{{\omega}Cat}$

Proof. The fact that the operation on $S^{*}$ defined above satisfy the laws
of ${\omega}$-category is obvious. The image $f(X)$ of a sub-face structure
$X$ of a positive face structure $S$ under a morphism $f:S\rightarrow T$ is a
sub-face structure of $T$. The association $X\mapsto f(X)$ is easily seen to
be an ${\omega}$-functor. $~{}~{}\Box$

Let $S$ be a positive face structure. We have a functor

$\Sigma^{S}:{\bf pFs}^{+/1}\downarrow S\longrightarrow{\bf Fs}^{+/1}$

such that

$\Sigma^{S}(f:B\rightarrow S)=B$

and a cocone

$\sigma^{S}:\Sigma^{S}\longrightarrow S$

such that

$\sigma^{S}_{(f:B\rightarrow S)}=f:\Sigma^{S}(f:B\rightarrow
S)=B\longrightarrow S$

We have

###### Lemma 6.4

The cocone
$\sigma^{S}:\Sigma^{S}\stackrel{{\scriptstyle}}{{\longrightarrow}}S$ is a
colimiting cocone in ${\bf Fs}^{+/1}$. Such colimiting cones are called
special colimits. Any special limit in ${\bf Fs}^{+/1}$ can be obtained via
some special pushouts and vice versa any special pushout can be obtained from
special limits. In particular, a functor from ${\bf Fs}^{+/1}$ preserves
special limits if and only if it preserves special pushouts.$~{}~{}\Box$

##  7 Normal positive face structures

Let $S$ be a $k$-normal positive face structure $S$. By ${\bf p}_{l}^{S}$ we
denote the unique element of the set $S_{l}-\delta(S_{l+1})$, for $l<k$.
Moreover, as we shall show below ${\bf p}_{k-1}\in\gamma(S_{k})$ and hence the
set $\\{x\in S_{k}:\gamma(x)={\bf p}_{k-1}\\}$ is not empty. We denote by
${\bf p}_{k}$ the $<^{+}$-maximal element of this set. We shall omit the
superscript $S$ if it does not lead to a confusion.

###### Lemma 7.1

Let $S$ be a $(k-1)$-principal positive face structure of dimension at least
$k$, $k>0$. Then

  1. 1.

$S_{l}=\delta^{(l)}(S_{k})\cup\gamma^{(l)}(S_{k})=\delta^{(l)}(S_{k})\cup\\{{\bf
p}_{l}\\}$, for $l<k$.

  2. 2.

$\delta(S_{l+1})=\delta^{(l)}(S_{k})$, for $l<k$.

  3. 3.

${\bf p}_{k}$ is $<^{-}$-largest element in $S_{k}-\delta(S_{k+1})$.

  4. 4.

$\gamma({\bf p}_{l})={\bf p}_{l-1}$, for $0<l\leq k$.

  5. 5.

$\delta({\bf p}_{l})=\delta(S_{l})-\gamma(S_{l})$, for $0<l<k$.

  6. 6.

$S_{l}=\delta^{(l)}({\bf p}_{k-1})\cup\gamma^{(l)}({\bf p}_{k-1})$, for
$l<k-2$.

Proof. Ad 1. If $H$ is a hypergraph of dimension greater than $l$ and
$\gamma(H_{l+1})\subseteq\delta(H_{l+1})$ then there is an infinite lower path
in $H_{l+1}$, i.e. $<^{H_{l},+}$ is not strict. Thus, if $S$ is a positive
face structure of dimension greater than $l$, we have
$\delta(S_{l+1})\subseteq\hskip-11.38109pt_{\not}\hskip 8.53581ptS_{l}$. A
positive face structure is normal iff this difference

$S_{l}-\delta(S_{l+1})$

is minimal possible (i.e. one-element set), for $l<k$. Thus, by the above, we
must have

$S_{l}=\delta(S_{l+1})\cup\gamma(S_{l+1})$ (1)

The first equation of the statement 1. we shall show by the downward induction
on $l$. Suppose that we have
$S_{l+1}=\delta^{(l)}(S_{k})\cup\gamma^{(l)}(S_{k})$ (for $l=k-2$ it is true
by the above). Then

$S_{l}=\delta^{(l)}(S_{k})\cup\gamma^{(l)}(S_{k})=$
$=\delta(\delta^{(l+1)}(S_{k})\cup\gamma^{(l+1)}(S_{k}))\cup\gamma(\delta^{(l+1)}(S_{k})\cup\gamma^{(l+1)}(S_{k}))=$
$=\delta\delta^{(l+1)}(S_{k})\cup\delta\gamma^{(l+1)}(S_{k})\cup\gamma\delta^{(l+1)}(S_{k})\cup\gamma\gamma^{(l+1)}(S_{k})=$
$=\delta^{(l)}(S_{k})\cup\delta\gamma^{(l)}(S_{k})\cup\gamma\delta^{(l)}(S_{k})\cup\gamma^{(l)}(S_{k})=$
$=\delta^{(l)}(S_{k})\cup\gamma^{(l)}(S_{k})$

where the last equation follows from Corollary 4.3.

The second equation of 1. is obvious for $l=k-1$. So assume that $l<k-1$. We
have

$\\{{\bf p}_{l}\\}=S_{l}-\delta(S_{l+1})=$
$=S_{l}-\delta(\delta^{(l+1)}(S_{k})\cup\gamma^{(l+1)}(S_{k}))=$
$=S_{l}-(\delta^{(l)}(S_{k})\cup\delta\gamma^{(l+1)}(S_{k}))=$
$S_{l}-\delta^{(l)}(S_{k}).$

Thus

$S_{l}=\delta^{(l)}(S_{k})\cup\\{{\bf p}_{l}\\}$

as required.

Ad 2. Let $l<k$ Then using 1. we have

$\delta^{(l)}(S_{k})\subseteq\delta(S_{l+1})\subseteq\hskip-11.38109pt_{\not}\hskip
8.53581pt\delta^{(l)}(S_{k})\cup\\{{\bf p}_{l}\\}$

Hence

$\delta^{(l)}(S_{k})=\delta(S_{l+1}).$

Ad 3. First we shall show that ${\bf p}_{k}\in S_{k}-\delta(S_{k+1})$. Suppose
contrary, that there is $\alpha\in S_{k+1}$ such that ${\bf
p}_{k}\in\delta(\alpha)$. Then $\gamma({\bf
p}_{k})\in\gamma\delta(\alpha)=\gamma\gamma(\alpha)\cup\iota(\alpha)$. If
$\gamma({\bf p}_{k})=\gamma\gamma(\alpha)$ then ${\bf
p}_{k}<^{+}\gamma(\alpha)$ i.e. $\gamma(\alpha)$ is $<^{+}$-smaller element
than ${\bf p}_{k}$ such that $\gamma(\gamma(\alpha))={\bf p}_{k-1}$. This
contradicts the choice of ${\bf p}_{k}$. If $\gamma({\bf
p}_{k})=\iota(\alpha)$ then there is $a\in\delta(\alpha)$ such that
$\gamma({\bf p}_{k})\in\delta(a)$. But this means that ${\bf
p}_{k-1}=\gamma({\bf p}_{k})\in\delta(S_{k})$ contradicting the choice of
${\bf p}_{k-1}\in S_{k-1}-\delta(S_{k})$. This shows that ${\bf p}_{k}\in
S_{k}-\delta(S_{k+1})$.

We need to prove that any maximal lower $(S_{k}-\delta(S_{k+1}))$-path ends at
${\bf p}_{k}$. By strictness, it is enough to show that if $x\in
S_{k}-\delta(S_{k+1})$ and $x\neq{\bf p}_{k}$ then there is $x^{\prime}\in
S_{k}-\delta(S_{k+1})$ such that $\gamma(x)\in\delta(x^{\prime})$. So fix
$x\in S_{k}-\delta(S_{k+1})$. If we were to have $\gamma(x)\in\iota(\beta)$
for some $\beta\in S_{k+1}$, then by Lemma 5.5 we would have
$x<^{+}\gamma(\beta)$. In particular, $x\in\delta(S_{k+1})$, contrary to the
assumption. Therefore $\gamma(x)\in S_{k-1}-\iota(S_{k+1})$. As $x,{\bf
p}_{k}\in S_{k}-\delta(S_{k+1})$, by $\gamma$-linearity we have
$\gamma(x)\neq\gamma({\bf p}_{k})={\bf p}_{k-1}$. Hence by 1. the set

$\Delta_{\gamma(x)}=\\{y\in S_{k}:\gamma(x)\in\delta(y)\\}$

is not empty. Let $x^{\prime}$ be the $<^{+}$-maximal element of this set. It
remains to show that $x^{\prime}\not\in\delta(S_{k+1})$. Suppose contrary,
that there is $\alpha\in S_{k+1}$ such that $x^{\prime}\in\delta(\alpha)$. As
$\gamma(x)\not\in\iota(S_{k+1})$ and $\gamma(x)\in\delta(x^{\prime})$ so
$\gamma(x)\not\in\iota(\alpha)$ and $\gamma(x)\neq\gamma\gamma(\alpha)$. Thus
$\gamma(x)\in\delta\gamma(\alpha)$. But this means that
$x^{\prime}<^{+}\gamma(\alpha)$ and $\gamma(\alpha)\in\Delta_{\gamma(x)}$.
This contradicts the choice of $x^{\prime}$. This ends the proof of 3.

Ad 4. $\gamma({\bf p}_{k})={\bf p}_{k-1}$ by definition. Fix $0<l<k$. As
$S_{l}=\delta(S_{l+1})\cup\\{{\bf p}_{l}\\}$, ${\bf p}_{l}$ is
$<^{+}$-greatest element in $S_{l}$. Assume that $\gamma({\bf p}_{l})\neq{\bf
p}_{l-1}$. Thus $\gamma({\bf p}_{l})<^{+}{\bf p}_{l-1}$. Let $x\in S_{l}$.
Then $x\leq{\bf p}_{l}$ and, by Lemma 5.9, $\gamma(x)\leq^{+}\gamma({\bf
p}_{l})<^{+}{\bf p}_{l-1}$. Thus ${\bf p}_{l-1}\not\in\gamma(S_{l})$. So
$\gamma(S_{l})\subseteq\delta(S_{l})$. But this is impossible in a positive
face structure as we noticed in the proof of 1. This ends 4.

Ad 5. Fix $l<k$. First we shall show that

$\delta({\bf p}_{l})\cap\gamma(S_{l})=\emptyset$ (2)

Let $z\in\gamma(S_{l})$, i.e. there is $a\in S_{l}$ such that $\gamma(a)=z$.
By 1. $a\leq^{+}{\bf p}_{l}$. By Lemma 5.7, there are $x\in\delta({\bf
p}_{l})$ and $y\in\delta(a)$ such that $x\leq^{+}y$. Hence
$x<^{+}\gamma(a)=z$. By Proposition 5.1, since $x\in\delta({\bf p}_{l})$ it
follows that $z\not\in\delta({\bf p}_{l})$. This shows (2).

By Lemma 5.19, we have

$\delta(S_{l})=\delta(S_{l}-\delta({S_{l+1}}))\cup\iota(S_{l+1})$ (3)

Since $\delta({\bf p}_{l})=\delta(S_{l}-\delta({S_{l+1}}))$ and
$\iota(S_{l+1})\subseteq\gamma(S_{l})$ we have by (2)

$\delta(S_{l}-\delta({S_{l+1}}))\cap\iota(S_{l+1})=\emptyset$ (4)

Next we shall show that

$\iota(S_{l+1})=\gamma(S_{l})\cap\delta(S_{l})$ (5)

The inclusion $\subseteq$ is obvious. Let
$x\in\gamma(S_{l})\cap\delta(S_{l})$. Hence there are $a,b\in S_{l}$ such that
$\gamma(a)=x\in\delta(b)$. We can assume that $a$ is $<^{+}$-maximal with this
property. As $a<^{-}b$, neither $a$ nor $b$ is equal to the $<^{+}$-greatest
element ${\bf p}_{l}\in S_{l}$. Therefore there is $\alpha\in S_{l+1}$ such
that $a\in\delta(\alpha)$. If we were to have
$x=\gamma(a)=\gamma\gamma(\alpha)$ then $\gamma(\alpha)$ would be a
$<^{+}$-greater element than $a$ with $\gamma(\gamma(\alpha))=x$. So
$\gamma(a)\neq\gamma\gamma(\alpha)$. Clearly, $x\in\gamma\delta(\alpha)$. By
globularity, $x\in\delta\delta(\alpha)$, as well. Thus $x\in\iota(\alpha)$,
and (5) is shown.

Using (2), (3), (4), and (5) we have

$\delta({\bf p}_{l})=\delta(S_{l}-\delta(S_{l+1}))=$
$=\delta(S_{l})-\iota(S_{l+1})=$
$=\delta(S_{l})-(\gamma(S_{l})\cap\delta(S_{l}))=$
$=\delta(S_{l})-\gamma(S_{l})$

as required.

Ad 6. By 1. and 2. it is enough to show

$\delta^{(l)}(S_{k-1})=\delta^{(l)}({\bf p}_{k-1}),$

for $l<k-2$. The inclusion $\supseteq$ is obvious.

Pick $x\in S_{k-1}$ We have an upper path $x,a_{1},\ldots,a_{r},{\bf
p}_{k-1}$. By Corollary 4.3, as $\gamma(a_{i})\in\delta(a_{i+1})$, we have

$\delta^{(l)}(a_{i})=\delta^{(l)}\gamma(a_{i})\subseteq\delta^{(l)}(\delta(a_{i+1}))=\delta^{(l)}(a_{i+1})$

for $i=0,\ldots,r-1$. Then, by transitivity of $\subseteq$ and again Corollary
4.3 we get

$\delta^{(l)}(x)\subseteq\delta^{(l)}(a_{1})\subseteq\delta^{(l)}(a_{r})\subseteq\delta^{(l)}(\gamma(a_{r}))=\delta^{(l)}({\bf
p}_{k-1}).$

This ends the proof of the inclusion $\subseteq$ and 6. $~{}\Box$

###### Lemma 7.2

Let $S$ be a positive face structure of dimension at least $k$. Then

  1. 1.

$S$ is $(k-1)$-principal iff $\;{\bf d}^{(k)}(S)$ is normal iff $\;{\bf
c}^{(k-1)}(S)$ is principal,

  2. 2.

if $S$ is normal, so is ${\bf d}(S)$,

  3. 3.

if $S$ is principal, so is ${\bf c}(S)$.

Proof. The whole Lemma is an easy consequence of Lemma 5.19. We shall show 1.
leaving 2. and 3. for the reader.

First note that all three conditions in 1. imply that,
$|S_{l}-\delta(S_{l+1}|=1$ for $l<k-2$. In addition to this they say:

  1. 1.

$S$ is $(k-1)$-principal iff $|S_{l}-\delta(S_{l+1})|=1$ for $l=k-2,k-1$.

  2. 2.

${\bf d}^{(k)}(S)$ is normal iff

    1. (a)

$|S_{k-1}-\delta(S_{k}-\gamma(S_{k+1}))|=1$, and

    2. (b)

$|S_{k-2}-\delta(S_{k-1})|=1$.

  3. 3.

${\bf c}^{(k-1)}(S)$ is principal iff

    1. (a)

$|(S_{k-1}-\iota(S_{k+1}))-\delta(S_{k}-\delta({S_{k+1}}))|=1$, and

    2. (b)

$|S_{k-2}-\delta(S_{k-1}-\iota({S_{k+1}}))|=1$.

So the equivalence of these conditions follows directly from Lemma 5.19.
$~{}~{}\Box$

Let $N$ be a $n$-normal positive face structure. We define a
$(n+1)$-hypergraph $N^{\bullet}$, that contains two additional faces: ${\bf
p}_{n+1}^{N^{\bullet}}$ of dimension $n+1$, and ${\bf p}_{n}^{N^{\bullet}}$ of
dimension $n$. We shall drop superscripts if it does not lead to confusions.
We also put

$\delta({\bf p}_{n+1})=N_{n},\;\;\;\;\gamma({\bf p}_{n+1})={\bf p}_{n},$
$\delta({\bf p}_{n})=\delta(N_{n})-\gamma(N_{n}),\;\;\;\;\gamma({\bf
p}_{n})={\bf p}_{n-1}(=\gamma(N_{n})-\delta(N_{n})).$

As $N$ is normal the $\gamma(N_{n})-\delta(N_{n})$ has one element so
$\gamma({\bf p}_{n})$ is well defined. This determines $N^{\bullet}$ uniquely.
$N^{\bullet}$ is called a simple extension of $N$.

Example. For a normal positive face structure $N$ like this

$x_{2}\vphantom{x_{1}}$$\vphantom{x_{2}}x_{1}$$f_{1}$$f_{2}$$f_{0}$$x_{3}\vphantom{x_{0}}$$\vphantom{x_{3}}x_{0}$

the hyper-graph $N^{\bullet}$ looks like this

$x_{2}\vphantom{x_{1}}$$\vphantom{x_{2}}x_{1}$$f_{1}$$f_{2}$$f_{0}$$x_{3}\vphantom{x_{0}}$$\vphantom{x_{3}}x_{0}$${\bf
p}^{N^{\bullet}}_{1}$$\Downarrow{\bf p}^{N^{\bullet}}_{2}$

We have

###### Proposition 7.3

Let $N$ be a normal positive face structure of dimension $n$. Then

  1. 1.

$N^{\bullet}$ is a principal positive face structure of dimension $n+1$.

  2. 2.

We have ${\bf d}(N^{\bullet})\cong N$, ${\bf c}(N^{\bullet})\cong({\bf
d}N)^{\bullet}$.

  3. 3.

If $N$ is a principal, then $N\cong({\bf d}N)^{\bullet}$.

  4. 4.

If $T$ is a positive sub-face structure of $N^{\bullet}$ then either
$T=N^{\bullet}$ or $T={\bf c}(N^{\bullet})$ or $T\subseteq N$.

Proof. Ad 1. We shall check globularity of the new added cells. The other
conditions are simple.

For ${\bf p}_{n+1}$, we have:

$\gamma\gamma({\bf p}_{n+1})=\gamma({\bf p}_{n})=$
$=\gamma(N_{n})-\delta(N_{n})=\gamma\delta({\bf p}_{n+1})-\delta\delta({\bf
p}_{n+1})$

and

$\delta\gamma({\bf p}_{n+1})=\delta({\bf p}_{n})=$
$=\delta(N_{n})-\gamma(N_{n})=\delta\delta({\bf p}_{n+1})-\gamma\delta({\bf
p}_{n+1}).$

So globularity holds for ${\bf p}_{n+1}$.

For ${\bf p}_{n}$, using Lemmas 7.1, 5.19 and normality of $N$, we have:

$\gamma\gamma({\bf p}_{n})=\gamma({\bf p}_{n-1})={\bf p}_{n-2}=$
$=\gamma(N_{n-1})-\delta(N_{n-1})=$
$=\gamma(N_{n-1}-\gamma(N_{n}))-\delta(N_{n-1}-\gamma(N_{n}))=$
$=\gamma(\delta(N_{n})-\gamma(N_{n}))-\delta(\delta(N_{n})-\gamma(N_{n}))=$
$=\gamma\delta({\bf p}_{n})-\delta\delta({\bf p}_{n})$

and similarly

$\delta\gamma({\bf p}_{n})=\delta({\bf p}_{n-1})=$
$=\delta(N_{n})-\gamma(N_{n})=$
$=\delta(\delta(N_{n})-\gamma(N_{n}))-\gamma(\delta(N_{n})-\gamma(N_{n}))=$
$=\delta\delta({\bf p}_{n})-\gamma\delta({\bf p}_{n})$

So globularity for ${\bf p}_{n}$ holds, as well.

Ad 2. The first isomorphism is obvious.

The faces of $(N^{\bullet})$, ${\bf c}(N^{\bullet})$, ${\bf d}N$, and $({\bf
d}N)^{\bullet}$ are as in the tables

$\begin{array}[]{|c|c|c|}dim&(N^{\bullet})&{\bf c}(N^{\bullet})\\\ \hline\cr
n+1&\\{{\bf p}_{n+1}^{N^{\bullet}}\\}&\emptyset\\\ \hline\cr
n&N_{n}\cup\\{{\bf p}_{n}^{N^{\bullet}}\\}&\\{{\bf p}_{n}^{N^{\bullet}}\\}\\\
\hline\cr n-1&N_{n-1}&N_{n-1}-(\gamma(N_{n})\cap\delta(N_{n}))\\\ \hline\cr
n-2&N_{n-2}&N_{n-2}\end{array}$

and

$\begin{array}[]{|c|c|c|}dim&{\bf d}N&({\bf d}N)^{\bullet}\\\ \hline\cr
n+1&\emptyset&\emptyset\\\ \hline\cr n&\emptyset&\\{{\bf p}_{n}^{({\bf
d}N)^{\bullet}}\\}\\\ \hline\cr
n-1&N_{n-1}-\gamma(N_{n})&(N_{n-1}-\gamma(N_{n}))\cup\\{{\bf p}_{n-1}^{({\bf
d}N)^{\bullet}}\\}\\\ \hline\cr n-2&N_{n-2}&N_{n-2}\end{array}$

We define the isomorphism $f:{\bf c}(N^{\bullet})\longrightarrow({\bf
d}N)^{\bullet}$ as follows

$f_{n}({\bf p}_{n+1}^{N^{\bullet}})={\bf p}_{n+1}^{({\bf d}N)^{\bullet}},$
$f_{n-1}(x)=\left\\{\begin{array}[]{ll}{\bf p}_{n-1}^{({\bf
d}N)^{\bullet}}&\mbox{if $x=\gamma({\bf p}_{n}^{N^{\bullet}})$,}\\\
x&\mbox{otherwise.}\end{array}\right.$

and $f_{l}=1_{N_{l}}$ for $l<n-1$. Clearly, all $f_{i}$’s are bijective. The
preservation of the domains and codomains is left for the reader.

3\. is left as an exercise.

Ad 4. If ${\bf p}_{n+1}\in T_{n+1}$ then $T=N^{\bullet}$. If ${\bf
p}_{n}\not\in T_{n}$ then $T\subseteq N$.

Suppose that ${\bf p}_{n+1}\not\in T_{n+1}$ but ${\bf p}_{n}\in T_{n}$. Since
$N^{\bullet}=[{\bf p}_{n+1}]$, by Lemma 6.1 it is enough to show that $T=[{\bf
p}_{n}]$. Clearly $[{\bf p}_{n}]\subseteq T$. As $[{\bf p}_{n}]_{l}=N_{l}$,
for $l<n-1$ we have $[{\bf p}_{n}]_{l}=T_{l}$, for $l<n-1$, as well.

Fix $x\in N_{n}$. As $x\in\delta({\bf p}_{n+1})$ and $\gamma({\bf
p}_{n+1})={\bf p}_{n}$, we have $x<^{N^{\bullet},+}{\bf p}_{n}$. So by
Corollary 5.11 $x\not\perp^{N^{\bullet},-}_{l}{\bf p}_{n}$, for any $l\leq n$.
Thus we cannot have $x\perp^{T,-}_{l}{\bf p}_{n}$, for any $l\leq n$, as well.
As $T$ is a positive face structure, again by Corollary 5.11, $x\not\in T$.
Since $x$ was an arbitrary element of $N_{n}$, we have $T_{n}=\\{{\bf
p}_{n}\\}=[{\bf p}_{n}]_{n}$.

It remains to show that $T_{n-1}=[{\bf p}_{n}]_{n-1}$. Suppose that $x\in
N_{n-1}-(\delta({\bf p}_{n})\cup\gamma({\bf p}_{n}))$. Then
$x<^{N,+}\gamma({\bf p}_{n})$ and hence
$x\not\perp^{N^{\bullet},-}_{l}\gamma({\bf p}_{n})$, for $l\leq n$. So $x$ and
$\gamma({\bf p}_{n})$ cannot be $<^{T,-}_{l}$ comparable, for $l\leq n$.
Since, as we have shown, $N_{n}\cap T_{n}=\emptyset$, it follows that $x$ and
$\gamma({\bf p}_{n})$ cannot be $<^{T,+}$ comparable. So by Lemma 5.11,
$x\not\in T_{n-1}$, i.e. $T_{n-1}=\delta({\bf p}_{n})\cup\gamma({\bf
p}_{n})=[{\bf p}_{n}]_{n-1}$. $~{}~{}\Box$

##  8 Decomposition of positive face structures

Let $T$ be a positive face structure, $X\subseteq T$ a subhypergraph of $T$,
$k\in{\omega}$, $a\in(T_{k}-\iota(T_{k+2}))$. We define two subhypergraphs of
$T$, $X^{\downarrow a}$ and $X^{\uparrow a}$, as follows:

$X^{\downarrow a}_{l}=\left\\{\begin{array}[]{ll}\\{\alpha\in
X_{l}:\gamma^{(k)}(\alpha)\leq^{+}a\\}&\mbox{for $l>k$,}\\\ \\{b\in
X_{k}:b\leq^{+}a\;{\rm or}\;\;b\not\in\gamma(X_{k+1})\\}&\mbox{for $l=k$}\\\
X_{l}&\mbox{for $l<k$.}\end{array}\right.$

$X^{\uparrow a}_{l}=\left\\{\begin{array}[]{ll}\\{\alpha\in
X_{l}:\gamma^{(k)}(\alpha)\not\leq^{+}a\\}&\mbox{for $l>k$,}\\\ \\{b\in
X_{k}:b\not<^{+}a\mbox{ or }b\not\in\delta(X_{k+1})\\}&\mbox{for $l=k$}\\\
X_{k-1}-\iota(X^{\downarrow a}_{k+1})&\mbox{for $l=k-1$}\\\ X_{l}&\mbox{for
$l<k-1$.}\end{array}\right.$

Intuitively, if $X$ is a positive face substructure, $X^{\downarrow a}$ is the
least positive face substructure of $X$ that contains faces ’smaller or equal’
$a$ and can be $k$-pre-composed with the ’rest’ to get $X$. $X^{\uparrow a}$
is this ’rest’ or in other words it is the largest positive face substructure
of $X$ that can be $k$-post-composed with $X^{\downarrow a}$ to get $X$ (or
largest positive face substructure of $X$ that do not contains faces ’smaller’
than $a$).

Examples. If $X$ is a hypergraph $a\in T$ then $X^{\downarrow a}$ is a
hypergraph as well. However, this is not the case with $X^{\uparrow a}$, if
$a\in\iota(T)$, as we can see below:

$\cdot\vphantom{\cdot}$$\vphantom{\cdot}\cdot$$a$$\Downarrow$$X$:$\cdot\vphantom{\cdot}$$\vphantom{\cdot}\cdot$$a$$\Downarrow$$X^{\uparrow
a}$:

Here $X=T$. The faces in the domain of the 2-dimensional face are not in
$X^{\uparrow a}$, i.e. $X^{\uparrow a}$ is not closed under $\delta$.

To see some real decompositions let fix a positive face structure $T$ as
follows:

$T$$\bullet\vphantom{}$$\bullet$$\vphantom{}y$$\bullet$$\Downarrow$$\Downarrow$$b$$a$$\bullet\vphantom{\bullet}$$\vphantom{\bullet}\bullet$$\bullet$$\Downarrow$$x\vphantom{}$$\vphantom{}\bullet$$\Downarrow$$\Downarrow$

Clearly $x,y,a,b\in T-\iota(T)$. Then

$T^{\downarrow
a}$$\bullet\vphantom{}$$\bullet$$\vphantom{}y$$\bullet$$b$$a$$\bullet\vphantom{\bullet}$$\vphantom{\bullet}\bullet$$\bullet$$\Downarrow$$x\vphantom{}$$\vphantom{}\bullet$$\Downarrow$

and

$T^{\uparrow
a}$$\bullet\vphantom{}$$\bullet$$\vphantom{}y$$\bullet$$\Downarrow$$\Downarrow$$a$$\vphantom{}\bullet$$x\vphantom{}$$\vphantom{}\bullet$$\Downarrow$

Moreover with

$X_{1}$$\bullet\vphantom{}$$\bullet$$\vphantom{}y$$\bullet$$\Downarrow$$\Downarrow$$a$$\vphantom{}\bullet$$x\vphantom{}$$\vphantom{}\bullet$$\Downarrow$

we have $X_{1}^{\uparrow b}=X_{1}$ and

$X_{1}^{\downarrow
b}$$\bullet\vphantom{}$$\bullet$$\vphantom{}y$$\bullet$$a$$\vphantom{}\bullet$$x\vphantom{}$$\vphantom{}\bullet$

i.e. $X_{1}^{\downarrow b}={\bf d}^{(1)}(X_{1})$. For

$X_{2}$$\bullet\vphantom{}$$\bullet$$\vphantom{}y$$\bullet$$\Downarrow$$\Downarrow$

we have $X_{2}^{\downarrow x}=X_{2}$ and $X_{2}^{\uparrow x}=\\{y\\}$.

We have

###### Lemma 8.1

Let $T$ be a positive face structure, $X\subseteq T$ a subhypergraph of $T$,
$a\in(T-\iota(T))$, $a\in X_{k}$. Then

  1. 1.

$X^{\downarrow a}$ and $X^{\uparrow a}$ are positive face structures;

  2. 2.

${\bf c}^{(k)}(X^{\downarrow a})={\bf d}^{(k)}(X^{\uparrow a})=X^{\downarrow
a}\cap X^{\uparrow a}$;

  3. 3.

${\bf d}^{(k)}(X^{\downarrow a})={\bf d}^{(k)}(X)$, ${\bf c}^{(k)}(X^{\uparrow
a})={\bf c}^{(k)}(X)$;

  4. 4.

$X=X^{\downarrow a}+_{k}X^{\uparrow a}=X^{\downarrow a}\cup X^{\uparrow a}$.

Proof. Ad 1. The verification that both $X^{\downarrow a}$ and $X^{\uparrow
a}$ are closed under $\gamma$ and $\delta$ is routine.

For any $k$, if $x,y\in X^{\downarrow a}_{k}$ then $x<^{+,X}y$ iff
$x<^{+,X^{\downarrow a}}y$. Similarly, for any $k$, if $x,y\in X^{\uparrow
a}_{k}$ then $x<^{+,X}y$ iff $x<^{+,X^{\uparrow a}}y$. Thus by Lemma 5.17 both
$X^{\downarrow a}$ and $X^{\uparrow a}$ positive face structures.

Ad 2. Let as spell in details both sides of the equation.

${\bf c}^{(k)}(X^{\downarrow a})$ is:

  1. 1.

${\bf c}^{(k)}(X^{\downarrow a})_{l}=\emptyset$, for $l>k$;

  2. 2.

${\bf c}^{(k)}(X^{\downarrow a})_{k}=\\\ (\\{b\in
X_{k}:b\leq^{+}a\\}\cup(X_{k}-\gamma(X_{k+1})))-\delta(\\{\alpha\in
X_{k+1}:\gamma(\alpha)\leq^{+}a\\})$;

  3. 3.

${\bf c}^{(k)}(X^{\downarrow a})_{k-1}=X_{k-1}-\iota(X^{\downarrow a}_{k+1})$;

  4. 4.

${\bf c}^{(k)}(X^{\downarrow a})_{l}=X_{l}$, for $l<k-1$.

and ${\bf d}^{(k)}(X^{\uparrow a})$ is:

  1. 1.

${\bf d}^{(k)}(X^{\uparrow a})_{l}=\emptyset$, for $l>k$;

  2. 2.

${\bf d}^{(k)}(X^{\uparrow a})_{k}=\\\ \\{b\in X_{k}:b\not<^{+}a\mbox{ or
}b\not\in\delta(X_{k+1})\\}-\gamma(X_{k+1}-\\{\alpha\in
X_{k+1}:\gamma(\alpha)\leq^{+}a\\})$;

  3. 3.

${\bf d}^{(k)}(X^{\uparrow a})_{k-1}=X_{k-1}-\iota(X^{\downarrow a}_{k+1})$;

  4. 4.

${\bf d}^{(k)}(X^{\uparrow a})_{l}=X_{l}$, for $l<k-1$.

Thus to show that ${\bf c}^{(k)}(X^{\downarrow a})={\bf d}^{(k)}(X^{\uparrow
a})$ we need to verify that ${\bf c}^{(k)}(X^{\downarrow a})_{k}={\bf
d}^{(k)}(X^{\uparrow a})_{k}$. As both sets are contained in $X_{k}$, we can
compare their complements. We have

$X_{k}-{\bf c}^{(k)}(X^{\downarrow
a})_{k}=\\{b\in\delta(X_{k+1}):b<^{+}a\\}\cup\gamma(X_{k+1}-\\{\alpha\in
X_{k+1}:\gamma(\alpha)\not\leq^{+}a\\})$

and

$X_{k}-{\bf d}^{(k)}(X^{\uparrow
a})_{k}=\\{b\in\gamma(X_{k+1}):b\not\leq^{+}a\\}\cup\delta(\\{\alpha\in
X_{k+1}\gamma(\alpha)\leq^{+}a\\}).$

But it easy to see that

$\\{b\in\delta(X_{k+1}):b<^{+}a\\}=\delta(\\{\alpha\in
X_{k+1}\gamma(\alpha)\leq^{+}a\\})$

and

$\gamma(X_{k+1}-\\{\alpha\in
X_{k+1}:\gamma(\alpha)\not\leq^{+}a\\})=\\{b\in\gamma(X_{k+1}):b\not\leq^{+}a\\}.$

The second equality uses the fact that $a\not\in\iota(T)$. Thus ${\bf
c}^{(k)}(X^{\downarrow a})_{k}={\bf d}^{(k)}(X^{\uparrow a})_{k}$, as
required.

Ad 3. To see that ${\bf c}^{(k)}(X^{\uparrow a})={\bf c}^{(k)}(X)$ it is
enough to note that $\iota(X_{k+1})=\iota(X^{\downarrow
a}_{k+1})\cup\iota(X^{\uparrow a}_{k+1})$. The equation ${\bf
d}^{(k)}(X^{\downarrow a})={\bf d}^{(k)}(X)$ is even simpler.

Ad 4. Obvious. $~{}~{}\Box$

###### Corollary 8.2

Let $T$ be a positive face structure, $k\in{\omega}$,
$a\in(T_{k}-\iota(T_{k+2}))$. Then the square

${\bf c}^{(k)}(T^{\downarrow a})\vphantom{T^{\uparrow a}}$$\vphantom{{\bf
c}^{(k)}(T^{\downarrow a})}T^{\uparrow a}$$T^{\downarrow
a}\vphantom{T}$$\vphantom{T^{\downarrow a}}T$

is a special pushout in ${\bf Fs}^{+/1}$.

Proof. Follows immediately from Lemmas 6.2 and 8.1. $~{}~{}\Box$

We need some notions and notations. Let $X$, $T$ be a positive face structures
$X\subseteq T$, $a\in(T-\iota(T))$. The decomposition $X=X^{\downarrow a}\cup
X^{\uparrow a}$ is said to be proper iff $size(X^{\downarrow
a}),size(X^{\uparrow a})<size(X)$. If the decomposition $X=X^{\downarrow
a}\cup X^{\uparrow a}$ is proper then $a$ is said to be a saddle face of $X$.
$Sd(X)$ is the set of saddle faces of $X$; $Sd(X)_{k}=Sd(X)\cap X_{k}$.

###### Lemma 8.3

Let $X$, $S$, $T$ be positive face structures, $X\subseteq T$, $l\in{\omega}$.
Then

  1. 1.

if $a\in(T_{l}-\iota(T))$ then $a\in Sd(X)$ iff there are $\alpha,\beta\in
X_{l+1}$ such that $\gamma(\alpha)\leq^{+}a$ and $\gamma(\beta)\not\leq^{+}a$;

  2. 2.

if ${\bf c}^{(k)}(S)={\bf d}^{(k)}(T)$ then

$size(S+_{k}T)_{l}=\left\\{\begin{array}[]{ll}size(S)_{l}+size(T)_{l}&\mbox{
if $l>k$,}\\\ size(T)_{l}&\mbox{ if $l\leq k$;}\end{array}\right.$

  3. 3.

$size(S)_{k}\geq 1$ iff $k\leq dim(S)$;

  4. 4.

if $a\in Sd(S)_{k}$ then $size(S)_{k+1}\geq 2$;

  5. 5.

$S$ is principal iff $Sd(S)$ is empty.

Proof. We shall show 5. The rest is easy.

If there is $a\in Sd(S)_{k}$ then by 2., 3. and Lemma 8.1 we have that
$size(S)_{k+1}=size(S^{\downarrow a})_{k+1}+size(S^{\uparrow a})_{k+1}\geq
1+1>1$. So in that case $S$ is not principal.

For the converse, assume that $S$ is not principal. Fix $k\in{\omega}$ such,
that $size(S)_{k+1}>1$. Thus there are $a,b\in S_{k+1}$, that $a\neq b$.
Suppose $\gamma(a)\in\iota(\alpha)$ for some $\alpha\in S_{k+2}$. Then by
Lemma 5.5, $a<^{+}\gamma(\alpha)$ contrary to the assumption on $a$. Hence
$a\in S-\iota(S)$ and for similar reasons $b\in S-\iota(S)$. We have
$a\not\perp^{+}b$ and, by pencil linearity, $\gamma(a)\neq\gamma(b)$. Then
either $\gamma(a)\not<^{+}\gamma(b)$ and then $\gamma(b)\in Sd(S)_{k}$ or
$\gamma(b)\not<^{+}\gamma(a)$ and then $\gamma(a)\in Sd(S)_{k}$. In either
case $Sd(S)$ is not empty, as required. $~{}\Box$

###### Lemma 8.4

Let $T$, $X$ be positive face structures, $X\subseteq T$, and $a,x\in
X-\iota(X)$, $k=dim(x)<dim(a)=m$.

  1. 1.

We have the following equations of positive face structures:

$X^{\downarrow x\downarrow a}=X^{\downarrow a\downarrow x},\;\;\;X^{\downarrow
x\uparrow a}=X^{\uparrow a\downarrow x},\;\;\;X^{\uparrow x\downarrow
a}=X^{\downarrow a\uparrow x},\;\;\;X^{\uparrow x\uparrow a}=X^{\uparrow
a\uparrow x},$

i.e. ’the decompositions of different dimension commute’.

  2. 2.

If $x\in Sd(X)$ then $x\in Sd(X^{\downarrow a})\cap Sd(X^{\uparrow a})$.

  3. 3.

Moreover, we have the following equations concerning domains and codomains

${\bf c}^{(k)}(X^{\downarrow x\downarrow a})={\bf c}^{(k)}(X^{\downarrow
x\uparrow a})={\bf d}^{(k)}(X^{\uparrow x\downarrow a})={\bf
d}^{(k)}(X^{\uparrow x\uparrow a})$ ${\bf c}^{(m)}(X^{\downarrow x\downarrow
a})={\bf d}^{(m)}(X^{\downarrow x\uparrow a}),\;\;\;\;{\bf
c}^{(m)}(X^{\uparrow x\downarrow a})={\bf d}^{(m)}(X^{\uparrow x\uparrow a}).$

  4. 4.

Finally, we have the following equations concerning compositions

$X^{\downarrow x\downarrow a}+_{m}X^{\downarrow x\uparrow a}=X^{\downarrow
x},\;\;\;\;X^{\uparrow x\downarrow a}+_{m}X^{\uparrow x\uparrow a}=X^{\uparrow
x}$ $X^{\downarrow x\downarrow a}+_{k}X^{\uparrow x\downarrow a}=X^{\downarrow
a},\;\;\;\;X^{\downarrow x\uparrow a}+_{k}X^{\uparrow x\uparrow a}=X^{\uparrow
a}.$

Proof. Simple check. $~{}\Box$

###### Lemma 8.5

Let $T$, $X$ be positive face structures, $X\subseteq T$, and $a,b\in
X-\iota(X)$, $dim(a)=dim(b)=m$.

  1. 1.

We have the following equations of positive face structures:

$X^{\downarrow a\downarrow b}=X^{\downarrow b\downarrow a},\;\;\;X^{\uparrow
a\uparrow b}=X^{\uparrow b\uparrow a},$

i.e. ’the decompositions in the same dimension and the same directions
commute’.

  2. 2.

Assume $a<^{+}b$. Then we have the following farther equations of positive
face structures:

$X^{\downarrow a}=X^{\downarrow a\downarrow b},\;\;\;X^{\uparrow
b}=X^{\uparrow a\uparrow b},\;\;\;X^{\downarrow b\uparrow a}=X^{\uparrow
a\downarrow b}.$

Moreover, if $a,b\in Sd(X)$ then $a\in Sd(X^{\downarrow b})$ and $b\in
Sd(X^{\uparrow a})$.

  3. 3.

Assume $a<^{-}_{l}b$, for some $l<m$. Then $X^{\uparrow b\downarrow a}$,
$X^{\uparrow a\downarrow b}$, are positive face structures, and

$X^{\downarrow a}+_{m}X^{\uparrow a\downarrow b}=X^{\downarrow
b}+_{m}X^{\uparrow b\downarrow a}$

Moreover, if $a,b\in Sd(X)$ then either there is $k$ such that $l-1\leq k<m$
and $\gamma^{(k)}(a)\in Sd(X)$ or $a\in Sd(X^{\uparrow b})$ and $b\in
Sd(X^{\uparrow a})$.

Proof. Simple check. $~{}\Box$

###### Lemma 8.6

Let $T$, $X$ be positive face structures, $X\subseteq T$, $dim(X)=n$, $l<n-1$,
$a\in Sd(X)_{l}$. Then

  1. 1.

$a\in Sd({\bf c}X)\cap Sd({\bf d}X)$;

  2. 2.

${\bf d}(X^{\downarrow a})=({\bf d}X)^{\downarrow a}$;

  3. 3.

${\bf d}(X^{\uparrow a})=({\bf d}X)^{\uparrow a}$;

  4. 4.

${\bf c}(X^{\downarrow a})=({\bf c}X)^{\downarrow a}$;

  5. 5.

${\bf c}(X^{\uparrow a})=({\bf c}X)^{\uparrow a}$.

Proof. The proof is again by a long and simple check. We shall check part of
5. We should consider separately cases: $l=n-2$, $l=n-3$, and $l<n-3$, but we
shall check the case $l=n-3$ only. The other cases can be also shown by
similar, but easier, check.

$({\bf c}X)^{\uparrow a}$ is:

  1. 1.

$({\bf c}X)^{\uparrow a}_{l}=\emptyset$, for $l\geq n$;

  2. 2.

$({\bf c}X)^{\uparrow a}_{n-1}=\\{x\in
X_{n-1}:\gamma^{(n-3)}(x)\not\leq^{+}a,\;\;x\not\in\delta(X_{n})\\}$;

  3. 3.

$({\bf c}X)^{\uparrow a}_{n-2}=\\{x\in
X_{n-2}:\gamma(x)\not\leq^{+}a,\;\;x\not\in\iota(X_{n})\\}$;

  4. 4.

$({\bf c}X)^{\uparrow a}_{n-3}=\\{x\in X_{n-3}:x\not<^{+}a\mbox{ or
}x\not\in\delta(X_{n-2}-\iota(X_{n}))\\}$;

  5. 5.

$({\bf c}X)^{\uparrow a}_{n-4}=X_{n-4}-\iota(\\{x\in
X_{n-2}:x\not\in\iota(X_{n}),\;\gamma(x)\leq^{+}a\\})$;

  6. 6.

$X^{\downarrow a}_{l}=X_{l}$, for $l<n-4$.

and ${\bf c}(X^{\uparrow a})$ is:

  1. 1.

${\bf c}(X^{\uparrow a})_{l}=\emptyset$, for $l\geq n$;

  2. 2.

${\bf c}(X^{\uparrow
a})_{n-1}=\\{x\\!\in\\!X_{n-1}:\gamma^{(n-3)}(x)\not\leq^{+}a\\}-\delta(\\{z\\!\in\\!X_{n}:\gamma^{(n-3)}(z)\not\leq^{+}a\\})$;

  3. 3.

${\bf c}(X^{\uparrow
a})_{n-2}=\\{x\\!\in\\!X_{n-2}:\gamma(x)\not\leq^{+}a\\}-\iota(\\{z\\!\in\\!X_{n}:\gamma^{(n-3)}(z)\not\leq^{+}a\\})$;

  4. 4.

${\bf c}(X^{\uparrow a})_{n-3}=\\{x\in X_{n-3}:x\not<^{+}a\mbox{ or
}x\not\in\delta(X_{n-2})\\}$;

  5. 5.

${\bf c}(X^{\uparrow a})_{n-4}=X_{n-4}-\iota(X^{\downarrow a}_{n-2})$;

  6. 6.

${\bf c}(X^{\uparrow a})_{l}=X_{l}$, for $l<n-4$.

We need to verify the equality $({\bf c}X)^{\uparrow a}_{l}={\bf
c}(X^{\uparrow a})_{l}$ for $l=n-1,\ldots,n-4$.

In dimension $n-1$, it is enough to show that if $x\in X_{n-1}$ and $z\in
X_{n}$ so that $\gamma^{(n-3)}(x)\not\leq^{+}a$ and $x\in\delta(z)$ then
$\gamma^{n-3}(z)\not\leq^{+}a$.

So assume that $x\in X_{n-1}$, $\gamma^{(n-3)}(x)\not\leq^{+}a$, $z\in X_{n}$
such that $x\in\delta(z)$. Hence $x\lhd^{+}\gamma(z)$. By Lemma 5.9.5
$\gamma^{(n-3)}(x)\leq^{+}\gamma^{(n-3)}(z)$. Therefore
$\gamma^{(n-3)}(z)\not\leq^{+}a$ (otherwise we would have
$\gamma^{(n-3)}(x)\not\leq^{+}a$), as required.

In dimension $n-2$, it is enough to show that if $x\in X_{n-2}$ and $z\in
X_{n}$ so that $x\not\leq^{+}a$ and $x\in\iota(z)$ then
$\gamma^{n-3}(z)\not\leq^{+}a$.

So assume that $x\in X_{n-2}$, $z\in X_{n}$ so that $x\not\leq^{+}a$ and
$x\in\iota(z)$. Hence $x\leq^{+}\gamma\gamma(z)$. By Lemma 5.9.5
$\gamma(x)\leq^{+}\gamma^{(n-3)}(z)$. Therefore
$\gamma^{(n-3)}(z)\not\leq^{+}a$, as required.

The equality in dimension $n-3$ follows immediately from Lemma 5.19.4.

To show that in dimension $n-4$, the above equation also holds we shall show
that

$\iota(X^{\downarrow a}_{n-2})\subseteq\iota(\\{x\in
X_{n-2}:x\not\in\iota(X_{n}),\;\gamma(x)\leq^{+}a\\})$

Note that, by Lemma 5.3.1, if $t\in X_{n-4}$ and $x\in X_{n-2}$, $y\in
X_{n-1}$, $t\in\iota(x)$ and $\gamma(x)\leq^{+}a$ and $x=\gamma(y)$ then there
is $x^{\prime}\in\delta(y)$ (i.e. $x^{\prime}\lhd^{+}x$ and hence
$\gamma(x^{\prime})\leq^{+}a$) such that $t\in\iota(x^{\prime})$.

Thus, as $<^{+}$ is well founded, from the above observation follows that, for
any $t\in X_{n-4}$ and $x\in X_{n-2}$ that $t\in\iota(x)$ and
$\gamma(x)\leq^{+}a$, there is $x^{\prime\prime}\leq^{+}x$ such that
$t\in\iota(x^{\prime\prime})$ and $x^{\prime\prime}\not\in\gamma(X)$. Then we
clearly have that $x^{\prime\prime}\not\in\iota(X)$ and
$\gamma(x^{\prime\prime})\leq^{+}a$, as required. $~{}\Box$

###### Lemma 8.7

Let $T,T_{1},T_{2}$ be positive face structures, $dim(T_{1}),dim(T_{2})>k$,
such that ${\bf c}^{(k)}(T_{1})={\bf d}^{(k)}(T_{2})$ and $T=T_{1}+_{k}T_{2}$.
Then ${\bf c}^{(k)}(T_{1})_{k}\cap\gamma(T_{1})\neq\emptyset$. For any
$a\in{\bf c}^{(k)}(T_{1})_{k}\cap\gamma(T_{1})$ we have $a\in Sd(T)_{k}$ and
either $T_{1}=T^{\downarrow a}$ and $T_{2}=T^{\uparrow a}$ or $a\in
Sd(T_{1})_{k}$, $T^{\downarrow a}=T_{1}^{\downarrow a}$ and $T^{\uparrow
a}=T_{1}^{\uparrow a}+_{k}T_{2}$.

Proof. By assumption $(T_{1})_{k+1}\neq\emptyset$ and
$(T_{2})_{k+1}\neq\emptyset$. So ${\bf
c}^{(k)}(T_{1})\cap\gamma(T_{1})\neq\emptyset$. Fix $a\in{\bf
c}^{(k)}(T_{1})\cap\gamma(T_{1})\neq\emptyset$. Then $T^{\downarrow
a}_{k+1}\neq\emptyset$. As $T^{\downarrow a}_{k+1}\cap(T_{2})_{k+1}=\emptyset$
we must have $a\in Sd(T)_{k}$.

Assume that $T_{1}\neq T^{\downarrow a}$. Then $T^{\downarrow
a}\subseteq\hskip-11.38109pt_{\not}\hskip 8.53581ptT_{1}$. Hence
$(T_{1})-(T^{\downarrow a})\neq\emptyset$. But this means that $a\in
Sd(T_{1})_{k}$. The verification that the equalities $T^{\downarrow
a}=T_{1}^{\downarrow a}$ and $T^{\uparrow a}=T_{1}^{\uparrow a}+_{k}T_{2}$
hold in this case is left as an exercise. $~{}\Box$

##  9 $S^{*}$ is a positive-to-one computad

###### Proposition 9.1

Let $S$ be a weak positive face structure. Then $S^{*}$ is a positive-to-one
computad.

Proof. The proof is by induction on dimension $n$ of the weak positive face
structure $S$. For $n=0,1$ the Proposition is obvious.

So assume that for any weak positive face structure $T$ of dimension $n$,
$T^{*}$ is a positive-to-one computad of dimension $n$, generated by faces in
$T$. Suppose that $S$ is a weak positive face structure of dimension $n+1$. We
shall show that $S^{*}$ is a computad generated by faces in $S$. Since
$S_{\leq n}$ is a weak positive face structure, by inductive assumption
$S_{\leq n}^{*}$ is the computad generated by faces in $S_{\leq n}$. So we
need to verify that for any ${\omega}$-functor $f:S_{\leq
n}^{*}\longrightarrow C$ to any ${\omega}$-category $C$, and any function
$|f|:S_{n+1}\longrightarrow C_{n+1}$ such, that for $a\in S_{n+1}$

$d_{C}(|f|(a))=f({\bf d}([a])),\;\;\;\;\;c_{C}(|f|(a))=f({\bf c}([a])),$

there is a unique ${\omega}$-functor $F:S^{*}\longrightarrow C$, such that

$F_{n+1}([a])=|f|(a),\;\;\;\;\;F_{\leq n}=f$

as in the diagram

$S_{\leq n}\vphantom{S_{\leq n}^{*}}$$\vphantom{S_{\leq n}}S_{\leq
n}^{*}$$[-]$$S_{n+1}\vphantom{S^{*}}$$\vphantom{S_{n+1}}S^{*}$$\delta$$\gamma$${\bf
d}$${\bf c}$$F$$f$$C$$[-]$$|f|$

We define $F_{n+1}$ as follows. For $X\in S^{*}_{n+1}$

$F_{n+1}(X)=\left\\{\begin{array}[]{ll}id_{f(X)}&\mbox{ if $dim(X)\leq n$,}\\\
|f|(a)&\mbox{ if $dim(X)=n+1$, $X$ is principal and $X=[a]$,}\\\
F_{n+1}(X^{\downarrow a});_{l}F_{n+1}(X^{\uparrow a})&\mbox{ if $dim(X)=n+1$,
$a\in Sd(X)_{l}$.}\end{array}\right.$

$;_{l}$ refers to the composition in $C$. Clearly $F_{k}=f_{k}$ for $k\leq n$.
The above morphism, if well defined, clearly preserves identities. We need to
verify, for $X\in S^{*}_{n+1}$ and $dim(X)=n+1$, three conditions:

  * I

$F$ is well defined, i.e. $F_{n+1}(X)=F_{n+1}(X^{\downarrow
a});_{l}F_{n+1}(X^{\uparrow a})$ does not depend on the choice of the saddle
face $a\in Sd(X)$;

  * II

$F$ preserves the domains and codomains i.e., $F_{n}(d(X))=d(F_{n+1}(X))$ and
$F_{n}(c(X))=c(F_{n+1}(X))$;

  * III

$F$ preserves compositions i.e.,
$F_{n+1}(X)=F_{n+1}(X_{1});_{k}F_{n+1}(X_{2})$ whenever $X=X_{1}+_{k}X_{2}$
and $dim(X_{1}),dim(X_{2})>k$.

We have an embedding $[-]:S_{\leq n}\longrightarrow S_{\leq n}^{*}$.

So let assume that for positive face substructures of $S$ of size less than
$size(X)$ the above assumption holds. If $size(X)_{n+1}=0$ or $X$ is principal
all three conditions are obvious. So assume that $X$ is not principal and
$dim(X)=n+1$. To save on notation we write $F$ for $F_{n+1}$.

Ad I. First we will consider two saddle faces $a,x\in Sd(X)$ of different
dimension $k=dim(x)<dim(a)=m$. Using Lemma 8.4 we have

$\displaystyle F(X^{\downarrow a});_{m}F(X^{\uparrow a})=$ $\displaystyle
ind.\;hyp.\;I$ $\displaystyle=(F(X^{\downarrow a\downarrow
x});_{k}F(X^{\downarrow a\uparrow x}));_{m}(F(X^{\uparrow a\downarrow
x});_{k}F(X^{\uparrow a\uparrow x}))=$ $\displaystyle MEL$
$\displaystyle=(F(X^{\downarrow a\downarrow x});_{m}F(X^{\uparrow a\downarrow
x}));_{k}(F(X^{\downarrow a\uparrow x});_{m}F(X^{\uparrow a\uparrow x}))=$
$\displaystyle=(F(X^{\downarrow x\downarrow a});_{m}F(X^{\downarrow x\uparrow
a}));_{k}(F(X^{\uparrow x\downarrow a});_{m}F(X^{\uparrow x\uparrow a}))=$
$\displaystyle ind.\;hyp.\;III$ $\displaystyle=F(X^{\downarrow
x});_{m}F(X^{\uparrow x})$

Now we will consider two saddle faces $a,b\in Sd(X)$ of the same dimension
$dim(a)=dim(b)=m$. We shall use Lemma 8.5. Assume that $a<^{-}_{l}b$, for some
$l<m$. If $\gamma^{(k)}(a)\in Sd(X)$, for some $k<m$, then this case reduces
to the previous one for two pairs $a,\gamma^{(k)}(a)\in Sd(X)$ and
$b,\gamma^{(k)}(a)\in Sd(X)$. Otherwise $a\in Sd(X^{\uparrow b})$ and $a\in
Sd(X^{\uparrow b})$ and we have

$\displaystyle F(X^{\downarrow a});_{k}F(X^{\uparrow a})=$ $\displaystyle
ind.\;hyp\;I$ $\displaystyle=F(X^{\downarrow a});_{k}(F(X^{\uparrow
a\downarrow b});_{k}F(X^{\uparrow a\uparrow b}))=$
$\displaystyle=(F(X^{\downarrow a});_{k}F(X^{\uparrow a\downarrow
b}));_{k}F(X^{\uparrow b\uparrow a})=$ $\displaystyle ind\;hyp\;III$
$\displaystyle=F(X^{\downarrow a};_{k}X^{\uparrow a\downarrow
b});_{k}F(X^{\uparrow b\uparrow a})=$ $\displaystyle=F(X^{\downarrow
b};_{k}X^{\uparrow b\downarrow a});_{k}F(X^{\uparrow b\uparrow a})=$
$\displaystyle ind\;hyp\;III$ $\displaystyle=(F(X^{\downarrow
b});_{k}F(X^{\uparrow b\downarrow a}));_{k}F(X^{\uparrow b\uparrow a})=$
$\displaystyle=F(X^{\downarrow b});_{k}(F(X^{\uparrow b\downarrow
a});_{k}F(X^{\uparrow b\uparrow a}))=$ $\displaystyle ind\;hyp\;I$
$\displaystyle=F(X^{\downarrow b});_{k}F(X^{\uparrow b})$

Finally, we consider the case $a<^{+}b$. We have

$\displaystyle F(X^{\downarrow a});_{k}F(X^{\uparrow a})=$ $\displaystyle
ind.\;hyp\;I$ $\displaystyle=F(X^{\downarrow a});_{k}(F(X^{\uparrow
a\downarrow b});_{k}F(X^{\uparrow a\uparrow b}))=$
$\displaystyle=F(X^{\downarrow b\downarrow a});_{k}(F(X^{\downarrow b\uparrow
a}));_{k}F(X^{\uparrow b})=$ $\displaystyle=(F(X^{\downarrow b\downarrow
a});_{k}F(X^{\downarrow b\uparrow a}));_{k}F(X^{\uparrow b})=$ $\displaystyle
ind\;hyp\;I$ $\displaystyle=F(X^{\downarrow b});_{k}F(X^{\uparrow b})$

This shows that $F(X)$ is well defined.

Ad II. We shall show that the domains are preserved. The proof that, the
codomains are preserved, is similar.

The fact that if $Sd(X)=\emptyset$ then $F$ preserves domains and codomains
follows immediately from the assumption on $f$ and $|f|$. So assume that
$Sd(X)\neq\emptyset$ and let $a\in Sd(X)$, $dim(a)=k$. We use Lemma 8.6. We
have to consider two cases $k<n$, and $k=n$.

If $k<n$ then

$\displaystyle F_{n}(d(X))=F_{n}(d(X^{\downarrow a}+_{k}X^{\uparrow a}))=$
$\displaystyle=F_{n}(d(X^{\downarrow a});_{k}d((X^{\uparrow a}))=$
$\displaystyle=F_{n}(d(X)^{\downarrow a};_{k}d(X)^{\uparrow a})=$
$\displaystyle ind\;hyp\;III$ $\displaystyle=F_{n}(d(X)^{\downarrow
a});_{k}F_{n}(d(X)^{\uparrow a})=$ $\displaystyle=F_{n}(d(X^{\downarrow
a}));_{k}F_{n}(d(X^{\uparrow a}))=$ $\displaystyle ind\;hyp\;II$
$\displaystyle=d(F_{n+1}(X^{\downarrow a}));_{k}d(F_{n+1}(X^{\uparrow a}))=$
$\displaystyle=d(F_{n+1}(X^{\downarrow a});_{k}F_{n+1}(X^{\uparrow a}))=$
$\displaystyle ind\;hyp\;I$ $\displaystyle=d(F_{n+1}(X))$

If $k=n$ then

$\displaystyle F_{n}(d(X))=F_{n}(d(X^{\downarrow a}+_{n}X^{\uparrow a}))=$
$\displaystyle=F_{n}(d(X^{\downarrow a}))=$ $\displaystyle ind\;hyp\;II$
$\displaystyle=d(F_{n+1}(X^{\downarrow a}))=$
$\displaystyle=d(F_{n+1}(X^{\downarrow a});_{n}F_{n+1}(X^{\uparrow a}))=$
$\displaystyle ind\;hyp\;I$ $\displaystyle=d(F_{n+1}(X))$

Ad III. Suppose that $X=X_{1};_{k}X_{2}$ and $dim(X)\leq n+1$. We shall show
that $F$ preserves this composition. If $dim(X_{1})=k$ then $X=X_{2}$,
$X_{1}={\bf d}^{(k)}(X_{2})$. We have

$F_{n+1}(X)=F_{n+1}(X_{2})=1^{(n+1)}_{F_{k}({\bf
d}^{(k)}(X_{2}))};_{k}F_{n+1}(X_{2})=$
$=1^{(n+1)}_{F_{k}(X_{1})};_{k}F_{n+1}(X_{2})=F_{n+1}(X_{1});_{k}F_{n+1}(X_{2})$

The case $dim(X_{2})=k$ is similar. So now assume that
$dim(X_{1}),dim(X_{2})>k$. We shall use Lemma 8.7. Fix $a\in{\bf
c}^{(k)}(X_{1})_{k}\cap\gamma(X_{1})$. So $a\in Sd(X)_{k}$. If
$X_{1}=X^{\downarrow a}$ and $X_{2}=X^{\uparrow a}$ then we have

$F(X)=F(X^{\downarrow a});_{k}F(X^{\uparrow a})=F(X_{1});_{k}F(X_{2}).$

If $a\in Sd(X_{1})_{k}$ then

$\displaystyle F(X)=F(X^{\downarrow a});_{k}F(X^{\uparrow a})=$ $\displaystyle
ind\;hyp\;II$ $\displaystyle=F(X^{\downarrow a});_{k}(F(X_{1}^{\uparrow
a});_{k}F(X_{2}))=$ $\displaystyle=(F(X_{1}^{\downarrow
a});_{k}F(X_{1}^{\uparrow a}));_{k}F(X_{2})=$ $\displaystyle ind\;hyp\;II$
$\displaystyle F(X_{1});_{k}F(X_{2})$

So in any case the composition is preserved.

This ends the proof of the Lemma. $~{}~{}\Box$

For $n\in{\omega}$, we have a functor

$(-)^{\sharp,n}:{\bf wFs}^{+/1}_{n}\longrightarrow{\bf Comma}^{+/1}_{n}$

such that, for $S$ in ${\bf wFs}^{+/1}_{n}$

$S^{\sharp,n}=(S_{n},S^{*}_{<n},[\delta],[\gamma])$

and for $f:S\rightarrow T$ in ${\bf wFs}^{+/1}_{n}$ we have

$f^{\natural,n}=(f_{n},(f_{<n})^{*}).$

###### Corollary 9.2

For every $n\in{\omega}$, the functor $(-)^{\sharp,n}$ is well defined, full,
faithful, and it preserves existing pushouts. Moreover, for $S$ in ${\bf
wFs}^{+/1}_{n}$ we have $S^{*}=\overline{S^{\sharp,n}}^{n}$.

Proof. The functor $\overline{(-)}^{n}:{\bf
Comma}^{+/1}_{n}\longrightarrow{{\bf Comp}^{+/1}_{n}}$ which is an equivalence
of categories is described in the Appendix.

Fullness and faithfulness of $(-)^{\sharp,n}$ is left for the reader. We shall
show simultaneously that for every $n\in{\omega}$, both functors

$(-)^{\sharp,n}:{\bf wFs}^{+/1}_{n}\longrightarrow{\bf Comma}^{+/1}_{n},\hskip
28.45274pt(-)^{*,n}:{\bf wFs}^{+/1}_{n}\longrightarrow{{\bf Comp}^{+/1}_{n}}$

preserve existing pushouts. For $n=0$ there is nothing to prove. For $n=1$
this is obvious. So assume that $n>1$ and that $(-)^{\sharp,n}$ preserves
existing pushouts. Let

$R\vphantom{T}$$\vphantom{R}T$$S\vphantom{S+_{R}T}$$\vphantom{S}S+_{R}T$

be a pushout in ${\bf wFs}^{+/1}_{n+1}$. Clearly its $n$-truncation is a
pushout in ${\bf wFs}^{+/1}_{n}$. Hence by inductive hypothesis it is
preserved by $(-)^{*,n}$. In dimension $n+1$, the functor $(-)^{\sharp,n+1}$
is an inclusion. Hence, in dimension $n+1$, this square is a pushout (of
monos) in $Set$. So the whole square

$R^{\sharp,n+1}\vphantom{T^{\sharp,n+1}}$$\vphantom{R^{\sharp,n+1}}T^{\sharp,n+1}$$S^{\sharp,n+1}\vphantom{(S+_{R}T)^{\sharp,n+1}}$$\vphantom{S^{\sharp,n+1}}(S+_{R}T)^{\sharp,n+1}$

is a pushout in ${\bf Comma}_{n+1}$, i.e. $(-)^{\sharp,n+1}$ preserves
pushouts. As $(-)^{*,n+1}$ is a composition of $(-)^{\sharp,n+1}$ with an
equivalence of categories it preserves the pushouts, as well. $~{}~{}\Box$

###### Corollary 9.3

The functor

$(-)^{*}:{\bf wFs}^{+/1}\longrightarrow{\bf Comp}^{+/1}$

is full and faithful and preserves special pushouts.

Proof. This follows from the previous Corollary and the fact that the functor
$\overline{(-)}^{n}:{\bf Comma}^{+/1}_{n}\longrightarrow{{\bf
Comp}^{+/1}_{n}}$ (see Appendix) is an equivalence of categories. $~{}~{}\Box$

Let $P$ be a positive-to-one computad, $a$ a $k$-cell in $P$. A description of
the cell $a$ is a pair

$<T_{a},\tau_{a}:T_{a}^{*}\longrightarrow P>$

where $T_{a}$ is a positive face structure and $\tau_{a}$ is a computad map
such that

$\tau_{a}(T_{a})=a.$

In the remainder of this section we shall define some specific positive face
structures that will be used later. First we define $\alpha^{n}$, for
$n\in{\omega}$. We put

$\alpha^{n}_{l}\;\;=\;\;\left\\{\begin{array}[]{ll}\emptyset&\mbox{ if
$l>n$}\\\ \\{2n\\}&\mbox{ if $l=n$}\\\ \\{2l+1,\,2l\\}&\mbox{ if $0\leq
l<n$}\end{array}\right.$

$d,\,c:\alpha^{n}_{l}\longrightarrow\alpha^{n}_{l-1}$

$d(x)=\\{2l-1\\}\;\;\;\;\;\;\;\;\;c(x)=2l-2$

for $x\in\alpha^{n}_{l}$, and $1\leq l\leq n$.

For example $\alpha^{4}$ can be pictured as follows:

${\bf 1}$${\bf 0}$${\bf 3}$${\bf 2}$${\bf 5}$${\bf 4}$${\bf 7}$${\bf 6}$${\bf
8}$

i.e. $8$ is the unique cell of dimension $4$ in $\alpha^{4}$ that has $7$ as
its domain and $6$ as its codomain, $7$ and $6$ have $5$ as its domain and $4$
as its codomain, and so on. Note that, for any $k\leq n$, we have

${\bf d}^{(k)}\alpha^{n}=\alpha^{k}={\bf c}^{(k)}\alpha^{n}.$

Let $n_{1}<n_{0},n_{2}$ and $n_{3}<n_{2},n_{4}$. We define the positive face
structures $\alpha^{n_{0},n_{1},n_{2}}$ and
$\alpha^{n_{0},n_{1},n_{2},n_{3},n_{4}}$ as the following colimits in ${\bf
Fs}^{+/1}$:

$\alpha^{n_{1}}\vphantom{\alpha^{n_{2}}}$$\vphantom{\alpha^{n_{1}}}\alpha^{n_{2}}$${\bf
d}^{(n_{1})}_{\alpha^{n_{2}}}$$\alpha^{n_{0}},\vphantom{\alpha^{n_{0},n_{1},n_{2}}}$$\vphantom{\alpha^{n_{0}},}\alpha^{n_{0},n_{1},n_{2}}$$\kappa_{1}$${\bf
c}^{(n_{1})}_{\alpha^{n_{0}}}$$\kappa_{2}$$\alpha^{n_{1}}\vphantom{\alpha^{n_{2}}}$$\vphantom{\alpha^{n_{1}}}\alpha^{n_{2}}$${\bf
d}^{(n_{1})}_{\alpha^{n_{2}}}$$\alpha^{n_{0}},\vphantom{\alpha^{n_{0},n_{1},n_{2},n_{3},n_{4}}}$$\vphantom{\alpha^{n_{0}},}\alpha^{n_{0},n_{1},n_{2},n_{3},n_{4}}$$\kappa_{1}$${\bf
c}^{(n_{1})}_{\alpha^{n_{0}}}$$\kappa_{2}$$\vphantom{\phantom{\alpha^{n_{2}}}}\alpha^{n_{3}}$${\bf
c}^{(n_{3})}_{\alpha^{n_{2}}}$$\vphantom{\phantom{\alpha^{n_{0},n_{1},n_{2},n_{3},n_{4}}}}\alpha^{n_{4}}$$\kappa_{3}$${\bf
d}^{(n_{3})}_{\alpha^{n_{4}}}$

We have

###### Proposition 9.4

The above colimits are preserved by the functor

$(-)^{*}:{\bf Fs}^{+/1}\longrightarrow{\bf Comp}^{+/1}.$

Moreover for any ${\omega}$-category $C$ we have bijective correspondences