K\setminus J}-\overline{\partial K\setminus J})}$
$\displaystyle=\left(\widetilde{\left((\widetilde{K\setminus
J}\sqcup\overline{\partial K\setminus J})\setminus\overline{\partial
K\setminus
J}\right)}\sqcup\overline{\overline{K_{J}}}-\overline{\overline{K_{J}}}\right)$
$\displaystyle\cong(\widetilde{\widetilde{K\setminus
J}}\sqcup\overline{\overline{J}}-\overline{\overline{J}})\cong(K-J),$

where we also used Lemma 3.3 and Proposition 1.11 for the last identity. ∎

Figure 14: A cc $K\in\mathsf{C}$ composed of two tetrahedra sharing a triangle
and seen as a 3-dimensional cobordism $(K-\emptyset),$ i.e. with empty ingoing
component. Its dual $\overline{(K-\emptyset)}$ is represented inside $K.$ In
this example the entire boundary of the dual is made of removed cells (i.e.
cells in the ingoing component). The cobordism $\overline{(K-\emptyset)}$ has
only two vertices not lying on the boundary corresponding to the dual of the
two tetrahedra in $K.$

Figure 15:  An example of a 2-dimensional cobordism that can be seen as a
discretization of the boundary of a cylinder. In this case the primal and dual
cobordisms are isomorphic, with the location of the ingoing and outgoing
components exchanged, the ingoing component being represented using cells with
white stripes. This is an example of an exactly collared cobordism.

###  3.2 Composition

Composing cobordisms will require the assumption of uniformity, simply defined
as follows for cobordisms.

######  Definition 3.7 (Uniform cobordism).

A cobordism $(K-J)\in\mathsf{Cob},$ will be called uniform if $(K,J^{\prime})$
is uniform for all connected components $J^{\prime}$ of $\partial K.$

The idea is then relatively simple: two uniform cobordisms $(K-J),$
$(K^{\prime}-L)$ such that $\partial K=J\sqcup L$ induce a sequence of the
type

$\displaystyle
M_{L}^{K}\vdash_{\pi_{L}^{K}}L(K)\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho_{L}^{K}}L\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{L}^{K^{\prime}}}L(K^{\prime})\dashv_{\pi_{L}^{K^{\prime}}}M_{L}^{K^{\prime}},$
(7)

which is geometric if the reductions $\rho_{L}^{K}$ and
$\rho_{L}^{K^{\prime}}$ are reflective (these maps were introduced in
Proposition 2.10). Such sequence is thus given the following name.

######  Definition 3.8 (Connecting sequence).

A geometric sequence of the form

$M\vdash_{\pi_{J}}J\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho_{J}}I\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{L}}L\dashv_{\pi_{L}}M^{\prime}$

is called a connecting sequence. More explicitly, the maps
$\pi_{J},\rho_{J},\pi_{L},\rho_{L}$ in such a connecting sequence satisfy

$\rho_{J}\leftthreetimes\pi_{J},\quad\rho_{L}\leftthreetimes\pi_{L},\quad\rho_{J}\mathbin{\top}\rho_{L}.$

One can then consider the dual of the connecting sequence (7). This produces a
slice sequence and allows to make use of the Correspondence Theorem to obtain
a slice. Using the duality map one the cells in the slice one can then ensure
that the cells in $K\cup K^{\prime}$ satisfy the axioms of a cc.

For the sake of completeness, we can formulate this in a somewhat more general
manner, using a freedom one has in choosing the (relative) reductions used in
the connecting sequence. This freedom is incorporated in the next definition.

######  Definition 3.9 (Connecting boundary components).

Let $K,H\in\mathsf{C}^{R}$ and $J,L\in\mathsf{B}^{R-1}$ be connected
components of $\partial K$ and $\partial H,$ respectively, such that $(K,J)$
and $(H,L)$ are uniform. Then we say that $K$ and $H$ connect through
$I\in\mathsf{B}^{R-1}$ if there exist cell complexes
$K^{\rho_{J}},H^{\rho_{L}}\in\mathsf{C}^{R}$ and relative reductions
$(I,K^{\rho_{J}})\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{J}}(J,K)$ and
$(I,H^{\rho_{L}})\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{L}}(L,H)$ such
that

$M_{J}^{K}\vdash_{\pi^{K}_{J}}J(K)\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho_{J}\circ\rho_{J}^{K}}I\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{L}\circ\rho_{L}^{H}}L(H)\dashv_{\pi_{L}}M_{L}^{H}$
(8)

is a connecting sequence ($\pi^{K}_{J},\rho_{J}^{K},\pi_{L}^{H}$ and
$\rho_{L}^{H}$ are defined in Proposition 2.10). In this case we say that
$\rho_{J}$ and $\phi_{L}$ are connecting $K$ and $H$ through $I.$

Connecting boundary components are then those that admit a lower bound with
respect to the partial order on relative cc induced by
$\mathrel{\raisebox{1.0pt}{$\prec$}}$ with the additional condition that the
corresponding reductions are reflective.

Two cc having a connecting boundary components can be associated a ranked
poset defined as their union in the following way, shown to be a cc in the
next corollary.

######  Definition 3.10 (Union of cc).

Let $K,H\in\mathsf{C}^{R}$ and $J,L\in\mathsf{B}^{R-1}$ be boundary components
of respectively $K$ and $H$ such that $(K,J)$ and $(H,L)$ are uniform. Let

$(K^{\rho_{J}},I)\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{J}}(K,J),\quad(H^{\rho_{L}},I)\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{L}}(H,L)$

be two relative reductions connecting $K$ and $H$ through $I$ and such that
$K^{\rho_{J}}\cap H^{\rho_{L}}=I.$ We define the union of $K$ and $H$ by
$\rho_{J}$ and $\rho_{L}$ to be the poset

$K\stackrel{{\scriptstyle\rho_{J}\leavevmode\nobreak\
\rho_{L}}}{{\cup}}H:=K^{\rho_{J}}\cup H^{\rho_{L}}$

with rank function

$\operatorname{rk}_{\rho_{J},\rho_{L}}=\operatorname{rk}_{K\stackrel{{\scriptstyle\rho_{J}\leavevmode\nobreak\
\rho_{L}}}{{\cup}}H}:=\operatorname{rk}_{K^{\rho_{J}}}+\operatorname{rk}_{H^{\rho_{L}}}-\operatorname{rk}_{I}.$

In the particular case were $J=I=L,$ i.e.
$\rho_{J}=\rho_{L}=\operatorname{id}_{I},$ we will simply write

$K\stackrel{{\scriptstyle\rho_{J}\leavevmode\nobreak\
\rho_{L}}}{{\cup}}H=K\stackrel{{\scriptstyle I}}{{\cup}}H.$

######  Corollary 3.11.

$(K\stackrel{{\scriptstyle\rho_{J}\leavevmode\nobreak\
\rho_{L}}}{{\cup}}H,\operatorname{rk}_{\rho_{J},\rho_{L}})$ as in Definition
3.10 is an element of $\mathsf{C}^{R}$ uniquely determined by $K,H$ and the
reductions $\rho_{J}:I\longrightarrow J,$ $\rho_{L}:I\longrightarrow L$ up to
cc-isomorphism.

###### Proof.

As pointed out in Definition 2.3, $K^{\prime}:=K^{\rho_{J}}$ and
$H^{\prime}:=H^{\rho_{L}}$ are uniquely determined by $K,H,\rho_{J}$ and
$\rho_{L}$, hence it remains to show that $K^{\prime}\cup
H^{\prime}\in\mathsf{C}^{R}.$ Notice first that
$\operatorname{rk}_{\rho_{J},\rho_{L}}|_{I}=\operatorname{rk}_{I}.$

The central argument of this proof is the following. By our assumptions, and
considering that we are using the same notation as in Definition 3.9, we
obtain the connecting sequence (8). As a consequence, and by Lemma 2.15, the
following sequence

$\overline{M_{J}^{K}}\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho^{\prime}_{J}}\overline{J(K)}\dashv_{\pi^{\prime}_{J}}\overline{I}\vdash_{\pi^{\prime}_{L}}\overline{L(H)}\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho^{\prime}_{L}}\overline{M_{L}^{H}}$

where

$\rho^{\prime}_{J}:=\overline{\pi_{J}^{K}},\quad\pi^{\prime}_{J}:=\overline{\rho_{J}\circ\rho_{J}^{K}},\quad\pi^{\prime}_{L}:=\overline{\rho_{L}\circ\rho_{L}^{H}},\quad\rho^{\prime}_{L}:=\overline{\pi_{L}^{H}},$

is a slice sequence. Therefore Theorem 2.22 provides us with a slice $S$ of
rank $R$ such that

$S=\overline{M_{J}^{K}}\sqcup(\pi_{J}^{\prime}\sqcup\pi^{\prime}_{L})(\overline{I})\sqcup\overline{M_{L}^{H}},\quad\partial
S=\overline{M_{J}^{K}}\sqcup\overline{M_{L}^{H}},\quad M^{S}=\overline{I}.$

Lemmas A.4 and 2.18 imply that there is an isomorphism of posets

$S\cong\overline{K_{I}^{\rho_{J}}}\sqcup\overline{I}\sqcup\overline{H_{I}^{\rho_{L}}}.$

Hence by Lemma 1.10 the duality map $x\longmapsto\overline{x}$ defines a poset
anti-isomorphism between $S$ and $K_{I}^{\rho_{J}}\sqcup I\sqcup
H_{I}^{\rho_{L}},$ i.e. a bijection between the two posets which reverses
inclusion relations. Moreover, for any element
$x\in(\pi_{J}^{\prime}\sqcup\pi^{\prime}_{L})(\overline{I})$ we have

$\operatorname{rk}_{S}(x)=\operatorname{rk}_{\overline{I}}(x)+1=(R-1)-\operatorname{rk}_{I}(\overline{x})=R-\operatorname{rk}_{\rho_{J},\rho_{L}}(\overline{x}),$

therefore
$\operatorname{rk}_{\rho_{J},\rho_{L}}(\overline{x})=R-\operatorname{rk}_{S}(x)$
for all $x\in S.$

By the same arguments as in the proof of Proposition 1.11 this implies that
the axioms of cc as well as all properties defining elements in $\mathsf{C}$
(i.e. graph-based, pure, non-branching, non-pinching, connected, cell-
connected) are satisfied for the cells in $K_{I}^{\rho_{J}}\sqcup I\sqcup
H_{I}^{\rho_{L}}$ as the same properties are true for the cells in $S.$ ∎

In particular, if $(K-J),(H-L)\in\mathsf{Cob}$ are uniform cobordisms such
that $\partial K=J\sqcup L$ and $K\cap H=L$ then

$(K\stackrel{{\scriptstyle L}}{{\cup}}H-J)\in\mathsf{Cob}.$

Hence the simpler case $J=I=L$ in Corollary 3.11 will be most important to us
for the remaining of this section, as the emphasis will be put primarily on
the structure of the resulting cc. However if we fix two cobordisms
$(K-J),(H-L)$ to start with as above then there remains some freedom in how
one can define their union: if $\alpha:(K,L)\longrightarrow(K_{\alpha},L)$ and
$\beta:(H,L)\longrightarrow(H_{\beta},L)$ are relative cc-isomorphisms (i.e.
relative cc-automorphisms) then $K_{\alpha}$ and $H_{\beta}$ also connect
through $L$ but in general $K\stackrel{{\scriptstyle L}}{{\cup}}H\not\cong
K_{\alpha}\stackrel{{\scriptstyle\alpha\leavevmode\nobreak\
\beta}}{{\cup}}H_{\beta}.$ If we consider the case where $L$ has a large class
of cc-automorphism, for example the case of a closed square lattice discussed
in 1.12 2), this freedom appears to be similar to a gauge symmetry, as it
amounts to choose the discrete analogue of a rotation of the space.

###  3.3 The category of causal cobordisms

Having defined a composition operation for cobordisms via the definition of
the union of two cc connecting through one of their boundary components, it
would be straightforward to define a category whose morphisms correspond to
cobordisms, and objects $\mathsf{B}_{\sqcup}^{R}$ for some rank $R.$ However
the dual $\overline{(K-J)}$ of a uniform cobordism $(K-J)$ is only uniform if
the relative cc $((\widetilde{K\setminus J}\sqcup\overline{\partial K\setminus
J}),\overline{K_{J}})$ is uniform, but there is no assumption guaranteeing
this to be true in general. The idea of this part is to restrict our attention
to a class of cobordisms having a causal structure generalizing that of CDT on
which both a composition operation and the duality map are defined. As a
result we obtain a category endowed with three non-trivial dualities (defined
as functors of categories).

In order to so, it is convenient to introduce what can be considered as
fundamental building blocks of geometric sequences that we call semi-
sequences. A semi-sequence corresponds to a geometric sequence

$M\vdash_{\pi}J^{\prime}\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho}J,$

for which we use the lighter notation
$M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}J.$ If
$M,J,J^{\prime}\in\mathsf{B}^{d}$ are as above then we say that the
corresponding semi-sequence $M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}J$ has
dimension $d$ and $J^{\prime}$ is called the transition of
$M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}J$ (although there is usually no
need to specify it in what follows).

Unless stated otherwise, we set the convention that if $(K,J)$ is a uniform
relative cc then the semi-sequence
$M_{J}^{K}\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}J$ stands for

$M_{J}^{K}\vdash_{\pi_{J}^{K}}J(K)\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho_{J}^{K}}J.$

It is also suited to borrow Dirac’s braket notation in order to talk about
states and dual states forming object of a category, as we do thereafter.

######  Definition 3.12 (Braket notation, sets of states $\mathcal{S},$
sequences of states $\mathcal{Q}$).

We introduce the following braket notation, where each ket
$|J_{1},J_{2},J_{3}\rangle$ and bra $\langle J_{1},J_{2},J_{3}|$ is labelled
by three elements $J_{i}\in\mathsf{B},\leavevmode\nobreak\ i=1,2,3,$
satisfying the following conditions:

  * •

$|M,L,M^{\prime}\rangle$ denotes the connecting state corresponding to the
connecting sequence

$M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}L\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv
M^{\prime},$

  * •

$\langle J,M,L|$ denotes the slice state corresponding to the slice sequence

$J\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv
M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}L,$

  * •

$\langle J^{\prime},M,J|N,L,N^{\prime}\rangle$ indicates that
$M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}J$ is equal to
$N\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}L,$

  * •

$|N^{\prime},J,N\rangle\langle L,M,L^{\prime}|$ indicates that
$J\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv N$ is equal to
$L\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv M.$

We define the set of states $\mathcal{S}$ to be the union of the set of
connecting states $\mathcal{S}^{\top}$ and the set of slice states
$\mathcal{S}^{\perp}.$ We define a sequence of states to be an expression
$s_{0}\dots s_{n}$ in which $s_{i}\in\mathcal{S},\leavevmode\nobreak\
i=0,\dots,n,$ and $n\in\mathbb{N}$ is the length of the sequence, satisfying
that $s_{0}\dots s_{n}$ defines a geometric sequence. We denote by
$\mathcal{Q}$ the set of sequences of states where for each state $s$ we
denote by $\operatorname{id}_{s}:=s$ the sequence of states of length 1
containing $s$ called the identity sequence on $s.$ If $\sigma:=s_{0}\dots
s_{n}\in\mathcal{Q}$ we define the image of $\sigma$ by
$\mathfrak{Im}(\sigma):=s_{0}$ and the domain of $\sigma$ by
$\mathfrak{D}(\sigma):=s_{n}.$

Two sequences of the form $\sigma=s_{0}\dots s_{n},\gamma=s_{n}\dots s_{m},$
where $n,m\in\mathbb{N},\leavevmode\nobreak\ n\leq m$ can be composed to form
the sequence

$\sigma\circ\gamma:=s_{0}\dots s_{m}.$

By Remark 1.26 all the labels of states belonging to a given sequence of
states are elements in $\mathsf{B}$ of the same rank. We define the set of
$d$-dimensional states $\mathcal{S}_{d}$ to be the set of states labelled by
elements in $\mathsf{B}^{d}.$ Similarly, the set of $d$-dimensional sequences
$\mathcal{Q}_{d}$ is defined as the set of sequences of states with labels in
$\mathsf{B}^{d}.$

######  Definition 3.13 (Category $\mathscr{C}_{d}$ of causal cobordism of
rank $d$).

Let $d\in\mathbb{N}^{*}.$ The category of $d$-dimensional causal cobordisms
$\mathscr{C}_{d}$ is defined by having the set of states $\mathcal{S}_{d-1}$
as its set of objects and set of morphisms from $a$ to $b$ defined by

$\mathsf{Hom}_{\mathscr{C}_{d}}(a,b)=\\{\sigma\in\mathcal{Q}\leavevmode\nobreak\
|\leavevmode\nobreak\ a=\mathfrak{D}(\sigma),\leavevmode\nobreak\
b=\mathfrak{Im}(\sigma)\\}.$

The composition of morphisms in $\mathscr{C}_{d}$ is defined via the
composition of sequences and the identity on $s\in\mathcal{S}_{d-1}$ is given
by $\operatorname{id}_{s}\in\mathsf{Hom}_{\mathscr{C}_{d}}(s,s).$

It is clear that $\mathscr{C}_{d}$ defines a category, hence it remains to
explain how it is related to cobordisms and slices.

Let us first specify that a cobordism $(S-L)$ is called a slice if $S$ is a
slice and we will denote the set of slices (seen as cobordisms) by
$\mathsf{Cob}_{s}.$

By the Correspondence Theorem 2.22, we can associate each slice state $\langle
J,M,L|$ with $(S-L)\in\mathsf{Cob}_{s}$ such that

$\partial S=J\sqcup L,\leavevmode\nobreak\ M^{S}\cong M,$

and where the slice $S$ corresponds, via the theorem, to the slice sequence

$J\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv
M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}L.$

Hence in this case we use the notation

$\langle S-L|:=\langle J,M,L|.$

If $\langle S-L|,\langle S^{\prime}-L^{\prime}|$ are slice states such that
there exists a connecting states $|M,L,M^{\prime}\rangle$ satisfying that

$\langle S-L|M,L,M^{\prime}\rangle\langle
S^{\prime}-L^{\prime}|\in\mathcal{Q}$

then this is equivalent to stating that $\partial S^{\prime}=L\sqcup
L^{\prime}$ and
$M\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}L\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv
M^{\prime}$ is the connecting sequence

$M^{S}\vdash_{\pi_{L}^{S}}L(S)\mathrel{\raisebox{1.0pt}{$\succ$}}_{\rho_{L}^{S}}L\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{L}^{S^{\prime}}}L(S^{\prime})\dashv_{\pi_{L}^{S^{\prime}}}M^{S^{\prime}}.$

Therefore Corollary 3.11 implies that $(S\stackrel{{\scriptstyle
L}}{{\cup}}S^{\prime}-L^{\prime})$ is a cobordism so that we define the bra

$\langle S\stackrel{{\scriptstyle L}}{{\cup}}S^{\prime}-L^{\prime}|:=\langle
S-L|M,L,M^{\prime}\rangle\langle S^{\prime}-L^{\prime}|,$

where $M=M^{S}$ and $M^{\prime}=M^{S^{\prime}}$ by the conditions of the
braket notation. More generally, we recursively define bras (and the
corresponding cobordisms) of the form $\langle K-J|$ with $K$ being a union of
slices $S_{1},\dots,S_{n}$ using the relation:

$\langle\cup^{n}_{i=1}S_{i}-L_{n}|:=\langle\cup_{i=1}^{n-1}S_{i}-L_{n-1}|M_{n-1},L_{n-1},M_{n}\rangle\langle
S_{n}-L_{n}|,$

for some connecting states $|M_{i},L_{i},M_{i+1}\rangle,\leavevmode\nobreak\
i=1,\dots,n-1.$

With the latter definitions at hand, we are able to give a precise meaning to
the notion of causal cobordisms.

######  Definition 3.14 (Causal cobordisms $\mathsf{Cob}_{c}$).

A cobordism $(K-J)\in\mathsf{Cob}$ is said to be causal if there exist slices
$S_{1},\dots,S_{n}$ such that

$(K-J)=(\cup_{i=1}^{n}S_{i}-L_{n})$

for some connecting states $|M_{i},L_{i},M_{i+1}\rangle,\leavevmode\nobreak\
i=1,\dots,n-1.$ The set of causal cobordisms is denoted by $\mathsf{Cob}_{c}.$
We will also call a bra of the form $\langle K-J|$ where
$(K-J)\in\mathsf{Cob}_{c}$ a causal cobordism.

Figure 15 in fact gives a simple example of causal cobordism.

For each $J\in\mathsf{B}$ we conventionally define an element called the empty
causal cobordism at $J$ denoted $\langle J-J|$ to be a causal cobordism with
boundary $J$ and rank $\operatorname{Rk}(J)+1.$ Empty cobordisms are simply
introduced to make some expressions of sequences of states more general. We
will for example use the bra $\langle J-J|$ in an expression of the form:

$|M,J,M^{\prime}\rangle\langle J-J|N,L,N^{\prime}\rangle\quad\text{to imply
that}\quad|M,J,M^{\prime}\rangle=|N,L,N^{\prime}\rangle.$

In order to more explicitly connect causal cobordisms with the related
category it is also convenient to introduce a notation indicating that a slice
is either the first or the last slice of a causal cobordism as follows. Let us
consider a causal cobordism $(K-J)\in\mathsf{Cob}_{c}$ given by

$(K-J)=(\cup_{i=1}^{n}S_{i}-L_{n}),$

for some slices $S_{1},\dots,S_{n}$ and connecting states
$|M_{i},L_{i},M_{i+1}\rangle,\leavevmode\nobreak\ i=1,\dots,n-1$ such that
$J=L_{n-1}.$ We will name $L$ the other boundary component of $K$ so that
$\partial K=J\sqcup L.$ We say that a slice $\langle S-J|=\langle
J^{\prime},M^{S},J|$ is the  ingoing slice of $\langle K-J|$ and use the
notation

$\langle J^{\prime},M^{S},J|\leq_{\text{in}}\langle K-J|,$

if $(S-J)=(S_{n}-L_{n}).$ Similarly, we say that a slice $\langle
S-L^{\prime}|=\langle L,M^{S},L^{\prime}|$ is the outgoing slice of $\langle
K-J|$ and use the notation

$\langle L,M^{S},L^{\prime}|\leq_{\text{out}}\langle K-J|,$

if $(S-L^{\prime})=(S_{1}-L_{1}).$

Using this notation, we can give expressions associated to morphisms in
$\mathscr{C}_{d}.$ These are of four different kinds depending on whether
their image and domain are connecting or slice states:

$\displaystyle\mathsf{Hom}_{\mathscr{C}_{d}}(|M,J,M^{\prime}\rangle,|N^{\prime},L,N\rangle):=\left\\{\leavevmode\nobreak\
|N^{\prime},L,N\rangle\langle K-J|M,J,M^{\prime}\rangle\leavevmode\nobreak\
:\leavevmode\nobreak\ \langle K-J|\in\mathsf{Cob}_{c}\right\\},$
$\displaystyle\mathsf{Hom}_{\mathscr{C}_{d}}(\langle
J^{\prime},M,J|,|N^{\prime},L,N\rangle):=\left\\{\leavevmode\nobreak\
|N^{\prime},L,N\rangle\langle K-J|\leavevmode\nobreak\
:\begin{array}[]{c}\langle K-J|\in\mathsf{Cob}_{c},\\\ \langle
J^{\prime},M,J|\leq_{\text{in}}\langle K-J|\end{array}\right\\},$
$\displaystyle\mathsf{Hom}_{\mathscr{C}_{d}}(|M,J,M^{\prime}\rangle,\langle
L^{\prime},N,L|):=\left\\{\leavevmode\nobreak\ \langle
K-J|M,J,M^{\prime}\rangle\leavevmode\nobreak\ :\begin{array}[]{c}\langle
K-J|\in\mathsf{Cob}_{c},\\\ \langle L^{\prime},N,L|\leq_{\text{out}}\langle
K-J|\end{array}\right\\},$
$\displaystyle\mathsf{Hom}_{\mathscr{C}_{d}}(\langle J^{\prime},M,J|,\langle
L^{\prime},N,L|):=\left\\{\leavevmode\nobreak\ \langle
K-J|\in\mathsf{Cob}_{c}\leavevmode\nobreak\ :\begin{array}[]{c}\langle
J^{\prime},M,J|\leq_{\text{in}}\langle K-J|,\\\ \langle
L^{\prime},N,L|\leq_{\text{out}}\langle K-J|\end{array}\right\\}.$

It is seen that only the morphisms in the last set
$\mathsf{Hom}_{\mathscr{C}_{d}}(\langle J^{\prime},M,J|,\langle
L^{\prime},N,L|)$ correspond exactly to the notion of causal cobordisms from
Definition 3.14. The other sets of morphisms include additional connecting
states corresponding to connecting sequences associated with one or both of
their boundary components. The morphisms of the first set
$\mathsf{Hom}_{\mathscr{C}_{d}}(|M,J,M^{\prime}\rangle,|N^{\prime},L,N\rangle)$
actually define the  sub-category of $d$-dimensional connecting states
$\mathscr{C}_{d}^{\top}$ with $\mathcal{S}^{\top}_{d}$ as its set of objects.
Similarly, one can single out the sub-category of $d$-dimensional slice states
$\mathscr{C}_{d}^{\perp}$ having $\mathcal{S}^{\perp}_{d}$ as objects and
morphisms defined by the last set $\mathsf{Hom}_{\mathscr{C}_{d}}(\langle
J^{\prime},M,J|,\langle L^{\prime},N,L|).$

######  Definition 3.15 (Duality map on $\mathcal{S}$ and $\mathcal{Q}$).

The dual of a sequence of states is defined via the following rules:

  * •

if $s=|M,L,M^{\prime}\rangle$ then
$\overline{s}=\langle\overline{M^{\prime}},\overline{L},\overline{M}|$ is the
slice state characterized by the slice sequence

$\overline{M^{\prime}}\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv\overline{L}\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}\overline{M},$

  * •

if $s=\langle J,M,L|$ then
$\overline{s}=|\overline{L},\overline{M},\overline{J}\rangle$ is the
connecting state characterized by the connecting sequence

$\overline{L}\vdash\mathrel{\raisebox{1.0pt}{$\succ$}}\overline{M}\mathrel{\raisebox{1.0pt}{$\prec$}}\dashv\overline{J},$

  * •

if $\sigma=\sigma_{1}\sigma_{2}$ is a sequence of states then

$\overline{\sigma}=\overline{\sigma_{2}}\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\sigma_{1}}.$

The definition of the duality map on $\mathcal{S}$ and $\mathcal{Q}$ for
example implies:

$\displaystyle\overline{\langle J,M,L|M,L,M^{\prime}\rangle}$
$\displaystyle=\overline{|M,L,M^{\prime}\rangle}\leavevmode\nobreak\
\leavevmode\nobreak\ \overline{\langle
J,M,L|}=\langle\overline{M^{\prime}},\overline{L},\overline{M}|\overline{L},\overline{M},\overline{J}\rangle.$
(9)

######  Corollary 3.16 (Duality map on $\mathscr{C}^{d}$).

Let $a,b\in\mathcal{S}^{d}$ and let $\sigma$ be a morphism in
$\mathscr{C}^{d}.$ We have the following equivalence:

$\sigma\in\mathsf{Hom}_{\mathscr{C}_{d}}(a,b)\quad\text{if and only
if}\quad\overline{\sigma}\in\mathsf{Hom}_{\mathscr{C}_{d}}(\overline{b},\overline{a}).$

In other words, $\mathscr{C}_{d}$ is self-dual as a category under the
contravariant functor $\mathtt{C}=\overline{(\cdot)}$ induced by the duality
map. It is also clear that $\mathtt{C}$ can be restricted to a functor between
the sub-categories $\mathscr{C}_{d}^{\top}$ and $\mathscr{C}_{d}^{\perp}.$

Let us furthermore note the following contravariant functor
$\mathtt{T}={(\cdot)}^{-1}$ in $\mathscr{C}_{d}.$ For any state
$s\in\mathcal{S},$ let the reverse state ${s}^{-1}$ be the state defined by
the same labels in reversed order, for example

${|M^{\prime},L,M\rangle}^{-1}=|M,L,M^{\prime}\rangle.$

Then $s\in\mathcal{S}^{\top}$ if and only if ${s}^{-1}\in\mathcal{S}^{\top}$
and similarly for $\mathcal{S}^{\perp}.$ We also define the reverse sequence
of states ${\sigma}^{-1}$ of a sequence $\sigma=\sigma_{1}\leavevmode\nobreak\
\sigma_{2}$ by

${\sigma}^{-1}={\sigma_{2}}^{-1}\leavevmode\nobreak\ {\sigma_{1}}^{-1},$

which evidently implies that

$\sigma\in\mathsf{Hom}_{\mathscr{C}_{d}}(a,b)\quad\text{if and only
if}\quad{\sigma}^{-1}\in\mathsf{Hom}_{\mathscr{C}_{d}}(b,a).$

Finally, it is worth mentioning a third (covariant) functor
$\mathtt{P}=(\cdot)^{*}$ defined as follows. The functor $\mathtt{P}$ acts on
states as a map $s\longmapsto s^{*},$ where $s^{*}$ is the state corresponding
to the sequence dual to the sequence characterizing $s$ labelled by the dual
of the labels of $s$ in the same order. For example,

$|M^{\prime},L,M\rangle^{*}=\langle\overline{M^{\prime}},\overline{L},\overline{M}|.$

The functor $\mathtt{P}$ acts on morphisms in $\mathscr{C}_{d}$ simply by
acting on each states of the corresponding sequence of states, i.e. if
$\sigma=\sigma_{1}\leavevmode\nobreak\ \sigma_{2}$ then

$\sigma^{*}=\sigma_{1}^{*}\leavevmode\nobreak\ \sigma_{2}^{*}.$

This in turn implies the equivalence:

$\sigma\in\mathsf{Hom}_{\mathscr{C}_{d}}(a,b)\quad\text{if and only
if}\quad\sigma^{*}\in\mathsf{Hom}_{\mathscr{C}_{d}}(a^{*},b^{*}).$

These functors satisfy
$\mathtt{T}^{2}=\mathtt{C}^{2}=\mathtt{P}^{2}=\operatorname{id}_{\mathscr{C}_{d}}$
and are in particular related by the relation
$\mathtt{C}=\mathtt{P}\leavevmode\nobreak\ \mathtt{T},$ as can be seen for the
specific example used in (9):

$\displaystyle\left({\left(\langle
J,M,L|M,L,M^{\prime}\rangle\right)}^{-1}\right)^{*}$
$\displaystyle=\left(|M^{\prime},L,M\rangle\langle M,L,J|\right)^{*}$
$\displaystyle=\langle\overline{M^{\prime}},\overline{L},\overline{M}|\overline{L},\overline{M},\overline{J}\rangle=\overline{\langle
J,M,L|M,L,M^{\prime}\rangle}.$

##  Appendix A Lemmas A.1 to A.8

######  Lemma A.1.

Let $x$ be a cell of a cc $K$ such that
$\left|\bigtriangledown\left(x\right)\right|\geq 2$ then

$x=\bigwedge\bigtriangledown\left(x\right)=\bigcap_{y\in\bigtriangledown\left(x\right)}y.$

###### Proof.

Since $\left|\bigtriangledown\left(x\right)\right|\geq 2$ Axiom iii) implies
$x^{\prime}:=\bigcap_{y\in\bigtriangledown\left(x\right)}y\in K.$ If
$y^{\prime}$ is a lower bound of $\bigtriangledown\left(x\right)$ then
$y^{\prime}\subset y$ for all $y\in\bigtriangledown\left(x\right)$ and
therefore $y^{\prime}\subset x^{\prime}.$ The latter implies

$x^{\prime}=\bigcap_{y\in\bigtriangledown\left(x\right)}y=\bigwedge\bigtriangledown\left(x\right).$

We also have that $x\subset x^{\prime}$ since $x$ is a lower bound of
$\bigtriangledown\left(x\right).$ If by contradiction $x^{\prime}\neq x$ then
by Axiom i) $\operatorname{rk}(x^{\prime})>\operatorname{rk}(x),$ hence
$\operatorname{rk}(\bigwedge\bigtriangledown\left(x\right))=\operatorname{rk}(x)+1.$
But if $y\in\bigtriangledown\left(x\right)$ then
$\bigwedge\bigtriangledown\left(x\right)\subset y$ and
$\operatorname{rk}(y)=\operatorname{rk}(\bigwedge\bigtriangledown\left(x\right))$
so that $y=\bigwedge\bigtriangledown\left(x\right)$ by Axiom i), which implies
$\left|\bigtriangledown\left(x\right)\right|=1$ and contradicts
$\left|\bigtriangledown\left(x\right)\right|\geq 2.$ ∎

######  Lemma A.2.

If $K$ is a non-singular cc such that there is no $K$-pinch in $\partial K$
then $\partial K$ is closed.

###### Proof.

Let $R=\operatorname{Rk}(K).$ It is sufficient to show that if $\partial K$
contains no $K$-pinch then

$\left|\bigtriangledown\left(x\right)\cap\partial K\right|=2,\quad\forall
x\in(\partial K)^{[R-2]}.$

We start by showing that $\left|\bigtriangledown\left(x\right)\cap\partial
K\right|\geq 2,$ which in particular implies that $\partial(\partial
K)=\emptyset.$ Since $x\in\partial K$ there is $y_{0}\in(\partial K)^{[R-1]}$
and $z_{0}\in K^{[R]}$ such that $x\subset y_{0}\subset z_{0}$ and by the
diamond property iv) there exists $y_{1}\in K^{[R-1]}$ such that
$\triangle\left(z_{0}\right)\cap\bigtriangledown\left(x\right)=\\{y_{0},y_{1}\\}.$
If $y_{1}\not\in\partial K$ then there exist $z_{1}\in
K^{[R]},\leavevmode\nobreak\ y_{2}\in K^{[R-1]}$ such that
$\bigtriangledown\left(y_{1}\right)=\\{z_{0},z_{1}\\}$ and
$\triangle\left(z_{1}\right)\cap\bigtriangledown\left(x\right)=\\{y_{1},y_{2}\\}.$
By repeating this argument we can construct a sequence of elements in
$K^{[R-1]},$ distinct by non-singularity of $K,$ until at some step $k$ we get
$y_{k}\in\partial K$ and this implies
$\\{y_{0},y_{k}\\}\subset\bigtriangledown\left(x\right)\cap\partial K.$ If by
contradiction we assume that there is more that two elements in
$\bigtriangledown\left(x\right)\cap\partial K$ then the previous construction
would show that it contains at least four elements and that the corresponding
two sequences of elements in $K^{[R]}$ sharing an element of $K^{[R-1]}$ are
not connected, implying that the local figure of $x$ is not strongly
connected. This would imply that $x\in\partial K$ is a $K$-pinch, a
contradiction with our hypothesis. ∎

######  Lemma A.3.

Let $K\in\mathsf{C}^{R}.$ Then $J\leq\partial K$ and
$J\in\mathsf{B}^{R-1}_{\sqcup}$ if and only if $J$ is a (possibly empty) union
of connected components of $\partial K.$

###### Proof.

Let $R=\operatorname{Rk}(K).$ A union of boundary components of $K$ is clearly
an element in $\mathsf{B}^{R-1}_{\sqcup},$ by Lemma A.2. In order to prove the
reverse implication it is sufficient to prove the case where $\partial K$ is
connected and to show that $J\neq\emptyset$ implies $J=\partial K.$ If
$J\in\mathsf{B}^{R-1}$ is non-empty then it is a closed $(R-1)$-cc. If by
contradiction there exists a vertex in $\partial K\setminus J,$ the
connectedness of $\partial K$ implies that this vertex is linked to a vertex
in $J$ via a path in $(\partial K)^{(1)}.$ Hence there is an edge
$e=\\{v,w\\}$ in $\partial K$ such that $v\in J$ and $w\in\partial K\setminus
J.$ Now we have that $\overline{J}\leq\overline{\partial K}$ and
$\overline{e}^{\partial K}$ (the dual of $e$ in $\partial K$) is a
$(R-2)$-cell of $\overline{\partial K}$ contained in $\overline{v}^{\partial
K},$ an $(R-1)$-cell of $\overline{J}.$ Therefore it implies that
$\overline{e}^{\partial K}=\overline{e}^{J}\in J^{[R-2]}$ but since
$\overline{J}$ is closed we have that $\overline{e}$ is contained in two
maximal cells of $J.$ Since $\overline{w}^{\partial K}$ is a maximal cell of
$\overline{\partial K}\setminus\overline{J}$ containing $\overline{e}^{J}$
this contradicts the non-singularity of $\overline{\partial K}.$ ∎

######  Lemma A.4.

Let $K$ be a local cell complex and $J\leq K.$ Then the map

$x\longmapsto\mathcal{E}_{J}^{x},\quad x\in K_{J}$ (10)

is an injective poset homomorphism such that the inverse is a poset
homomorphism.

###### Proof.

We first notice that since $K$ is cell-connected we have that
$\mathcal{E}^{x}_{J}\neq\emptyset$ for all $x\in K_{J}.$ In order to prove
that (10) is injective, one considers $x,y\in K_{J}$ such that
$\mathcal{E}_{J}^{x}=\mathcal{E}_{J}^{y}$ and show that $x=y.$ As consequence
of the definition of collar we have
$\mathcal{E}_{J}^{x}=\mathcal{E}_{J}^{y}=\mathcal{E}_{J}^{x\cap y}$ hence we
can without loss of generality assume that $x\subset y.$

We then proceed by induction on $r=\operatorname{rk}(x)\geq 1.$ If
$\operatorname{rk}(x)=1$ and $x\subsetneq y$ then by Axiom ii) we can assume
that $\operatorname{rk}(y)=\operatorname{rk}(x)+1.$ This means that $y$ is a
2-cell which, as discussed before, corresponds to a cycle of edges. One can
then easily show that there must be a edge in $\mathcal{E}_{J}^{y}$ other than
$x$ and since $\mathcal{E}_{J}^{x}={x}$ this contradict
$\mathcal{E}_{J}^{x}=\mathcal{E}_{J}^{y}.$

Assume by induction that the claim is true for all $x,y$ such that $x\subset
y$ and $\operatorname{rk}(x)\leq r-1.$ Take $x\in K_{J}^{[r]}$ and $y\in
K_{J}$ such that $\mathcal{E}_{J}^{x}=\mathcal{E}_{J}^{y}$ and $x\subsetneq
y.$ Like before we can without loss of generality assume that
$\operatorname{rk}(y)=r+1.$ By induction we have that
$\mathcal{E}_{J}^{w}\subsetneq\mathcal{E}_{J}^{x}$ for all $w\subsetneq x,$
and, using this inclusion with $w=x\cap x^{\prime},$ this implies in
particular that $\mathcal{E}_{J}^{x^{\prime}}\not\subset\mathcal{E}_{J}^{x}$
for all $x^{\prime}\in K_{J}\cap K_{x}$ (since
$\mathcal{E}_{J}^{x^{\prime}}\neq\emptyset$ by locality of $K$). Since
$\mathcal{E}_{J}^{x}\neq\emptyset$ there exists
$w\in\triangle\left(x\right)\cap K_{J}.$ By the diamond property iv), there
exists $x^{\prime}\in K^{[r]}$ such that
$\bigtriangledown\left(w\right)\cap\triangle\left(y\right)=\\{x,x^{\prime}\\}.$
But $x^{\prime}\in K_{J}$ since so does $w$ and $w\subset x^{\prime},$
therefore
$\emptyset\neq\mathcal{E}_{J}^{x^{\prime}}\subset\mathcal{E}_{J}^{y}=\mathcal{E}_{J}^{x}.$
This is a contradiction with our induction hypothesis and therefore $x=y.$

The definition of $\mathcal{E}_{J}^{x}$ and the injectivity proven above imply
that if $\mathcal{E}_{J}^{x}\subset\mathcal{E}_{J}^{y}$ then

$\mathcal{E}_{J}^{(x\cap
y)}=\mathcal{E}_{J}^{x}\cap\mathcal{E}_{J}^{y}=\mathcal{E}_{J}^{x}$

and therefore $x=x\cap y$ or equivalently $x\subset y$ and this concludes the
proof. ∎

######  Lemma A.5.

The barycentric subdivision of a cc $K$ is a subdivision of $K,$ i.e.
$\widecheck{K}\mathrel{\raisebox{1.0pt}{$\prec$}}K.$

###### Proof.

Define $\rho:\widecheck{K}\longrightarrow K$ by
$\rho(\\{x_{0}\subsetneq\dots\subsetneq x_{r}\\}):=x_{r},$ clearly a
surjective poset homomorphism and injective on
$\rho^{-1}(K^{[0]})=\\{\\{v\\}\leavevmode\nobreak\ |\leavevmode\nobreak\ v\in
K^{[0]}\\}.$

Let $\\{x_{0}\subsetneq\dots\subsetneq x_{r}\\}\in\widecheck{K}$ and $y\in K$
such that $\rho(\\{x_{0}\subsetneq\dots\subsetneq x_{r}\\})\subset y$ then by
ii) there exists a cell of $\widecheck{K}$ of the form
$\sigma=\\{x_{0}\subsetneq\dots\subsetneq x_{r}\subset\dots\subset y\\}$ such
that $\operatorname{rk}_{\widecheck{K}}(\sigma)=\operatorname{rk}_{K}(y)$ and
$\rho(\sigma)=y$ and therefore $\rho$ also satisfies condition (r2) of a
reduction. For condition (r3), if
$r:=\operatorname{rk}_{\widecheck{K}}(\\{x_{0}\subsetneq\dots\subsetneq
x_{r}\\})=\operatorname{rk}_{K}(x_{r})+1$ then there exists a unique $0\leq
i\leq r$ such that
$\operatorname{rk}_{K}(x_{i})=\operatorname{rk}_{K}(x_{i+1})-2$ and therefore

$\displaystyle\left|\bigtriangledown\left(\\{x_{0}\subsetneq\dots\subsetneq
x_{r}\\}\right)\cap\rho^{-1}(x_{r})\right|$
$\displaystyle=\left|\\{\\{x_{0}\subsetneq\dots\subsetneq x_{i}\subsetneq
y\subsetneq x_{i+1}\subsetneq\dots\subsetneq x_{r}\\}\leavevmode\nobreak\
|\leavevmode\nobreak\
y\in\bigtriangledown\left(x_{i}\right)\cap\triangle\left(x_{i+1}\right)\\}\right|$
$\displaystyle=\left|\bigtriangledown\left(x_{i}\right)\cap\triangle\left(x_{i+1}\right)\right|=2.$

Finally, for condition (r4), if
$r:=\operatorname{rk}_{\widecheck{K}}(\\{x_{0}\subsetneq\dots\subsetneq
x_{r}\\})=\operatorname{rk}_{K}(x_{r})$ then $\operatorname{rk}(x_{i})=i$ for
all $0\leq i\leq r.$ Hence for all
$x_{r+1}\in\bigtriangledown\left(x_{r}\right),$ we have

$\displaystyle\left|\bigtriangledown\left(\\{x_{0}\subsetneq\dots\subsetneq
x_{r}\\}\right)\cap\rho^{-1}(x_{r+1})\right|=\left|\\{\\{x_{0}\subsetneq\dots\subsetneq
x_{r+1}\\}\\}\right|=1.$

∎

######  Lemma A.6.

Let $(K,J)$ be a pure relative cc satisfying (U) and let $x\in J(K).$ Then for
all $x^{\prime}\in K_{J}^{-}[x]$ and all maximal cells
$w\in\mathbf{B}_{J}(x^{\prime})$ we have

$\operatorname{rk}_{J}^{K}(x):=\operatorname{rk}_{K}(x^{\prime})-1=\operatorname{rk}_{K}(w).$

Moreover, for all $x^{\prime}\in K_{J}^{-}[x]$ we have that
$K_{J}(x)=K_{J}(x^{\prime}\cap J^{[0]})$ and

$K_{J}^{-}(x)=K_{J}^{-}[x]=K_{J}^{-}(x^{\prime}\cap J^{[0]}),$

which implies that for any given $y^{\prime}\in K_{J}(x)$ there exists
$x^{\prime}\in K_{J}^{-}[x]$ such that $x^{\prime}\subset y^{\prime}.$

###### Proof.

First, it is clear that $K_{J}[x]\subset K_{J}(x).$ An element $x^{\prime}\in
K_{J}^{-}[x]$ satisfies that $x\subset x^{\prime},$ which implies the
inclusion

$K_{J}(x)\supset K_{J}(x^{\prime}\cap J^{[0]}).$

The reverse inclusion is also valid by the following argument. Note that an
element $y^{\prime}\in K_{J}(x)$ satisfies

$x^{\prime}_{J}=x\subset y^{\prime}_{J}$

and (U) applied to $(K,J)$ gives

$x^{\prime}\cap J^{[0]}\subset y^{\prime}\cap J^{[0]},$

which in particular implies that $y\in K_{J}(x^{\prime}\cap J^{[0]}).$
Therefore we indeed have $K_{J}(x)=K_{J}(x^{\prime}\cap J^{[0]}).$

Moreover, a minimal element in $K_{J}[x]$ is also a minimal element in
$K_{J}(x).$ This is a consequence of the fact that if $x^{\prime}\in K_{J}[x]$
and $y^{\prime}\in K_{J}(x)$ satisfy $y^{\prime}\subset x^{\prime}$ then

$x\subset y^{\prime}_{J}\subset x^{\prime}_{J}=x$

and therefore $y^{\prime}_{J}=x$ i.e. $y\in K_{J}[x].$ As a consequence of the
results above, the purity of $(K,J)$ implies that for all $x^{\prime}\in
K_{J}^{-}[x]$ we have

$K_{J}^{-}[x]\subset K_{J}^{-}(x)=K_{J}^{-}(x^{\prime}\cap J^{[0]}).$

Therefore if $x^{\prime}\in K_{J}^{-}[x]$ then $x^{\prime}\in
K_{J}^{-}(x^{\prime}\cap J^{[0]})$ and since $J\cap x^{\prime}$ is pure, Axiom
ii) implies that a maximal cell $w$ of $J\cap x^{\prime}$ satisfies

$\operatorname{rk}_{K}(w)=\operatorname{rk}_{K}(x^{\prime})-1.$

Suppose by contradiction that there exists $w^{\prime}\in
K_{J}^{-}(x)\setminus K_{J}^{-}[x]$ and let $x^{\prime}\in K_{J}^{-}[x].$ Then
$w^{\prime}\cap x^{\prime}$ cannot be an element of $K_{J}$ as it would in
particular contradict the minimality of $x^{\prime},$ hence $w^{\prime}\cap
x^{\prime}\in J.$ But this implies that $J^{[0]}\cap w^{\prime}=J^{[0]}\cap
x^{\prime}$ which by condition (U) implies $x=x^{\prime}_{J}=w^{\prime}_{J}.$
This contradicts our assumptions and concludes the proof. ∎

######  Lemma A.7.

If $(K,J)$ is uniform and $x\in J$ then there exists a unique cell $x_{J}\in
J(K)$ such that

$K_{J}^{-}[x_{J}]=K_{J}^{-}(x)\quad\text{and}\quad\operatorname{rk}_{J(K)}(x_{J})=\operatorname{rk}_{J}^{K}(x).$

###### Proof.

In order to prove the first statement, it is sufficient to show that if
$y,y^{\prime}$ are distinct elements in $K_{J}^{-}(x)$ then
$y_{J}=y^{\prime}_{J}.$ Since $y$ and $y^{\prime}$ are minimal elements in
$K_{J}(x)$ then $y\cap y^{\prime}$ cannot be contained in $K_{J},$ so $y\cap
y^{\prime}\in J.$ Hence $J^{[0]}\cap y=J^{[0]}\cap y^{\prime}$ and condition
(U) implies that $y_{J}=y^{\prime}_{J}.$

Also, if $y\in K_{J}^{-}(x)$ then by definition
$\operatorname{rk}_{J}^{K}(x)=\operatorname{rk}_{K}(y)-1$ and by Lemma A.6 we
have $\operatorname{rk}_{K}(y)=\operatorname{rk}_{K}(w)+1$ for all maximal
cells $w$ in $\mathbf{B}_{J}(y).$ The second statement then follows since
$y\in K_{J}^{-}[x_{J}]$ by the first statement and therefore

$\operatorname{rk}_{J(K)}(x_{J})=\operatorname{rk}_{J(K)}(y_{J})=\operatorname{rk}_{K}(w)=\operatorname{rk}_{J}^{K}(x).$

∎

######  Lemma A.8.

Let $K\in\mathsf{C}$ and let $J^{\prime}$ be a connected component of
$\partial K.$ If $\rho:J\longrightarrow J^{\prime}$ is a reduction (where $J$
is disjoint from $K$) then the map from $K_{J^{\prime}}$ to
$\mathcal{P}\left(\left(K^{[0]}\setminus(J^{\prime})^{[0]}\right)\sqcup
J^{[0]}\right)$ defined by

$y^{\prime}\longmapsto\left(y^{\prime}\setminus(J^{\prime})^{[0]}\right)\sqcup\left(\bigcup_{x\in
J,\leavevmode\nobreak\ \rho(x)\subset y^{\prime}}x\right)$ (11)

is an injective poset homomorphism such that the inverse is also a poset
homomorphism.

###### Proof.

Let us denote by $\phi$ the map (11), which clearly is a poset homomorphism
since so is $\rho.$

The injectivity of $\phi$ is a direct consequence of condition (r1) and the
observation that any vertex $v\in\rho^{-1}((J^{\prime})^{[0]}\cap y^{\prime})$
satisfies that $\phi(v)\subset y^{\prime}$ hence $v$ is included in
$\left(\bigcup_{x\in J,\leavevmode\nobreak\ \rho(x)\subset
y^{\prime}}x\right).$

From the last observation one also get that the inverse of $\phi$ is a poset
homomorphism since if $x^{\prime},y^{\prime}\in K_{J^{\prime}}$ satisfy that
$\phi(x^{\prime})\subset\phi(y^{\prime})$ then it implies that
$x^{\prime}\setminus(J^{\prime})^{[0]}\subset
y^{\prime}\setminus(J^{\prime})^{[0]}$ and

$\rho^{-1}((J^{\prime})^{[0]}\cap
x^{\prime})\subset\rho^{-1}((J^{\prime})^{[0]}\cap y^{\prime}).$

Therefore (r1) implies that $(J^{\prime})^{[0]}\cap
x^{\prime}\subset(J^{\prime})^{[0]}\cap y^{\prime}$ and we can indeed conclude
that $x^{\prime}\subset y^{\prime}.$ ∎

##  Appendix B Proof of Proposition 1.24

We will show that $\mathrel{\raisebox{1.0pt}{$\prec$}}$ is a partial order on
the set of closed cc. The proof applies also to $\vdash$ using the same
arguments while reversing inclusion relations between cells and replacing
vertices with maximal cells.

###### Proof.

The relation $\mathrel{\raisebox{1.0pt}{$\prec$}}$ is clearly reflexive. The
main part of this proof focuses on showing that
$\mathrel{\raisebox{1.0pt}{$\prec$}}$ is transitive. For this, we assume that
$J\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{1}}K,$
$K\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho_{2}}L$ and check the five
conditions for $\rho_{2}\circ\rho_{1}.$

Condition (r1) for $\rho_{2}\circ\rho_{1}$ is checked easily: if $y\in
L^{[0]}$ we have $\left|\rho_{2}^{-1}(y)\right|=1$ since $\rho_{2}$ is a
reduction, i.e. $v=\rho_{2}^{-1}(y)\in K^{[0]}$ and we have
$\left|\rho_{1}^{-1}(v)\right|=1$ since $\rho_{1}$ is a reduction, hence
$\left|(\rho_{2}\circ\rho_{1})^{-1}(y)\right|=1.$

Proving condition (r2) for $\rho_{2}\circ\rho_{1}$ is relatively direct as
well. If $y\in L^{[r]},x\in J$ are such that $\rho_{2}\circ\rho_{1}(x)\subset
y$ then since $\rho_{2}$ is a reduction there exists
$w^{\prime}\in\rho_{2}^{-1}(y)^{[r]}$ such that $\rho_{1}(x)\subset
w^{\prime}.$ Also, since $\rho_{1}$ is a reduction there exists
$w\in\rho^{-1}(w^{\prime})^{[r]}\subset(\rho_{2}\circ\rho_{1})^{-1}(y)^{[r]}$
such that $x\subset w$ and this proves that

$A(\rho_{2}(\rho_{1}(x)))^{[r]}\subset\rho_{2}(\rho_{1}(A(x)^{[r]})).$

We will show condition (r4) for $\rho_{2}\circ\rho_{1}$ and finish the proof
of transitivity with condition (r3) below. If we assume that
$\operatorname{rk}_{J}(x)=\operatorname{rk}_{L}(\rho_{2}(\rho_{1}(x))),$ it
implies that $\operatorname{rk}_{K}(\rho_{1}(x))=\operatorname{rk}_{J}(x)$
since the rank is increasing under subdivision as observed in Remark 1.26.
Hence conditions (r4) for $\rho_{1}$ and $\rho_{2}$ imply respectively the two
following conditions:

$\left|\bigtriangledown\left(x\right)\cap\rho_{1}^{-1}(y)\right|=1\quad\text{for
all }y\in\bigtriangledown\left(\rho_{1}(x)\right);$ (12)

$\left|\bigtriangledown\left(\rho_{1}(x)\right)\cap\rho_{2}^{-1}(y)\right|=1\quad\text{for
all }y\in\bigtriangledown\left(\rho_{2}(\rho_{1}(x))\right).$ (13)

Using that

$f(A\cap f^{-1}(B))=f(A)\cap B$ (14)

for a function $f:S\longrightarrow T$ between two sets $S,T$ and $A\subset
S,\leavevmode\nobreak\ B\subset T$ and the fact that condition (r2) implies
$\bigtriangledown\left(\rho_{1}(x)\right)\subset\rho_{1}(\bigtriangledown\left(x\right)),$
we get the following inequality for all
$y\in\bigtriangledown\left(\rho_{2}(\rho_{1}(x))\right):$

$\displaystyle\left|\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(y)\right|$
$\displaystyle\geq\left|\rho_{1}(\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(y))\right|$
$\displaystyle=\left|\rho_{1}(\bigtriangledown\left(x\right))\cap\rho_{2}^{-1}(y)\right|$
$\displaystyle\geq\left|\bigtriangledown\left(x\right)\cap\rho_{2}^{-1}(y)\right|=1,$

where the last equality is due to condition (12).

Moreover if we suppose that for some
$y\in\bigtriangledown\left(\rho_{2}(\rho_{1}(x))\right)$ there are two
distinct elements

$w,w^{\prime}\in\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(y)$

then since
$\bigtriangledown\left(\rho_{2}(\rho_{1}(x)\right)\subset\rho_{2}(\bigtriangledown\left(\rho_{1}(x)\right))$
by condition (r2), there exists
$y^{\prime}\in\bigtriangledown\left(\rho_{1}(x)\right)\cap\rho_{2}^{-1}(y)$
and this element is unique by condition (13). This implies that

$\\{w,w^{\prime}\\}\subset\left|\bigtriangledown\left(x\right)\cap\rho_{1}^{-1}(y^{\prime})\right|$

which contradicts condition (12) and therefore we have proven (r4) for
$\rho_{2}\circ\rho_{1}.$

Finally, we show condition (r3) for $\rho_{2}\circ\rho_{1}$ as follows. If
$x\in J$ satisfies

$\operatorname{rk}_{J}(x)=\operatorname{rk}_{L}(\rho_{2}(\rho_{1}(x)))-1$

then we can have the two following possible cases.

The first possible case is

$rk_{J}(x)=\operatorname{rk}_{K}(\rho_{1}(x)),\quad\operatorname{rk}_{K}(\rho_{1}(x))=\operatorname{rk}_{L}(\rho_{2}(\rho_{1}(x)))-1,$

which respectively implies condition (12) as well as condition (r3) for
$\rho_{2},$ i.e.

$\left|\bigtriangledown\left(\rho_{1}(x)\right)\cap\rho_{2}^{-1}(\rho_{2}(\rho_{1}(x)))\right|=2.$
(15)

Every element
$w\in\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(\rho_{2}(\rho_{1}(x)))$
satisfies $\rho(x)\subset\rho(w)$ and
$\rho_{2}(\rho_{1}(w))=\rho_{2}(\rho_{1}(x)),$ hence

$r+1=\operatorname{rk}_{J}(w)\leq\operatorname{rk}_{K}(\rho_{1}(w))\leq\operatorname{rk}_{L}(\rho_{2}(\rho_{1}(w)))=\operatorname{rk}_{L}(\rho_{2}(\rho_{1}(x)))=r+1.$

Therefore $\operatorname{rk}_{K}(\rho_{1}(w))=r+1$ and
$\rho_{1}(w)\in\bigtriangledown\left(\rho_{1}(x)\right).$ Condition (12)
implies that $\rho_{1}^{-1}(\rho_{1}(w))=\\{w\\}$ for every such $w$ and using
(15) we get

$\left|\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(\rho_{2}(\rho_{1}(x)))\right|=\left|\bigtriangledown\left(\rho_{1}(x)\right)\cap\rho_{2}^{-1}(\rho_{2}(\rho_{1}(x)))\right|=2.$

The second possible case is

$rk_{J}(x)=\operatorname{rk}_{K}(\rho_{1}(x))-1,\quad\operatorname{rk}_{K}(\rho_{1}(x))=\operatorname{rk}_{L}(\rho_{2}(\rho_{1}(x))).$

This case implies condition (r3) for $\rho_{1}:$

$\left|\bigtriangledown\left(x\right)\cap\rho_{1}^{-1}(\rho_{1}(x))\right|=2.$
(16)

If
$w\in\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(\rho_{2}(\rho_{1}(x)))$
then $\rho(x)\subset\rho(w)$ and
$\operatorname{rk}_{K}(\rho(w))=r+1=\operatorname{rk}_{K}(\rho_{1}(x))$ by the
same argument as above, therefore $\rho_{1}(w)=\rho_{1}(x).$ This implies that

$\bigtriangledown\left(x\right)\cap(\rho_{2}\circ\rho_{1})^{-1}(\rho_{2}(\rho_{1}(x)))=\bigtriangledown\left(x\right)\cap\rho_{1}^{-1}(\rho_{1}(x))$

and we get the desired result by (16). This concludes the proof that
$\rho_{2}\circ\rho_{1}$ is a reduction and therefore
$\mathrel{\raisebox{1.0pt}{$\prec$}}$ is transitive.

At last, the relation $\mathrel{\raisebox{1.0pt}{$\prec$}}$ is antisymmetric
by the following argument. Suppose
$J\mathrel{\raisebox{1.0pt}{$\prec$}}_{\phi}K$ and
$K\mathrel{\raisebox{1.0pt}{$\prec$}}_{\gamma}J$ (which by Remark 1.26 implies
$\operatorname{Rk}(J)=\operatorname{Rk}(K)$) and define
$\phi_{r}:=\phi|^{K^{[r]}}$ and $\gamma_{r}:=\gamma|^{J^{[r]}}.$ We will prove
by induction on $r$ that $\phi_{r}:J^{[r]}\longrightarrow K^{[r]}$ is a
bijection for all $0\leq r\leq R$ implying that $\phi$ is an isomorphism by
Lemma 1.6. For the case $r=0$ we know that $\phi_{0}$ and $\gamma_{0}$ are
surjective and this implies that $\left|J^{[0]}\right|=\left|K^{[0]}\right|.$
Hence $\phi_{0}$ and $\gamma_{0}$ are also injective and therefore bijective.
Suppose that $\phi_{s}:J^{[s]}\longrightarrow K^{[s]}$ and
$\gamma_{s}:K^{[s]}\longrightarrow J^{[s]}$ are bijective for all $s=0,1\dots
r-1.$ As noted in Remark 1.25 since $\phi$ is a reduction we have
$\phi^{-1}(K^{[r]})\subset J^{(r)}$ and $J^{(r-1)}=\phi^{-1}(K^{(r-1)}).$
Hence $\phi^{-1}(K^{[r]})\subset J^{[r]}$ and
$\left|K^{[r]}\right|\leq\left|J^{[r]}\right|.$ By the same argument for
$\gamma$ we also get the reverse inequality and therefore
$\phi_{r}:J^{[r]}\longrightarrow K^{[r]}$ and
$\gamma_{r}:K^{[r]}\longrightarrow J^{[r]}$ are indeed bijections. ∎

##  Appendix C Proof of Proposition 2.10

In this appendix we show the following result: let $K\in\mathsf{C}^{R+1}$ and
$J\leq\partial K,$ $J\in\mathsf{B}^{R}$ such that $(K,J)$ is uniform and
local. The two following statements hold.

  1. 1)

We have $J\mathrel{\raisebox{1.0pt}{$\prec$}}J(K)$ as the poset homomorphism
$\rho_{J}^{K}:J\longrightarrow J(K)$ defined by

$\rho_{J}^{K}(x)=x_{J}$

is a reduction.

  2. 2)

We have $M_{J}^{K}\vdash J(K)$ as the poset homomorphism
$\pi_{J}^{K}:M_{J}^{K}\longrightarrow J(K)$ defined by

$\pi_{J}^{K}(\mathcal{E}^{x}_{J})=x_{J}$

is a collapse.

###### Proof.

In order to prove the point 1) we first prove that
$\rho_{J}^{K}:J\longrightarrow J(K)$ is a reduction. The map $\rho_{J}^{K}$ is
clearly a surjective poset homomorphism. Condition (r1) follows from the fact
that if a vertex $v\in J^{[0]}$ satisfies $v_{J}\in J(K)^{[0]}$ then we have
$v_{J}=\\{v\\}.$

Condition (r2) is obtained in the following way. Suppose $x\in J$ satisfies

$x_{J}\subset y\in J(K)^{[r]}.$

For $y^{\prime}\in K_{J}^{-}[y]$ we then have that
$\operatorname{Rk}(\mathbf{B}_{J}(y^{\prime}))=r$ and
$x\in\mathbf{B}_{J}(y^{\prime}).$ By purity of $(K,J)$ there exists an
$r$-cell $x^{\prime}$ in $\mathbf{B}_{J}(y^{\prime})$ such that $x\subset
x^{\prime}.$ If we name $w=x_{J}^{\prime}$ then Lemma A.7 implies that

$K_{J}^{-}[w]=K_{J}^{-}(x^{\prime}).$

Since $x^{\prime}$ is a maximal cell of $\mathbf{B}_{J}(y^{\prime})$ we have
that $y^{\prime}\in K_{J}^{-}(x^{\prime})$ and by the previous relation this
implies

$y=y^{\prime}_{J}=w=x^{\prime}_{J},$

which proves condition (r2) for $\rho_{J}^{K}.$

We then turn to condition (r3). Let $x\in J$ such that
$\operatorname{rk}_{J}(x)=\operatorname{rk}_{J(K)}(x_{J})-1$ and let
$x^{\prime}\in K_{J}^{-}[x_{J}].$ Then
$\operatorname{rk}_{K}(x^{\prime})=\operatorname{rk}_{J(K)}(x_{J})+1,$ hence
$\operatorname{rk}_{K}(x^{\prime})=\operatorname{rk}_{K}(x)-2.$ Since
$x\subset x^{\prime},$ Axiom iv) for $K$ provides us with two cells
$y,y^{\prime}$ such that

$\bigtriangledown\left(x\right)\cap\triangle\left(x^{\prime}\right)=\\{y,y^{\prime}\\}.$

By minimality of $x^{\prime},$ we have that $y$ and $y^{\prime}$ are in $J$
and

$x^{\prime}\in K_{J}^{-}(y)\cap K_{J}^{-}(y^{\prime})=K_{J}^{-}[y_{J}]\cap
K_{J}^{-}[y^{\prime}_{J}],$

where the equality is due to Lemma A.7. Also, if
$w\in\bigtriangledown\left(x\right)$ satisfies $w_{J}=x_{J}$ then $w\subset
x^{\prime}$ because $x^{\prime}\in K_{J}^{-}[x_{J}].$ Since
$\operatorname{rk}_{J}(w)=\operatorname{rk}_{J}(x)+1=\operatorname{rk}_{K}(x^{\prime})-1,$
we have $w\in\triangle\left(x^{\prime}\right)$ and therefore
$w\in\\{y,y^{\prime}\\}.$ This completes the proof that

$\left|\bigtriangledown\left(x\right)\cap(\rho_{J}^{K})^{-1}(x_{J})\right|=2.$

Then condition (r4) can be proven by taking $x\in J$ such that
$\operatorname{rk}_{J}(x)=\operatorname{rk}_{J(K)}(x_{J}).$ This implies that
$x$ is a maximal cell of $\mathbf{B}_{J}(x^{\prime})$ for all $x^{\prime}\in
K_{J}^{-}[x_{J}].$ If we take $y\in\bigtriangledown\left(x_{J}\right),$
$y^{\prime}\in K_{J}^{-}[y]$ and $x^{\prime}\in K_{J}^{-}[x_{J}]$ such that
$x^{\prime}\subset y^{\prime}$ then we obtain

$\operatorname{rk}_{K}(y^{\prime})=\operatorname{rk}_{J(K)}(y)+1=\operatorname{rk}_{K}(x)+2.$

Therefore there exists $x^{\prime\prime}\in\bigtriangledown\left(x\right)$
such that

$\bigtriangledown\left(x\right)\cap\triangle\left(y^{\prime}\right)=\\{x^{\prime},x^{\prime\prime}\\}.$

If $x^{\prime\prime}\in K_{J}$ then $x^{\prime}\cap
J^{[0]}=x^{\prime\prime}\cap J^{[0]}=x$ which implies $y^{\prime}\cap
J^{[0]}=x$ but this contradicts the fact that
$\operatorname{Rk}(\mathbf{B}_{J}(y^{\prime}))=\operatorname{rk}_{J}(x)+1.$
Therefore $x^{\prime\prime}\in J,$ hence
$x^{\prime\prime}_{J}=y^{\prime}_{J}=y$ and we have

$\bigtriangledown\left(x\right)\cap(\rho_{J}^{K})^{-1}(y)=\\{x^{\prime\prime}\\}.$

Next we prove that $\pi_{J}^{K}:M_{J}^{K}\longrightarrow J(K)$ is a collapse.
Condition (c1) is clear since $\left|\bigtriangledown\left(z\right)\right|=1$
for all $z\in J^{[R-1]}$ and therefore $K_{J}^{-}[z]=K_{J}^{-}(z)$ contains a
unique cell $z^{\prime}$ which is a maximal cell of $K,$ in other words

$(\pi_{J}^{K})^{-1}(z)=\\{\mathcal{E}_{J}^{z^{\prime}}\\}.$

Condition (c2) is a consequence of Lemma A.6, since if $x\in J(K)$ and
$\mathcal{E}_{J}^{y^{\prime}}\in M_{J}^{K}$ are such that $x\subset
y_{J}^{\prime}$ then there exists $x^{\prime}\in K_{J}^{-}[x]$ such that
$x^{\prime}\subset y^{\prime}.$ Hence we have
$\operatorname{rk}_{M_{J}^{K}}(\mathcal{E}_{J}^{x^{\prime}})=\operatorname{rk}_{J(K)}(x)$
and Lemma A.4 implies that
$\mathcal{E}_{J}^{x^{\prime}}\subset\mathcal{E}_{J}^{y^{\prime}}.$

In order to prove condition (c3), we take $\mathcal{E}_{J}^{x^{\prime}}\in
M_{J}^{K}$ such that

$\operatorname{rk}_{M_{J}^{K}}(\mathcal{E}_{J}^{x^{\prime}})=\operatorname{rk}_{J(K)}(x^{\prime}_{J})+1=\operatorname{rk}_{K}(x^{\prime})-1.$

If $w\in\mathbf{B}_{J}(x^{\prime})$ is maximal cell then Axiom iv) for $K$
gives

$\bigtriangledown\left(w\right)\cap\triangle\left(x^{\prime}\right)=\\{y,y^{\prime}\\},$

for $y,y^{\prime}\in K.$ The maximality of $w$ then implies that
$\\{y,y^{\prime}\\}\subset K_{J}.$ We therefore have that

$w\subset y\cap y^{\prime}\in J,$

which by Axiom i) implies that $w=y\cap y^{\prime}.$ Therefore
$y_{J}=y^{\prime}_{J}=x_{J}$ and we indeed have

$\left|\triangle\left(\mathcal{E}_{J}^{x^{\prime}}\right)\cap(\pi^{-1})(x)\right|=2.$

Finally, we can prove condition (c4) in the following way. Suppose that
$\mathcal{E}_{J}^{x^{\prime}}\in M_{J}^{K}$ satisfies
$\operatorname{rk}_{M_{J}^{K}}(\mathcal{E}_{J}^{x^{\prime}})=\operatorname{rk}_{J(K)}(x^{\prime}_{J}).$
If $w$ is a maximal cell in $\mathbf{B}_{J}(x^{\prime})$ we have that
$\operatorname{rk}_{K}(w)=\operatorname{rk}_{K}(x^{\prime})-1,$ i.e.
$w\in\triangle\left(x^{\prime}\right).$ By Lemma A.7 we also have that

$x^{\prime}\in K_{J}^{-}(w)=K_{J}^{-}[w_{J}].$

If we fix an element $y\in\triangle\left(x\right)$ where
$x:=x^{\prime}_{J}=w_{J}$ then by Lemma A.6 there exists $y^{\prime}\in
K_{J}^{-}[y]$ such that $y^{\prime}\subset x^{\prime}$ and

$\operatorname{rk}_{K}(y^{\prime})=\operatorname{rk}_{J(K)}(y)+1=\operatorname{rk}_{J(K)}(x)=\operatorname{rk}_{K}(x^{\prime})-1.$

This implies that $y^{\prime}\in\triangle\left(x^{\prime}\right),$ i.e.
$\mathcal{E}_{J}^{y^{\prime}}\in\triangle\left(\mathcal{E}_{J}^{x^{\prime}}\right)$
by Lemma A.4. Moreover if
$\mathcal{E}_{J}^{w^{\prime}}\in\triangle\left(\mathcal{E}_{J}^{x^{\prime}}\right)\setminus\\{\mathcal{E}_{J}^{y^{\prime}}\\}$
satisfies $w^{\prime}_{J}=y$ then by minimality of $y^{\prime}$ we have
$y^{\prime}\cap w^{\prime}\in J.$ Hence it must be that
$\mathbf{B}_{J}((y^{\prime}\cap w^{\prime}))$ has a unique maximal cell $u$
and

$\\{y^{\prime},w^{\prime}\\}\subset\bigtriangledown\left(u\right)\cap\triangle\left(x^{\prime}\right).$

This is a contradiction with Axiom iv) for $K$ since
$\mathbf{B}_{J}(x^{\prime})$ is a pure cc such that

$\operatorname{Rk}(\mathbf{B}_{J}(x^{\prime}))=\operatorname{rk}_{K}(x^{\prime})-1=\operatorname{rk}_{K}(y^{\prime})=\operatorname{rk}_{K}(u)+1$

and therefore there exists a maximal cell $w$ of $\mathbf{B}_{J}(y^{\prime})$
such that
$w\in\bigtriangledown\left(u\right)\cap\bigtriangledown\left(x^{\prime}\right)\setminus\\{y^{\prime},w^{\prime}\\}.$
Thus we can indeed conclude that

$\triangle\left(\mathcal{E}_{J}^{x^{\prime}}\right)\cap(\pi_{J}^{K})^{-1}(y)=\\{\mathcal{E}_{J}^{y^{\prime}}\\}.$

∎

##  Appendix D Proof of Proposition 2.13

In this appendix we show the following result: let $K\in\mathsf{C}$ and let
$J^{\prime}$ be a connected component of $\partial K$ such that
$(K,J^{\prime})$ is uniform. If $\rho:J\longrightarrow J^{\prime}$ is a
reduction such that $\rho\circ\rho_{J^{\prime}}^{K}:J\longrightarrow
J^{\prime}(K)$ is compatible with
$\pi_{J^{\prime}}^{K}:M_{J^{\prime}}^{K}\longrightarrow J^{\prime}(K)$ and $J$
is disjoint from $K$ then the cc $K^{\rho}$ (introduced in Definition 2.3) is
an element in $\mathsf{C}$ and $J$ is a connected component of $\partial
K^{\rho},$ in other words

$(K^{\rho},J)\mathrel{\raisebox{1.0pt}{$\prec$}}_{\rho}(K,J^{\prime})$

defines a relative reduction. Moreover, we have that
$\rho_{J}^{K^{\rho}}=\rho\circ\rho_{J^{\prime}}^{K}$ and $(K^{\rho},J)$ is
uniform.

###### Proof.

The transitivity of $\mathrel{\raisebox{1.0pt}{$\prec$}}$ (Proposition 1.24)
implies that $\rho\circ\rho_{J^{\prime}}^{K}$ is also a reduction hence we can
without loss of generality assume that
$\rho_{J^{\prime}}^{K}=\operatorname{id}_{J^{\prime}(K)},$ i.e.
$J^{\prime}=J^{\prime}(K)$ and $\rho=\rho_{J^{\prime}}^{K}\circ\rho.$

Define $(J^{\prime})^{c}:=\mathbf{B}_{K}(K^{[0]}\setminus(J^{\prime})^{[0]}),$
i.e. the sub-cc of $K$ containing the cells in $K$ not intersecting
$J^{\prime}.$ Let $\phi$ be the map defined by (11) which by Lemma A.8 defines
a poset isomorphism

$\phi:K_{J^{\prime}}\longrightarrow\phi(K_{J^{\prime}}).$

We define $(K^{\rho},\operatorname{rk}_{K^{\rho}})$ by

$\displaystyle K^{\rho}$ $\displaystyle:=(J^{\prime})^{c}\sqcup
J\sqcup\phi(K_{J^{\prime}})$ $\displaystyle\operatorname{rk}_{K^{\rho}}$
$\displaystyle:=\operatorname{rk}_{K}|_{(J^{\prime})^{c}}+\operatorname{rk}_{J}+\operatorname{rk}_{K}|_{K_{J^{\prime}}}\circ\phi^{-1}.$

It then follows that $K_{J}^{\rho}=\phi(K_{J^{\prime}}).$ Also, since
$J^{\prime}=\pi_{J^{\prime}}^{K}(M_{J^{\prime}}^{K})$ as
$\rho_{J^{\prime}}^{K}=\operatorname{id}_{J^{\prime}}$ we have that

$\text{for all }y^{\prime}\in
K_{J^{\prime}},(y^{\prime})_{J^{\prime}}=y^{\prime}\cap({J^{\prime}})^{[0]}.$
(17)

Let’s now check the axioms of cc for $K^{\rho}.$ To make our notations more
explicit, for a cell $x$ in a cc $L$ we will denote by
$\bigtriangledown_{L}\left(x\right)$ the set of co-faces of $x$ in $L$ and
similarly for faces. Since these axioms are satisfied for $K$ they are
automatically satisfied when all cells involved are in $J$ or
$(J^{\prime})^{c}\sqcup\phi(K_{J})$ by the poset isomorphism property of
$\phi.$ This allows us to only check the axioms in the case of inclusions of
cells between the sets $J$ and $(J^{\prime})^{c}\sqcup\phi(K_{J}).$

To check Axiom i) we only need to consider the case where $x\in J,$ $y\in
K^{\rho}_{J}$ and $x\subsetneq y.$ In this case we have that
$\rho(x)\subsetneq\phi^{-1}(y)$ and therefore

$\operatorname{rk}_{K^{\rho}}(x)\leq\operatorname{rk}_{K}(\rho(x))<\operatorname{rk}_{K}(\phi^{-1}(y))=\operatorname{rk}_{K^{\rho}}(y).$

To check Axiom iii) for the non-direct cases amounts to show that $x\cap y\in
K^{\rho}$ for $x\in K_{J}^{\rho}\sqcup J$ and $y\in K_{J}^{\rho}.$ If $x\in
K_{J}^{\rho}$ and $x^{\prime},y^{\prime}\in K_{J^{\prime}}$ are such that
$\phi(x^{\prime})=x$ and $\phi(y^{\prime})=y$ then we have two possible cases:
either
$\mathcal{E}_{J^{\prime}}^{x^{\prime}}\cap\mathcal{E}_{J^{\prime}}^{y^{\prime}}\neq\emptyset$
or
$\mathcal{E}_{J^{\prime}}^{x^{\prime}}\cap\mathcal{E}_{J^{\prime}}^{y^{\prime}}=\emptyset.$
In the first case one has that $x^{\prime}\cap y^{\prime}\in K_{J^{\prime}}$
and

$\phi(x^{\prime}\cap y^{\prime})=\phi(x^{\prime})\cap\phi(y^{\prime})=x\cap
y\in K^{\rho}_{J}.$

For the case
$\mathcal{E}_{J^{\prime}}^{x^{\prime}}\cap\mathcal{E}_{J^{\prime}}^{y^{\prime}}=\emptyset,$
then by property (17), we have that

$x^{\prime}\cap
y^{\prime}=(x^{\prime})_{J^{\prime}}\cap(y^{\prime})_{J^{\prime}}\in
J^{\prime}(K)=J^{\prime}.$

By Proposition 2.10 we know that $\pi_{J^{\prime}}^{K}$ is a collapse, hence
by (c2) we can pick $x^{\prime}_{0}\in\mathbf{B}(x^{\prime})^{[r+1]}$ and
$y^{\prime}_{0}\in\mathbf{B}(y^{\prime})^{[r+1]}$ where $r=rk(x^{\prime}\cap
y^{\prime})=\operatorname{rk}_{M_{J^{\prime}}^{K}}(\mathcal{E}_{J^{\prime}}^{x^{\prime}_{0}})=\operatorname{rk}_{M_{J^{\prime}}^{K}}(\mathcal{E}_{J^{\prime}}^{y^{\prime}_{0}}).$
By Lemma A.6 this implies that $x^{\prime}_{0},y^{\prime}_{0}\in
K^{-}_{J^{\prime}}[x^{\prime}\cap y^{\prime}],$ and since
$\mathcal{E}_{J^{\prime}}^{x^{\prime}}\cap\mathcal{E}_{J^{\prime}}^{y^{\prime}}=\emptyset,$
we also have that
$\mathcal{E}_{J^{\prime}}^{x_{0}^{\prime}}\cap\mathcal{E}_{J^{\prime}}^{y_{0}^{\prime}}=\emptyset.$
The latter implies that $\left|(\pi_{J^{\prime}}^{K})^{-1}(x^{\prime}\cap
y^{\prime})\right|>1$ which by condition 1) of compatibility implies that
there exists $x\in J$ such that $\rho^{-1}(x^{\prime}\cap
y^{\prime})=\\{w_{0}\\}.$ Therefore we obtain

$\phi(x^{\prime})\cap\phi(y^{\prime})=\bigcup_{w\in\rho^{-1}(B(x^{\prime}\cap
y^{\prime}))}w=w_{0}\in J.$

For the case where $x\in J$ and if $y^{\prime}=\phi^{-1}(y)\in K_{J^{\prime}}$
then one has to show that

$x\cap\left(\bigcup_{w\in J\leavevmode\nobreak\ :\leavevmode\nobreak\
\rho(w)\subset y^{\prime}}w\right)\in J.$

Let $w_{0}:=(y^{\prime})_{J^{\prime}},$ then $\\{w\in J\leavevmode\nobreak\
|\leavevmode\nobreak\ \rho(w)\subset y^{\prime}\\}=\rho^{-1}(w_{0}),$ hence we
need to show that there exists a unique maximal cell in
$\mathbf{B}(x)\cap\rho^{-1}(w_{0}).$ Suppose by contradiction that
$w_{1},w_{2}$ are two distinct such maximal cells. Since we have $w_{i}\subset
x$ for $i=1,2,$ we have that $w_{1}\vee w_{2}\neq\overline{\emptyset}.$
Condition 2) of compatibility then implies that

$\rho_{J^{\prime}}^{K}(w_{1}\vee w_{2})=\rho(w_{1})\vee\rho(w_{2})\subset
w_{0},$

where the inclusion is due to the fact that $\rho(w_{i})\subset w_{0}$ for
$i=1,2.$ But this implies that $w_{1}\vee
w_{2}\in\rho^{-1}(\mathbf{B}(w_{0}))$ and contradicts the maximality of
$w_{1}$ and $w_{2}.$ This finishes the proof of Axiom iii) for $K^{\rho}.$

Axiom ii) can be inferred from the case $J\ni x\subsetneq y\in K^{\rho}_{J}$
which is proven as follows. By the poset isomorphism property of $\phi$ we
have that $\rho(x)\subset\phi^{-1}(y)\in K_{J^{\prime}}.$ Therefore by Axiom
ii) for $K$ we can suppose without loss of generality that
$\operatorname{rk}_{K}(y)=\operatorname{rk}_{K}(\rho(x))+1.$ If
$\operatorname{rk}_{J}(x)=\operatorname{rk}_{K}(\rho(x))$ then we have that
$y\in\bigtriangledown_{K^{\rho}}\left(x\right)$ as desired. If
$\operatorname{rk}_{J}(x)<\operatorname{rk}_{K}(\rho(x))$ then, as noted in
Remark 1.25, there exists $w\in\rho^{-1}(\rho(x))$ such that
$\operatorname{rk}_{J}(w)=\operatorname{rk}_{K}(\rho(x)).$ Therefore using ii)
for $J$ we can find a co-face of $x$ contained in $w$ which is then also
contained in $y.$

For Axiom iv) it is also sufficient to focus on the case where $x\in J,y\in
K_{J}^{\rho},$ $x\subsetneq y$ and

$\operatorname{rk}_{K^{\rho}}(x)=\operatorname{rk}_{K^{\rho}}(y)-2.$

Since $\operatorname{rk}_{K^{\rho}}(y)=\operatorname{rk}_{K}(\phi^{-1}(y)),$
this leaves two possible cases: either we have
$\operatorname{rk}_{K}(\rho(x))=\operatorname{rk}_{K^{\rho}}(x)$ or
$\operatorname{rk}_{K}(\rho(x))=\operatorname{rk}_{K^{\rho}}(x)+1.$ In the
first case where
$\operatorname{rk}_{K}(\rho(x))=\operatorname{rk}_{K^{\rho}}(x),$ Axiom iv)
for $K$ implies that

$\left|\bigtriangledown_{K}\left(\rho(x)\right)\cap\triangle_{K}\left(\phi^{-1}(y)\right)\right|=2.$

To conclude this first case, it remains to show that there is a bijection
between
$\bigtriangledown_{K}\left(\rho(x)\right)\cap\triangle_{K}\left(\phi^{-1}(y)\right)$
and
$\bigtriangledown_{K^{\rho}}\left(x\right)\cap\triangle_{K^{\rho}}\left(y\right).$
If
$w^{\prime}\in\bigtriangledown_{K}\left(\rho(x)\right)\cap\triangle_{K}\left(\phi^{-1}(y)\right)$
then either $w^{\prime}\in K_{J^{\prime}}$ or $w^{\prime}\in J^{\prime}.$
Since $w^{\prime}\in K_{J^{\prime}}$ if and only if
$\phi(w^{\prime})\in\bigtriangledown_{K^{\rho}}\left(x\right)\cap\triangle_{K^{\rho}}\left(y\right)$
and $\phi$ is bijection, we are good for this case. If $w^{\prime}\in
J^{\prime}$ then
$w^{\prime}\in\bigtriangledown_{J^{\prime}}\left(\rho(x)\right)$ and (r4)
implies that the map

$\bigtriangledown_{J}\left(x\right)\cap\rho^{-1}(\bigtriangledown_{K}\left(\rho(x)\right))\ni
w\longmapsto\rho(w)\in\bigtriangledown_{J^{\prime}}\left(\rho(x)\right)$

is a bijection, and this concludes the case
$\operatorname{rk}_{K}(\rho(x))=\operatorname{rk}_{K^{\rho}}(x).$

In the second case where
$\operatorname{rk}_{K}(\rho(x))=\operatorname{rk}_{K^{\rho}}(x)+1$ condition
(r3) implies that

$\left|\bigtriangledown_{J}\left(x\right)\cap\rho^{-1}(\rho(x))\right|=2.$

Moreover, if
$w\in\bigtriangledown_{K^{\rho}}\left(x\right)\cap\triangle_{K^{\rho}}\left(y\right)$
then $\rho(w)\subset\phi^{-1}(y)$ and

$\operatorname{rk}_{J}(x)+1=\operatorname{rk}_{J}(w)\leq\operatorname{rk}_{K}(\rho(w))<\operatorname{rk}_{K}(y)=\operatorname{rk}_{J}(x)+2.$

Therefore $\operatorname{rk}_{K}(\rho(w))=\operatorname{rk}_{K}(\rho(x))$ and
$\rho(w)=\rho(x).$ Thus
$w\in\bigtriangledown_{J}\left(x\right)\cap\rho^{-1}(\rho(x))$ and this also
proves that

$\left|\bigtriangledown_{K^{\rho}}\left(x\right)\cap\triangle_{K^{\rho}}\left(y\right)\right|=2,$

which concludes the proof of iv) and shows that $K^{\rho}$ is a cc

The cc $K^{\rho}$ is graph-based if $K^{\rho}_{J}$ is graph-based, which is
the case since $e\in(K^{\rho}_{J})^{[1]}$ if and only if
$e^{\prime}:=\phi^{-1}(e)\in K^{[1]}_{J^{\prime}}$ and therefore

$\left|e\right|=\left|e^{\prime}\setminus(J^{\prime})^{[0]}\sqcup\rho^{-1}(J^{[0]}\cap
e)\right|=2$

by condition (r1).

The previous argument also implies that $K^{\rho}_{J}$ is cell-connected.
Since $K$ is connected and cell-connected, this shows that $K^{\rho}$ is
connected and cell-connected.

One has that $K^{\rho}$ is clearly pure and
$R:=\operatorname{Rk}(K^{\rho})=\operatorname{Rk}(K).$ And $K^{\rho}$ is non-
branching if $\left|\bigtriangledown\left(y^{\prime}\right)\right|=2$ for all
$y^{\prime}\in(K^{\rho}_{J})^{[R-1]},$ which is directly verified as a
consequence of the poset isomorphism property of $\phi.$

A $K^{\rho}$-pinch cannot occur in $(J^{\prime})^{c}$ nor $K^{\rho}_{J}$ since
the poset isomorphism property of $\phi$ would produce a $K$-pinch. If $x\in
J$ then $x\subsetneq y\in K^{\rho}$ if and only if
$\rho(x)\subsetneq\phi^{-1}(y)\in K$ hence we have

$M_{\mathbf{B}(x)}^{K^{\rho}}\cong K_{x}^{\rho}\cong K_{\rho(x)}\cong
M^{K}_{\mathbf{B}(\rho(x))},$

where $\cong$ means isomorphic as ranked posets, the second isomorphism being
$\phi^{-1}.$ Therefore there also cannot be a $K^{\rho}$-pinch in $J$ since
this would contradict that $K$ has no $K$-pinch on $J^{\prime}.$ This in
addition shows that $K^{\rho}\in\mathsf{C}.$

By Remark 1.26 we have
$\operatorname{Rk}(J)=\operatorname{Rk}(J^{\prime})=R-1$ and every cell in $J$
is included in a cell $y\in J^{[R-1]}.$ By Remark 1.25 such a $y\in J^{[R-1]}$
satisfies that $\rho(y)\in(J^{\prime})^{[R-1]}.$ Therefore we have

$\left|\bigtriangledown_{K^{\rho}}\left(y\right)\right|=\left|\bigtriangledown_{K}\left(\rho(y)\right)\right|=1,$

hence $J$ is indeed a connected component of $\partial K^{\rho}.$

We have that $\rho_{J}^{K^{\rho}}=\rho$ by the following arguments. If
$v^{\prime}$ is the vertex in $e\cap(J^{\prime})^{[0]}$ for
$e\in\mathcal{E}^{\prime}_{J^{\prime}}$ then by (r1) we have that
$\rho^{-1}(\\{v^{\prime}\\})$ contains exactly one vertex $v$ from $J.$ We
will therefore consider the vertices $v^{\prime}$ and $v$ as identical to
simplify the discussion. With this convention, a cell $y\in J^{\prime}(K)$ can
be seen as having its vertices in $J.$ Moreover, if $y^{\prime}\in
K_{J^{\prime}}^{-}[y]$ then $\phi(y^{\prime})\in(K^{\rho})_{J}^{-}(x)$ for all
$x\in J$ such that $\rho(x)\subset y.$ Since we also assume that
$J^{\prime}=J^{\prime}(K)$ and
$\rho_{J^{\prime}}^{K}=\operatorname{id}_{J^{\prime}}$ we have that
$y=y_{J^{\prime}}$ for all $y\in J^{\prime}.$ Therefore, if we take $x\in J$
and $y^{\prime}\in K_{J^{\prime}}^{-}[\rho(x)_{J^{\prime}}]$ then

$x_{J}=\phi(y^{\prime})_{J}=y^{\prime}_{J^{\prime}}=\rho(x)_{J^{\prime}}=\rho(x)$

and this proves that $\rho=\rho_{J}^{K^{\rho}}.$ This also directly implies
that $J(K^{\rho})=J^{\prime}(K)$ and that $(K^{\rho},J)$ is uniform. ∎

##  Appendix E Proof of Theorem 3.5

This section simply contains the proof of Theorem 3.5 stating that if
$(K-J)\in\mathsf{Cob}^{R}$ then:

  * •

$\overline{(K-J)}\in\mathsf{Cob}^{R},$

  * •

$\overline{(K-J)}$ is exactly collared,

  * •

$\overline{K_{J}}\in\mathsf{B}^{R-1}_{\sqcup}$ and
$\partial\overline{(K-J)}=\overline{\partial K\setminus
J}\sqcup\overline{K_{J}}.$

###### Proof.

Let $R=\operatorname{Rk}(K).$ We split the proof into three statements 1a, 1b
and 2 as follows.

Suppose that $(K-J)\in\mathsf{Cob}.$

  1. 1.

Then we have $K^{\prime}:=\widetilde{K\setminus J}\sqcup\overline{\partial
K\setminus J}\in\mathsf{C},$

    1. (a)

for the special case $J=\emptyset,$

    2. (b)

and for $J$ arbitrary as a consequence of the following statement: if
$K^{\prime}\in\mathsf{C}$ and $J_{0}$ is a connected component of $\partial
K\setminus J$ such that $(K,J\sqcup J_{0})$ is local then

$K^{\prime}_{0}:=\widetilde{K\setminus(J\sqcup J_{0})}\sqcup\overline{\partial
K\setminus(J\sqcup J_{0})}\in\mathsf{C};$

  2. 2.

If we define $J^{\prime}:=\overline{\partial K\setminus J},$ then
$J^{\prime}\in\mathsf{B}^{R-1}_{\sqcup},$ $(K^{\prime},J^{\prime})$ is exactly
collared and

$\partial K^{\prime}=J^{\prime}\sqcup\overline{K_{J}}.$

To prove the point 1a, we show that
$K^{\prime}:=\widetilde{K}\sqcup\overline{\partial K}$ satisfies Axioms i),
iii), ii) and iv) and is a local non-singular non-pinching cell complex of
rank $R.$

Axiom i) is a consequence of Lemma 3.3 for the cells of $K^{\prime}$ included
in $\widetilde{K}$ and is a consequence of Lemma 1.10 for the cells included
in $\overline{\partial K}.$ To prove iii), we take $x^{\prime},y^{\prime}\in
K^{\prime}$ such that $x^{\prime}\cap y^{\prime}\neq\emptyset.$ If
$x^{\prime}=\widetilde{x}$ and $y^{\prime}=\widetilde{y}$ are included in
$\widetilde{K}$ then there exists a cell $w\supset x\cup y$ of minimal rank,
and by Lemma 3.3 we get $\widetilde{w}=\widetilde{x}\cap\widetilde{y}$ (using
the same arguments as in the proof of Proposition 1.11). If
$x^{\prime}=\overline{x}$ and $y^{\prime}=\overline{y}$ are cells in
$\overline{\partial K}$ then iii) is a consequence of Proposition 1.11. If say
$x^{\prime}=\widetilde{x}\in\widetilde{K}$ and
$y^{\prime}=\overline{y}\in\overline{\partial K}$ then

$\widetilde{x}\cap\overline{y}=(\overline{x}^{K}\cup\overline{x}^{\partial
K})\cap\overline{y}^{\partial K}=\overline{x}^{\partial
K}\cap\overline{y}^{\partial K},$

so $\overline{x}^{\partial K}\cap\overline{y}^{\partial K}\neq\emptyset.$
Therefore $x,y\in\partial K$ and

$\overline{x}^{\partial K}\cap\overline{y}^{\partial K}=\overline{w}^{\partial
K},$

where $w=\overline{\overline{x}\cap\overline{y}}\in\partial K.$ Axiom ii) is
satisfied for elements in $\widetilde{K\setminus\partial K}$ or in
$\overline{\partial K}$ as a consequence of Axiom ii) for $K$ and $\partial K$
and by Lemma 3.3 and Proposition 1.11 respectively. The case
$\overline{x}\subsetneq\widetilde{y}$ is shown by finding a cell
$x^{\prime}\in K^{\prime}$ included in $\widetilde{y}$ and having
$\overline{x}$ as a face by the following arguments. We have

$\overline{x}^{\partial
K}\subsetneq(\overline{y}^{K}\cup\overline{y}^{\partial K})$

which in particular implies $\overline{x}^{\partial
K}\subset\overline{y}^{\partial K}.$ This in turn implies that
$\overline{y}^{\partial K}\neq\emptyset$ and thus $y\in\partial K.$ If
$\overline{x}\subsetneq\overline{y}$ then the existence of $x^{\prime}$ is
given by Axiom ii) for $\overline{\partial K}.$ If $\overline{x}=\overline{y}$
then $x=y$ and since
$\operatorname{rk}_{K^{\prime}}(\widetilde{y})=\operatorname{rk}_{\overline{\partial
K}}(\overline{y})+1$ then $\overline{x}$ is a face of $\widetilde{y}$ and
$x^{\prime}=\widetilde{y}$ satisfies the desired conditions. The diamond
property iv) is proven in a similar way. The only non-direct case to check is
when $\overline{x}\in\overline{\partial K},$ $\widetilde{y}\in\widetilde{K},$
$\overline{x}\subset\widetilde{y}$ and

$\operatorname{rk}_{K^{\prime}}(\overline{x})=\operatorname{rk}_{K^{\prime}}(\widetilde{y})-2.$
(18)

By the same argument as before, $y\in\partial K$ and (18) implies that $y$ is
a face of $x.$ We also have that $\widetilde{x}$ is a coface of $\overline{x}$
and a face of $\widetilde{y}$ by (18). Since $y\subset x$ this implies that

$\\{\overline{y},\widetilde{x}\\}\subset\bigtriangledown\left(\overline{x}\right)\cap\triangle\left(\widetilde{y}\right).$
(19)

If
$z\in\bigtriangledown\left(\overline{x}\right)\cap\triangle\left(\widetilde{y}\right)$
then either $z\in\overline{\partial K}$ and

$\overline{x}\subsetneq z\subset\overline{y}$

which implies $z=\overline{y}$ or $z\in\widetilde{K}$ and

$\widetilde{x}\subset z\subsetneq\widetilde{y}$

which implies $z=\widetilde{x}.$ Therefore the inclusion in (19) is in fact an
equality and this shows Axiom iv) as well as that $K^{\prime}$ is a cc.

We then prove that $K^{\prime}$ is cell-connected. $K^{\prime}$ is graph-based
since the non-singularity of $K$ implies that the edges of $K^{\prime}$ are
either composed of two maximal cells of $K$ or of one maximal cell $y$ of
$\partial K$ and the unique maximal cell of $K$ containing $y.$ The cc
$K^{\prime}$ is cell-connected by the following argument. If
$x^{\prime}\in\widetilde{K}$ is a cell of the form
$x^{\prime}=\overline{x}^{K}$ where $x\in K\setminus\partial K$ then
$x^{\prime}$ is connected since $K$ is non-pinching. If
$x^{\prime}=\widetilde{x}$ where $x\in\partial K$ then $x^{\prime}$ is
connected if and only if there is and edge $e^{\prime}$ of $K^{\prime}$ with a
vertex in $\overline{x}^{K}$ and in $\overline{x}^{\partial K},$ as both of
these sets are connected since $K$ is non-pinching. Let $y$ be a maximal cell
of $\partial K$ containing $x.$ The cell $y$ is then included in a maximal
cell $z$ of $K,$ we can then take $e^{\prime}=\\{y,z\\}.$ Cells in
$\overline{\partial K}$ are connected since $\partial K$ is non-pinching and
this shows that $K^{\prime}$ is cell-connected.

The cc $K^{\prime}$ is connected by the following argument. If
$v_{1}^{\prime},v_{2}^{\prime}$ are vertices of $K^{\prime}$ then by purity of
$K$ these vertices are included in maximal cells $z_{1}^{\prime}$ and
$z_{2}^{\prime}$ of $\widetilde{K}.$ These maximal cells are duals to vertices
$v_{1},v_{2}$ of $K$ which are connected by a path in $K^{(1)}.$ This path
provides a finite sequence of maximal cells of $\widetilde{K}$ starting with
$z_{1}^{\prime}$ and ending with $z_{2}^{\prime}$ such that each element in
the sequence shares a sub-maximal cell with the next element in the sequence
and since the cells in $\widetilde{K}$ are connected this ensures that there
is a path linking $v_{1}^{\prime}$ to $v_{2}^{\prime}$ in $\widetilde{K}.$

The following arguments show that $K^{\prime}$ is a non-pinching non-singular
cell complex of rank $R.$ In order to show that $K^{\prime}$ is non-pinching,
we first note that $\partial K^{\prime}=\overline{\partial K}$ therefore there
is no $\partial K^{\prime}$-pinch in $K^{\prime}$ since $\partial K$ is cell-
connected. Let us assume by contradiction that there is a $K^{\prime}$-pinch
in $K^{\prime},$ i.e. a cell $x^{\prime}\in K^{\prime}$ such that
$\overline{K^{\prime}}^{(1)}\cap\overline{x^{\prime}}$ is not connected. If we
take $x\in K$ such that $\widetilde{x}=x^{\prime},$ we can see that a path
between two vertices $v,w$ in $x$ corresponds to a path between two vertices
$\overline{\widetilde{v}}$ and $\overline{\widetilde{w}}$ in
$\overline{K^{\prime}}^{(1)}\cap\overline{x^{\prime}}.$ Therefore if
$\overline{K^{\prime}}^{(1)}\cap\overline{x^{\prime}}$ is not connected then
so is $K^{(1)}\cap x,$ which contradicts the assumption that $K$ is cell-
connected. Every cell in $K^{\prime}$ is included in a $R$-cell since this is
clear for the cells in $\widetilde{K}$ and every $(R-1)$-cell
$\overline{v}\in\overline{\partial K}$ is contained in
$\widetilde{v}\in(K^{\prime})^{[R]},$ which is then pure of dimension $R.$
Finally $K^{\prime}$ is also non-singular since $K$ is graph-based.

We then turn to the point 1b. Proving this statement amounts to verify that
all the properties shown for the point 1a are still valid when restricted to
cells not included in the sets $\widetilde{J_{0}}\sqcup\overline{J_{0}}.$
Axioms i), ii) and iv) are simply a consequence of the cc structure of
$K^{\prime}$ and the fact that whenever $x^{\prime},y^{\prime}\in
K^{\prime}_{0}$ are two cells such that $x^{\prime}\subset y^{\prime}$ and
$z^{\prime}\in K^{\prime}$ satisfies $x^{\prime}\subset z^{\prime}\subset
y^{\prime}$ then $z^{\prime}\in K^{\prime}_{0}.$ Axiom iii) is also verified
for $K^{\prime}_{0}$ since if $x^{\prime},y^{\prime}\in K^{\prime}_{0}$ then
$x^{\prime}\cap y^{\prime}$ contains no maximal cell of $J_{0}$ and therefore
cannot be included in $\widetilde{J_{0}}\sqcup\overline{J_{0}}.$ We have that
$K^{\prime}_{0}$ is clearly graph-based since so is $K^{\prime}$ and the set
of edges of $K^{\prime}_{0}$ is a subset of the set of edges of $K^{\prime}.$
Also $K_{0}^{\prime}$ is cell-connected since so is $K^{\prime}.$ The purity
of $K^{\prime}_{0}$ is verified if we can show that for every cell
$x^{\prime}$ of $K^{\prime}_{0}$ contained in a maximal cell of $K^{\prime}$
of the form $\widetilde{v}$ where $v\in J_{0}^{[0]},$ we have that
$x^{\prime}$ is also contained in a maximal cell of the form $\widetilde{w},$
where $w\in K^{[0]}\setminus(J_{0}^{[0]}\sqcup J^{[0]}).$ This is ensured by
the non-degeneracy of $(K,J_{0}\sqcup J)$ together with the connectedness of
the cells in $K_{0}^{\prime}.$ The cc $K^{\prime}_{0}$ is also non-singular
since the set of $(R-1)$-cells of $K_{0}^{\prime}$ is a subset of the set of
$(R-1)$-cells of $K^{\prime}$ and their set of cofaces in $K_{0}^{\prime}$ is
identical to their set of cofaces in $K^{\prime}.$

Now comes a key point of this proof where the locality of $(K,J)$ enters
crucially: showing that $K^{\prime}_{0}$ is non-pinching. First we can see
that there is no $K^{\prime}_{0}$-pinch in $K_{0}^{\prime}.$ Indeed
$K^{\prime}$ has no $K^{\prime}$-pinch and
$\mathcal{G}_{\overline{K^{\prime}_{0}}}\cap\widetilde{x^{\prime}}=\mathcal{G}_{\overline{K^{\prime}}}\cap\widetilde{x^{\prime}}$
for all $x^{\prime}\in K^{\prime}_{0}$ since if two maximal cells of
$K^{\prime}_{0}$ contain $x^{\prime}$ and share a sub-maximal cell
$y^{\prime}\in K^{\prime}$ then $y^{\prime}$ is also in $K^{\prime}_{0}.$ It
is therefore sufficient to show that $\partial K^{\prime}_{0}$ contains no
$\partial K^{\prime}_{0}$-pinch. For this we first prove the following
relation:

$\partial K_{0}^{\prime}=\overline{\partial K\setminus(J\sqcup
J_{0})}\sqcup\overline{K_{J}}\sqcup\overline{K_{J_{0}}}.$ (20)

In order to prove (20) we first observe that $\overline{\partial
K\setminus(J\sqcup J_{0})}\leq\partial K_{0}^{\prime}$ as a consequence of the
following: every vertex $v\in(\partial K\setminus(J\sqcup J_{0})^{[0]}$
satisfies that $y^{\prime}:=\overline{v}^{\partial K}\in(\partial
K\setminus(J\sqcup J_{0}))^{[R-1]}$ and $\widetilde{v}$ is the only $R$-cell
of $K^{\prime}$ containing $y^{\prime}.$ Since $(K,J\sqcup J_{0})$ is non-
singular, every $(R-1)$-cell of $K_{0}^{\prime}$ not in $\overline{\partial
K\setminus(J\sqcup J_{0})}$ is of the form $y^{\prime}=\overline{e}^{K}$ for
some $e\in K^{[1]}$ such that $\left|e\cap(J\sqcup J_{0})^{[0]}\right|\leq 1.$
Hence $\left|\bigtriangledown\left(y^{\prime}\right)\right|=1$ if and only if
$e\in K_{(J\sqcup J_{0})}$ and this proves (20).

We can now turn back to showing that there is no $\partial
K^{\prime}_{0}$-pinch on $\partial K^{\prime}_{0}.$ Since we assume that
$K^{\prime}\in\mathsf{C},$ the relation (20) implies that it is sufficient to
prove that $\overline{K_{J_{0}}}$ is non-pinching. By Lemma A.4, we have that
the map $(M_{J_{0}}^{K})\ni\mathcal{E}_{J_{0}}^{x}\mapsto x\in K_{J_{0}}$ is
an isomorphism of posets. By Proposition 1.11,
$(\overline{M_{J_{0}}^{K}},\operatorname{rk}_{\overline{M_{J_{0}}^{K}}})$ is a
closed cc and for all $x\in K_{J_{0}}$ we have

$\operatorname{rk}_{\overline{M_{J_{0}}^{K}}}(\overline{\mathcal{E}_{J_{0}}^{x}})=R-1-\operatorname{rk}_{M_{J_{0}}^{K}}(\mathcal{E}_{J_{0}}^{x})=R-\operatorname{rk}_{K}(x)=\operatorname{rk}_{\overline{K}}(\overline{x})=:\operatorname{rk}_{\overline{K_{J_{0}}}}(\overline{x}).$

Hence we have that
$(\overline{K_{J_{0}}},\operatorname{rk}_{\overline{K_{J_{0}}}})$ is a cc and
$\overline{K_{J_{0}}}\cong\overline{M_{J_{0}}^{K}}.$ Therefore
$\overline{K_{J_{0}}}$ is non-pinching since $M_{J_{0}}^{K}$ is cell-connected
by locality of $(K-J).$ We can then conclude that the point 1 is verified for
all possible $J.$

The point 2 is shown using (20) (with $J_{0}=\emptyset$) and observing that
$J^{\prime}:=\overline{\partial K\setminus J}$ is an element of
$\mathsf{B}^{R-1}_{\sqcup}$ since it is the dual of a union of local non-
pinching closed cell complexes. Finally $\overline{(K-J)}$ is exactly collared
since, using the notation $L=\partial K\setminus J,$ the map

$\\{\\{y,z\\}\leavevmode\nobreak\ |\leavevmode\nobreak\ L^{[R-1]}\ni y\subset
z\in K^{[R]}\\}\ni\\{y,z\\}\longmapsto y\in L^{[R-1]}$

is a bijection and we have the isomorphism $\cong$ in the following
derivation:

$\displaystyle M_{\overline{L}}^{\widetilde{K\setminus J}\sqcup\overline{L}}$
$\displaystyle=\\{\mathcal{E}_{\overline{L}}^{\widetilde{x}}\leavevmode\nobreak\
|\leavevmode\nobreak\ x\in L\\}$
$\displaystyle=\\{\\{\overline{e}\in\widetilde{K}^{[1]}_{\overline{L}}\leavevmode\nobreak\
|\leavevmode\nobreak\ \overline{e}\subset\widetilde{x}\\}\leavevmode\nobreak\
|\leavevmode\nobreak\ x\in L\\}$
$\displaystyle=\\{\\{\\{y,z\\}\leavevmode\nobreak\ |\leavevmode\nobreak\ y\in
L^{[R-1]},\leavevmode\nobreak\ z\in K^{[R]}_{L}\leavevmode\nobreak\
:\leavevmode\nobreak\ x\subset y\subset z\\}\leavevmode\nobreak\
|\leavevmode\nobreak\ x\in L\\}$
$\displaystyle\cong\\{\overline{x}^{L}\leavevmode\nobreak\
|\leavevmode\nobreak\ x\in L\\}=\overline{L}.$

∎

##### Acknowledgements

The author was supported by the ERC SG CONSTAMIS.

## References

  * [ADJ97] J. Ambjørn, B. Durhuus, and T. Jonsson.  Quantum geometry: a statistical field theory approach.  Cambridge Monographs on Mathematical Physics. Cambridge University Press, Cambridge, 1997.
  * [Asc96] M. Aschbacher.  Combinatorial cell complexes.  In Progress in algebraic combinatorics (Fukuoka, 1993), volume 24 of Adv. Stud. Pure Math., pages 1–80. Math. Soc. Japan, Tokyo, 1996.
  * [Ati88] M. Atiyah.  Topological quantum field theories.  Inst. Hautes Études Sci. Publ. Math., (68):175–186 (1989), 1988\.
  * [Bas10] T. Basak.  Combinatorial cell complexes and Poincaré duality.  Geom. Dedicata, 147:357–387, 2010.
  * [Bry02] J. L. Bryant.  Piecewise linear topology.  In Handbook of geometric topology, pages 219–259. North-Holland, Amsterdam, 2002.
  * [BS10] J. Baez and M. Stay.  Physics, topology, logic and computation: A rosetta stone.  In New Structures for Physics, pages 95–172. Springer Berlin Heidelberg, 2010.
  * [BvN36] G. Birkhoff and J. von Neumann.  The logic of quantum mechanics.  Ann. of Math. (2), 37(4):823–843, 1936.
  * [CG01] M. L. Dalla Chiara and R. Giuntini.  Quantum logic, 2001.
  * [DJ15] B. Durhuus and T. Jonsson.  Exponential bounds on the number of causal triangulations.  Comm. Math. Phys., 340(1):105–124, 2015.
  * [Fra04] P. Di Francesco.  2d quantum gravity, matrix models and graph combinatorics, 2004.
  * [Fre05] L. Freidel.  Group field theory: An overview.  Int. J. Theor. Phys., 44(10):1769–1783, 2005.
  * [FT19] D. S. Freed and C. Teleman.  Topological dualities in the ising model.  2019\.  To appear in Geometry and Topology.
  * [GR12] R. Gurau and J. P. Ryan.  Colored tensor models—a review.  SIGMA Symmetry Integrability Geom. Methods Appl., 8:Paper 020, 78, 2012.
  * [Jau68] J. M. Jauch.  Foundations of quantum mechanics.  Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1968\.
  * [Lol20] R. Loll.  Quantum gravity from causal dynamical triangulations: a review.  Classical Quantum Gravity, 37(1):013002, 50, 2020.
  * [Mac63] G. W. Mackey.  The mathematical foundations of quantum mechanics: A lecture-note volume.  W. A. Benjamin, Inc., New York-Amsterdam, 1963.
  * [McK64] H. P. McKean, Jr.  Kramers-Wannier duality for the 2-dimensional Ising model as an instance of Poisson’s summation formula.  J. Mathematical Phys., 5:775–776, 1964.
  * [MS02] P. McMullen and E. Schulte.  Abstract regular polytopes, volume 92 of Encyclopedia of Mathematics and its Applications.  Cambridge University Press, Cambridge, 2002.
  * [RS72] C. P. Rourke and B. J. Sanderson.  Introduction to piecewise-linear topology.  Springer-Verlag, New York-Heidelberg, 1972.
  * [Sav21] M. Savoy.  Combinatorial cell complexes: Duality, reconstruction and causal cobordisms.  PhD Thesis, 2021.
  * [Seg04] G. Segal.  The definition of conformal field theory.  In Topology, geometry and quantum field theory, volume 308 of London Math. Soc. Lecture Note Ser., pages 421–577. Cambridge Univ. Press, Cambridge, 2004.
  * [Tho54] R. Thom.  Quelques propriétés globales des variétés différentiables.  Comment. Math. Helv., 28:17–86, 1954.