# Non-Hausdorff Manifolds via Adjunction Spaces

David O’Connell

(Okinawa Institute of Science and Technology
1919-1 Tancha, Onna-son, Okinawa 904-0495, Japan)

###### Abstract

In this paper we will introduce and develop a theory of adjunction spaces
which allows the construction of non-Hausdorff topological manifolds from
standard Hausdorff ones. This is done by adjoining Hausdorff manifolds along
homeomorphic open submanifolds, but leaving the boundaries of these subspaces
unidentified. It is shown that all non-Hausdorff manifolds may be constructed
in this manner. We also provide a partial characterisation of the maximal
Hausdorff submanifolds of a given non-Hausdorff manifold.

###### Contents

  1. 1 Topological Adjunction Spaces
    1. 1.1 The Binary Case
    2. 1.2 Generalised Adjunction Spaces
    3. 1.3 Subsystems and Subspaces
  2. 2 Preservation of Properties
    1. 2.1 Countability
    2. 2.2 Connectedness
    3. 2.3 Compactness and Paracompactness
    4. 2.4 Separation and Metrizability
  3. 3 Non-Hausdorff Manifolds
    1. 3.1 Adjoined Manifolds
    2. 3.2 Non-Hausdorff Manifolds and $H$-Submanifolds
    3. 3.3 Representation of Non-Hausdorff Manifolds
    4. 3.4 Partitions of Unity
  4. 4 Characterisation of Canonical Embeddings
    1. 4.1 Initial Steps in the Characterisation
    2. 4.2 Some Preliminary Results
    3. 4.3 The Binary and Finite Cases
    4. 4.4 The Countable Case
  5. 5 Conclusion
  6. A Appendices
    1. A.1 Proofs from Section 1
      1. A.1.1 Proof of Lemma 1.4
      2. A.1.2 Proof of Lemma 1.6
    2. A.2 Some Results from General Topology
    3. A.3 Proof of Paracompactness Preservation
    4. A.4 Proof of Lemma 4.3

## Introduction

Adjunction spaces are a relatively well-known concept in general topology. The
configuration is fairly simple – take two topological spaces, specify which
regions of both that you would like to glue together, and then identify them.
Typically, this to-be-glued region is taken to be a topologically-closed
subset of one of the component spaces. This is for several reasons, the most
relevant of which being that the closedness of this region directly relates to
the possibility of gluing (topological) manifolds along their boundaries,
giving rise to the connected sum operation.

However, if you instead take this to-be-glued region to be topologically open,
oddities occur in the manifold context – there is no guarantee that gluing two
manifolds along an open region will result in another manifold. In fact, under
certain assumptions, the resulting structure will be a non-Hausdorff manifold.
Intuitively speaking, this is because the gluing may result in “doubled
points” on the boundary of the glued regions.

Surprisingly, there is little to no known literature connecting adjunction
spaces with non-Hausdorff manifolds. In this paper we will take steps towards
remedying this. We will introduce a theory of adjunction spaces that is
sufficiently general as to serve as a formal foundation to the construction of
non-Hausdorff manifolds. Amongst many other things, we will see that gluing
two Hausdorff manifolds will indeed result in a non-Hausdorff manifold, and
moreover every non-Hausdorff manifold arises in this manner.111This is a
recent result of Placek and Luc, see [1] for their proof.

This paper is structured as follows: in the first section we will introduce
the standard theory of adjunction spaces, and generalise it to the setting of
gluing arbitrarily-many topological spaces together. In Section 2, we will
further study these generalised adjunction spaces by asking which properties
from Topology are preserved in the gluing process. In the next section, we
turn our eye to the setting of adjoining manifolds. We will combine the
results of Section 2 with the currently-existing literature of non-Hausdorff
manifolds. Finally, in Section 4 we will study and partially characterise the
Hausdorff submanifolds that a non-Hausdorff might possess.

Throughout this paper we will assume basic familiarity with General Topology,
i.e. familiarity up to the undergraduate level, though we will remind the
reader of relevant facts where necessary.

##  1 Topological Adjunction Spaces

In this section we introduce a generalised approach to adjunctions of
topological spaces. We will identify conditions under which arbitrarily-many
topological spaces can be glued together to form a new, well-defined space.
Moreover, we will discuss conditions under which the component spaces can be
embedded into the glued space. The focus is mainly on the situation in which
topological spaces are glued along open subspaces, since this will be an
important precursor to our discussions of non-Hausdorff manifolds in Section
3.

For the sake of pedagogy, we will first recall the standard notion of an
adjunction space, that is, the case in which two topological spaces are glued
together. Throughout this paper, we will refer to these as binary adjunction
spaces. After a brief analysis of this binary instance of adjunctions, we will
move on to generalise to the setting of arbitrarily-many spaces.

###  1.1 The Binary Case

The idea behind a binary adjunction space is fairly simple – we take two
topological spaces, say $X$ and $Y$, and specify the parts which are to be
glued to each other. The specification is made formal by using a subspace $A$
of $X$ and a continuous map $f:A\rightarrow Y$. The intuition is depicted in
Figure 1, and the formal definition is as follows.

######  Definition 1.1.

Let $X,Y$ be topological spaces and $A\subseteq X$. Given a continuous
function $f:A\rightarrow Y$, we define the adjunction space of $X$ and $Y$
under $f$ as:

$X\cup_{f}Y:=\faktor{X\sqcup Y}{\sim}$

where $\sim$ is the reflexive transitive closure of the relation that
identifies $(a,1)$ and $(f(a),2)$ for each $a\in A$. The topology on
$X\cup_{f}Y$ induced from the quotient of the disjoint union topology on
$X\sqcup Y$ is called the adjunct topology, and we will denote this topology
by $\tau_{A}$.

We will casually refer to the spaces $X$ and $Y$ as component spaces of
$X\cup_{f}Y$, and we will denote the points in $X\cup_{f}Y$ by $[z,Z]$, where
$Z$ is a component space, and $z$ is a point in $Z$. It is worth explictly
describing the nature of these points.

######  Remark 1.2.

By construction, the elements of the adjunction space are equivalence classes,
of which there are four possible types:

  * •

$[x,1]=\\{(x,1)\\}$ if $x$ lies in $X\backslash A$,

  * •

$[y,2]=\\{(y,2)\\}$ if $y$ lies in $Y\backslash f(A)$,

  * •

$[a,1]=\\{(f(a),2)\\}\cup\\{(a^{\prime},1)\ |\ a^{\prime}\in f^{-1}(f(a))\\}$
for each $a$ in $A$, and

  * •

$[y,2]=\\{(y,2)\\}\cup\\{(a,1)\ |\ a\in f^{-1}(y)\\}$ for each $y$ in
$f(A)$.222Here we adopt the convention of abbreviating the equivalence class
$[(x,\cdot)]$ to $[x,\cdot]$.

In particular, the equivalence relation in Definition 1.1 ensures that
$[a,1]=[f(a),2]$ for all $a$ in $A$.

We also remark that in the case that the subspace $A$ is empty, we recover the
definition of the disjoint union, that is, $X\cup_{f}Y=X\sqcup Y$.

Observe that Definition 1.1 closely resembles the standard definition of
adjunction spaces found in, say Munkres [2, pg. 224], except that now we drop
the requirement that $A$ be a closed subset of $X$.

$X$$Y$$A$$f(A)$$X\backslash Y$$Y\backslash X$$f$ Figure 1: The configuration
of a binary adjunction space

Categorically speaking, the construction of $X\cup_{f}Y$ can be described with
the following diagram,

${A}$${X}$${Y}$${X\sqcup
Y}$${X\cup_{f}Y}$$\scriptstyle{f}$$\scriptstyle{\iota_{A}}$$\scriptstyle{\varphi_{1}}$$\scriptstyle{\phi_{X}}$$\scriptstyle{\varphi_{2}}$$\scriptstyle{\phi_{Y}}$$\scriptstyle{q}$

where the maps $\phi_{X}$ and $\phi_{Y}$ emerge as the compositions of the
canonical inclusions $\varphi_{i}$ and the quotient map $q$ associated to the
equivalence relation $\sim$, that is, $\phi_{X}=q\circ\varphi_{1}$ and
$\phi_{Y}=q\circ\varphi_{2}$. Throughout this paper, we will refer to the maps
$\phi_{X}$ and $\phi_{Y}$ as canonical maps. We may abbreviate the above with
the following (commutative) diagram.

${A}$${X}$${Y}$${X\cup_{f}Y}$$\scriptstyle{\iota_{A}}$$\scriptstyle{f}$$\scriptstyle{\phi_{X}}$$\scriptstyle{\phi_{Y}}$

We also make the following useful observation, which is an immediate
consequence of Definition 1.1.

######  Proposition 1.3.

Let $U$ be a subset of $X\cup_{f}Y$. Then $U$ is open in $X\cup_{f}Y$ iff both
$\phi^{-1}_{X}(U)$ and $\phi^{-1}_{Y}(U)$ are open in their respective spaces.

The above result can then be used to show that the adjunction space
$X\cup_{f}Y$ is the pushout of the diagram

${A}$${X}$${Y}$$\scriptstyle{f}$$\scriptstyle{\iota_{A}}$

in the category TOP. The following result confirms this fact.

######  Lemma 1.4.

Let $Z$ be a topological space in which $\psi_{X}:X\rightarrow Z$ and
$\psi_{Y}:Y\rightarrow Z$ are continuous maps such that
$\psi_{X}(a)=\psi_{Y}\circ f(a)$ for each $a\in A$. Then there is a unique
continuous map $g:X\cup_{f}Y\rightarrow Z$ that makes the following diagram
commute.

${A}$${X}$${Y}$${X\cup_{f}Y}$${Z}$$\scriptstyle{f}$$\scriptstyle{\iota_{A}}$$\scriptstyle{\phi_{X}}$$\scriptstyle{\psi_{X}}$$\scriptstyle{\phi_{Y}}$$\scriptstyle{\psi_{Y}}$$\scriptstyle{g}$

###### Proof.

We define $g$ by:

$g([x,i])=\begin{cases}\psi_{X}(x)&\textrm{if}\ i=1\\\
\psi_{Y}(x)&\textrm{if}\ i=2\end{cases}.$

Observe first that the map $g$ is well-defined: if $[x,1]=[y,2]$, then
$f(x)=y$ and thus:

$g([x,1])=\psi_{X}(x)=\psi_{Y}\circ f(x)=\psi_{Y}(y)=g([y,2]).$

We now show that $g$ is continuous. Let $U\subseteq Z$ be open and consider
the preimage $g^{-1}(U)$. According to Prop. 1.3, it suffices to show that
both $\phi_{X}^{-1}(g^{-1}(U))$ and $\phi_{Y}^{-1}(g^{-1}(U))$ are open
subsets of $X$ and $Y$, respectively. It is not hard to see that
$\phi_{X}^{-1}(g^{-1}(U))=\psi_{X}^{-1}(U)$, which is open in $X$ since we
assumed $\psi_{X}$ was continuous. A similar situation holds for $Y$, from
which we can conclude that $g^{-1}(U)$ is open in $X\cup_{f}Y$, and thus $g$
is continuous.

To see that $g$ is unique, suppose we have some other continuous map
$g^{\prime}:X\cup_{f}Y\rightarrow Z$ such that
$\psi_{X}=g^{\prime}\circ\phi_{X}$ and $\psi_{Y}=g^{\prime}\circ\phi_{Y}$.
Consider an element $[x,1]$ in $X\cup_{f}Y$. Then
$g([x,1])=g\circ\phi_{X}(x)=\psi_{X}(x)=g^{\prime}\circ\phi_{X}(x)=g^{\prime}([x,1])$
and thus $g([x,1])=g^{\prime}([x,1])$ for all $x\in X$. The argument for $Y$
is similar, and from this we can conclude that $g=g^{\prime}$. ∎

It will be useful to know under what circumstances $X$ and $Y$ topologically
embed into $X\cup_{f}Y$ under the canonical maps $\phi_{X}$ and
$\phi_{Y}$.333Recall that a topological embedding is an injective, continuous
map between topological spaces that acts as a homeomorphism once restricted to
its image. The following lemma summarises the situation.444Note that in Lemma
1.5 we have omitted certain well-known results for the case when $A$ is
assumed to be closed, since these are not useful for the purposes of this
paper.

######  Lemma 1.5 (Basic facts about canonical maps).

Let $X\cup_{f}Y$ be an adjunction space.

  1. 1.

$\phi_{Y}$ is always an injection.

  2. 2.

If $f$ is an injection, then so is $\phi_{X}$.

  3. 3.

If $A$ is open, then $\phi_{Y}$ is an open map.

  4. 4.

If $f$ is injective and open map, then $\phi_{X}$ is open.

  5. 5.

$\phi_{Y}$ is always a topological embedding.

  6. 6.

If $f$ is an injective, open map, then $\phi_{X}$ is a topological embedding.

###### Proof.

See Appendix A1.1. ∎

We can combine the above results to obtain the followsing useful corollary.

######  Corollary 1.6.

If $A$ is an open subset of $X$ and $f$ is an injective, open map, then the
canonical maps $\phi_{X}$ and $\phi_{Y}$ are both open topological embeddings.

The above Corollary suggests that the canonical map $\phi_{X}$ will be as
structure-preserving as $f$ is. This agrees with the correct intuition
regarding binary adjunction spaces – by construction it is the space $X$ that
you are gluing to $Y$, and it may be the case that some features of the to-be-
glued region $A$ are spoiled in the process. Indeed, as an extreme example, we
could glue all of $A$ to a single point in $Y$. On the other hand, if $f$ is a
topological embedding, the entire structure of $A$ will be preserved in the
gluing process, and thus so is $X$.

It will be useful to know whether a given topological property is preserved
under the adjunction construction, and specifically how much is required in
order to ensure preservation. The following lemma teases some results
regarding the preservation of various properties. We state these without
proof, since they follow from basic facts about general topology, and the
general versions of these results are proved in Section 2.

######  Lemma 1.7.

Let $X\cup_{f}Y$ be an adjunction space.

  1. 1.

If $\mathcal{B}^{X}$ and $\mathcal{B}^{Y}$ are bases for $X$ and $Y$
respectively, and $\phi_{X}$ and $\phi_{Y}$ are open maps, then the collection

$\mathcal{B}=\\{\phi_{X}(U)\ |\ U\in\mathcal{B}^{X}\\}\cup\\{\phi_{Y}(V)\ |\
V\in\mathcal{B}^{Y}\\}$

forms a basis for the topology $\tau_{A}$.

  2. 2.

If $X$ and $Y$ are connected and $A$ is non-empty, then $X\cup_{f}Y$ is
connected.

  3. 3.

If $X$ and $Y$ are compact, then so is $X\cup_{f}Y$.

###  1.2 Generalised Adjunction Spaces

We will now set about generalising the binary adjunction spaces to the setting
in which multiple spaces can be adjoined. It is reasonable to suggest that one
can glue together several topological spaces by successively performing the
binary construction. However, we will skip this inductive construction for now
and instead consider the case in which multiple topological spaces are glued
together simultaneously.

As a motivating example, suppose that we have three topological spaces
$X_{1},X_{2}$ and $X_{3}$ that we would like to adjoin. In order to join all
three spaces together, we need information telling us how to glue the spaces
pairwise, namely for all $i,j=1,2,3$, we need a collection of subsets $A_{ij}$
of $X_{i}$ and attaching maps $f_{ij}:A_{ij}\rightarrow X_{j}$. Drawing
inspiration from Definition 1.1, we would like there to be some topological
space, which we will tentatively denote by $\bigcup_{\mathcal{F}}X_{i}$, that
quotients the disjoint union $\bigsqcup_{i=1}^{3}X_{i}$ via some relation
which identifies each $A_{ij}$ their image under the associated gluing map
$f_{ij}$. Following Lemma 1.4, we would again like the resulting space to be
the colimit (i.e. the generalised pushout) of the diagram formed from $X_{i}$,
$A_{ij}$ and $f_{ij}$. The case for three space is depicted below.

${X_{1}}$${A_{12}}$${A_{31}}$${\bigcup_{\mathcal{F}}X_{i}}$${X_{2}}$${X_{3}}$${A_{23}}$$\scriptstyle{\phi_{1}}$$\scriptstyle{f_{12}}$$\scriptstyle{\iota_{A_{12}}}$$\scriptstyle{\iota_{A_{31}}}$$\scriptstyle{f_{31}}$$\scriptstyle{\phi_{2}}$$\scriptstyle{\phi_{3}}$$\scriptstyle{f_{23}}$$\scriptstyle{\iota_{A_{23}}}$

This can be generalised by fixing some index set $I$ to enumerate the
component spaces that we would like to glue together, and defining a triple of
sets $\mathcal{F}:=(\textbf{X},\textbf{A},\textbf{F})$, where:

  * •

the set X is a collection of topological spaces $X_{i}$,

  * •

the set A is a collection of sets $A_{ij}$ such that $A_{ij}\subseteq X_{i}$
for all $j\in I$, and

  * •

the set F is a collection of continuous maps $f_{ij}:A_{ij}\rightarrow X_{j}$.

We will now define what it means for such an $\mathcal{F}$ to be an adjunction
system.

######  Definition 1.8.

The tuple $\mathcal{F}=(\textbf{X},\textbf{A},\textbf{F})$, is called an
adjunction system if it satisfies the following conditions for all $i,j\in I$.

  1. A1)

$A_{ii}=X_{i}$ and $f_{ii}=id_{X_{i}}$

  2. A2)

$A_{ji}=f_{ij}(A_{ij})$, and $f_{ij}^{-1}=f_{ji}$

  3. A3)

$f_{ik}(a)=f_{jk}\circ f_{ij}(a)$ for each $a\in A_{ij}\cap A_{ik}$.

Spelled out, the conditions A1 to A3 amount to guaranteeing that

  1. 1.

we glue all of $X_{i}$ to itself in a trivial manner,

  2. 2.

if we glue $X_{i}$ to some $X_{j}$ along the subspace $A_{ij}$, then we also
glue $X_{j}$ to $X_{i}$ along the subspace $f_{ij}(A_{ij})$, and

  3. 3.

if we glue $X_{i}$ to some $X_{j}$ along $A_{ij}$, and we then glue $X_{j}$ to
some $X_{k}$ along (part of) the subspace $f_{ij}(A_{ij})$, then we also
$X_{i}$ to $X_{k}$ along $A_{ik}$,

respectively.555The conditions listed above are nothing new – they resemble
familiar gluing constructions from differential geometry such as the
(re)construction of fibre bundles from local trivialisations. We remark that
condition A2 ensures that each $f_{ij}$ is a homeomorphism from $A_{ij}$ to
$A_{ji}$.

Given an adjunction system $\mathcal{F}$, we can then define the adjunction
space subordinate to $\mathcal{F}$, denoted $\bigcup_{\mathcal{F}}X_{i}$, as
the topological space obtained from quotienting the disjoint union

$\bigsqcup_{i}X_{i}:=\\{(x,i)\ |\ x\in X_{i}\\}$

under the relation $\cong$, where $(x,i)\cong(y,j)$ iff $f_{ij}(x)=y$. In
fact, as the following result confirms, the conditions of Definition 1.8 are
precisely what is needed to ensure the equivalence relation $\cong$ is well-
defined.

######  Proposition 1.9.

The relation $\cong$ described above is an equivalence relation.

###### Proof.

Reflexivity and symmetry follow immediately from conditions A1 and A2
respectively. For transitivity, suppose that there are three elements
$(x,i),(y,j)$ and $(z,k)$ such that $(x,i)\cong(y,j)\cong(z,k)$. Then by
definition $f_{ij}(x)=y$ and $f_{jk}(y)=z$. By condition A2, it follows that
$y\in A_{ji}$ and $f_{ji}(y)=x$. Then $y\in A_{ji}\cap A_{jk}$, hence by A3 it
follows that

$z=f_{jk}(y)=f_{ik}\circ f_{ji}(y)=f_{ik}(x),$

thus $(x,i)\cong(z,k)$. ∎

The above result shows that as a set, the adjunction space
$\bigcup_{\mathcal{F}}X_{i}$ is a collection of equivalence classes of the
form

$[x,i]=\\{[y,j]\ |\ f_{ij}(x)=y\\},$

where here we have simplified our notation slightly, by using the index $i$
instead of the component space $X_{i}$. As in the previous section, we define
the topology on $\bigcup_{\mathcal{F}}X_{i}$ to be the quotient topology
induced from the disjoint union topology $\bigsqcup_{i}X_{i}$ by the quotient
map $q$ associated to the equivalence relation $\cong$. We again have a
collection of canonical maps
$\phi_{i}:X_{i}\rightarrow\bigcup_{\mathcal{F}}X_{i}$ defined as the
compositions $q\circ\varphi_{i}$, where $\varphi_{i}$ are the canonical maps
from the component spaces $X_{i}$ into the disjoint union
$\bigsqcup_{i}X_{i}$. There is also a generalisation of Proposition 1.3, which
again follows immediately by construction.

######  Proposition 1.10.

A subset $U$ of $\bigcup_{\mathcal{F}}X_{i}$ is open in the adjunction
topology iff $\phi_{i}^{-1}(U)$ is open in $X_{i}$ for all $i$ in $I$.

We also have the following result, which confirms that the adjunction space is
indeed the colimit of the diagram formed from the adjunction system
$\mathcal{F}$.

######  Lemma 1.11.

Let $\psi_{i}:X_{i}\rightarrow Y$ be a collection of continuous maps from each
$X_{i}$ to some topological space $Y$, such that for every $i,j\in I$ it is
the case that $\psi_{i}(x)=\psi_{j}(f_{ij}(x))$. Then there is a unique
continuous map $g:\bigcup_{\mathcal{F}}X_{i}\rightarrow Y$ and
$\psi_{i}=g\circ\phi_{i}$ for all $i$ in $I$.

###### Proof.

We define the map $g$ by $g([x,i])=\psi_{i}(x)$, that is,
$g=\psi_{i}\circ\phi_{i}^{-1}$. To see that this defines a function, we need
to confirm that $g$ preserves equivalence classes. Suppose that $[x,i]=[y,j]$,
i.e. $x=f_{ij}(y)$. Then:

$g([x,i])=\psi_{i}(x)=\psi_{j}(f_{ij}(x))=\psi_{j}(y)=g([y,j])$

as required. We now show that $g$ is continuous. Let $U$ be open in $Y$, and
consider the set $g^{-1}(U)=\phi_{i}\circ\psi_{i}^{-1}(U)$. Recall the set
$g^{-1}(U)$ is open in $\bigcup_{\mathcal{F}}X_{i}$ iff for each $i\in I$, the
set $\phi_{i}^{-1}(g^{-1}(U))$ is open in $X_{i}$. Observe that:

$\phi_{i}^{-1}(g^{-1}(U))=\phi_{i}^{-1}\circ\phi_{i}\circ\psi_{i}^{-1}(U)=\psi_{i}^{-1}(U)$

which is open since $\psi_{i}$ is continuous. It follows that $g^{-1}(U)$ is
open in $\bigcup_{\mathcal{F}}X_{i}$, and thus $g$ is continuous. To see that
$g$ is unique, we can use a similar argument to that in 1.4. ∎

In contrast to Lemma 1.5, each $\phi_{i}$ is always continuous and injective –
continuity follows immediately from the definition of canonical maps, and
injectivity follows as a consequence of condition A2. Moreover, we also have
the following result, which is an analogue of Corollary 1.6.

######  Lemma 1.12.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space. If each $A_{ij}$ is
an open subset of $X_{i}$, then each $\phi_{i}$ is an open topological
embedding.

###### Proof.

Fix some $\phi_{i}$. By construction $\phi_{i}$ is injective and continuous,
so it suffices to show that $\phi_{i}$ is an open map. So, let $U$ be an open
subset of $X_{i}$, and consider $\phi_{i}(U)$. By Prop. 1.10 this set is open
in $\bigcup_{\mathcal{F}}X_{i}$ iff for every $j\in I$, the preimage
$\phi_{j}^{-1}\circ\phi_{i}(U)$ is open in $X_{j}$. Observe that
$\phi_{j}^{-1}\circ\phi_{i}(U)=f_{ij}(U\cap A_{ij})$.666Indeed – if
$\phi_{j}(y)\in\phi_{i}(U)$, then $\phi_{j}(y)=\phi_{i}(u)$ for some $u\in U$.
Thus $[y,j]=[u,i]$, hence $(y,j)\cong(u,i)$, and $f_{ij}(u)=y$. Thus $y\in
f_{ij}(U\cap A_{ij})$. Conversely, if $y\in f_{ij}(U\cap A_{ij})$ then there
is some $u\in U\cap A_{ij}$ such that $f_{ij}(u)=y$. Hence $(y,j)\cong(u,i)$,
so $[y,j]=[u,i]$ and thus $\phi_{j}(y)=\phi_{i}(u)$, i.e.
$y\in\phi_{j}^{-1}\circ\phi_{i}(u)$. Since $U$ is open in $X_{i}$, the set
$U\cap A_{ij}$ is open in $A_{ij}$ (equipped with the subspace topology). By
construction $f_{ij}:A_{ij}\rightarrow X_{j}$ is a topological embedding. As
such, $f_{ij}:A_{ij}\rightarrow A_{ji}$ is a homeomorphism, thus the set
$f_{ij}(U\cap A_{ij})$ is open in $A_{ji}$ (again equipped with the subspace
topology). By assumption $A_{ji}$ is an open subspace, thus $f_{ij}(U\cap
A_{ij})$ is also open in $X_{j}$.777Recall that in an open subspace $U$ of
$X$, $V\subset U$ is open in $A$ iff $V=A\cap W$ where $W$ and $A$ are open in
$X$, so $V$ is the finite intersection of sets open in $X$, thus is also open
in $X$. It follows that $\phi_{j}^{-1}\circ\phi_{i}(U)$ is open in $X_{j}$ for
all $j$. Since $U$ was arbitrary, we may conclude that $\phi_{i}$ is an open
map. The result then follows from the fact that every continuous, open
injective map is an open topological embedding. ∎

We finish this section by introducing our first generalised adjunction space,
which will serve as an illustrative example going forward.

$\mathbb{R}_{n}$$\mathbb{R}_{3}$$\mathbb{R}_{2}$$\mathbb{R}_{1}$$\vdots$$0$$0$$0$$0$$\vdots$$[0,i]$
Figure 2: The construction of the $n$-branched real line.

######  Example 1.13 (The $n$-branched real line).

Take $I$ to be an indexing set of size $n$, and consider the triple
$\mathcal{F}=(\textbf{X},\textbf{A},\textbf{f})$, where:

  * •

each $X_{i}$ equal to a copy of the real line equipped with the standard
topology,

  * •

each $A_{ij}$ equal to the set $(-\infty,0)$ for all $i,j$ distinct, and

  * •

each $f_{ij}:A_{ij}\rightarrow X_{j}$ is the identity map.

Once we require that each $A_{ii}$ equals $X_{i}$ empty set and each $f_{ii}$
equals the identity map, the triple $\mathcal{F}$ forms an adjunction system.
The adjunction space subordinate to $\mathcal{F}$ will be a collection of
$n$-many copies of the real line all glued to each other along the negative
numbers, as pictured in Figure 2. Observe that the origins of each $X_{i}$ are
not identified, thus there are $n$-many distinct equivalence classes $[0,i]$
in $\bigcup_{\mathcal{F}}\mathbb{R}_{i}$.

###  1.3 Subsystems and Subspaces

There is a natural sense in which subspaces can be defined. Suppose that we
have an adjunction space $\bigcup_{\mathcal{F}}X_{i}$ built from the
adjunction system $\mathcal{F}$. It follows immediately from Definition 1.8
that any subset $J$ of the underlying indexing set $I$ will yield another
adjunction system $\tilde{\mathcal{F}}$, which can be obtained by simply
forgetting the parts of $\mathcal{F}$ that are not contained in $J\subset I$.
As such, there is an associated adjunction space
$\bigcup_{\tilde{\mathcal{F}}}X_{j}$, that consists of gluing only the
component spaces $X_{j}$ where $j$ lies in the subset $J$.

In order to distinguish objects in $\bigcup_{\tilde{\mathcal{F}}}X_{j}$ from
objects in the full adjunction space, we will use the double-bracketed
notation $\llbracket\cdot\rrbracket$ to denote objects in the sub-adjunction
space. In particular, we will use

$\llbracket x,j\rrbracket:=\\{(y,k)\ |\ k\in J\ \wedge\ f_{jk}(x)=y\\}$

to denote the members of $\bigcup_{\tilde{\mathcal{F}}}X_{j}$. Furthermore, we
will denote the canonical maps of the sub-adjunction space by $\chi_{i}$’s.

Conveniently, the universal property of the adjunction space
$\bigcup_{\tilde{\mathcal{F}}}X_{j}$ yields the existence of a continuous map
$g:\bigcup_{\tilde{\mathcal{F}}}X_{j}\rightarrow\bigcup_{\mathcal{F}}X_{i}$,
which will send each equivalence class $\llbracket x,i\rrbracket$ of
$\bigcup_{\tilde{\mathcal{F}}}X_{j}$ into the associated equivalence class
$[x,i]$ of $\bigcup_{\mathcal{F}}X_{i}$. At the level of the topological
spaces $X_{i}$ that are begin glued together, this continuous function $g$
will send each $X_{i}$ of $\mathcal{F}^{\prime}$ into its copy that sits
inside $\bigcup_{\mathcal{F}}X_{i}$. Note that in general this map will not
necessarily preserve the structure of the component spaces, that is, the map
$g$ will not act as a homeomorphism between $\chi_{j}(X_{j})$ and
$\phi_{j}(X_{j})$. That being said, this happens to be the case, provided that
the gluing maps $f_{ij}$ and gluing regions $A_{ij}$ behave as in Lemma 1.12.
Indeed, we have the following result on the matter.

######  Theorem 1.14.

Let $\mathcal{F}$ be an adjunction system in which each gluing region $A_{ij}$
is an open subset of $X_{i}$ and each $f_{ij}:A_{ij}\rightarrow X_{j}$ is an
open topological embedding. For any sub-adjunction system
$\tilde{\mathcal{F}}\subseteq\mathcal{F}$, the continuous function
$g:\bigcup_{\tilde{\mathcal{F}}}X_{j}\rightarrow\bigcup_{\mathcal{F}}X_{i}$
defined by $\llbracket x,i\rrbracket\mapsto[x,i]$ is an open topological
embedding.

###### Proof.

Since we already know that the function $g$ is continuous, in order to prove
the result it suffices to show that $g$ is both injective and open. For
injectivity, let $\llbracket x,j\rrbracket$ and $\llbracket y,k\rrbracket$ be
two distinct points in $\bigcup_{\tilde{\mathcal{F}}}X_{j}$. By construction
the equivalence class $\llbracket x,j\rrbracket$ contains all elements of the
form $(f_{jl}(x),l)$. Since $\llbracket y,k\rrbracket$ is distinct from
$\llbracket x,j\rrbracket$, we may conclude that $f_{ij}(x)$ does not equal
$y$, and thus $[x,j]\neq[y,k]$ as well. This confirms that $g$ is injective.

Suppose now that $U$ is some open subset of the sub-adjunction space
$\bigcup_{\mathcal{F}^{\prime}}X_{j}$. By Prop 1.11, $U$ is open in the
adjunction space iff the preimages under all of the canonical maps are open,
i.e. $\chi_{j}^{-1}(U)$ is open in $X_{j}$. By assumption, all of the
canonical maps $\phi_{i}$ of the larger adjunction space
$\bigcup_{\mathcal{F}}X_{i}$ are open topological embeddings, thus the images
$\phi_{j}\circ\chi_{j}^{-1}(U)$ are open in the spaces $\phi_{j}(X_{j})$.
Since these are open subspaces of $\bigcup_{\mathcal{F}}X_{i}$, we may
conclude that each $\phi_{j}\circ\chi_{j}^{-1}(U)$ is an open subset of
$\bigcup_{\mathcal{F}}X_{i}$. It follows that the union

$\bigcup_{j\in J}\phi_{j}\circ\chi_{j}^{-1}(U)$

is open in $\bigcup_{\mathcal{F}}X_{i}$. However, this set is precisely equal
to the image $g(U)$. Since we chose $U$ arbitrarily, it follows that $g$ is
open and therefore $g$ is an open topological embedding. ∎

##  2 Preservation of Properties

In this section we will survey the types of properties that might be preserved
under the gluing of topological spaces. In some cases, the continuity of the
canonical maps $\phi_{i}$ is enough to preserve certain properties (e.g.
compactness of subsets), though in general more stringent requirements are
needed. Given the scope of this paper, that is, the goal of adjoining smooth
manifolds along homeomorphic open submanifolds, we will pay special attention
to the topological properties surrounding the theory of manifolds. As such, we
will consider the countability axioms, connectedness, compactness, and the
separation axioms.

All definitions in this section pertaining to General Topology are taken from
Munkres’ excellent book [2], unless otherwise specified. Furthermore,
throughout this section we will fix an adjuction system
$\mathcal{F}=(\textbf{X},\textbf{A},\textbf{f})$, with no additional
assumptions on its structure.

###  2.1 Countability

We begin by discussing the prospect of transferring countability properties
from the component spaces $X_{i}$ into the adjunction space
$\bigcup_{\mathcal{F}}X_{i}$ subordinate to $\mathcal{F}$.

Before considering any such properties, we need to first identify conditions
under which bases of the component spaces $X_{i}$ can be “glued” to form a
basis of the adjunction space. It turns out that this cannot be performed
unrestrictedly – in order to ensure that every basis element remains open we
need the additional requirement that every canonical map $\phi_{i}$ be open.

######  Lemma 2.1.

Suppose that for each $i$ in $I$, the collection $\mathcal{B}_{i}$ forms a
basis for $X_{i}$. If each $\phi_{i}$ is an open map, then the collection
$\mathcal{B}=\\{\phi_{i}(B)\ |\ B\in\mathcal{B}_{i}\\}$ forms a basis for
$\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

By assumption each $\phi_{i}(B)$ is open in $\bigcup_{\mathcal{F}}X_{i}$, so
it suffices to show that every open set of $\bigcup_{\mathcal{F}}X_{i}$ can be
represented as union of elements of $\mathcal{B}$. Let $U$ be some open subset
of $\bigcup_{\mathcal{F}}X_{i}$, and consider the subset $\phi_{i}^{-1}(U)$
for some fixed $i$ (which is open in $X_{i}$ since $\phi_{i}$ is continuous).
Since $\mathcal{B}_{i}$ forms a basis for $X_{i}$, the set $\phi_{i}^{-1}(U)$
can be represented as some union of open sets:

$\phi_{i}^{-1}(U)=\bigcup_{\alpha\in\mathcal{A}_{i}}B^{i}_{\alpha},$

where $\mathcal{A}_{i}$ is some $i$-dependent indexing set and each
$B^{i}_{\alpha}\in\mathcal{B}_{i}$. It then follows that:

$\phi_{i}(X_{i})\cap
U=\phi_{i}(\phi_{i}^{-1}(U))=\phi_{i}\left(\bigcup_{\alpha\in\mathcal{A}_{i}}B^{i}_{\alpha}\right)=\bigcup_{\alpha\in\mathcal{A}_{i}}\phi_{i}(B^{i}_{\alpha})$

and, since the $\phi_{i}(X_{i})$ cover $\bigcup_{\mathcal{F}}X_{i}$, we have

$U=U\cap\bigcup_{i\in I}\phi_{i}(X_{i})=\bigcup_{i\in
I}\left(U\cap\phi_{i}(X_{i})\right)=\bigcup_{i\in
I}\bigcup_{\alpha\in\mathcal{A}_{i}}B^{i}_{\alpha},$

that is, $U$ can be represented as a union of elements of $\mathcal{B}$, as
required. ∎

Now that we have discussed the transference of bases from the $X_{i}$ into the
adjunction space $\bigcup_{\mathcal{F}}X_{i}$, we can begin to consider some
countability properties. We start with first-countability.

Recall that a topological space $X$ is said to have a countable basis at a
point $x$ if there is a countable collection $\mathcal{B}$ of open
neighbourhoods of $x$ such that each open set $U$ containing $x$ intersects at
least one of the members of $\mathcal{B}$. We then say that $X$ is first-
countable if every point in $X$ has a countable basis. As one might expect, it
is possible to transfer these countable bases of points in $X_{i}$ into the
adjunction space, provided the canonical maps are open. The following result
confirms this.

######  Lemma 2.2.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which each
$\phi_{i}$ is an open map. If each $X_{i}$ is first-countable, then so is the
adjunction space $\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

Let $[x,i]$ be some arbitrary point in $\bigcup_{\mathcal{F}}X_{i}$. By
assumption, the space $X_{i}$ is first-countable, so there is some countable
collection $\mathcal{B}$ of open neighbourhoods of $x$ such that every open
set $U$ in $X_{i}$ satisfies $U\cap\bigcup\mathcal{B}\neq\emptyset$. We will
now show that the collection $\phi_{i}(\mathcal{B}):=\\{\phi_{i}(B)\ |\
B\in\mathcal{B}\\}$ forms a countable basis at the point $[x,i]$.

Observe first that by assumption, the sets $\phi_{i}(B)$ are all open in the
adjunction space. Now, suppose that $V$ is an open subset of
$\bigcup_{\mathcal{F}}X_{i}$ that contains $[x,i]$. Since $V$ is open, and
$\phi_{i}$ is continuous, the preimage $\phi_{i}^{-1}(V)$ is an open subset of
$X$ that contains $x$. Since $X$ is first-countable, there is some
$B\in\mathcal{B}$ such that $B\cap\phi_{i}^{-1}(V)\neq\emptyset$. Thus

$\phi_{i}\left(B\cap\phi_{i}^{-1}(V)\right)=\phi_{i}(B)\cap\phi_{i}(\phi_{i}^{-1}(V))=\phi_{i}(B)\cap
V$

is non-empty, from which the result follows. ∎

We will now consider the Lindelöf property. Recall that a topological space is
said to be Lindelöf if every open cover has a countable subcover.888Here by
“subcover” we simply mean a subset of the open cover that is still a cover.
Our next result confirms that the Lindelöf property is preserved, assuming a
reasonable restriction on the number of spaces being glued together.

######  Lemma 2.3.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which the indexing
set $I$ is countable. If every $X_{i}$ is Lindelöf, then so is
$\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

Let $\mathcal{U}$ be an open cover of $\bigcup_{\mathcal{F}}X_{i}$. For each
$i$ in $I$, we can use the continuity of the canonical map $\phi_{i}$ to pull
back $\mathcal{U}$ to an open cover

$\phi_{i}^{-1}(\mathcal{U}):=\\{\phi_{i}^{-1}(U)\ |\ U\in\mathcal{U}\\}$

of $X_{i}$. Since by assumption each $X_{i}$ is Lindelöf, it follows that
there is some countable subcover $\mathcal{V}_{i}$ of $\phi_{i}^{-1}(U)$. We
can now consider the collection

$\mathcal{V}:=\bigcup_{i\in I}\\{U\in\mathcal{U}\ |\
\phi_{i}^{-1}(U)\in\mathcal{V}_{i}\\},$

which is clearly a subcover of $\mathcal{U}$. Moreover, since $I$ is
countable, so is $\mathcal{V}$. ∎

Recall that a topological space is said to be separable if it contains a
countable, dense subset. In similar spirit to Lemma 2.3, this property is
preserved in the adjunction process, provided that we only glue together
countably-many spaces.

######  Lemma 2.4.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which the indexing
set $I$ is countable. If each $X_{i}$ is separable, then so is
$\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

For each $i$ in $I$, denote by $Z_{i}$ the countable dense subset of each
$X_{i}$. We show that the set

$Z:=\bigcup_{i\in I}\phi_{i}(Z_{i})$

is a countable dense subset of the adjunction space
$\bigcup_{\mathcal{F}}X_{i}$. This set is manifestly countable, so it suffices
to verify its density. Before doing so, we make the observation that
$\phi_{i}(Z_{i})$ is the continuous image of a dense subset, so is dense in
the subspace $\phi_{i}(X_{i})$.999See Appendix A.2 for the proof of this. In
order to show that $Cl(Z)$ equals the whole adjunction space, pick some
arbitrary element $z$ of $\bigcup_{\mathcal{F}}X_{i}$. Since the subspaces
$\phi_{i}(X_{i})$ cover $\bigcup_{\mathcal{F}}X_{i}$, it must be the case that
$z=[x,i]$ for some $x\in X_{i}$. Let $U$ be any open neighbourhood of $[x,i]$.
Then $U\cap\phi_{i}(X_{i})$ is an open neighbourhood in $\phi_{i}(X_{i})$.
Since $\phi_{i}(Z_{i})$ is a dense subset of $\phi_{i}(X_{i})$, it follows
that $U\cap\phi_{i}(Z_{i})$ is non-empty. Hence $U\cap Z$ is also non-empty,
from which it follows that $z\in Cl(Z)$, and consequently $Z$ is dense in
$\bigcup_{\mathcal{F}}X_{i}$. ∎

We finish this section by discussing the strongest notion of countability that
we will need when discussing topological manifolds: second countability.
Recall that a topological space is called second-countable whenever it has a
countable basis. It is well-known that second-countability implies first-
countability, Lindelöf and separability in the senses introduced
above.101010In fact, one needs to assume at least the axiom of countable
choice in order to prove these implications. We have a corresponding
preservation result for second-countability. As a slight anticlimax, the
result actually follows trivially from Lemma 2.1.

######  Corollary 2.5.

Suppose each $X_{i}$ is second-countable. If $I$ is countable and each
$\phi_{i}$ is an open map, then $\bigcup_{\mathcal{F}}X_{i}$ is also second-
countable.

###### Proof.

We use the collection $\mathcal{B}$ as in Lemma 2.1. Since $I$ is countable,
the basis $\mathcal{B}$ will be a countable union of countable sets, i.e. a
countable basis for $\bigcup_{\mathcal{F}}X_{i}$. ∎

###  2.2 Connectedness

Now that we have covered some of the countability axioms, we will move on to a
discussion connectedness in adjunction spaces. We note that a deeper theory of
connected adjunction spaces is lurking in the background, though we will turn
a blind eye to this and instead focus only on the prospect of preserving
connectedness.

Recall that a topological space $X$ is called connected if there are no
disjoint, non-empty pair of open subsets $U,V$ covering $X$. There are several
equivalent characterisations of connectedness, we will make use of the
following two well-known facts:

  1. 1.

$X$ is connected iff there are no non-empty proper clopen subsets, and

  2. 2.

$X$ is connected iff every continuous function from $X$ to $\\{0,1\\}$
(equipped with the discrete topology) is constant.

By construction, if all of the gluing regions $A_{ij}$ in $\mathcal{F}$ are
empty, then the adjunction space $\bigcup_{\mathcal{F}}X_{i}$ will reduce to
the disjoint union $\bigsqcup_{i\in I}X_{i}$, which is clearly disconnected.
So, even if the component spaces $X_{i}$ are connected to begin with, there is
no guarantee that the resulting adjunction space remains connected. For our
first result, we shall see that a rather strong requirement ensures
preservation of connectedness.

######  Lemma 2.6.

If each $X_{i}$ is connected and each $A_{ij}$ is non-empty, then
$\bigcup_{\mathcal{F}}X_{i}$ is connected.

###### Proof.

Let $\psi:\bigcup_{\mathcal{F}}X_{i}\rightarrow\\{0,1\\}$ be a continuous map,
and suppose towards a contradiction that $\psi$ is not constant, i.e. suppose
there are elements $[x,i]$ and $[y,j]$ of the adjunction space
$\bigcup_{\mathcal{F}}X_{i}$ such that $\psi([x,i])=0$ and $\psi([y,j])=1$.
Since $\phi_{i}$ and $\phi_{j}$ are continuous, it follows that the maps
$\psi_{i}:=\psi\circ\phi_{i}:X_{i}\rightarrow\\{0,1\\}$ and
$\psi_{j}:=\psi\circ\phi_{j}$ are continuous maps. By assumption $X_{i}$ and
$X_{j}$ are both connected, so it follows that $\psi_{i}$ and $\psi_{j}$ are
both constant. Observe that $\psi_{i}(x)=\psi([x,i])=0$, hence it follows that
$\psi_{i}(X_{i})=0$, and by a similar argument, $\psi_{j}(X_{j})=1$. However,
by assumption the set $A_{ij}$ is non-empty. Then for any element $a$ in
$A_{ij}$, we have that:

$\psi_{j}(f_{ij}(a))=\psi\circ\phi_{j}\circ f_{ij}(a)=\psi\circ\phi_{i}(a)=0$

which contradicts $\psi_{j}(X_{j})=1$. We can conclude that the map $\psi$ is
constant, and thus $\bigcup_{\mathcal{F}}X_{i}$ is indeed connected. ∎

The above result is slightly overkill – it is also possible to obtain a
connected adjunction space in which some (but not all) of the gluing regions
$A_{ij}$ are empty – see Figure 3 for the idea. This observation is discussed
more in Section 5.

We will now move on to the setting in which the component spaces are path-
connected. Recall that a path with endpoints in a topological space $X$ is a
continuous map $\gamma:[0,1]\rightarrow X$, where here the unit interval is
endowed with the subspace topology induced from the standard topology on
$\mathbb{R}$. We say that two points $x,y$ in $X$ are path-connected if there
exists a path $\gamma:[0,1]\rightarrow X$ such that $\gamma(0)=x$ and
$\gamma(1)=y$. A topological space $X$ is then called path-connected if any
two points in $X$ can be connected by a path.

With that out of the way, we will now set about proving that path-
connectedness is preserved in the adjunction process (provided again, that
each $A_{ij}$ is non-empty). In order to do so, we first need the following
intermediary result.

######  Proposition 2.7.

Let $X$ be a topological space. If $A$ and $B$ are path-connected subsets of
$X$ such that $A\cap B\neq\emptyset$, then the union $A\cup B$ is also path-
connected.

###### Proof.

Consider two points $a,b$ in $A\cup B$. Without loss of generality, we may
assume that $a\in A$ and $b\in B$ (since in any other case, the result follows
trivially). By assumption the sets $A$ and $B$ are not disjoint, which means
that there is some $x\in A\cap B$. Since $A$ is path-connected, there is a
path $\gamma_{1}$ connected $a$ to $x$. Since $B$ is path-connected, there is
a path $\gamma_{2}$ connecting $x$ to $b$. We can glue these paths
together,111111Specifically, we define the path $\gamma_{3}$ by:
$\gamma_{3}(t)=\begin{cases}\gamma_{1}(2t)&t\in[0,\frac{1}{2}]\\\
\gamma_{2}(2t-1)&t\in[\frac{1}{2},1]\end{cases}$ and observe that $\gamma_{3}$
is well-defined and continuous. to form a path that connects $a$ to $b$
through $x$. ∎

$A_{12}$$X_{1}$$X_{2}$$A_{23}$$X_{3}$ Figure 3: An adjunction space which is
connected, but in which one of the gluing regions (i.e. $A_{13}$) is empty.

We then have the following result, which is a natural continuation of the
previous argument.

######  Lemma 2.8.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which every
$A_{ij}$ is non-empty. If each $X_{i}$ is path-connected, then so is
$\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

Let $[x,i]$ and $[y,j]$ be arbitrary elements of $\bigcup_{\mathcal{F}}X_{i}$.
Since the canonical maps $\phi_{i}$ are continuous, and continuous maps
preserve path-connectedness, it follows that the images $\phi_{i}(X_{i})$ and
$\phi_{j}(X_{j})$ are both path-connected subsets of
$\bigcup_{\mathcal{F}}X_{i}$. Moreover, by assumption we have that $A_{ij}$ is
non-empty, hence the intersection $\phi_{i}(X_{i})\cap\phi_{j}(X_{j})$ is also
non-empty. Thus we may apply Proposition 2.7 to conclude that there is a path
connecting $[x,i]$ to $[y,j]$. Since we chose $[x,i]$ and $[y,j]$, we are
done. ∎

Again, we remark that there most likely exist some weaker conditions that will
still guarantee the preservation of path-connectedness, though this is beyond
the scope of this paper.

###  2.3 Compactness and Paracompactness

We will now consider compactness and paracompactness in adjunction spaces. As
with the case of countability seen in Section 2.1, (para)compactness may be
preserved in the adjunction process provided some restrictions on the size of
the indexing set $I$. We start with compactness.

Recall that a topological space is called compact if every open cover has a
finite subcover. In general, it is not true that the adjunction of infinitely-
many compact spaces is again compact. As a counterexample, consider countably-
many unit intervals $[0,1]$, each glued to the next along its endpoints. The
resulting structure is will be homeomorphic to the real line and thus not
compact. That being said, there is preservation of compactness provided the
indexing set $I$ is finite.

######  Lemma 2.9.

If $I$ is finite and each $X_{i}$ is compact, then so is
$\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

We can use that compactness is preserved in finite disjoint unions and under
quotients, as in Lemma 1.7.3, together with [3, Thm 3.2.3]. ∎

We remark that in general, the canonical maps $\phi_{i}$ are not proper, that
is, the preimages of compact subsets of $\bigcup_{\mathcal{F}}X_{i}$ are not
necessarily compact in the component spaces $X_{i}$. To see this, consider
again the $n$-branched real line as in Example 1.13. The subset $[-1,1]$ is a
compact subset of $\mathbb{R}_{i}$. By the previous lemma, the image
$\phi_{i}([-1,1])$ is compact in the space
$\bigcup_{\mathcal{F}}\mathbb{R}_{i}$. Consider now any $\phi_{j}$ where $j$
is distinct from $i$. Then the preimage of this compact subset is:

$\phi_{j}^{-1}\circ\phi_{i}([-1,1])=f_{ij}([-1,0))=[-1,0)$

which is not compact in $\mathbb{R}_{j}$.

We will now move on to paracompactness. Recall that a topological space $X$ is
called paracompact if for every open cover $\mathcal{U}$ of $X$ there exists
another open cover $\mathcal{V}$ of $X$ that is

  * •

a refinement of $\mathcal{U}$, meaning that each $V$ in $\mathcal{V}$ is
contained in some $U$ in $\mathcal{U}$, and

  * •

locally-finite, meaning that every point $x$ in $X$ has some open
neighbourhood $W_{x}$ that intersects finitely-many members of $\mathcal{V}$.

As with compact spaces, in general it is not true that adjunctions of
paracompact spaces will remain paracompact. As an illustration of this,
consider the $\aleph_{0}$-branched real line, which is constructed
analoguously to the $n$-branched real line of Example 1.13, except that now we
consider the indexing set $I$ to be countably infinite. Any neighbourhood of a
copy $[0,i]$ of the origin will not be locally-finite.

It would be nice if we could obtain a simple argument akin to Lemma 2.9 for
the preservation of paracompactness under finite adjunctions. However, there
is the immediate issue that continuous images of paracompact spaces need not
be paracompact. This means that a more involved argument is needed.
Unfortunately, the case of open adjunction spaces is rather broad, so we will
need to make some additional assumptions on the nature of the gluing process.
In particular, we will need to assume that all of the gluing regions $A_{ij}$
and $A_{ji}$ have homeomorphic boundaries, which will simplify the situation
significantly.

We can use the above lemma to identify some conditions under which
paracompactness can be preserved in the adjunction process. Note that in the
following result (and throughout this paper), we adopt the convention of
indexing the topological closure, interior and boundary with the space that we
are considering, for instance $\partial^{X}(U)$ denotes the $X$-relative
boundary of the subset $U$. With this in mind, we can make explicit what we
mean by two gluing regions to have homeomorphic boundaries. This is done in
the formulation of the following result.

######  Lemma 2.10.

Suppose $\bigcup_{\mathcal{F}}X_{i}$ is an adjunction space in which all
$A_{ij}$ are open, and each gluing map $f_{ij}:A_{ij}\rightarrow A_{ji}$ can
be extended to a homeomorphism $\tilde{f}:Cl^{X_{i}}(A_{ij})\rightarrow
Cl^{X_{j}}(A_{ji})$. Let $\mathcal{U}_{i}$ be a locally-finite open cover of
$X_{i}$. If each $\phi_{i}$ is an open map and the indexing set $I$ is finite,
then the collection

$\mathcal{U}:=\\{\phi_{i}(U)\ |\ U\in\mathcal{U}_{i},\ i\in I\\}$

is a locally-finite open cover of the adjunction space
$\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

(Sketch). Let $[x,i]$ be an arbitrary point in $\bigcup_{\mathcal{F}}X_{i}$.
We would like to construct an open neighbourhood $W$ of $[x,i]$ that
intersects only finitely-many members of $\mathcal{U}$. This is done by
treating each $X_{j}$ separately, and proceeding via a case distinction
depending on where $x$ lies in $X_{i}$: either $x\in A_{ij}$, or $x\in
Int^{X_{i}}(X_{i}\backslash A_{ij})$, or $x\in\partial^{X_{i}}(A_{ij})$. In
any case one can construct an open neighbourhood $W_{j}$ of $[x,i]$ that
intersects only finitely-many of the members of $\mathcal{U}$ that come from
either $\mathcal{U}_{i}$ or $\mathcal{U}_{j}$. We then let $W:=\bigcap_{j\neq
i}W_{j}$ to obtain a witness for the local-finiteness of $\mathcal{U}$ at the
point $[x,i]$. The painstaking details can be found in Appendix A.3. ∎

Using the above result, we can obtain a preservation theorem for
paracompactness in a fairly straightforward manner.

######  Theorem 2.11.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which all $A_{ij}$
are open, and each $f_{ij}:A_{ij}\rightarrow A_{ji}$ can be extended to a
homeomorphism $\tilde{f}:Cl^{X_{i}}(A_{ij})\rightarrow Cl^{X_{j}}(A_{ji})$. If
all the spaces $X_{i}$ are paracompact and $I$ is finite, then
$\bigcup_{\mathcal{F}}X_{i}$ is paracompact.

###### Proof.

Let $\mathcal{U}$ be some open cover of $\bigcup_{\mathcal{F}}X_{i}$. Consider
the cover

$\mathcal{V}:=\bigcup_{i\in I}\\{\phi_{i}(V_{\alpha})\ |\
V_{\alpha}\in\mathcal{V}_{i}\\},$

where $\mathcal{V}_{i}$ is a locally-finite refinement of the open cover
$\phi_{i}^{-1}(\mathcal{U})$ in $X_{i}$. By Lemma 2.10, we have that
$\mathcal{V}$ is a locally-finite cover of $\bigcup_{\mathcal{F}}X_{i}$.

To see that $\mathcal{V}$ is a refinement of $\mathcal{U}$, consider some
arbitrary set $V$ in $\mathcal{V}$. By construction, $V$ must be of the form
$\phi_{i}(V_{\alpha})$ for some $V_{\alpha}$ in some $\mathcal{V}_{i}$. Since
$\mathcal{V}_{i}$ is a refinement of $\phi_{i}^{-1}(\mathcal{U})$, there is
some $\phi_{i}^{-1}(U)$ such that $V_{\alpha}\subseteq\phi_{i}^{-1}(U)$. It
will then be the case that $V=\phi_{i}(V_{\alpha})$ is contained in $U$, as
required. ∎

###  2.4 Separation and Metrizability

We will finish this section by discussing the issue of separation in
adjunction spaces. We focus on the $T_{1}$ and $T_{2}$ axioms, since the non-
preservation of separation is already apparent at the level of $T_{2}$.

Recall that a topological space is called $T_{1}$ if every pair of distinct
points have open neighbourhoods containing one but not the other. For our
first result, we show that the adjunction of open adjunction spaces preserves
the $T_{1}$ property.

######  Lemma 2.12.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space formed from a
collection of $T_{1}$ spaces. If the canonical maps $\phi_{i}$ are all open,
then $\bigcup_{\mathcal{F}}X_{i}$ is $T_{1}$.

###### Proof.

Let $a$ and $b$ be two distinct points in $\bigcup_{\mathcal{F}}X_{i}$. There
are two cases to consider:121212These are indeed the only two possibilities:
if it is not the case that $a,b\in\phi_{i}(X_{i})$, then $a\in\phi_{i}(X_{i})$
implies that $b\notin\phi_{i}(X_{i})$, and $b\in\phi_{j}(X_{j})$ implies that
$a\notin\phi_{j}(X_{j})$.

  1. Case 1:

$a$ and $b$ both lie in $\phi_{i}(X)$ for some $i\in I$. Then $a=[x_{1},i]$
and $b=[x_{2},i]$. By assumption there are two open subsets $U_{1}$ and
$U_{2}$ of $X_{i}$ such that $x_{1}\in U_{1}$ and $x_{1}\notin U_{2}$, and
$x_{2}\in U_{2}$ and $x_{2}\notin U_{1}$. Since $\phi_{X}$ is an open map, the
sets $\phi_{i}(U_{1})$ and $\phi_{i}(U_{2})$ will be open neighbourhoods of
$[x_{1},i]$ and $[x_{2},i]$ respectively. It is still the case that
$[x_{1},i]\in\phi_{i}(U_{1})$ and $[x_{1},i]\notin\phi_{i}(U_{2})$, and
$[x_{2},i]\in\phi_{i}(U_{2})$ and $[x_{2},i]\notin\phi_{i}(U_{1})$.

  2. Case 2:

$a$ lies in $\phi_{i}(X_{i})\backslash\phi_{j}(X_{j})$ and $b$ lies in
$\phi_{j}(X_{j})\backslash\phi_{i}(X_{i})$, or vice-versa. In this case, we
may use the fact that the canonical maps are open and pick the open
neighbourhoods to be $\phi_{X}(X)$ and $\phi_{Y}(Y)$.

In either case, we have the existence of two open sets that separate the
points $a$ and $b$. It follows that $\bigcup_{\mathcal{F}}X_{i}$ is indeed
$T_{1}$. ∎

The next separation axiom is the Hausdorff property. Recall that a topological
space $X$ is called Hausdorff (also known as $T_{2}$) whenever any two
distinct points can be separated by disjoint open sets. Clearly Hausdorffness
implies $T_{1}$.

As previously mentioned, the prospect of Hausdorff preservation is not as
simple as the $T_{1}$ case. Consider the $2$-branched real line (cf. 1.13), as
pictured in Figure 4. In this case, two two copies of the origin are not
identified, and consequently they will violate the Hausdorff property. Indeed
– any open neighbourhoods of these origins will necessarily intersect along
the negative numbers.131313That being said, there are some instances in which
gluing along open subspaces can give rise to a Hausdorff space. For instance,
if we consider $X=Y=\mathbb{R}$, with $A=(0,1)$ and $f:A\rightarrow Y$ a
homeomorphism, then the resulting adjunction space $X\cup_{f}Y$ will be
equivalent to the real line. Generally speaking, the Hausdorff violation of an
adjunction space will occur when there is a sequence in some open $A_{ij}$
that simultaneously converges to two points in the adjunction space.141414Such
a sequence in the $2$-branched real line would be $a_{n}=[-\frac{1}{n},i]$,
which exists within the identified negative reals, and converges to both
$[0,i]$. The following result makes this more precise.

######  Lemma 2.13.

Suppose that $X,Y$ are connected and Hausdorff, and $A$ is a non-empty, proper
subset of $X$ that is open. If the map $f:A\rightarrow Y$ can be extended to a
homeomorphism $\tilde{f}:Cl^{X}(A)\rightarrow Cl^{Y}(f(A))$, then the
adjunction space $X\cup_{f}Y$ is not Hausdorff.

###### Proof.

Since $X$ is connected, the only clopen subsets are the empty set and $X$
itself. Thus, since $A$ is non-empty, proper subset of $A$ that is open, it
cannot be the case that $A$ is clopen. This amounts to saying that the
$X$-boundary $\partial^{X}(A)$ is non-empty. Since $A$ is open, we have that

$\partial^{X}(A):=Cl^{X}(A)\backslash Int^{X}(A)=Cl^{X}(A)\backslash
A\neq\emptyset.$

Let $x$ be some element of $\partial^{X}(A)$. By assumption $f$ can be
extended to a homeomorphism $\tilde{f}:Cl^{X}(A)\rightarrow Cl^{Y}(f(A))$. In
particular, we have that
$\tilde{f}(\partial^{X}(A))=\partial^{Y}(f(A))$.151515This is actually an
instance of a more-general result: let $f:X\rightarrow Y$ be a homeomorphism
between two topological spaces, and $A$ some subset of $X$. Then
$f(\partial(A))=\partial^{Y}(f(A))$. See Appendix AX for the proof of this.
Thus the element $x$ is mapped to some $y:=\tilde{f}(x)$ in the boundary
$\partial^{Y}(f(A))$.

We now show that the points $[x,1]$ and $[y,2]$ are Hausdorff-inseparable in
the adjunction space $X\cup_{f}Y$. Let $U$ be some open neighbourhood of
$[x,1]$ and $V$ be some open neighbourhood of $[y,2]$. Since $\phi_{X}$ is
continuous, the set $\phi_{X}^{-1}(U)$ is an open neighbourhood of $x$ in $X$.
Since $x\in Cl^{X}(A)$, the set $U\cap Cl^{X}(A)$ is an open neighbourhood of
$x$ in the subspace topology $Cl^{X}(A)$. By assumption $\tilde{f}$ is a
homeomorphism, so the image $\tilde{f}(U\cap Cl^{X}(A))$ is an open
neighbourhood of $y$ in $Cl^{Y}(f(A))$. The set $\phi_{Y}^{-1}(V)\cap
Cl^{Y}(f(A))$ is also an open neighbourhood of $y$ in $Cl^{Y}(f(A))$. We can
intersect these two sets to conclude that

$\tilde{f}(U\cap Cl^{X}(A))\cap\phi_{Y}^{-1}(V)\cap Cl^{Y}(f(A))$

is also an open neighbourhood of $y$ in $Cl^{Y}(f(A))$. Since $y$ lies in the
$Y$-relative boundary of $f(A)$, the above set necessarily intersects $f(A)$.
Let $z$ be any element of the set

$\tilde{f}(U\cap Cl^{X}(A))\cap\phi_{Y}^{-1}(V)\cap Cl^{Y}(f(A))\cap f(A),$

and consider the element $[z,2]$ in $X\cup_{f}Y$. Then $[z,2]$ will lie in
both $U$ and $V$. ∎

We have the immediate corollary for generalised adjunction spaces.

$\mathbb{R}_{2}$$\mathbb{R}_{1}$$0_{2}$$0_{1}$$[0_{i},i]$ Figure 4: The
Hausdorff violation of the $2$-branched real line. Here the points $0_{1}$ and
$0_{2}$ are not identified, and are thus Hausdorff inseparable.

######  Corollary 2.14.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which each $A_{ij}$
is open and each $f_{ij}$ can be extended to a homeomorphism
$\tilde{f}_{ij}:Cl^{X_{i}}(A_{ij})\rightarrow Cl^{X_{j}}(A_{ji})$. If each
$X_{i}$ is Hausdorff and connected, then $\bigcup_{\mathcal{F}}X_{i}$ is not
Hausdorff.

###### Proof.

We can argue as in the previous lemma, except that now we pick two points
$[x,i]$ and $[\tilde{f}_{ij}(x),j]$ where $x\in Cl^{X_{i}}(A_{ij})$. ∎

In the previous two results, we obtained Hausdorff-violating pairs by mapping
points of the boundaries of $A_{ij}$ into the adjunction space under the
canonical maps $\phi_{i}$. We will now show that every Hausdorff-violating
pair arises in this manner, provided that the spaces $X_{i}$ are first-
countable. In order to prove this, we will need to make use of the following
well-known facts regarding first-countable spaces.

######  Proposition 2.15.

Let $X$ be a first-countable topological space. Then

  1. 1.

For any subset $A$ of $X$, a point $x$ lies in $Cl(A)$ iff there is a sequence
in $A$ that converges to $x$ in $X$.

  2. 2.

$X$ is Hausdorff iff every convergent sequence has a unique limit.

These two results can be used to give a useful characterisation of the
Hausdorff-violating points of an adjunction space. But, before we prove our
result we will first introduce some auxillary notation that will be useful
throughout the remainder of this paper.161616This notation is due to Hajicek,
who first introduced it in [4] during a discussion of non-Hausdorff
spacetimes.

Given a topological space $(X,\tau)$, we consider the binary relation
$\textsf{Y}\subseteq X\times X$ defined for distinct $x,y$ as follows:

$x\textsf{Y}y\ \textrm{iff}\ \forall U,V\left((U\in\tau\wedge V\in\tau\wedge
x\in U\wedge y\in V)\rightarrow U\cap V\neq\emptyset\right),$

that is, $x\textsf{Y}y$ is shorthand for saying that $x$ and $y$ form a
Hausdorff-violating pair. As a slight aside, observe that the relation Y is
manifestly irreflexive and symmetric. However, it is not necessarily
transitive, as the following example illustrates.

######  Example 2.16.

Let $\mathcal{F}$ be a tertiary adjunction system in which $X_{i}=\mathbb{R}$,
with $A_{12}=(-\infty,0)$, $A_{23}=(0,\infty)$, and $A_{13}=\emptyset$, and
all $f$-maps equal to the identity. The resulting space is depicted in Figure
5. Here we have an adjoined manifold in which there are three distinct copies
of the origin. By Corollary 2.14 we have that
$[0_{1},1]\textsf{Y}[0_{2},2]\textsf{Y}[0_{3},3]$, but any images
$\phi_{i}(U)$ and $\phi_{j}(V)$ where $U,V$ are open intervals around $0_{1}$
and $0_{3}$ respectively, will Hausdorff-separate $[0_{1},1]$ and $[0_{3},3]$.

$0_{1}$$0_{2}$$0_{3}$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$0_{i}$$\cdots$$\cdots$$\cdots$$\cdots$
Figure 5: An adjunction space in which the relation Y is not transitive. Here
there are three distinct copies of the origin in which $0_{1}\textsf{Y}0_{2}$
and $0_{2}\textsf{Y}0_{3}$, but $0_{1}$ and $0_{3}$ can be separated by
disjoint open sets.

Given some subset $W$ of $X$, we denote by $\textsf{Y}^{W}$ the set of all
points that form a Hausdorff-violating pair with some member of $W$, that is:

$\textsf{Y}^{W}:=\\{x\in X\ |\ \exists w(w\in W\wedge x\textsf{Y}w)\\}.$

Using this notation, we can now characterise the space of Hausdorff-violating
points for each image $\phi_{i}(X_{i})$ in our adjunction space. We have the
following result.

######  Theorem 2.17.

Let $\bigcup_{\mathcal{F}}X_{i}$ be a connected adjunction space formed from
first-countable, Hausdorff spaces, in which the gluing maps are all open
topological embeddings and each $A_{ij}$ is open in $X_{i}$. If each
$f_{ij}:A_{ij}\rightarrow X_{j}$ can be extended to a homeomorphism
$\tilde{f}:Cl^{X_{i}}(A_{ij})\rightarrow Cl^{X_{j}}(A_{ji})$ then in the
adjunction space we have the equality

$\partial(\phi_{i}(X_{i}))=\textsf{Y}^{\ \phi_{i}(X_{i})}$

for all $i$ in $I$.

###### Proof.

Before proving the equality, we make the observation that by Lemma 1.12 the
canonical maps $\phi_{i}$ are open topological embeddings. Thus we may apply
Lemma 2.2 to conclude that the adjunction space $\bigcup_{\mathcal{F}}X_{i}$
is first-countable.

For the inclusion from left-to-right, let $z$ be some arbitrary element in
$\partial(\phi_{i}(X_{i}))$. Since $\bigcup_{\mathcal{F}}X_{i}$ is connected
and $\phi_{i}(X_{i})$ is open, the point $z$ cannot lie in $\phi_{i}(X_{i})$.
Thus $z=[y,j]$ for some $j\neq i$. Since $\bigcup_{\mathcal{F}}X_{i}$ is
first-countable, by Prop. 2.15.1 there exists some sequence in
$\phi_{i}(X_{i})$ that converges to $[y,j]$. Since $\phi_{j}(X_{j})$ is open,
without loss of generality we may assume that this sequence lies in the
intersection $\phi_{i}(A_{ij})$. It follows that $[y,j]$ is in the boundary
$\partial(\phi_{i}(A_{ij}))$, and since $\phi_{j}$ is a homeomorphism, it is
also the case that $y\in\partial^{X_{j}}(A_{ji})$. By assumption, we can use
the homeomorphism of boundaries $\tilde{f}_{ji}$ to map $y$ to some element
$x$ in $A_{ij}$. Then $[y,j]\textsf{Y}[x,i]$ in $\bigcup_{\mathcal{F}}X_{i}$.
It follows that $[y,j]\in\textsf{Y}^{X_{i}}$, and thus
$\partial(\phi_{i}(X_{i}))\subseteq\textsf{Y}^{X_{i}}$.

For the converse inclusion, suppose that $[y,j]$ is some element of
$\textsf{Y}^{X_{i}}$. Thus there is some element $[x,i]$ in $\phi_{i}(X_{i})$
such that $[y,j]\textsf{Y}[x,i]$. Since $\bigcup_{\mathcal{F}}X_{i}$ is a
first-countable space, by Prop. 2.15.2 there exists some sequence in
$\bigcup_{\mathcal{F}}X_{i}$ that converges to both $[x,i]$ and $[y,j]$.
Without loss of generality we may assume that this sequence lies in the
intersection $\phi_{i}(A_{ij})$. It follows that $[y,j]$ lies in the closure
$Cl(\phi_{i}(X_{i}))$ and thus $[y,2]\in\partial(\phi_{i}(X_{i}))$, as
required. ∎

The reader may convince themselves of the following equivalence of boundaries:

$\partial(\phi_{i}(X_{i}))=\partial(\phi_{i}(A_{ij}))\backslash\phi_{i}(X_{i})=\bigcup_{j\neq
i}\phi_{j}(\partial^{j}A_{ji}).$

As an application of the above theorem we thus have the equality

$\textsf{Y}^{\left(\bigcup_{\mathcal{F}}X_{i}\right)}=\bigcup_{i\in
I}\textsf{Y}^{\phi_{i}(X_{i})}=\bigcup_{i\in I}\partial(\phi_{i}(A_{ij})).$

From this we may conclude that Hausdorff violation in an open adjunction space
occurs precisely at the (images of) boundaries of the gluing regions $A_{ij}$.

#### Metrizability

It is well-known that every metrizable space is Hausdorff. We can apply the
contrapositive of this fact to the setting of Theorem 2.17 to conclude that
any such adjunction space is not metrizable – intuitively, any Hausdorff-
violating pairs ought to be distance zero away from each other.

Although the adjunction spaces of Corollary 2.14 are not metrizable, they can
be locally metrizable, meaning that every point has an open neighbourhood that
is homeomorphic to a metric space. Indeed – suppose we have a adjunction space
as in Corollary 2.14 such that each $X_{i}$ is additionally a metric space. In
this case, each of the images $\phi_{i}(X_{i})$ will be homeomorphic to
$X_{i}$, thus we can take the open neighbourhoods to be the $\phi_{i}(X_{i})$.
Moreover, such adjunction spaces will also be locally Hausdorff.

##  3 Non-Hausdorff Manifolds

We have seen in the previous section that various topological features may,
under certain assumptions, be preserved in the adjunction process. In this
section, we will refine our attention to the case of gluing manifolds
together. In order to fix terminology, we will call a topological space $M$ a
generalised manifold171717This terminology is standard within the literature.
See [5], and [1] for instances and discussions of generalised manifolds. if it
is both second-countable and locally-Euclidean, meaning that every point in
$M$ has a neighbourhood that is homeomorphic to some Euclidean space of fixed
dimension. In the case that $M$ also satisfies the Hausdorff property, we will
simply refer to $M$ as a manifold, and if $M$ happens to violate the Hausdorff
property, we will call $M$ a non-Hausdorff manifold. Clearly the class of
generalised manifolds can be partitioned into manifolds and non-Hausdorff
manifolds.

We will soon identify several conditions that are required in order to ensure
that the resulting space is a non-Hausdorff topological manifold. Amongst
other things, we will show that in actual fact all non-Hausdorff manifolds can
be represented as an adjunction space of Hausdorff manifolds. In order to show
that this representation holds true, we will need an auxiliary way in which to
refer to generalised manifolds formed from an adjunction space. We will call
these adjoined manifolds. Our first task is to make precise what we mean by
this.

###  3.1 Adjoined Manifolds

Recall that a topological space $X$ is locally-Euclidean if every point in $X$
has a neighbourhood that is homeomorphic to some Euclidean space
$\mathbb{R}^{n}$ of fixed dimension. In symbols, this means that at each point
$x$ in $X$, one can find a tuple $(U,\varphi)$, where $U$ is an open
neighbourhood of $x$ and $\varphi:U\rightarrow\mathbb{R}^{n}$ is an open
embedding. The following result shows that this property can be transferred to
an adjunction space.181818Note that Lemma 3.2 won’t work if we assume that the
gluing regions $A_{ij}$ are topologically closed. As an example – consider
$X=Y=\mathbb{R}$, with $A=(\infty,0]$ and $f=id_{\mathbb{R}}$. The resulting
space will not be locally-Euclidean. See [5] for suggestive depictions of this
space.

######  Lemma 3.1.

Let $\bigcup_{\mathcal{F}}X_{i}$ be an adjunction space in which each $X_{i}$
is locally-Euclidean. If each $\phi_{i}$ is an open topological embedding,
then $\bigcup_{\mathcal{F}}X_{i}$ is also locally-Euclidean.

###### Proof.

Let $[x,i]$ be a point in $\bigcup_{\mathcal{F}}X_{i}$. Then by construction
$x$ lies in $X_{i}$. Since $X_{i}$ is locally-Euclidean, there is a chart
$(U,\varphi)$ of $X_{i}$ at $x$. We show that
$(\phi_{i}(U),\varphi\circ\phi_{i}^{-1})$ is a chart for
$\bigcup_{\mathcal{F}}X_{i}$ at $[x,i]$. By assumption $\phi_{i}$ is open,
thus $\phi_{i}(U)$ is an open subset of the adjunction space. Since we assumed
that $\phi_{i}$ is a topological embedding, it is homeomorphic onto its image.
Then the restriction of $\phi_{i}$ to an open subset is also a homeomorphism,
as is its inverse. Thus the map
$\varphi\circ\phi_{i}^{-1}:\phi_{i}(U)\rightarrow\varphi(U)$ is a composition
of homeomorphisms, thus is also a homeomorphism. It follows that
$(\varphi(U),\varphi\circ\phi_{i}^{-1})$ is a chart for
$\bigcup_{\mathcal{F}}X_{i}$ at $[x,i]$. ∎

We can now combine Lemma 3.1 with the results of Section 2 to obtain the
following theorem, which forms the backbone of the definition of an adjoined
manifold. Note that in order to emphasise that we are working with manifolds,
in what follows we will denote our adjunction spaces by
$\bigcup_{\mathcal{F}}M_{i}$.

######  Theorem 3.2.

Let $\bigcup_{\mathcal{F}}M_{i}$ be an adjunction space formed from Hausdorff
topological manifolds in which:

  1. 1.

each $A_{ij}$ is an open submanifold of $M_{i}$,

  2. 2.

each $f_{ij}$ is an open topological embedding, and

  3. 3.

the indexing set $I$ is countable.

Then the adjunction space $\bigcup_{\mathcal{F}}X_{i}$ is a generalised
topological manifold.

###### Proof.

The proof follows immediately from Lemma 1.12, Corollary 2.5 and Lemma 3.1. ∎

We will refer to any topological space satisfying the antecedent conditions of
Theorem 3.3 as an adjoined manifold. Note that we may apply Lemma 2.6 as we
please to obtain connected adjoined manifolds. As well as this, we may also
invoke the results in the discussion following Lemma 2.14 to conclude that
adjoined manifolds are typically non-Hausdorff, and in the case that each pair
of gluing regions $A_{ij}$ and $A_{ji}$ have connected, homeomorphic
boundaries, we have that the Hausdorff-violating points are characterised by
the equalities

$\partial(\phi_{i}(M_{i}))=\textsf{Y}^{M_{i}}.$

In fact, much more is true of these canonical images $\phi_{i}(X_{i})$ –
Section 4 is devoted to expanding on this.

###  3.2 Non-Hausdorff Manifolds and $H$-Submanifolds

As previously mentioned, a non-Hausdorff manifold is a generalised manifold
$M$ in which there exist at least two points $x$ and $y$ such that
$x\textsf{Y}y$. These are relatively unstudied compared to their Hausdorff
cousins. In varying contexts, some mention of non-Hausdorff manifolds can be
found in [6, 7, 8, 4, 9, 5, 10, 1].

There has been interest in the literature in studying the possible Hausdorff
submanifolds of a given non-Hausdorff manifold. In fact, this evolved into the
refined notion of $H$-submanifolds. The definition of these subspaces is
originally due to Hajicek, who in [4] introduced the term whilst studying the
prospect of non-Hausdorff spacetimes. The definition is as follows.

######  Definition 3.3.

Given a non-Hausdorff manifold $M$, a subset $V$ of $M$ is called an
$H$-submanifold if $V$ is open, Hausdorff and connected, and is maximal with
respect to these properties.

We remark that since $H$-submanifolds of a manifold $M$ are defined to be open
submanifolds, they are necessarily of the same dimension as $M$.

In the context of adjoined manifolds, in order to preserve manifold structures
during the adjunction process, we demanded stringent conditions on the to-be-
glued regions $A_{ij}$ and the gluing maps $f_{ij}$. It follows from an
application of Lemma 1.12 that the canonical maps $\phi_{i}$, which take each
Hausdorff manifold $M_{i}$ into the adjunction space
$\bigcup_{\mathcal{F}}M_{i}$, now act as open topological embeddings. This
means that the $M_{i}$ effectively live inside the adjunction space as
Hausdorff submanifolds. In fact, more is true – we will prove in Section 4
that the images $\phi_{i}(M_{i})$ are actually $H$-submanifolds.

At first glance, it is somewhat difficult to give examples of $H$-submanifolds
of an arbitrary generalised manifold. Fortunately, Hajicek provides a useful
characterisation in terms of Hausdorff-violating points.191919We note that the
original formulation (found as [4, Thm. 2]) of this result is in terms of
smooth manifolds. But, the proof contains no obstructions to passing into the
more general setting of topological non-Hausdorff manifolds.

######  Theorem 3.4 (Hajicek).

Given a non-Hausdorff manifold $M$, a subset $V$ is an $H$-submanifold of $M$
if and only if

$\partial V=Cl(\textsf{Y}^{V}),$

where $Y^{V}:=\\{x\in M\ |\ \exists y\in V(x\textsf{Y}y)\\}$.

We also have the following enjoyable lemma, which will be useful in the
remainder of this paper.

######  Proposition 3.5.

Let $M$ be a non-Hausdorff manifold. The collection of $H$-submanifolds forms
an open cover of $M$.

###### Proof.

(Sketch). For any point $x$ in $M$, we consider the family $\mathcal{H}$ of
open, connected, Hausdorff spaces containing $x$, viewed as a poset under set-
theoretic inclusion. Any chain of elements of $\mathcal{H}$ has upper-bound
given by their union, thus we may apply Zorn’s Lemma to conclude that there
exists at least one maximal open, connected, Hausdorff subspace $H_{x}$ of $M$
containing $x$. The open cover is then given by $\\{H_{x}\ |\ x\in M\\}$. For
the full argument, see [4, Thm. 1]. ∎

###  3.3 Representation of Non-Hausdorff Manifolds

We are now in the position to argue that all non-Hausdorff manifolds are
actually adjoined manifolds. We should note that this idea is in no sense
original – this is an adaptation of a recent theorem of Placek and Luc [1],
which we will rephrase in a slightly more formal setting.202020Moreover, this
proof strategy has been employed in [7], where various arguments of this type
are used when discussing the vector bundles over a given adjoined manifold.

######  Theorem 3.6.

Let $M$ be a non-Hausdorff manifold. Then $M$ is homeomorphic to an adjunction
space $\bigcup_{\mathcal{F}}X_{i}$.

###### Proof.

Consider the family $M_{i}$ consisting of all $H$-submanifolds of $M$. We know
from Prop. 3.5 that the $X_{i}$ form an open cover of $M$. Since $M$ is
second-countable, in particular $M$ is Lindelöf. Thus without loss of
generality we may assume that the $M_{i}$ form a countable open cover of $M$.
We can then define an adjunction system
$\mathcal{F}:=(\textbf{M},\textbf{A},\textbf{f})$ as follows:

  * •

$\textbf{M}=\\{M_{i}\\}_{i\in I}$,

  * •

$A_{ij}:=M_{i}\cap M_{j}$, and

  * •

$f_{ij}:A_{ij}\rightarrow M_{j}$ is the identity map.

Clearly this collection satisfies the conditions of Definition 1.8 and the
criteria of Theorem 3.2. Thus the associated adjunction space
$\bigcup_{\mathcal{F}}M_{i}$ is well-defined, and by Theorem 3.2 it is an
adjoined manifold. We can invoke the universal property of adjunction spaces
(Prop. 1.11) to see that there exists a unique continuous map $g$ such that we
have the following commutative diagram for every pair of $M_{i}$’s:

${A_{ij}}$${M_{i}}$${X_{j}}$${\bigcup_{\mathcal{F}}X_{i}}$${M}$$\scriptstyle{id_{A_{ij}}}$$\scriptstyle{\phi_{i}}$$\scriptstyle{\iota_{i}}$$\scriptstyle{\phi_{j}}$$\scriptstyle{\iota_{j}}$$\scriptstyle{g}$

where the $\iota_{\cdot}$ are the inclusion maps, and

$g:\bigcup_{\mathcal{F}}M_{i}\rightarrow M\ \textrm{is given by}\ [x,i]\mapsto
x.$

This map is clearly open – this follows from the fact that $\phi_{i}$ and
$\iota_{i}$ are open maps for all $i$. Moreover, $f$ is a bijection – the
inverse is given by $f^{-1}(x)=[x,i]$ where $i$ is the index of any $M_{i}$
that contains $x$. By construction, this map is well-defined. Thus we have
obtained a bijective, continuous, open map from $\bigcup_{\mathcal{F}}M_{i}$
to $M$. ∎

###  3.4 Partitions of Unity

We finish this section with a discussion of partitions of unity in non-
Hausdorff manifolds. Recall first the standard definition of a partition of
unity subordinate to an open cover.212121This definition is a slight
adaptation of the standard (i.e. Hausdorff) definition found in, say [2, pg.
258] or [6, pg. 43].

######  Definition 3.7.

Let $M$ be a generalised manifold, and $\mathcal{U}=\\{U_{i}\ |\ i\in I\\}$ be
an open cover of $M$. A partition of unity subordinate to $\mathcal{U}$ is an
indexed family of continuous maps $\psi_{i}:M\rightarrow[0,1]$ such that:

  1. 1.

the support $Supp(\psi_{i}):=\\{p\in M\ |\ \psi_{i}(p)\neq 0\\}$ is contained
with $U_{i}$ for all $i$ in $I$,

  2. 2.

the family of supports $\\{Supp(\psi_{i})\ |\ i\in I\\}$ is locally-
finite,222222Recall that a family of subsets $C_{i}$ of a topological space
$X$ is locally-finite if every point $x$ in $X$ has an open neighbourhood that
intersects only finitely-many $C_{i}$. and

  3. 3.

$\sum_{i\in I}\psi_{i}(p)=1$ for all $p$ in $M$.

The strength of partitions of unity is that they allow the construction of
global objects by pasting together locally-defined ones. Within (Hausdorff)
differential geometry, partitions of unity are used in the construction of
bump functions, in extensions of smooth functions, in the definition of
integration, in the proof of the Fundamental Theorem of Riemannian Geometry,
and so on.

There are some well-known results regarding the equivalence of various
topological properties surrounding partitions of unity. These can be
summarised as follows.

######  Theorem 3.8 (Equivalence Results for Partition of Unity).

Let $X$ be a topological space.

  1. 1.

If $X$ is $T_{1}$, then $X$ is Hausdorff and paracompact iff there exists a
partition of unity subordinate to every open cover.

  2. 2.

If $X$ is locally-Euclidean and Hausdorff, then $X$ is second-countable iff
$X$ is paracompact and has countably-many connected components.

  3. 3.

If $X$ is a Hausdorff manifold, then $X$ is paracompact and admits a partition
of unity subordinate to every open cover.

###### Proof.

For the proofs of (1) and (2), see [6]. ∎

Given item (1) in the above theorem, one can immediately conclude that in
every non-Hausdorff manifold, there exists at least some open open cover that
admits no partition of unity subordinate to it. In fact, we can describe such
an open cover explicitly.

######  Theorem 3.9.

Every non-Hausdorff manifold $M$ has an open cover $\mathcal{U}$ that does not
admit a partition of unity.

###### Proof.

By the results of the previous section, without loss of generality we may
assume that $M=\bigcup_{\mathcal{F}}M_{i}$, where $M_{i}$ is some countable
cover of $H$-submanifolds of $M$. Since $H$-submanifolds are open, we can
consider the open cover

$\mathcal{U}:=\\{M_{i}\ |\ i\in I\\}.$

Since $M$ is non-Hausdorff, there exists at least two points $a,b$ in $M$ such
that $a\textsf{Y}b$. Since the images $\phi_{i}(M_{i})$ cover $M$, these two
points are of the form $a=[x,i]$ and $b=[y,j]$ for some $x\in M_{i}$ and $y\in
M_{j}$, with $i\neq j$.

Suppose towards a contradiction that there exists some partition of unity
$\\{\psi_{i}\ |\ i\in I\\}$ subordinate to the cover $\mathcal{U}$, and
consider the two maps $\psi_{i}$ and $\psi_{j}$. We will arrive at a
contradiction by identifying a point $c$ nearby both $a$ and $b$ in which the
sum $\sum\psi_{i}(z)$ exceeds $1$. In order to do this, we need to exploit the
local-finiteness of the $Supp(\psi_{i})$.

By local-finiteness, there exists some open neighbourhood $U_{a}$ of $a$ such
that the set $\\{\psi_{i}\ |\ Supp(\psi_{i})\cap U_{a}\\}$ is finite, of
cardinality $n$. Similarly, we may conclude the existence of some
neighbourhood $U_{b}$ of $b$ that intersects only finitely-many of the
supports of the $\psi_{i}$. For convenience, we will re-index as follows:

  * •

let $\psi_{\alpha_{1}},...,\psi_{\alpha_{n}}$ be the $\psi_{i}$ whose supports
intersect $U_{a}$, and

  * •

let $\psi_{\beta_{1}},...,\psi_{\beta_{m}}$ be the $\psi_{i}$ whose supports
intersect $U_{b}$.

Using this notation, we have that:

$\sum_{i\in I}\psi_{i}(a)=\psi_{\alpha_{1}}(a)+...+\psi_{\alpha_{n}}(a)=1,$

that is, only the $\alpha_{i}$’s contribute to the sum at $a$. We have an
analogous result for $b$.

Consider now the subsets $A_{i}$ and $B_{j}$ of the unit interval given by:

$A_{i}:=\left(\psi_{\alpha_{i}}(a)+\frac{1}{2n+1},1\right),\ \textrm{and}\
B_{j}:=\left(\psi_{\beta_{i}}(b)+\frac{1}{2m+1},1\right).$

It follows from the continuity of the $\psi$’s and the local finiteness
discussed above that the intersections

$\bigcap_{i=1}^{n}\psi_{\alpha_{i}}(A_{i})\ \textrm{and}\
\bigcap_{j=1}^{m}\psi_{\beta_{j}}(B_{j})$

are open neighbourhoods of $a$ and $b$, respectively.

We are now in the position to obtain our element $c$. Observe that since
$a\textsf{Y}b$, all open neighbourhoods of these elements must intersect. So,
let $c$ be any element of the intersection

$\left(\bigcap_{i=1}^{n}\psi_{\alpha_{i}}(A_{i})\right)\cap\left(\bigcap_{j=1}^{m}\psi_{\beta_{j}}(B_{j})\right).$

We then have:

$\displaystyle\sum_{i\in I}\psi_{i}(c)$
$\displaystyle=\left(\sum_{i=1}^{n}\psi_{\alpha_{i}}(c)\right)+\left(\sum_{j=1}^{m}\psi_{\beta_{j}}(c)\right)$
$\displaystyle\geq\left(\sum_{i=1}^{n}\psi_{\alpha_{i}}(a)+\frac{1}{2n+1}\right)+\left(\sum_{j=1}^{m}\psi_{\beta_{j}}(b)+\frac{1}{2m+1}\right)$
$\displaystyle=\left(\sum_{i=1}^{n}\psi_{\alpha_{i}}(a)\right)-\frac{n}{2n+1}+\left(\sum_{j=1}^{m}\psi_{\beta_{j}}(b)\right)+\frac{m}{2m+1}$
$\displaystyle=\left(1-\frac{n}{2n+1}\right)+\left(1-\frac{m}{2m+1}\right)$
$\displaystyle>\frac{1}{2}+\frac{1}{2},$

which contradicts item (3) in Definition 3.7.

∎

Observe that in the above result, we only used the images $\phi_{i}(M_{i})$ to
guarantee the existence of an open cover. The real issue was with the non-
Hausdorffness of $M$ – for any open neighbourhoods around a Hausdorff-
violating pair $a\textsf{Y}b$ in $M$ we can find a third point arbitrarily
close to both $a$ and $b$ that forces any attempted partition of unity to have
sum greater than one. As such, we can really pick any open cover
$\mathcal{U}$, and reproduce the argumentation above in order to deduce the
following dire situation.

######  Theorem 3.10.

Let $M$ be a non-Hausdorff manifold. Then every open cover $\mathcal{U}$ of
$M$ does not admit a partition of unity subordinate to it.

Although beyond the scope of this paper, we remark that the above Theorem will
likely result in some non-trivial obstructions when attempting to recreate the
standard theory of differential geometry in the non-Hausdorff regime.

##  4 Characterisation of Canonical Embeddings

In the previous section we saw that all non-Hausdorff (topological) manifolds
are actually adjoined manifolds. This was shown by considering an open cover
of $H$-submanifolds that could be trivially glued together to recover the
original manifold. In this section, we will switch gears, and instead study
the topological properties of the subspaces $\phi_{i}(M_{i})$ of a given
adjoined manifold $\bigcup_{\mathcal{F}}M_{i}$. Inspired by Theorem 3.2, we
will refer to these spaces $\phi_{i}(M_{i})$ as the canonical embeddings.
Throughout this section we will work towards determining which topological
properties uniquely characterise these canonical embeddings.

###  4.1 Initial Steps in the Characterisation

In order to characterise the canonical embeddings $\phi_{i}(M_{i})$, it makes
sense to first explicate their topological properties. Observe that by Theorem
2.17 the canonical embeddings satisfy the property

$\partial(\phi_{i}(M_{i}))=\textsf{Y}^{\phi_{i}(M_{i})}.$

It follows from Hajicek’s condition (Thm. 3.4) that all of the canonical
embeddings $\phi_{i}(M_{i})$ are $H$-submanifolds of
$\bigcup_{\mathcal{F}}M_{i}$.

However, an important observation (initiated by Müller [11]) is that in
general there may exist other $H$-submanifolds distinct from all of the images
$\phi_{i}(M_{i})$. We will now consider an important illustration of this.

######  Theorem 4.1.

There exists a non-Hausdorff manifold $M$ in which there are infinitely-many
$H$-submanifolds.

###### Proof.

Consider the binary adjunction space $M$ formed from two copies of
$\mathbb{R}^{2}$, in which the gluing region $A$ is given by

$A:=\\{(x,y)\in\mathbb{R}^{2}\ |\ y<0\\},$

that is, $A$ is an open half-plane, and the gluing map
$f:A\rightarrow\mathbb{R}^{2}$ is the identity map,232323This can be seen as a
two-dimensional analogy of the $2$-branched real line. We need this because
our $H$-submanifold requires the boundary of the gluing region $A_{ij}$ to be
at least one dimensional. as pictured in Figure 6. Observe that the Hausdorff-
violating points of $M$ are precisely the images of the $x$-axes under the
maps $\phi_{i}$.

In order to define our $H$-submanifold (which we will denote by $V$), we first
define some particular subspaces of the copies of $\mathbb{R}^{2}$. Let:

  * •

$U_{1}$ be the complement of the non-negative $x$ axis in the first copy of
$\mathbb{R}^{2}$, and

  * •

$U_{2}$ be the complement of the non-positive $x$-axis in the second copy of
$\mathbb{R}^{2}$.

The idea behind $U_{1}$ and $U_{2}$ is depicted in the left-hand image in
Figure 6. Observe that $U_{1}$ and $U_{2}$ are open submanifolds of their
respective copies of $\mathbb{R}^{2}$.

By construction of $M$, the canonical maps $\phi_{i}$ are open. With this in
mind, we define the subset $V$ of $M$ as follows:

$V:=\phi_{1}(U_{1})\cup\phi_{2}(U_{2}).$

Since $V$ is open in $M$, we may endow $V$ with an open submanifold structure.
Moreover, $V$ is clearly connected, and $V$ is Hausdorff by construction,
since it only contains one half of each copy of the $x$-axis.

We will now argue that $V$ is an $H$-submanifold of $M$. There are two ways to
do this – we can argue via maximality, or we can employ Hajicek’s condition.
We will do the latter. The topological boundary of $V$ can be described
explicitly as:

$\partial V=\phi_{1}(\mathbb{R}^{2}_{1}\backslash
U_{1})\cup\phi_{2}(\mathbb{R}^{2}_{2}\backslash U_{2}),$

that is, $\partial V$ consists of the images of the red axes in Figure 6 under
the maps $\phi_{i}$.

The Hausdorff-violating points $\textsf{Y}^{V}$ will almost coincide with
$\partial V$, with the exception of the two copies of the origin, which are
both excluded from $V$, thus also do not appear in $\textsf{Y}^{V}$. However,
when we take the topological closure of $\textsf{Y}^{V}$, we include these two
copies of the origin. We may conclude from this that the equality $\partial
V=Cl(\textsf{Y}^{V})$ holds. It follows that $V$ is an $H$-submanifold, as
required. Moreover, we can obtain infinitely-many other $H$-submanifolds by
choosing Y-related points on the $x$-axes different from origin(s). ∎

The above result confirms that generally there will exist infinitely-many
$H$-submanifolds of a given adjoined manifold. This means that the canonical
embeddings are in general not the unique $H$-submanifolds of a given adjoined
manifold. As such, a stronger property is needed.

In order to work towards such a property, we can consider the ways in which
the $H$-submanifold $V$ of Theorem 4.1 differs from the canonical embeddings.
Actually, there are several ways: $V$ contains elements from both parts of the
set-theoretic difference of both copies of $\mathbb{R}^{2}$, $V$ happens to be
a dense subset of $M$, etc. Although these are valid observations, they are
dependent on the fact that we are dealing with a binary adjunction space, and
one may entertain possible counterexamples in higher orders which possess
$H$-submanifolds violating these properties.

As it turns out, there is another distinguishing feature – the origin of both
copies of $\mathbb{R}^{2}$ are excluded from $V$. This means that both origins
are members of the topological boundary $\partial V$, without forming
Hausdorff-violating points for $V$. As we saw in the proof of 4.1, we have
that:

$\partial V=Cl(\textsf{Y}^{V}),\ \textrm{but}\ \partial V\neq\textsf{Y}^{V}.$

By Theorem 2.17, we know that this equality holds for all $\phi_{i}(M_{i})$.
It is thus natural to conjecture that the images $\phi_{i}(X)$ are the unique
$H$-submanifolds with this property. As it turns out, we have the following
result for the case that $I$ is finite.

$f(A)$$\mathbb{R}^{2}$$A$$\mathbb{R}^{2}$$M$ Figure 6: The construction of $M$
in the proof of Theorem 6. Here the complements of the sets $U_{i}$ are
displayed in red.

######  Theorem 4.2.

Let $\mathcal{F}$ be an adjunction system for an adjoined manifold, in which
$I$ is finite and each of the gluing maps $f_{ij}:A_{ij}\rightarrow M_{j}$ can
be extended to a homeomorphism $\tilde{f}_{ij}:Cl^{M_{i}}(A_{ij})\rightarrow
Cl^{M_{j}}(A_{ji})$. Then, in the adjoined manifold
$\bigcup_{\mathcal{F}}M_{i}$, the canonical embeddings $\phi_{i}(M_{i})$ are
the unique $H$-submanifolds satisfying the equality:
$\partial(\phi_{i}(M_{i}))=\textsf{Y}^{\phi_{i}(M_{i})}$.

The remainder of this section is dedicated to proving this result. In order to
do so, we will first need to consider some consequences of the assumptions at
play.

###  4.2 Some Preliminary Results

In Section 4.3 we will prove Theorem 4.2 by an inductive argument on the size
of the adjunction space. In order to perform this induction, we will need some
preliminary results that make explicit use of the additional conditions at
play in Theorem 4.2. As such, throughout this section we fix an adjoined
manifold

$M:=\bigcup_{\mathcal{F}}M_{i}$

that satisfies the criteria of Theorem 4.2. Our first result is the following,
the proof of which can be found in Appendix A.4.

######  Lemma 4.3.

If $V$ is a subset of $M$ such that $\partial V=\textsf{Y}^{V}$, then for
every $i,j\in I$, the set
$\textsf{Y}^{V}\cap\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$ is
open in $\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$ (equipped
with the subspace topology).

The above result has the following corollary, which will be very useful in
Section 4.3.

######  Corollary 4.4.

Let $V$ be an $H$-submanifold of $\bigcup_{\mathcal{F}}M_{i}$ such that
$\partial V=\textsf{Y}^{V}$. Then for any connected boundary
$\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$, if
$\textsf{Y}^{V}\cap\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$ is
non-empty, then
$\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)\subseteq\textsf{Y}^{V}$.

###### Proof.

Since $\textsf{Y}^{V}=\partial V$, the set $\textsf{Y}^{V}$ is closed in the
whole space $\bigcup_{\mathcal{F}}M_{i}$. It follows from Lemma 4.3 that
$\textsf{Y}^{V}\cap\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$ is
clopen. Since $\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$ is
connected and by assumption
$\textsf{Y}^{V}\cap\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)$ is
non-empty, it follows that
$\textsf{Y}^{V}\cap\partial^{\phi_{i}(M_{i})}\left(\phi_{i}(A_{ij})\right)=\textsf{Y}^{V}$
as required. ∎

The next result says that the equality $\partial V=\textsf{Y}^{V}$ may be
preserved in any adjunction spaces built from subsystems of $\mathcal{F}$.
Recall that in Section 1.3 we denoted objects in the sub-adjunction space with
double-brackets $\llbracket\cdot\rrbracket$.

######  Lemma 4.5.

Let $V$ be a subset of $\bigcup_{\mathcal{F}}M_{i}$ such that $\partial
V=\textsf{Y}^{V}$. If $V$ is contained within the image some sub-adjunction
space $\bigcup_{\tilde{\mathcal{F}}}M_{j}$ under the canonical inclusion of
Theorem 1.14, then $\partial\llbracket V\rrbracket=\textsf{Y}^{\llbracket
V\rrbracket}$ in $\bigcup_{\tilde{\mathcal{F}}}M_{j}$ as well.

###### Proof.

For the inclusion from left-to-right, suppose that $\llbracket x,i\rrbracket$
is a point in $\partial\llbracket V\rrbracket$. Since $g$ is an open
embedding, it follows that $g(\llbracket x,i\rrbracket)=[x,i]$ is in the
boundary $\partial V$. By assumption $\partial V=\textsf{Y}^{V}$, so there
exists some $[y,j]$ in $V$ such that $[x,i]\textsf{Y}[y,j]$ in the larger
adjunction space. Since $[y,j]$ is a point in $V$, its preimage $\llbracket
y,j\rrbracket$ makes sense as a point in $\llbracket V\rrbracket$. We will
show that $\llbracket x,i\rrbracket\textsf{Y}\llbracket y,j\rrbracket$. Let
$U_{x}$ and $U_{y}$ be neighbourhoods of $\llbracket x,i\rrbracket$ and
$\llbracket y,j\rrbracket$ respectively. Without loss of generality, we may
assume that $U_{y}$ is contained in $\llbracket V\rrbracket$. By Theorem 1.14,
the map $g$ is open, thus $g(U_{x})$ and $g(U_{y})$ are open neighbourhoods of
$[x.i]$ and $[y,j]$ in $\bigcup_{\mathcal{F}}M_{i}$. Since
$[x,i]\textsf{Y}[y,j]$, these two neighbourhoods overlap, and any element in
such an overlap will pullback to an element in the intersection $U_{x}\cap
U_{y}$. Since we chose $U_{x}$ and $U_{y}$ to be arbitrary neighbourhoods of
$\llbracket x,i\rrbracket$ and $\llbracket y,j\rrbracket$, we may conclude
that $\llbracket x,i\rrbracket\textsf{Y}\llbracket y,j\rrbracket$. It follows
that $\partial\llbracket V\rrbracket\subseteq\textsf{Y}^{\llbracket
V\rrbracket}$.

For the converse inclusion, suppose now that $\llbracket x,i\rrbracket$ is a
point in $\textsf{Y}^{\llbracket V\rrbracket}$. By definition, this means that
there is some point $\llbracket y,j\rrbracket$ in $\llbracket V\rrbracket$
that cannot be separated from $\llbracket x,i\rrbracket$. We know from Theorem
1.14 that the function $g$ mapping the sub-adjunction space
$\bigcup_{\tilde{\mathcal{F}}}M_{j}$ into the full space
$\bigcup_{\mathcal{F}}M_{i}$ is an open topological embedding. Moreover,
Hausdorff-violation is preserved under open maps, so we may conclude that
$\llbracket x,i\rrbracket\textsf{Y}\llbracket y,j\rrbracket$ implies that
$[x,i]\textsf{Y}[y,j]$. By assumption, in the full adjunction space we have
that $\textsf{Y}^{V}=\partial V$, so we may conclude that $[x,i]\in\partial
V$. That $\llbracket x,i\rrbracket$ lies in $\partial(\llbracket V\rrbracket)$
then follows from the fact that boundaries pull-back to boundaries under open
topological embeddings. ∎

It is well-known that path-connectedness implies connectedness in all
topological spaces. Our next result provides a restricted converse, the proof
of which can be found in Appendix A.2.

######  Proposition 4.6.

If $A$ is a connected, open subset of a generalised manifold $M$, then $A$ is
also path-connected.

We finish this section with an extremely useful lemma that will underpin our
later arguments. Roughly speaking, this lemma gives us a way to describe open
subsets of a given $H$-submanifold.

######  Lemma 4.7 (Extension Lemma).

Suppose that $V$ is an $H$-submanifold of $M$, and consider an open, Hausdorff
and connected subset $U$. If $V\cap U\neq\emptyset$, with
$U\cap\textsf{Y}^{V}=\emptyset$ then $U\subseteq V$.

###### Proof.

To see that $U\cup V$ is Hausdorff, suppose that we have two elements $a,b\in
U\cup V$. In the case that both $a$ and $b$ both lie in $V$ (or $U$), then the
Hausdorffness of $V$ (or $U$) ensures that $a$ and $b$ can be separated.
Moreover, if $a\in V$ and $b\in U$, then $a\textsf{Y}b$ would imply that
$b\in\textsf{Y}^{V}$, contradicting our assumption that
$U\cap\textsf{Y}^{V}=\emptyset$. It follows that $U\cup V$ is Hausdorff.
Moreover, $U\cup V$ is a union of overlapping, open, connected subsets, hence
it is also open and connected. Thus $U\cup V$ is a Hausdorff, connected, open
submanifold of $M$. It follows from the maximality of $V$ that $U\cup V=V$,
from which the result follows. ∎

###  4.3 The Binary and Finite Cases

Now that we have some useful lemmas out of the way, we can prove Theorem 4.2.
In order to do so, we will proceed by induction on the size of $I$.

#### The Binary Case

We start with the case of a binary adjunction space. As it turns out, we have
done the majority of the work already – there are several ways in which the
results of Section 4.2 guarantee the uniqueness result that we seek. We only
pick one such argument in the following result.

######  Theorem 4.8.

Let $V$ be an $H$-submanifold of the adjoined manifold $M_{1}\cup_{f}M_{2}$.
If the gluing map $f:A\rightarrow Y$ can be extended to a homeomorphism of
boundaries $\tilde{f}:Cl^{M_{1}}(A)\rightarrow Cl^{M_{2}}(f(A))$, and $A$ has
connected boundary, then:

$\partial V=\textsf{Y}^{V}\ \textrm{iff}\ V=\phi_{1}(M_{1})\ \textrm{or}\
\phi_{2}(M_{2}).$

###### Proof.

The direction from right-to-left has already been proved as Theorem 2.17. So,
let $V$ be some $H$-submanifold in which $\partial V=\textsf{Y}^{V}$, and
suppose towards a contradiction that $V$ is distinct from both
$\phi_{1}(M_{1})$ and $\phi_{2}(M_{2})$. Thus $V$ must contain elements from
both $\phi_{1}(M_{1})\backslash\phi_{2}(M_{2})$ and
$\phi(Y)\backslash\phi_{1}(M_{1})$. Since $V$ is a connected locally-Euclidean
space, by Lemma 4.6 $V$ is path-connected. Thus, let $\gamma$ be a path in $V$
that connects an element $x$ from $\phi_{1}(M_{1})\backslash\phi_{2}(M_{2})$
to some element $y$ in $\phi(Y)\backslash\phi_{1}(M_{1})$. Intuitively
speaking, this path $\gamma$ has to exit the difference
$\phi_{1}(M_{1})\backslash\phi_{2}(M_{2})$ at some point, thus has to pass
through the boundary $\partial^{M_{1}}((\phi_{1}(A))$. It follows from
Corollary 4.4 that $\partial^{M_{1}}((\phi_{1}(A))\subseteq V$. We may then
apply Lemma 4.7 to conclude that $\phi_{1}(M_{1})=V$, a contradiction. ∎

$\phi_{1}(M_{1})$$x$$\phi_{2}(M_{2})$$y$$\phi_{1}(A)$$\phi_{1}(A)$$\phi_{1}(M_{1})\backslash\phi_{2}(M_{2})$$\phi_{2}(M_{2})\backslash\phi_{1}(M_{1})$$\gamma$$\gamma$
Figure 7: The idea behind the proof of Theorem 4.8. We create a path between
two points in $V$ that are in both sides of the relative complements
$\phi_{1}(M_{1})\backslash\phi_{2}(M_{2})$ and
$\phi_{2}(M_{2})\backslash\phi_{1}(M_{1})$. We can then use the Extension
Lemma to conclude that $V$ contains either all of $\phi_{1}(M_{1})$ or all of
$\phi_{2}(M_{2})$.

#### The Finite Case

Now that we have proved the binary version of Theorem 4.2, we can set about
performing an inductive argument for the finite case. In fact, we will prove a
slightly stronger result, from which Theorem 4.2. will follow as a corollary.

######  Theorem 4.9.

Let $M:=\bigcup_{\mathcal{F}}M_{i}$ be an adjoined manifold satisfying the
criteria of Theorem 4.2. If $V$ is an $H$-submanifold of $M$ that is contained
within finitely-many $\phi_{i}(M_{i})$ and $\partial V=\textsf{Y}^{V}$, then
$V=\phi_{i}(M_{i})$ for some $i$ in $I$.

###### Proof.

We will prove the result by induction on the size of the family of
$\phi_{i}(M_{i})$ containing $V$. Note that we have already done the majority
of the leg-work for the base case. Suppose that $V$ is an $H$-submanifold of
$\bigcup_{\mathcal{F}}M_{i}$ that is contained within two canonical images,
say $V\subseteq\phi_{1}(M_{1})\cup\phi_{2}(M_{2})$. We can consider the sub-
adjunction space formed from $M_{1}$ and $M_{2}$. By Lemma 4.5, the
restriction of $V$ will be an $H$-submanifold of the binary adjunction space
$M_{1}\cup_{f_{12}}M_{2}$ in which the equality $\partial V=\textsf{Y}^{V}$
still holds true. We may thus apply Theorem 4.8 to conclude that $V$ equals
either $\phi_{1}(M_{1})$ or $\phi_{2}(M_{2})$.

Inductive Step: Suppose that for any collection of $n$-many $M_{i}$, if
$V\subseteq\bigcup_{j=i_{1}}^{i_{n}}\phi_{j}(M_{j})$ then $V=\phi_{j}(M_{j})$
for some $j\in\\{i_{1},...,i_{n}\\}$. Consider now a collection of
$(n+1)$-many $\phi_{i}(M_{i})$, and suppose that $V$ is a subset of the union
of these $\phi_{i}(M_{i})$’s. Suppose towards a contradiction that $V$ is
distinct from each $\phi_{i}(M_{i})$. Then both

$V\not\subseteq\bigcup_{j=i_{1}}^{i_{n}}\phi_{j}(M_{j})\ \textrm{and}\
V\not\subseteq\phi_{i_{n+1}}(M_{i_{n+1}})$

follow from the induction hypothesis and maximality, respectively. The first
of the above expressions means that there exists some element $u$ of $V$ that
lies in the set-theoretic difference

$\phi_{i_{n+1}}(M_{i_{n+1}})\backslash\left(\bigcup_{j=i_{1}}^{i_{n}}\phi_{j}(M_{j})\right).$

Without loss of generality, we may suppose that $u$ lies in the interior

$Z:=Int\left(\phi_{i_{n+1}}(M_{i_{n+1}})\backslash\left(\bigcup_{j=i_{1}}^{i_{n}}\phi_{j}(M_{j})\right)\right),$

since otherwise, $u$ lies on the boundary of this set, and we can find an
element of the interior by using the fact that $V$ is open. It follows from an
application of the Extension Lemma that the entirety of $Z$ is contained
within $V$.

Let $U$ be the maximal subset of $V\cap\phi_{i_{n+1}}(M_{i_{n+1}})$ that is
open and connected, and contains the interior $Z$. Note that $U$ exists by an
application of Zorn’s Lemma, in a manner similar to that of Proposition 3.5.
We will arrive at a contradiction by showing that $U$ is a non-trivial clopen
subset of $V$, which contradicts $V$ as connected. It is clear by construction
that $U$ is an open, non-empty subset of $V$, and moreover $U$ is also a
proper subset of $V$ since by assumption
$V\not\subseteq\phi_{i_{n+1}}(M_{i_{n+1}})$. We will now show that $U$ is
closed in $V$.

Let $v$ be some element of $Cl^{V}(U)$, and suppose towards a contradiction
that $v$ does not lie in $U$. We will obtain two contradictions by proceeding
via case distinction.

  * Case 1:

$v\in\phi_{i_{n+1}}(M_{i_{n+1}})$. Pick any open neighbourhood
$W\subseteq\phi_{i_{n+1}}(M_{i_{n+1}})$ of $v$. Then the set $W\cap V$ is an
open neighbourhood of $v$ in $V$. Since $v$ lies in the $V$-relative closure
of $U$, the set $W\cap V\cap U$ is non-empty. Without loss of generality we
may suppose that $W\cap V$ is a connected open neighbourhood of $v$ in
$V$.242424Since $V$ is locally-Euclidean, it is also locally-connected, thus
the connected component of $W\cap V$ containing $v$ is still open in $V$.
Since the sets $U$ and $W\cap V$ are both open, connected, non-disjoint
subsets of $V$, their union $U\cup(W\cap V)$ is a connected, open subset of
$V\cap\phi_{i_{n+1}}(M_{i_{n+1}})$ that contains $Z$ as a subset. Since $U$ is
the maximal subset of $V\cap\phi_{i_{n+1}}(M_{i_{n+1}})$ with this property,
it follows that $W\cap V\subseteq U$, thus $v\in U$, which contradicts our
assumption that $v\notin U$.

  * Case 2:

$v\notin\phi_{i_{n+1}}(M_{i_{n+1}})$. Then $v$ lies in some (possibly many)
$M_{j}$ where $j\leq n$. We first show that for any such $j$, it is the case
that $v\in\partial^{\phi_{j}(M_{j})}(M_{j}\cap M_{i_{n+1}})$. Indeed – let $W$
be any open neighbourhood of $v$. Since $v\in V$, the set $W\cap V$ is an open
neighbourhood of $v$ in the subspace $V$. Since $v\in\partial^{V}U$, the set
$W\cap V$ has non-empty intersection with $U$. Since
$U\subseteq\phi_{i_{n+1}}(M_{i_{n+1}})$ and $W\cap V\subseteq W$, it is also
the case that $W\cap\phi_{i_{n+1}}(M_{i_{n+1}})=\emptyset$. It follows that

$v\in
M_{j}\cap\partial\left(\phi_{i_{n+1}}(M_{i_{n+1}})\right)=\partial^{\phi_{j}(M_{j})}(\phi_{i}(A_{ji_{n+1}})).$

Since $v\in V$, it follows that the set
$\partial^{\phi_{j}(M_{j})}(\phi_{i}(A_{ji_{n+1}}))\cap\textsf{Y}^{V}$ is non-
empty, and thus
$\partial^{\phi_{i_{n+1}}(M_{i_{n+1})}}(\phi_{i}(A_{ji_{n+1}}))\subseteq\textsf{Y}^{V}$
by Corollary 4.4. Since $U\subseteq V$, we have that
$U\cap\partial^{\phi_{i_{n+1}}(M_{i_{n+1})}}(\phi_{i}(A_{ji_{n+1}}))=\emptyset$.
We then have the following string of equivalences:

$\displaystyle U$ $\displaystyle=U\cap\phi_{i_{n+1}}(M_{i_{n+1}})$
$\displaystyle=U\cap\left(Int(\phi_{i_{n+1}}(M_{i_{n+1}}\backslash
M_{j}))\cup\phi_{i}(A_{ji_{n+1}})\cup\partial^{n+1}(\phi_{i}(A_{ji_{n+1}}))\right)$
$\displaystyle=\left(U\cap Int(\phi_{i_{n+1}}(M_{i_{n+1}}\backslash
M_{j}))\right)\cup\left(U\cap(\phi_{i}(A_{ji_{n+1}}))\right)\cup\left(U\cap\partial^{n+1}(\phi_{i}(A_{ji_{n+1}}))\right)$
$\displaystyle=\left(U\cap Int(\phi_{i_{n+1}}(M_{i_{n+1}}\backslash
M_{j}))\right)\cup\left(U\cap(\phi_{i}(A_{ji_{n+1}}))\right),$

and thus $U$ may be represented as a union of two non-empty disjoint $V$-open
sets, from which we may conclude that $U$ is disconnected in $V$. This
contradicts our assumption that $U$ is a connected subset of $M_{i_{n+1}}\cap
V$.

In either case, we arrive at a contradiction. We may conclude from this that
$v$ lies in $U$, and thus, since we chose $v$ arbitrarily, $Cl^{V}(U)\subseteq
U$. It follows that $U$ is closed in $V$, as required.

We thus constructed a subset $U$ of $V$ that is non-empty, clopen and proper.
It follows that $V$ is not connected, which contradicts $V$ as an
$H$-submanifold. We may thus conclude that either

$V\subseteq\bigcup_{j=i_{1}}^{i_{n}}\phi_{j}(M_{j})\ \textrm{or}\ V\subseteq
M_{i_{n+1}}.$

In the first case, we may apply the induction hypothesis to conclude that
$V\subseteq\phi_{i}(M_{i})$ for some $i=i_{1},...,i_{n},i_{n+1}$. The desired
equality then follows from the maximality of the canonical images. ∎

It should be clear that the above result allows us to prove Theorem 4.2.
Indeed, we simply take the special case that $I$ is finite, and Theorem 4.2
follows as a corollary.

###  4.4 The Countable Case

Given the results of the previous section, it is natural to wonder whether
there is some form of transfinite induction that can be performed in order to
lift the result of Theorem 4.2 into a setting in which $I$ is allowed to be
countable. Unfortunately, this is the not possible, since it is possible to
construct an adjoined manifold in which the indexing set $I$ is countably
infinite, and in which there is an $H$-submanifold $V$ in which $\partial
V=\textsf{Y}^{V}$, yet $V\neq\phi_{i}(M_{i})$ for all $i$ in $I$. Indeed,
consider the triple $\mathcal{F}:=(\textbf{X},\textbf{A},\textbf{f})$, where:

  * •

each $M_{i}$ equals the real line equipped with the standard topology,

  * •

$A_{ij}=\begin{cases}(-\infty,i)&\textrm{if}\ i<j\\\ (-\infty,j)&\textrm{if}\
i>j\\\ \mathbb{R}&\textrm{if}\ i=j\end{cases}$, and

  * •

each $f_{ij}$ is the identity map on the appropriate domain.

The reader may verify that this collection $\mathcal{F}$ does indeed form an
adjunction system. The resulting adjunction space, which we denote by $N$,
will be a countable collection of real lines, successively splitting in two at
each natural number, as pictured in Figure 8.

$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\cdots$$\vdots$
Figure 8: The idea behind the construction of our counterexample $N$.

We will now show that our adjunction space $N$ contains an $H$-submanifold $V$
that is distinct from each copy of the real line embedded in $M$, yet
nevertheless retains the property $\partial V=\textsf{Y}^{V}$.

######  Lemma 4.10.

Let $N$ be defined as above. The subspace $V$ defined by

$V=\bigcup_{i,j\in I}\phi_{i}(A_{ij})$

is an $H$-submanifold of $N$ in which $\partial V=\textsf{Y}^{V}$.

###### Proof.

Observe first that $V$ is a union of open, connected, pairwise intersecting
subsets of $M$, so $V$ is open and connected. Moreover, since $V$ is the union
of all the $A_{ij}$, and these are Hausdorff, $V$ is also Hausdorff. For
maximality, we can simply apply Hajicek’s criterion. Observe that the boundary
of $V$ is:

$\partial(V)=\partial\left(\bigcup_{i,j\in
I}\phi_{i}(A_{ij})\right)=\bigcup_{i,j\in I}\partial(A_{ij}),$

which is clearly equal to the Hausdorff-violating points $\textsf{Y}^{V}$. ∎

We may thus conclude that in the countable case, some additional property is
required in order to characterise the canonical embeddings of an adjoined
manifold.

##  5 Conclusion

In this paper we have introduced a particular notion of adjunction spaces.
These spaces are sufficiently general as to allow the construction and
characterisation of non-Hausdorff topological manifolds. Specifically, it was
shown in Section 3 that all non-Hausdorff manifolds can be constructed by
gluing together at most countably-many Hausdorff topological manifolds.
Moreover, we saw in Section 4 that these glued-together Hausdorff manifolds
may be the unique submanifolds whose topological boundary coincides with
Hausdorff-violating points.

Although the results of this paper are encouraging, one still gets the feeling
that there is so much still to be done. There is a plethora of results in the
smooth category that need to be studied and outlined – questions of smooth
structures, bundles, calculus on non-Hausdorff manifolds, and so on. Of
course, we will not discuss these here. Some results already exist, and the
interested reader is invited to read the relevant sections of [7]. For now, we
will finish this paper with some potential future developments in the
topological regime.

#### Relationship of Adjunctions with Other Constructions

There are several standard topological constructions – subspaces, product
spaces, quotient spaces, function spaces, to name but a few. Although strictly
outside the scope of this paper, it would be an interesting exercise to
determine how well these operations commute with the adjunction of two spaces.
For instance, in the case of product spaces, we could ask if

$(X\times Z)\cup_{f}(Y\times Z)\cong_{homeo}\left(X\cup_{f}Y\right)\times Z.$

#### Characterisation of Canonical Embeddings in the Countable Case

As mentioned in Section 4.4, in the case of adjoined manifolds formed from
countably-many Hausdorff manifolds, the property $\partial V=\textsf{Y}^{V}$
is not enough to uniquely characterise the canonical embeddings. Some
additional condition is needed, and it is not entirely obvious what that
condition is.

#### Refined Connectedness

We saw in Section 2.2 that it is possible to adjoin connected topological
spaces in such a way that the resulting space remains connected. Given the
scope of this paper, this was done briefly and a discussion of deeper results
was brushed under the carpet. Specfically – in Lemma 2.6 we required that each
gluing region $A_{ij}$ was non-empty, but then remarked that this condition is
slightly too strong, with Figure 3 depicting the intuition for a weaker
condition. Drawing inspiration from this Figure, one guess at the refined
connectivity condition would be to consider a binary relation $C\subseteq
I\times I$ on the indexing set of $\mathcal{F}$, defined as follows:

$iCj\ \textrm{iff}\ A_{ij}\neq\emptyset.$

Clearly the relation $C$ is reflexive and symmetric by conditions A1 and A2 of
Definition 1.8, respectively. Thus the pair $(I,C)$ forms an undirected graph.
However, in general the relation $C$ is not transitive, as Figure 3
illustrates.

It should be clear that at least in the case of finite indexing set $I$, the
connectedness of $C$ viewed as a graph implies the connectedness of the
topological space $\bigcup_{\mathcal{F}}X_{i}$. Indeed – one can consider the
inductive construction of $\bigcup_{\mathcal{F}}X_{i}$ à la Section 1.3, and
successively apply Lemma 2.6 at each stage of the construction. For the
infinite case, perhaps more work is needed.

#### The Algebraic Topology of Adjunction Spaces

During the discussion of connectedness in Section 2.2, it would have been
natural to continue onto a study of the preservation of simple-connectedness.
There are many ways to phrase the definition of a simply-connected space, but
an interesting way is to say that the first fundamental group $\pi_{1}(X)$ is
trivial. This gives rise to an interesting general question – what is the
relationship between the homology groups of an adjunction space and its
component spaces? It is likely that, at least for the first fundamental group,
and application of the Seifert-van Kampen theorem is enough to preserve the
simple-connectedness of the component spaces. However, the general theory of
(co)homology remains unconsidered.

In summary, this paper is only a brief survey of some basic topological
results in the theory of generalised adjunction spaces. The theory of General
Topology is vast, and no doubt there is always something new to find and study
under the gluing operation.

##### Acknowledgment

This paper was made possible with funding provided by the Okinawa Institute of
Science and Technology. Naturally, I am indebted to Yasha Neiman’s Quantum
Gravity unit for acting as a nice soundboard for my ideas. Moreover, I’d like
to thank KP Hart for his help getting this paper out there, and of course Tim
Henke for the ever-stimulating discussions.

##  Appendix A Appendices

Throughout this paper we have occasionally used results whose proofs would
obscure the point of the overall narrative. This is where all these extra
proofs live.

###  A.1 Proofs from Section 1

####  A.1.1 Proof of Lemma 1.4

######  Proposition A.1.

$\phi_{X}$ is injective iff $f$ is injective.

###### Proof.

Suppose that $f$ is injective, and let $x,x^{\prime}\in X$ be distinct.
Without loss of generality we can assume that $x,x^{\prime}$ are elements of
$A$, since otherwise the result is trivial252525If $x\in X\backslash A$ then
$\phi_{X}(x)=[x,1]=\\{(x,1)\\}$. Since $f$ is injective it has a right
inverse, i.e. a map $f^{-1}:f(A)\rightarrow A$ such that $f^{-1}(f(a))=a$ for
each $a$ in $A$. Following the description of the equivalence classes given
above, for each $a\in A$, the mapping under $\phi_{X}$ is given by:

$\phi_{X}(a)=[a,1]=\\{(a,1),(f(a),2)\\}$

It is not hard to see that any distinct element $a^{\prime}$ cannot be mapped
to this equivalence class.

For the converse direction, suppose that $f$ is not injective. Then there is
some $a,a^{\prime}$ such that $f(a)=f(a^{\prime})$. Hence:

$\phi_{X}(a)=[a,1]=\\{(a,1),(a^{\prime},1),(f(a),2)\\}=[a,2]=\phi_{X}(a^{\prime})$

It follows that $\phi_{X}$ is also not injective. By contraposition our result
follows. ∎

######  Lemma A.2.

If $A$ is open in $X$ then $\phi_{Y}$ is an open map.

###### Proof.

Let $U$ be open in $Y$. From the previous proposition we know that
$\phi_{Y}(U)$ is open in $X\cup_{f}Y$ iff both $\phi_{X}^{-1}(\phi_{Y}(U))$
and $\phi_{Y}^{-1}(\phi_{Y}(U))$ are open in $X$ and $Y$ respectively. The
latter is open, since $\phi_{Y}$ is always an injection, we have that
$\phi_{Y}^{-1}(\phi_{Y}(U))=U$. So, it suffices to show that
$\phi_{X}^{-1}(\phi_{Y}(U))$ is an open subset of $X$. It is not hard to
see262626Something like: $a\in f^{-1}(U)$ implies $f(a)\in U$ so
$\phi_{X}(a)=\phi_{Y}(f(a))$ implies that $\phi_{X}(a)\in\phi_{Y}(U)$, hence
$a\in\phi_{X}^{-1}(\phi_{Y}(U))$. Conversely, $x\in\phi_{X}^{-1}(\phi_{Y}(U))$
implies that $\phi_{X}(x)\in\phi_{Y}(U)$, so there is some $u\in U$ such that
$\phi_{X}(x)=\phi_{Y}(u)$, and this is only the case if $x\in A$ and $f(x)=u$,
i.e. $x\in f^{-1}(u)\subseteq f^{-1}(U)$. that:

$\phi_{X}^{-1}(\phi_{Y}(U))=\\{x\in X\ |\ \phi_{X}(x)\in\phi_{Y}(U)\\}=\\{a\in
A\ |\ f(a)\in U\\}$

Put differently, $\phi_{X}^{-1}(\phi_{Y}(U))=f^{-1}(U)$. Since $f$ is a
continuous map and $U$ open, the set $f^{-1}(U)$ is open in $A$. Since $A$ is
equipped with the subspace topology, this is the case iff $f^{-1}(U)=A\cap V$
where $V$ is some open set in $X$. By assumption, $A$ is open in $X$ and thus
$f^{-1}(U)$ is the intersection of open subsets of $X$, so is also open in
$X$. It follows that $\phi_{Y}(U)$ is open in $X\cup_{f}Y$. Since $U$ was
arbitrary, we can conclude that $\phi_{Y}$ is an open map. ∎

######  Lemma A.3.

If $A$ is open in $X$ and $f$ is an open map, then $\phi_{X}$ is an open map.

###### Proof.

Let $U$ be an open subset of $X$, and consider $\phi_{X}(U)\subset
X\cup_{f}Y$. As in the previous lemma, in order to show that $\phi_{X}(U)$ is
open in the adjunction space, it suffices to show that the preimages of
$\phi_{X}(U)$ under $\phi_{X}$ and $\phi_{Y}$ are open in their respective
spaces.

Consider first $\phi_{X}^{-1}(\phi_{X}(U))$. Observe272727If
$x\in\phi_{X}^{-1}(\phi_{X}(U))$ then $\phi_{X}(x)=\phi_{X}(u)$ for some $u\in
U$. If $x\notin U$, this can only be the case if $x$ and $u$ are in $A$ and
$f(x)=f(u)$, i.e. $x\in f^{-1}(f(u))$. Since $u\in A\cap U$, we get that $x\in
f^{-1}(f(A\cap U))$. Conversely, $x\in f^{-1}(f(A\cap U))$ implies that
$f(x)=f(u)$ for some $u\in A\cap U$. Hence $[x,1]=\\{(f(x),2)\\}\cup\\{(a,1)\
|\ f(a)=f(x)\\}=[u,1]$, hence $\phi_{X}(x)=\phi_{X}(u)$ and thus
$x\in\phi_{X}^{-1}(\phi_{X}(u))\subseteq\phi_{X}^{-1}(\phi_{X}(U))$. that:

$\displaystyle\phi_{X}^{-1}(\phi_{X}(U))$ $\displaystyle=\\{x\in X\ |\
\phi_{X}(x)\in\phi_{X}(U)\\}$ $\displaystyle=U\cup\\{a\in A\ |\ \exists u\in
U\cap A(f(u)=f(a)\\}$ $\displaystyle=U\cup f^{-1}(f(A\cap U))$

We show that $f^{-1}(f(A\cap U))$ is open in $X$. Observe that since $U$ is
open in $X$, the set $A\cap U$ is open in $A$. By assumption, $f$ is an open
map, hence $f(A\cap U)$ is open in $Y$. Since $f$ is continuous,
$f^{-1}(f(A\cap U))$ is open in $A$, i.e. $f^{-1}(f(A\cap U))=A\cap V$ for
some $V$ open in $X$. By assumption, $A$ is open in $X$ and thus
$f^{-1}(f(A\cap U))$ is open in $X$. We can conclude from all this that
$\phi_{X}^{-1}(\phi_{X}(U))$ is open in $X$.

Consider now the set $\phi_{Y}^{-1}(\phi_{X}(U))$. Observe282828If
$y\in\phi_{Y}^{-1}(\phi_{X}(U))$ then $\phi_{Y}(y)=\phi_{X}(u)$ for some $u\in
U$, which can only be the case if $u\in A$ and $u\in f^{-1}(y)$. Hence
$f(u)=y$ and $y\in f(A\cap U)$. Conversely, $y\in f(A\cap U)$ implies there is
some $u\in A\cap U$ such that $f(u)=y$. In particular, $u\in A$, hence
$\phi_{X}(u)=\phi_{Y}(f(u))=\phi_{Y}(y)$. So since $\phi_{Y}$ is injective,
$y=\phi_{Y}^{-1}(\phi_{X}(u))\subseteq\phi_{Y}^{-1}(\phi_{X}(U))$. that:

$\displaystyle\phi_{Y}^{-1}(\phi_{X}(U))$ $\displaystyle=\\{y\in Y\ |\ \exists
u(u\in U\wedge\phi_{Y}(y)=\phi_{X}(u))\\}$ $\displaystyle=\\{y\in Y\ |\
\exists u(u\in U\cap A\wedge f(u)=y)\\}$ $\displaystyle=f(A\cap U)$

Since $U$ is open in $X$, the set $U\cap A$ is open in $A$. By assumption, $f$
is an open map, hence $f(A\cap U)$ is an open set in $Y$.

We can conclude that $\phi_{X}(U)$ is open in $X\cup_{f}Y$, and thus
$\phi_{X}$ is an open map. ∎

######  Lemma A.4.

If $f$ is injective and open, then $\phi_{X}$ is open.

###### Proof.

Let $U\subseteq X$ and consider $\phi_{X}(U)$. Since $f$ is injective, it
follows from Prop. A.1 that $\phi_{X}$ is also injective, and thus
$\phi_{X}^{-1}(\phi_{X}(U))=U$, which is clearly open in $X$. Consider now
$\phi_{Y}^{-1}(\phi_{X}(U))$. We can as in the previous lemma to conclude that
$\phi_{Y}^{-1}(\phi_{X}(U))$ is open. It follows that $\phi_{X}$ is an open
map. ∎

######  Lemma A.5.

$\phi_{Y}$ is always a topological embedding.

###### Proof.

Since $\phi_{Y}$ is injective, the restriction to its image is a bijection.
Similarly, the restriction of a continuous map to its image is also
continuous292929Let $f:X\rightarrow Y$ be any continuous map between
topological spaces, and consider $f|_{X}:X\rightarrow f(X)$. Let $U\subseteq
f(X)$ be open in the subspace topology. Then $U=f(X)\cap V$ where $V$ is open
in $Y$. The preimage is then $(f|_{X})^{-1}(U)=f^{-1}(f(X)\cap V)=X\cap
f^{-1}(V)=f^{-1}(V)$ which is open in $X$ since $f$ is continuous. and thus it
suffices to show that $\phi_{Y}$ is homeomorphic onto it’s domain, i.e. the
map $\phi_{Y}:Y\rightarrow\phi_{Y}(Y)$ is open.

Let $U$ be an open subset of $Y$. Since $\phi_{Y}(Y)$ is equipped with the
subspace topology, in order to show that $\phi_{Y}(U)$ is open in
$\phi_{Y}(Y)$, we need to show that $\phi_{Y}(U)=\phi_{Y}(Y)\cap V$, where $V$
is some open subset of $X\cup_{f}Y$. Observe that $U$ is open in $Y$ and $f$
is continuous, so the preimage $f^{-1}(U)$ is open in $A$. Since $A$ is
equipped with the subspace topology, this means that $f^{-1}(U)=A\cap W$,
where $W$ is some open set in $X$.

We now argue that the set $\phi_{X}(W)\cup\phi_{Y}(U)$ suffices as our choice
for $V$. Note first that:

$\displaystyle\phi_{Y}(Y)\cap(\phi_{X}(W)\cup\phi_{Y}(U))$
$\displaystyle=(\phi_{Y}(Y)\cap\phi_{X}(W))\cup(\phi_{Y}(Y)\cap\phi_{Y}(U))$
$\displaystyle=(\phi_{Y}(Y)\cap\phi_{X}(W))\cup\phi_{Y}(U)$

Let $z\in\phi_{X}(W)\cap\phi_{Y}(Y)$. Then $z=\phi_{X}(w)$ for some $w\in W$,
and $z=\phi_{Y}(y)$ for some $y\in Y$. This can only be the case if $w\in A$
and $f(w)=y$. Since $w\in A$, it follows that $w\in A\cap W=f^{-1}(U)$, hence
$f(w)\in U$, and thus $z=\phi_{Y}(y)=\phi_{Y}(f(w))\in U$. We can conclude
that $\phi_{X}(W)\cap\phi_{Y}(Y)\subseteq U$, and thus
$(\phi_{Y}(Y)\cap\phi_{X}(W))\cup\phi_{Y}(U)=\phi_{Y}(U)$ as required.

We now show that the set $\phi_{X}(W)\cup\phi_{Y}(U)$ is open in $X\cup_{f}Y$.
Observe first that:

$\displaystyle\phi_{X}^{-1}(\phi_{X}(W)\cup\phi_{Y}(U))$
$\displaystyle=\phi_{X}^{-1}(\phi_{X}(W))\cup\phi_{X}^{-1}(\phi_{Y}(Y))$
$\displaystyle=(W\cup f^{-1}(U))\cup f^{-1}(U)$ $\displaystyle=W$
$\displaystyle\phi_{Y}^{-1}(\phi_{X}(W)\cup\phi_{Y}(U))$
$\displaystyle=\phi_{Y}^{-1}(\phi_{X}(W))\cup\phi_{Y}^{-1}(\phi_{Y}(Y))$
$\displaystyle=f(W\cap A)\cup U$ $\displaystyle=U$

Since $W$ and $U$ are open sets of $X$ and $Y$ respectively, it follows from
Prop. 1.3 that $\phi_{X}(W)\cup\phi_{Y}(U)$ is open in $X\cup_{f}Y$. We may
thus conclude that $\phi_{Y}(U)=\phi_{Y}(Y)\cap(\phi_{X}(W)\cup\phi_{Y}(U))$
is open in the subspace topology $\phi_{Y}(Y)$. Since we picked an arbitrary
open set $U$ of $Y$, it follows that the map
$\phi_{Y}:Y\rightarrow\phi_{Y}(Y)$ is an open map, and thus is a homeomorphism
as required. ∎

######  Lemma A.6.

If $f$ is an injective open map, then $\phi_{X}$ is a topological embedding.

###### Proof.

Note first that since $f$ is by assumption an injection, it follows from Prop.
A.1 that $\phi_{X}$ is also injective. Since $\phi_{X}$ is continuous, it is
continuous once restricted to its image. So, it suffices to show that
$\phi_{X}$ is homeomorphic onto its image.

Let $U\subseteq X$ be an open set. We want a set $V$ such that
$\phi_{X}(U)=\phi_{X}(X)\cap V$. The set $V:=\phi_{X}(U)\cup\phi_{Y}(f(U\cap
A))$ will suffice. Observe first that:

$\phi_{X}(X)\cap(\phi_{X}(U)\cup\phi_{Y}(f(U\cap
A)))=...=\phi_{X}(U)\cup(\phi_{X}(X)\cap\phi_{Y}(f(U\cap A)))$

Let $z\in\phi_{X}(X)\cap\phi_{Y}(f(U\cap A))$. Then there is some $x\in X$ and
some $y\in f(U\cap A)$ such that $\phi_{X}(x)=z=\phi_{Y}(y)$. This can only be
the case if $f(x)=y$ and since $y\in f(U\cap A)$, it must be the case that
$x\in U$. It follows that $z\in\phi_{X}(U)$ and thus
$\phi_{X}(X)\cap\phi_{Y}(f(U\cap A))$ is a subset of $U$, whence
$\phi_{X}(U)=\phi_{X}(X)\cap V$.

We now show that $V$ is open in $X\cup_{f}Y$, i.e. $\phi_{X}^{-1}(V)$ and
$\phi_{Y}^{-1}(V)$ are open in $X$ and $Y$ respectively.

Consider first $\phi_{X}^{-1}(V)$. Observe:

$\displaystyle\phi_{X}^{-1}(\phi_{X}(U)\cup\phi_{Y}(f(U\cap A)))$
$\displaystyle=\phi_{X}^{-1}(\phi_{X}(U))\cup\phi_{X}^{-1}(\phi_{Y}(f(U\cap
A)))$ $\displaystyle=U\cup f^{-1}(f(A\cap U))$ $\displaystyle=U\cup(A\cap U)$
$\displaystyle=U$

Where in the second and third lines we use that $\phi_{X}$ and $f$ are
injective. Since $U$ is open, we may conclude that $\phi_{X}^{-1}(V)$ is open.
Now observe that:

$\displaystyle\phi_{Y}^{-1}(\phi_{X}(U)\cup\phi_{Y}(f(U\cap A)))$
$\displaystyle=\phi_{Y}^{-1}(\phi_{X}(U))\cup\phi_{Y}^{-1}(\phi_{Y}(f(U\cap
A)))$ $\displaystyle=f(A\cap U)\cup f(A\cap U)$ $\displaystyle=f(A\cap U)$

Since $U$ is open, the set $A\cap U$ is an open set the subspace topology on
$A$. Since we assumed that $f$ is an open map, it follows that the set
$f(A\cap U)=\phi_{Y}^{-1}(V)$ is open in $Y$.

We may conclude that $V$ is open in $X\cup_{f}Y$, and since
$\phi_{X}(U)=V\cap\phi_{X}(X)$, it follows that $\phi_{X}(U)$ is open in
$\phi_{X}(X)$ equipped with the subspace topology. As such, $\phi_{X}$ is an
open map onto $\phi_{X}(X)$, and thus $\phi_{X}:X\rightarrow X\cup_{f}Y$ is a
topological embedding. ∎

####  A.1.2 Proof of Lemma 1.6

######  Proposition A.7.

Let $\mathcal{B}^{X}$ and $\mathcal{B}^{Y}$ be bases for $X$ and $Y$
respectively. If $\phi_{X}$ and $\phi_{Y}$ are open maps, then the collection
$\mathcal{B}=\\{\phi_{X}(U)\ |\ U\in\mathcal{B}^{X}\\}\cup\\{\phi_{Y}(V)\ |\
V\in\mathcal{B}^{Y}\\}$ forms a basis for the adjunction topology $\tau_{A}$.

###### Proof.

It suffices to show that $\mathcal{B}$ consists of open sets, and that every
open set in $X\cup_{f}Y$ can be represented as a union of elements of
$\mathcal{B}$. To see that $\mathcal{B}$ consists of open sets, we can appeal
to our assumption that the canonical maps $\phi_{X}$ and $\phi_{Y}$ are open.
Since every element $U$ of $\mathcal{B}^{X}$ is open in $X$, it follows that
$\phi_{X}(U)$ is open in $X\cup_{f}Y$. The case for $Y$ is similar. Thus
$\mathcal{B}$ consists of open sets.

Suppose now that $W$ is an open set in $X\cup_{f}Y$. By Prop. 1.3, the sets
$\phi_{X}^{-1}(W)$ and $\phi_{Y}^{-1}(W)$ are open in $X$ and $Y$,
respectively. As such, these preimages can be represented as unions of basis
elements, i.e.

$\phi_{X}^{-1}(W)=\bigcup_{i\in I}U_{i}\ \ \ \textrm{and}\ \ \
\phi_{Y}^{-1}(W)=\bigcup_{j\in J}V_{j}$

where each $U_{i}\in\mathcal{B}^{X}$ and each $V_{j}\in\mathcal{B}^{Y}$. Since
for any303030Since $X\cup_{f}Y=\phi_{X}(X)\cup\phi_{Y}(Y)$, any subset $Z$ of
$X\cup_{f}Y$ can be represented as
$Z=(Z\cap\phi_{X}(X))\cup(Z\cap\phi_{Y}(Y))$. Since $\phi_{X}$ is a surjective
map once restricted to its image, it has a right-inverse $\phi_{X}^{-1}$. In
this case, $\phi_{X}(\phi^{-1}(W))=W$ for any subset $W\subseteq\phi_{X}(X)$.
In particular, $Z\cap\phi_{X}(X)$ is a subset of $\phi_{X}(X)$, so
$Z\cap\phi_{X}(X)=\phi_{X}(\phi_{X}^{-1}(Z\cap\phi_{X}(X)))=...=\phi_{X}(\phi_{X}^{-1}(Z))\cap\phi_{X}(\phi_{X}^{-1}(\phi_{X}(X)))=\phi_{X}(\phi_{X}^{-1}(Z))\cap\phi_{X}(X)=\phi_{X}(\phi_{X}^{-1}(Z))$.
The case for $\phi_{Y}$ is similar, and we then get
$Z=(Z\cap\phi_{X}(X))\cup(Z\cap\phi_{Y}(Y))=\phi_{X}(\phi_{X}^{-1}(Z))\cup\phi_{Y}(\phi_{Y}^{-1}(Z))$.
subset $Z\subseteq X\cup_{f}Y$ it is the case that
$Z=\phi_{X}(\phi_{X}^{-1}(Z))\cup\phi_{Y}(\phi_{Y}^{-1}(Z))$, it follows that:

$\displaystyle W$
$\displaystyle=\phi_{X}(\phi_{X}^{-1}(W))\cup\phi_{Y}(\phi_{Y}^{-1}(W))$
$\displaystyle=\phi_{X}\left(\bigcup_{i\in
I}U_{i}\right)\cup\phi_{Y}\left(\bigcup_{j\in J}V_{j}\right)$
$\displaystyle=\left(\bigcup_{i\in
I}\phi_{X}(U_{i})\right)\cup\left(\bigcup_{j\in J}\phi_{Y}(V_{j})\right)$

and thus $W$ is a union of members of $\mathcal{B}$. We can conclude that
$\mathcal{B}$ is indeed a basis. ∎

######  Lemma A.8.

If $X,Y$ are connected and $A$ is non-empty, then $X\cup_{f}Y$ is connected.

###### Proof.

Suppose towards a contradiction that $X\cup_{f}Y$ is disconnected, i.e. there
exist disjoint open sets $U,V$ of $X\cup_{f}Y$ such that $U,V$ are both non-
empty and cover $X\cup_{f}Y$. Consider $\phi_{X}^{-1}(U)$ and
$\phi_{X}^{-1}(V)$. By assumption, these sets are open in $X$, and since pre-
images preserve disjointedness313131Simple enough -
$\phi_{X}^{-1}(U)\cap\phi_{X}^{-1}(V)=\phi_{X}^{-1}(U\cap
V)=\phi_{X}^{-1}(\emptyset)=\emptyset$, they are disjoint. It is also the case
that they cover323232$\phi^{-1}(U)\cup\phi_{X}(V)=\phi_{X}^{-1}(U\cup V)=X$
$X$, in order to preserve the connectedness of $X$, it must be the case that
either $\phi_{X}^{-1}(U)$ or $\phi_{X}^{-1}(V)$ is empty. Without loss of
generality suppose that $\phi_{X}^{-1}(U)=\emptyset$. It follows from
$\phi_{X}^{-1}(V)=X$ that $\phi_{X}(X)\subseteq V$ and, since
$\phi_{X}(A)=\phi_{Y}(f(A))$, that $f(A)\subseteq\phi_{Y}^{-1}(V)$. Since we
have assumed that $A$ is non-empty, this means that $\phi_{Y}^{-1}(V)$ is also
non-empty. However, in order to preserve the connectedness of $Y$, it cannot
be the case that $\phi_{Y}^{-1}(U)$ is also non-empty. We have thus arrived at
our contradiction - if $\phi_{Y}^{-1}(U)=\emptyset$, then
$\phi_{Y}(Y)\subseteq V$, and thus
$X\cup_{f}Y=\phi_{X}(X)\cup\phi_{Y}(Y)\subset V$, contradicting $U$ as a non-
empty set333333This proof is a bit weak at the end - we want to say that
$X\cup_{f}Y=V$ and thus $U=\emptyset$.. ∎

######  Lemma A.9.

If both $X$ and $Y$ are compact, then so is $X\cup_{f}Y$.

###### Proof.

It is well-known that the disjoint union of two (or, finitely-many) compact
spaces is again compact, and that the quotient of a compact space is
compact343434This is a corollary of [3, Thm. 3.2.3.].. In our context, this
means that $X\sqcup Y$ is compact and thus $X\cup_{f}Y$ is also compact. ∎

###  A.2 Some Results from General Topology

######  Lemma A.10.

Let $f:X\rightarrow Y$ be a continuous map. If $D$ is a dense subset of $X$,
then the image $f(D)$ is dense in the subspace $f(X)$.

###### Proof.

Suppose $f:X\rightarrow Y$ is a continuous map and $D$ is some dense subset of
$X$. To see that $f(D)$ is dense in $f(X)$, let $y$ be any point in $f(X)$ and
consider some open neighbourhood $U$ of $y$ in $f(X)$. Since $f$ is
continuous, the preimage $f^{-1}(U)$ is an open neighbourhood of $x$ in $X$.
Since $D$ is dense, it is the case that $f^{-1}(U)\cap D$ is non-empty. Any
element in this intersection will get mapped into the set $U\cap f(D)$ under
$f$. It follows that $y$ lies in the $f(X)$-relative closure of $f(D)$, and
thus, since we chose $y$ arbitrarily, we may conclude that $f(X)=Cl(f(D))$,
that is, $f(D)$ is dense in $f(X)$. ∎

######  Proposition A.11.

If $A$ is a connected, open subset of $M$, then $A$ is also path-connected.

###### Proof.

The proof of this result revolves around the same result for Euclidean space
(see [12, Prop 12.25]). Let $a\in A$ and let $X$ be the set of all members of
$A$ that are path-connected to $a$. Suppose towards a contradiction that
$X\neq A$. We show that both $X$ and $A\backslash X$ are open.

Let $b\in X$, and consider a chart $(U,\varphi)$ of $M$ at $b$. We show that
$U\cap A\subseteq X$. Let $c\in U\cap A$. Since $\varphi(U)$ is homeomorphic
to $\mathbb{R}^{n}$, in particular it is connected and open. As such, we can
use [12, Prop 12.25] to conclude that there is a path
$\gamma:[0,1]\rightarrow\varphi(U)$ in $\mathbb{R}^{n}$ connecting
$\varphi(b)$ and $\varphi(c)$. Since $\varphi$ is a homeomorphism, in
particular it’s inverse map $\varphi^{-1}$ is continuous, thus the composition
$\varphi^{-1}\circ\gamma:[0,1]\rightarrow U$ defines a path in $M$ that
connects $b$ and $c$. Since we assumed that $b\in X$, there is also a path
connecting $a$ to $b$. We can adjoin these two paths to conclude that $a$ and
$c$ are path-connected. Since $c$ was an arbitrary member of $U\cap A$, we may
conclude that $U\cap A\subset X$. Since $b$ was arbitrary, it follows that $X$
is open.

We now show that the complement $Y:=A\backslash X$ is open. Let $b\in Y$, and
again consider the set $A\cap U$, where $(U,\varphi)$ is a chart for $M$ at
$b$. Suppose towards a contradiction that $U\cap A$ and $X$ intersect. Then
there is some element $c\in U\cap A$ that is path-connected to $a$. However,