# Decomposition space theory

Boldizsár Kalmár

##  1\. Introduction

In these notes we give a brief introduction to decomposition theory and we
summarize some classical and well-known results. The main question is that if
a partitioning of a topological space (in other words a _decomposition_) is
given, then what is the topology of the quotient space. The main result is
that an _upper semi-continuous_ decomposition yields a homeomorphic
decomposition space if the decomposition is _shrinkable_ (i.e. there exist
self-homeomorphisms of the space which shrink the partitions into arbitrarily
small sets in a controllable way). This is called _Bing shrinkability
criterion_ and it was introduced in [Bi52, Bi57]. It is applied in major
$4$-dimensional results: in the disk embedding theorem and in the proof of the
$4$-dimensional topological Poincaré conjecture [Fr82, FQ90, BKKPR]. It is
extensively applied in constructing approximations of manifold embeddings in
dimension $\geq 5$, see [AC79] and Edwards’s cell-like approximation theorem
[Ed78]. If a decomposition is shrinkable, then a decomposition element has to
be _cell-like_ and _cellular_. Also the quotient map is approximable by
homeomorphisms. A cell-like map is a map where the point preimages are similar
to points while a cellular map is a map where the point preimages can be
approximated by balls. There is an essential difference between the two types
of maps: ball approximations always give cell-like sets but in a smooth
manifold for a cell-like set $C$ the complement has to be simply connected in
a nbhd of $C$ in order to be cellular. Finding conditions for a decomposition
to be shrinkable is one of the main goal of the theory. For example, cell-like
decompositions are shrinkable if the non-singleton decomposition elements have
codimension $\geq 3$, that is any maps of disks can be made disjoint from them
[Ed16]. In many constructions Cantor sets (a set of uncountably many points
that cutting out from the real line we are left with a manifold) arise as
limits of sequences of sets defining the decomposition. The interesting fact
is that a limit Cantor set can be non-standard and it can have properties very
different from the usual middle-third Cantor set in $[0,1]$. An example for
such a non-standard Cantor set is given by Antoine’s necklace but many other
explicit constructions are studied in the subsequent sections. The present
notes will cover the following: upper semi-continuous decompositions, defining
sequences, cellular and cell-like sets, examples like Whitehead continuum,
Antoine’s necklace and Bing decomposition, shrinkability criterion and near-
homeomorphism, approximating by homeomorphisms and shrinking countable upper
semi-continuous decompositions. We prove for example that every cell-like
subset in a $2$-dimensional manifold is cellular, that Antoine’s necklace is a
wild Cantor set, that in a complete metric space a usc decomposition is
shrinkable if and only if the decomposition map is a near-homeomorphism and
that every manifold has collared boundary.

##  2\. Decompositions

A neighborhood (nbhd for short) of a subset $A$ of a topological space $X$ is
an open subset of $X$ which contains $A$.

######  Definition 2.1.

Let $X$ be a topological space. A set $\mathcal{D}\subset\mathcal{P}(X)$ is a
_decomposition_ of $X$ if the elements of $\mathcal{D}$ are pairwise disjoint
and $\bigcup\mathcal{D}=X$. An element of $\mathcal{D}$ which consists of one
single point is called a _singleton_. A non-singleton decomposition element is
called _non-degenerate_. The elements of $\mathcal{D}$ are the _decomposition
elements_. The set of non-degenerate elements is denoted by
$\mathcal{H}_{\mathcal{D}}$.

If $f\colon X\to Y$ is an arbitrary (not necessarily continuous) map between
the topological spaces $X$ and $Y$, then the set

$\\{f^{-1}(y):y\in Y\\}$

is a decomposition of $X$. A decomposition defines an equivalence relation on
$X$ as usual, i.e. $a,b\in X$ are equivalent iff $a$ and $b$ are in the same
element of $\mathcal{D}$.

######  Definition 2.2.

If $\mathcal{D}$ is a decomposition of $X$, then the _decomposition space_
$X_{\mathcal{D}}$ is the space $\mathcal{D}$ with the following topology: the
subset $U\subset\mathcal{D}$ is open exactly if $\pi^{-1}(U)$ is open. Here
$\pi\colon X\to\mathcal{D}$ is the _decomposition map_ which maps each $x\in
X$ into its equivalence class.

In other words $X_{\mathcal{D}}$ is the quotient space with the quotient
topology and

$\pi\colon X\to X_{\mathcal{D}}$

is just the quotient map. Recall that by well-known statements
$X_{\mathcal{D}}$ is compact, connected and path-connected if $X$ is compact,
connected and path-connected, respectively. Obviously $\pi$ is continuous.

######  Proposition 2.3.

The decomposition space is a $T_{1}$ space if the decomposition elements are
closed.

###### Proof.

We have to show that the points in the space $X_{\mathcal{D}}$ are closed. If
$U$ is a point complement in $X_{\mathcal{D}}$, then $\pi^{-1}(U)$ is the
complement of a decomposition element, which is open so $U$ is also open. ∎

We would like to construct and study such decompositions which have especially
nice properties concerning the behavior of the sequences of decomposition
elements.

######  Definition 2.4.

Let $f\colon X\to\mathbb{R}$ be a function. It is _upper semi-continuous_
(resp.  _lower semi-continuous_) if for every $x\in X$ and $\varepsilon>0$
there is a nbhd $V_{x}$ such that $f(V_{x})\subset(-\infty,f(x)+\varepsilon)$
(resp. $f(V_{x})\subset(f(x)-\varepsilon,\infty))$.

For us, upper semi-continuous functions will be important. They are such
functions, where a sequence $f(x_{n})$ can have only smaller or equal values
than $f(x)+\varepsilon_{n}$ as $x_{n}\rightarrow x$, where
$\varepsilon_{n}\geq 0$ and $\varepsilon_{n}\rightarrow 0$. Let
$f\colon\mathbb{R}\to\mathbb{R}$ be an upper semi-continuous, positive
function and consider the following decomposition of $\mathbb{R}^{2}$. Take
the vertical segments of the form

(2.1) $A_{x}=\\{(x,y):y\in[0,f(x)]\\}$

for each $x\in\mathbb{R}$. Together with the points in $\mathbb{R}^{2}$ which
are not in these segments (these points are the so-called singletons) this
gives a decomposition of $\mathbb{R}^{2}$. This has an interesting property:
let $y\in[0,f(x)]$ for some $x\in\mathbb{R}$ and let $(x_{n})\in\mathbb{R}$ be
a sequence (which is not necessarily convergent). If every nbhd of the point
$(x,y)$ intersects all but finitely many segments $A_{x_{n}}$, then the points
$(u,v)\in\mathbb{R}^{2}$ each of whose nbhds intersects all but finitely many
$A_{x_{n}}$ are in $A_{x}$ as well, see Figure 1. The set of the points
$(u,v)$ is called the _lower limit_ of the sequence $A_{x_{n}}$. In other
words, if an $A_{x}$ intersects the lower limit of a sequence $A_{x_{n}}$,
then all the lower limit is a subset of $A_{x}$. More generally we have the
following.

Figure 1. The graph of an upper semi-continuous function $f$ and some segments
$A_{x}$. If the segments $A_{x_{n}}$ “converge” to a segment $A_{x}$, then
$f(x_{n})$ converges to a number $\leq f(x)$.

######  Definition 2.5.

Let $A_{n}$ be a sequence of subsets of the space $X$. The _lower limit_ of
$A_{n}$ is the set of the points $p\in X$ each of whose nbhds intersects all
but finitely many $A_{n}$. It is denoted by $\liminf A_{n}$. The _upper limit_
of $A_{n}$ is the set of the points $p\in X$ each of whose nbhds intersects
infinitely many $A_{n}$s. It is denoted by $\limsup A_{n}$.

Note that $\liminf A_{n}\subset\limsup A_{n}$ is always true. In the previous
example the sets $A_{x_{n}}$ could approach the set $A_{x}$ only in a manner
determined by the function $f$. This leads to the following general
definition.

######  Definition 2.6.

Let $\mathcal{D}$ be a decomposition of a space $X$ such that all elements of
$\mathcal{D}$ are closed and compact and they can converge to each other only
in the following way: if $A\in\mathcal{D}$, then for every nbhd $U$ of $A$
there is a nbhd $V$ of $A$ with the property $V\subset U$ such that if some
element $B\in\mathcal{D}$ intersects $V$, then $B\subset U$, i.e. the set $B$
is completely inside the nbhd $U$. Then $\mathcal{D}$ is an _upper semi-
continuous decomposition_ (_usc_ decomposition for short). If all the
decomposition elements are closed but not necessarily compact, then we say it
is a _closed upper semi-continuous decomposition_.

For example, the decomposition defined in (2.1) is usc.

######  Lemma 2.7.

Let $\mathcal{D}$ be a decomposition of the space $X$ such that each
decomposition element is closed. The following are equivalent:

  1. (1)

$\mathcal{D}$ is a closed usc decomposition,

  2. (2)

for every $D\in\mathcal{D}$ and every nbhd $U$ of $D$ there is a saturated
nbhd $W\subset U$ of $D$, that is an open set $W$ which is a union of
decomposition elements,

  3. (3)

for each open subset $U\subset X$, the set $\cup\\{D\in\mathcal{D}:D\subset
U\\}$ is open,

  4. (4)

for each closed subset $F\subset X$, the set $\cup\\{D\in\mathcal{D}:D\cap
F\neq\emptyset\\}$ is closed,

  5. (5)

the decomposition map $\pi\colon X\to X_{\mathcal{D}}$ is a closed map.

###### Proof.

Suppose $\mathcal{D}$ is usc and $U$ is a nbhd of $D$. Let $W$ be the union of
all decomposition elements which are subsets of $U$. Then $D\subset W$
obviously and $W$ is open because if $x\in W$, then $x\in D^{\prime}$ for some
decomposition element $D^{\prime}\subset W$ and $D^{\prime}\subset U$, so by
definition $D^{\prime}\subset V\subset U$ for a nbhd $V$ but the nbhd $V$ of
$x$ is in $W$ since all the decomposition elements intersecting $V$ have to be
in $U$, which means they are in $W$ as well. This shows that (1) implies (2).
Suppose (2) holds. If $U$ is an open set, then for each decomposition element
$D\subset U$ a saturated nbhd $W$ of $D$ is also in $U$ and also in
$\cup\\{D\in\mathcal{D}:D\subset U\\}$. This means that the set
$\cup\\{D\in\mathcal{D}:D\subset U\\}$ is a union of open sets, which proves
(3). We have that (3) and (4) are equivalent because we can take the
complement of a given closed set $F$ or an open set $U$. We have that (4) and
(5) are equivalent: from (4) we can show (5) by taking an arbitrary closed set
$F\subset X$, then $\cup\\{D\in\mathcal{D}:D\cap F\neq\emptyset\\}$ is closed,
its complement is a saturated open set whose $\pi$-image is open so $\pi(F)$
is closed. If we suppose (5), then for a closed set $F\subset X$ the set

$\pi^{-1}(\pi(F))=\cup\\{D\in\mathcal{D}:D\cap F\neq\emptyset\\}$

is closed so we get (4). Finally (3) implies (1): if $D\in\mathcal{D}$ and $U$
is a nbhd of $D$, then let $V$ be the open set
$\cup\\{D\in\mathcal{D}:D\subset U\\}$, this is a nbhd of $D$, it is in $U$
and if a $D^{\prime}\in\mathcal{D}$ intersects $V$, then it is in $V$ and
hence also in $U$. ∎

There is also the notion of _lower semi-continuous decomposition_ : a
decomposition $\mathcal{D}$ of a metric space is lower semi-continuous if for
every element $A\in\mathcal{D}$ and for every $\varepsilon>0$ there is a nbhd
$V$ of $A$ such that if some decomposition element $B$ intersects $V$, then
$A$ is in the $\varepsilon$-nbhd of $B$. A decomposition of a metric space is
_continuous_ if it is upper and lower semi-continuous, see Figure 2. We will
not study decompositions which are only lower semi-continuous.

Figure 2. A lower semi-continuous, an upper semi-continuous and a continuous
decomposition. In each of the cases the non-degenerate decomposition elements
are line segments, which converge to other line segments. The dots indicate
convergence. Only the non-singleton decomposition elements are sketched. The
lower semi-continuous decomposition consists of decomposing the area under the
graph of a lower semi-continuous function into vertical line segments, there
are no singletons among the decomposition elements and the decomposed space
itself is not closed. The upper semi-continuous and continuous decompositions
are decompositions of the rectangle. Only the upper semi-continuous
decomposition has singletons.

######  Theorem 2.8.

Let $X$ be a $T_{3}$ space and $\mathcal{D}$ is a closed usc decomposition. If
$A_{n}\in\mathcal{D}$ is a sequence of decomposition elements and
$A\in\mathcal{D}$ are such that $A\cap\liminf A_{n}\neq\emptyset$, then
$\limsup A_{n}\subset A$.

###### Proof.

Suppose there is a point $x\in A$ such that $x\in\liminf A_{n}$ as well. By
contradiction suppose that $\limsup A_{n}\nsubseteq A$, this means that a
point $y\in\limsup A_{n}$ is such that $y\notin A$. Since $y\in D$ for a
decomposition element, we get $D\neq A$ so $D$ is disjoint from the
decomposition element $A$. The space $X$ is $T_{3}$, the sets $D$ and
$\\{x\\}$ are closed so there is a nbhd $U$ of $D$ and a nbhd $V$ of $x$ which
are disjoint from each other. We also have a nbhd $W\subset U$ of $D$ which is
a union of decomposition elements by Lemma 2.7. Since $x\in\liminf A_{n}$, we
have that for an integer $k$ the sets $A_{k},A_{k+1},\ldots$ intersect $V$.
The nbhd $W$ is saturated, this implies that a decomposition element does not
intersect both of $W$ and $V$. So $A_{k},A_{k+1},\ldots$ are disjoint from
$W$. This contradicts to that $W$ is a nbhd of $y$ and so infinitely many
$A_{n}$ has to intersect $W$ because $y\in\limsup A_{n}$. ∎

An other example for a usc decomposition is the equivalence relation on
$S^{n}$ defined by $x\sim-x$. Here the decomposition elements are not
connected and the decomposition space is the projective space
${\mathbb{R}}{P}^{n}$. Or another example is the closed usc decomposition of
$\mathbb{R}^{2}$, where the two non-singleton decomposition elements are the
two arcs of the graph of the function $x\mapsto 1/x$, all the other
decomposition elements are singletons. The decomposition space is homeomorphic
to

$A\cup_{\varphi}B\cup_{\psi}A^{\prime},$

where $A$ and $A^{\prime}$ are open disks, each of them with one additional
point in its frontier denoted by $a$ and $a^{\prime}$ respectively. The space
$B$ is an open disk with two additional points $b,b^{\prime}$ in its frontier
and the gluing homeomorphisms are $\varphi\colon\\{a\\}\to\\{b\\}$ and
$\psi\colon\\{a^{\prime}\\}\to\\{b^{\prime}\\}$. If a decomposition is given,
then we would like to understand the decomposition space as well.

######  Proposition 2.9.

The decomposition space of a closed usc decomposition of a normal space is
$T_{4}$.

###### Proof.

We have to show that if $\mathcal{D}$ is a usc decomposition of a normal space
$X$, then any two disjoint closed sets in the space $X_{\mathcal{D}}$ can be
separated by open sets. Let $A,B$ be disjoint closed sets in
$X_{\mathcal{D}}$. Then $\pi^{-1}(A)$ and $\pi^{-1}(B)$ are disjoint closed
sets and by being $X$ normal and by Lemma 2.7 they have disjoint saturated
nbhds $U_{1}$ and $U_{2}$. Taking $\pi(U_{1})$ and $\pi(U_{2})$ we get
disjoint nbhds of $A$ and $B$. The decomposition elements are closed so
$X_{\mathcal{D}}$ is $T_{1}$, which finally implies that $X_{\mathcal{D}}$ is
$T_{4}$. ∎

If a space $X$ is not normal, then it is easy to define such closed usc
decomposition, where the decomposition space is even not $T_{2}$. Take two
disjoint closed sets $A,B$ in $X$ which can not be separated by open sets. For
example the direct product of the Sorgenfrei line with itself is not normal
and choose the points with rational and irrational coordinates in the
antidiagonal respectively, to have two closed sets $A$ and $B$. These two sets
are the two non-singleton elements of the decomposition $\mathcal{D}$, other
elements are singletons. Then $\mathcal{D}$ is closed usc but
$X_{\mathcal{D}}$ is not $T_{2}$ because $\pi(A)$ and $\pi(B)$ can not be
separated by open sets.

######  Definition 2.10.

Let $\mathcal{D}$ be a decomposition of the space $X$. A decomposition is
_finite_ if it has only finitely many non-degenerate elements and _countable_
if it has countably many non-degenerate elements. A decomposition is
_monotone_ if every decomposition element is connected. If $X$ is a metric
space, then a decomposition is _null_ if the decomposition elements are
bounded and for every $\varepsilon>0$ there is only a finite number of
elements whose diameter is greater than $\varepsilon$.

######  Proposition 2.11.

Let $\mathcal{D}$ be a decomposition and suppose that all elements are closed.
If $\mathcal{D}$ is finite, then it is a closed usc decomposition.

###### Proof.

Let $C\subset X$ be a closed subset, then $\pi^{-1}(\pi(C))$ is closed because
it is the finite union of the closed set $C$ and the non-degenerate elements
which intersect $C$. Then by Lemma 2.7 (4) the statement follows. ∎

######  Proposition 2.12.

If $\mathcal{D}$ is a closed and null decomposition of a metric space, then it
is usc.

###### Proof.

Denote the metric by $d$. All the decomposition elements are compact because
they are bounded. Let $U$ be a nbhd of a $D\in\mathcal{D}$, then there is an
$\varepsilon>0$ such that the $\varepsilon$-nbhd of $D$ is in $U$. Since
$\mathcal{D}$ is null, there are only finitely many decomposition elements
$D_{1},\ldots,D_{n}$ whose diameter is greater than $\varepsilon/4$ and
$D_{i}\neq D$. Let $\delta$ be the minimum of $\varepsilon/4$ and the
distances between $D$ and the $D_{i}$s. If $D^{\prime}\in\mathcal{D}$ is such
that the distance between $D^{\prime}$ and $D$ is less than $\delta$, then
$D^{\prime}$ is in the $\varepsilon$-nbhd of $D$: there are $x\in D$ and $y\in
D^{\prime}$ such that $d(x,y)<\delta$ so for every $a\in D^{\prime}$

$\inf\\{d(a,b):b\in D\\}\leq d(a,y)+d(y,x)+\inf\\{d(x,b):b\in
D\\}=d(a,y)+d(y,x)\leq\\\ {\mathrm{diam}}\thinspace
D^{\prime}+\delta\leq\varepsilon/2,$

which means that $D^{\prime}$ is in the $\varepsilon$-nbhd of $D$ so
$D^{\prime}\subset U$. ∎

######  Proposition 2.13.

Let $\mathcal{D}$ be a usc decomposition of a space $X$.

  1. (1)

If $X$ is $T_{2}$, then $X_{\mathcal{D}}$ is $T_{2}$ as well.

  2. (2)

If $X$ is regular, then $X_{\mathcal{D}}$ is $T_{3}$.

###### Proof.

The decomposition elements are compact so every $\pi^{-1}(a)$ and
$\pi^{-1}(b)$ for different $a,b\in X_{\mathcal{D}}$ can be separated by open
sets. The statement follows easily. ∎

######  Proposition 2.14.

Let $\mathcal{D}$ be a usc decomposition of a $T_{2}$ space $X$. The
decomposition $\mathcal{D}^{\prime}$ whose elements are the connected
components of the elements of $\mathcal{D}$ is a monotone usc decomposition.

###### Proof.

Take an element $D^{\prime}\in\mathcal{D}^{\prime}$ and denote by $D$ the
decomposition element in $\mathcal{D}$ which contains $D^{\prime}$. Suppose
$D\neq D^{\prime}$. Then $D-D^{\prime}$ is closed in $D$ so it is closed in
$X$. Let $U$ be a nbhd of $D^{\prime}$. Then there exists a nbhd
$U^{\prime}\subset U$ of $D^{\prime}$ which is disjoint from a nbhd
$U^{\prime\prime}$ of the closed set $D-D^{\prime}$. By the usc property we
can find a nbhd $V$ of $D$ such that $V\subset U^{\prime}\cup
U^{\prime\prime}$ and if a $C\in\mathcal{D}$ intersects $V$, then $C\subset
U^{\prime}\cup U^{\prime\prime}$. If $C^{\prime}\in\mathcal{D}^{\prime}$
intersects $V\cap U^{\prime}$, then the element $C\in\mathcal{D}$ which
contains $C^{\prime}$ as a connected component intersects $V$ hence $C\subset
U^{\prime}\cup U^{\prime\prime}$. Since $U^{\prime}$ and $U^{\prime\prime}$
are disjoint, the component $C^{\prime}$ of $C$ is in $U^{\prime}$ because it
intersects $U^{\prime}$. We got that $C^{\prime}\subset U$. ∎

For example, it follows that the decomposition of a compact $T_{2}$ space $X$
whose elements are the connected components of the space is a usc
decomposition. To see this, at first take the decomposition $\mathcal{D}$,
where $\mathcal{H}_{\mathcal{D}}=\\{X\\}$ and hence the decomposition has no
singletons. This is usc so we can apply the previous proposition.

######  Proposition 2.15.

If $X$ is a metric space and $\mathcal{D}$ is its usc decomposition, then
$X_{\mathcal{D}}$ is metrizable. If $X$ is separable, then $X_{\mathcal{D}}$
is also separable.

###### Proof.

By [St56] if there is a continuous closed map $f$ of a metric space onto a
space $Y$ such that for every $y\in Y$ the closed set
$f^{-1}(y)-{\mathrm{int}}\thinspace f^{-1}(y)$ is compact, then $Y$ is
metrizable. But for every $y\in X_{\mathcal{D}}$ the set $\pi^{-1}(y)$ and so
its closed subset $\pi^{-1}(y)-{\mathrm{int}}\thinspace\pi^{-1}(y)$ are
compact hence $X_{\mathcal{D}}$ is metrizable. Moreover if $X$ is separable,
then there is a countable subset $S\subset X$ intersecting every open set,
which gives the countable set $\pi(S)$ intersecting every open set in
$X_{\mathcal{D}}$. ∎

##  3\. Examples and properties of decompositions

Usually, we are interested in the topology of the decomposition space if a
decomposition of $X$ is given. Especially those situations are stimulating
where the decomposition space turns out to be homeomorphic to $X$.

Let $X=\mathbb{R}$ and let $\mathcal{D}$ be a decomposition such that
$\mathcal{H}_{\mathcal{D}}$ consists of countably many disjoint compact
intervals. Then this is a usc decomposition: any open interval
$U\subset\mathbb{R}$ contains at most countably many compact intervals of
$\mathcal{H}_{\mathcal{D}}$ and the infimum of the left endpoints of these
intervals could be in $U$ or it could be the left boundary point of $U$.
Similarly, we have this for the right endpoints. In all cases the union of the
decomposition elements being in $U$ is open. For an arbitrary open set
$U\subset\mathbb{R}$ we have the same, this means we have a usc decomposition.
Later we will see, that the decomposition space $X_{\mathcal{D}}$ is
homeomorphic to $\mathbb{R}$. Moreover the decomposition map $\pi\colon X\to
X_{\mathcal{D}}$ is approximable by homeomorphisms, which means there are
homeomorphisms from $\mathbb{R}$ to $\mathbb{R}$ arbitrarily close to $\pi$ in
the sense of uniform metric. For example, let $X=\mathbb{R}$ and consider the
infinite Cantor set-like construction by taking iteratively the middle third
compact intervals in the interval $[0,1]$. These are countably many intervals
and define the decomposition ${\mathcal{D}}$ so that the non-degenerate
elements are these intervals. We can obtain this decomposition ${\mathcal{D}}$
by taking the connected components of $[0,1]-\mbox{Cantor set}$ and then
taking the closure of them. This is usc and we will see that the decomposition
space is $\mathbb{R}$.

If $X=\mathbb{R}^{2}$, then an analogous decomposition is that
$\mathcal{H}_{\mathcal{D}}$ consists of countably many compact line segments.
More generally, let $\mathcal{H}_{\mathcal{D}}$ be countably many _flat_ arcs,
that is such subsets $A$ of $\mathbb{R}^{2}$ for which there exist self-
homeomorphisms $h_{A}$ of $\mathbb{R}^{2}$ mapping $A$ into the standard
compact interval $\\{(x,0)\in\mathbb{R}^{2}:0\leq x\leq 1\\}$. Such a
decomposition is not necessarily usc, for example take the function
$f\colon[0,1)\to\mathbb{R}$, $f(x)=1+x$, and the sequence $x_{n}=1-1/n$.
Define the decomposition by
$\mathcal{H}_{\mathcal{D}}=\\{(x_{n},y):y\in[0,f(x_{n})],n\in\mathbb{N}\\}$
and the singletons are all the other points of $\mathbb{R}^{2}$. Then
$\mathcal{H}_{\mathcal{D}}$ consists of countably many straight line segments
but this decomposition is not usc: consider the point $(1,3/2)\in\mathcal{D}$
and its $\varepsilon$-nbhds for small $\varepsilon>0$. These intersect
infinitely many non-degenerate decomposition elements but none of the elements
is a subset of any of these $\varepsilon$-nbhds. The decomposition space is
not $T_{2}$: the points $\pi((1,y))$, where $0\leq y\leq 2$, cannot be
separated by disjoint nbhds because the sequence $\pi((x_{n},0))$ converges to
all of them.

However, if $\mathcal{D}$ is such a decomposition of $\mathbb{R}^{2}$ that
$\mathcal{H}_{\mathcal{D}}$ consists of countably many flat arcs and further
we suppose that $\mathcal{D}$ is usc, then the decomposition space
$X_{\mathcal{D}}$ is homeomorphic to $\mathbb{R}^{2}$ and again $\pi$ can be
approximated by homeomorphisms, we will see this later.

We get another interesting example by taking a smooth function with finitely
many critical values on a closed manifold $M$. Then the decomposition elements
are defined to be the connected components of the point preimages of the
function. This is a monotone decomposition ${\mathcal{D}}$ and it is usc
because the decomposition map $\pi\colon M\to M_{\mathcal{D}}$ is a closed
map: in $M$ a closed set is compact, its $\pi$-image is compact as well and
$M_{\mathcal{D}}$ is $T_{2}$ because it is a graph [Iz88, Re46, Sa20] so this
$\pi$-image is also closed.

If $X$ is $3$-dimensional, then the possibilites increase tremendously. This
is illustrated by the following surprising statement.

######  Proposition 3.1.

For every compact metric space $Y$ there exists a monotone usc decomposition
of the compact ball $D^{3}$ such that $Y$ can be embedded into the
decomposition space.

###### Proof.

Recall that by the Alexandroff-Hausdorff theorem the Cantor set in the $[0,1]$
interval can be mapped surjectively and continuously onto every compact metric
space. Let $T$ be a tetrahedron in $D^{3}$, denote two of its non-intersecting
edges by $e$ and $f$. Identify these edges linearly with $[0,1]$ and let
$C_{1}$ and $C_{2}$ be the Cantor sets in $e$ and $f$, respectively. For
$i=1,2$ denote the existing surjective maps of $C_{i}$ onto $Y$ by
$\psi_{i}\colon C_{i}\to Y$. For every $x\in Y$ take the union of all the line
segments in $T$ connecting all the points of $\psi_{1}^{-1}(x)$ to all the
points of $\psi_{2}^{-1}(x)$. Denote this subset of $T$ by $D_{x}$, see Figure
3. They are compact and connected for all $x\in Y$ and they are pairwise
disjoint because all the lines in $T$ connecting points of $e$ and $f$ are
pairwise disjoint. So we have a monotone usc decomposition with
$\mathcal{H}_{\mathcal{D}}=\\{D_{x}:x\in Y\\}$. Define the embedding $i$ of
$Y$ into $D^{3}_{\mathcal{D}}$ by $i(x)=\pi(\psi_{1}^{-1}(x))$. This map is
injective, closed because $\pi$ is closed and continuous because $\psi_{1}$ is
closed. ∎

$f$$e$

Figure 3. The tetrahedron $T$, the edges $e$ and $f$ and a set $D_{x}$
pictured in blue.

To see further examples in $\mathbb{R}^{3}$ let us introduce some notions.

######  Definition 3.2 (Defining sequence).

Let $X$ be a connected $n$-dimensional manifold. A _defining sequence_ for a
decomposition of $X$ is a sequence

$C_{1},C_{2},\ldots,C_{n},\ldots$

of compact $n$-dimensional submanifolds-with-boundary in $X$ such that
$C_{n+1}\subset\mathrm{int}\thinspace C_{n}$. The decomposition elements of
the defined decomposition are the connected components of
$\cap_{n=1}^{\infty}C_{n}$ and the other points of $X$ are singletons.

Obviously a decomposition defined in this way is monotone. The set
$\cap_{n=1}^{\infty}C_{n}$ is closed and compact so its connected components
are closed and compact as well. Also the space $\cap_{n=1}^{\infty}C_{n}$ is
$T_{2}$ hence its decomposition to its connected components is usc. Then
adding all the points of $X-\cap_{n=1}^{\infty}C_{n}$ to this decomposition as
singletons results our decomposition. This is usc: the only thing which is not
completely obvious is that in a nbhd of an added point the conditions being
usc are satisfied or not. But $\cap_{n=1}^{\infty}C_{n}$ is closed, its
complement is open so every such singleton has a nbhd disjoint from
$\cap_{n=1}^{\infty}C_{n}$.

######  Proposition 3.3.

If all $C_{n}$ in a defining sequence is connected, then
$\cap_{n=1}^{\infty}C_{n}$ is connected.

###### Proof.

Let $C$ denote the non-empty set $\cap_{n=1}^{\infty}C_{n}$. Suppose $C$ is
not connected, this means there are disjoint closed non-empty subsets
$A,B\subset C$ such that $A\cup B=C$. These $A$ and $B$ are closed in the
ambient manifold $X$ as well, so there exist disjoint nbhds $U$ of $A$ and $V$
of $B$ in $X$. It is enough to show that for some $n\in\mathbb{N}$ we have
$C_{n}\subset U\cup V$, because then $C_{n}\cap U\neq\emptyset$, $C_{n}\cap
V\neq\emptyset$ imply that $C_{n}$ is not connected, which is a contradiction.
If we suppose that for every $n\in\mathbb{N}$ we have $C_{n}\cap(X-(U\cup
V))\neq\emptyset$, then for every $n$ we have
$C_{n}\cap(X-U)\cap(X-V)\neq\emptyset$, i.e. the closed set $F=(X-U)\cap(X-V)$
and each element of the nested sequence $C_{1},C_{2},\ldots$ satisfy

$C_{n}\cap F\neq\emptyset.$

Of course

$C_{n+1}\cap F\subset C_{n}\cap F$

which implies that

$F\cap C=F\cap(\cap_{n=1}^{\infty}C_{n})=\cap_{n=1}^{\infty}(C_{n}\cap
F)\neq\emptyset$

because all $C_{n}\cap F$ is closed in the compact space $C_{1}$. But $F\cap
C\neq\emptyset$ contradicts to $C\subset U\cup V$. ∎

The $\pi$-image of the union of non-degenerate elements of a decomposition
associated to a defining sequence is closed and also totally disconnected
because if $\cap_{n=1}^{\infty}C_{n}$ is not connected, then all the pairs of
decomposition elements have disjoint saturated nbhds which yield disjoint
nbhds of their $\pi$-image.

###  3.1. The Whitehead continuum

One of the most famous such decomposition is related to the so called
Whitehead continuum. Its defining sequence consists of solid tori embedded
into each other in such a way that $C_{i+1}$ is a thickened Whitehead double
of the center circle of $C_{i}$, see Figure 4. The intersection
$\cap_{i=1}^{\infty}C_{i}$ is a compact subset of $\mathbb{R}^{3}$, this is
the Whitehead continuum, which we denote by $\mathcal{W}$. The decomposition
consists of the connected components of $\mathcal{W}$ and the singletons in
the complement of them. If the diameters $d_{i}$ of the meridians of the tori
$C_{i}$ converges to $0$ as $i$ goes to $\infty$, then $\mathcal{W}$
intersects the vertical sheet $S$ in Figure 4 in a Cantor set: $C_{i}\cap S$
is equal to $2^{i-1}$ copies of disks of diameter $d_{i}$ nested into each
other. The intersection $S\cap(\cap_{i=1}^{\infty}C_{i})$ is then a Cantor
set. The Whitehead continuum $\mathcal{W}$ is connected because the $C_{i}$
tori are connected but it is not path-connected. We will see later that the
decomposition space $\mathbb{R}^{3}_{\mathcal{W}}$ is not homeomorphic to
$\mathbb{R}^{3}$ but taking its direct product with $\mathbb{R}$ we get
$\mathbb{R}^{4}$. An important property of $\mathbb{R}^{3}-\mathcal{W}$ is
that it is a contractible $3$-manifold, which is not homeomorphic to
$\mathbb{R}^{3}$.

For understanding further properties of this decomposition, we are going to
define some notions.

$S$

Figure 4. A sketch of the defining sequence of the Whitehead decomposition.
The first figure shows the solid torus $C_{1}$ and the Whitehead double of its
center circle. The second figure shows the Whitehead double of the center
circle of $C_{2}$. The torus $C_{2}$ is not shown but we get it by thickening
the Whitehead double in $C_{1}$. Then thicken the knot in the second figure
(so we get the solid torus $C_{3}$) and take its center circle. Take the
Whitehead double of this circle and so we get the knot embedded in $C_{3}$ in
the third figure. In the third figure we can see the intersection of $C_{3}$
with a vertical sheet $S$, which is four small disks. This vertical sheet $S$
intersects the Whitehead continuum in a Cantor set.

######  Definition 3.4 (Cellular set, cell-like set).

Let $X$ be an $n$-dimensional manifold and $C\subset X$ be a subset of $X$.
The set $C$ is _cellular_ if there is a sequence
$B_{1},B_{2},\ldots,B_{n},\ldots$ of closed $n$-dimensional balls in $X$ such
that $B_{n+1}\subset\mathrm{int}\thinspace B_{n}$ and
$C=\cap_{n=1}^{\infty}B_{n}$. A compact subset $C$ of a topological space $X$
is _cell-like_ if for every nbhd $U$ of $C$ there is a nbhd $V$ of $C$ in $U$
such that the inclusion map $V\to U$ is homotopic in $U$ to a constant map.
Similarly, a decomposition is called cellular or cell-like if each of its
decomposition elements is cellular or cell-like, respectively.

For example the “topologist’s sine curve” in $\mathbb{R}^{2}$ is cellular. A
cellular set is compact and also connected but not necessarily path-connected.
It is also easy to see that every compact contractible subset of a manifold is
cell-like. Also a compact and contractible metric space is cell-like in
itself. A cell-like set $C$ is connected because if there were two open
subsets $U_{1}$ and $U_{2}$ in $X$ separating some connected components of
$C$, then it would be not possible to contract any nbhd $V\subset U_{1}\cup
U_{2}$ of $C$ to one single point.

######  Proposition 3.5.

The set $\mathcal{W}$ is cell-like but not cellular.

###### Proof.

Let $U$ be a nbhd of $\mathcal{W}$. Then there is an $n$ such that
$C_{i}\subset U$ for all $i\geq n$. Let $V$ be such a small tubular nbhd of
$C_{n+1}$ which is inside $C_{n}$. Then since the Whitehead double of the
center circle of $C_{n}$ is null-homotopic in the solid torus $C_{n}$, the
thickened Whitehead double $C_{n+1}$ and its nbhd $V$ are also null-homotopic
in $C_{n}$, hence the map $V\to U$ is homotopic in $U$ to a constant map.

######  Lemma 3.6.

The $3$-manifold $S^{3}-\mathcal{W}$ is not simply connected at infinity.

###### Proof.

We have to show that there is a compact subset $C\subset S^{3}-\mathcal{W}$
such that for every compact set $D\subset S^{3}-\mathcal{W}$ containing $C$
the induced homomorphism

$\varphi\colon\pi_{1}(S^{3}-\mathcal{W}-D)\to\pi_{1}(S^{3}-\mathcal{W}-C)$

is not the zero homomorphism. Let $C$ be the closure of $S^{3}-C_{1}$. If $D$
is a compact set in $S^{3}-\mathcal{W}$ containing $C$, then $S^{3}-D$ is a
nbhd of $\mathcal{W}$ in $C_{1}$. Then there is an $n$ such that $C_{i}\subset
S^{3}-D$ for all $i\geq n$. Consider the commutative diagram

$\begin{CD}\pi_{1}(S^{3}-C_{n}-D)@>{}>{}>\pi_{1}(S^{3}-C_{n}-C)\\\
@V{}V{}V@V{}V{\alpha}V\\\
\pi_{1}(S^{3}-\mathcal{W}-D)@>{\varphi}>{}>\pi_{1}(S^{3}-\mathcal{W}-C).\end{CD}$

By [NW37] the generator of the group $\pi_{1}(S^{3}-C_{n}-C)$ represented by
the meridian of the torus $C_{n}$ is mapped by $\alpha$ into a generator of
$\pi_{1}(S^{3}-\mathcal{W}-C)$. Since this meridian also represents an element
of $\pi_{1}(S^{3}-C_{n}-D)$, we get that $\varphi$ is not the zero
homomorphism. ∎

Let us continue the proof of Proposition 3.5. If $\mathcal{W}$ is cellular,
then there are $B_{1},B_{2},\ldots,B_{n},\ldots$ closed $n$-dimensional balls
in $S^{3}$ such that $B_{n+1}\subset\mathrm{int}\thinspace B_{n}$ and
$\mathcal{W}=\cap_{n=1}^{\infty}B_{n}$. This would imply that
$S^{3}-\mathcal{W}$ is simply connected at infinity because if $C\subset
S^{3}-\mathcal{W}$ is a compact set, then take a $B_{n}\subset S^{3}-C$ and a
loop in $\mathrm{int}B_{n}-\mathcal{W}$, then there is a
$B_{m}\subset\mathrm{int}B_{n}$ not containing this loop and the loop in null-
homotopic in $\mathrm{int}B_{n}-B_{m}$ because
$\pi_{1}(\mathrm{int}B_{n}-B_{m})=0$. Hence we obtain that $S^{3}-\mathcal{W}$
is not cellular. ∎

With more effort we could show that $S^{3}-\mathcal{W}$ is contractible so it
is homotopy equivalent to $\mathbb{R}^{3}$ but by the previous statement it is
not homeomorphic to $\mathbb{R}^{3}$. It is known that the set
$\mathcal{W}\times\\{0\\}$ is cellular in $\mathbb{R}^{3}\times\mathbb{R}$ and
the decomposition space of the decomposition of
$\mathbb{R}^{3}\times\mathbb{R}$ whose only non-degenerate element is
$\mathcal{W}\times\\{0\\}$ is homeomorphic to $\mathbb{R}^{4}$. This fact is
the starting point of the proof of the $4$-dimensional Poincaré conjecture.

Being cell-like often does not depend on the ambient space. To understand
this, we have to introduce a new notion.

######  Definition 3.7 (Absolute nbhd retract).

A metric space $Y$ is an _absolute nbhd retract_ (or _ANR_ for short) if for
an arbitrary metric space $X$ and its closed subset $A$ every map $f$ from $A$
to $Y$ extends to a nbhd of $A$. In other words, the nbhd $U$ and the dashed
arrow exist in the following diagram and make the diagram commutative.

$A$$X$$U$$Y$$f$$\subseteq$$\subseteq$$\subseteq$

This is equivalent to say that for every metric space $Z$ and embedding
$i\colon Y\to Z$ such that $i(Y)$ is closed there is a nbhd $U$ of $i(Y)$ in
$Z$ which retracts onto $i(Y)$, that is $r|_{i(Y)}=\mathrm{id}_{i(Y)}$ for
some map $r\colon U\to i(Y)$. It is a fact that every manifold is an ANR.

The property of cell-likeness is independent of the ambient space until that
is an ANR as the following statement shows.

######  Proposition 3.8.

If $C\subset X$ is a compact cell-like set in a metric space $X$, then the
embedded image of $C$ in an arbitrary ANR is also cell-like.

###### Proof.

Suppose $e\colon C\to Y$ is an embedding into an ANR $Y$. We have to show that
$e(C)$ is cell-like. Let $U$ be a nbhd of $e(C)$. Since $Y$ is ANR, there is a
nbhd $\tilde{V}$ of $C$ in $X$ such that $e$ extends to an
$\tilde{e}\colon\tilde{V}\to Y$. Let $V\subset X$ be the open set
$\tilde{V}\cap\tilde{e}^{-1}(U)$, it is a nbhd of $C$. There is a nbhd $W$ of
$C$ such that $C\subset W\subset V$ and there is a homotopy of the inclusion
$W\subset V$ to the constant in $V$ since $C$ is cell-like, denote this
homotopy by $\varphi\colon W\times[0,1]\to V$. Then $\varphi|_{C\times[0,1]}$
is a homotopy of the inclusion $C\subset V$ to the constant. Take

$\tilde{e}\circ\varphi|_{C\times[0,1]}\circ(e^{-1}|_{e(C)}\times\mathrm{id}_{[0,1]}),$

this is a homotopy of the inclusion $e(C)\subset U$ to the constant in $U$.
The space $e(C)\times[0,1]$ is compact in $Y\times[0,1]$ and the homotopy maps
it into $Y$, which is ANR. This implies that there is a nbhd $\tilde{U}\subset
U$ of $e(C)$ such that the inclusion $\tilde{U}\subset U$ is homotopic to
constant in $U$. ∎

For example, this shows that a compact and contractible metric space is cell-
like if we embed it into any ANR. In practice, we do not consider cell-like
sets as subsets in some ambient space but rather as compact metric spaces
which are cell-like if we embed them into an arbitrary ANR.

It is clear that every cellular set $C$ is cell-like because in every nbhd $U$
of $C$ some open ball is contractible. Also, we have seen that the Whitehead
continuum is cell-like but not cellular. In order to compare cell-like and
cellular sets we introduce the notion of cellularity criterion.

######  Definition 3.9 (Cellularity criterion).

A subset $Y\subset X$ satisfies the _cellularity criterion_ if for every nbhd
$U$ of $Y$ there is a nbhd $V$ of $Y$ such that $V\subset U$ and every loop in
$V-Y$ is null-homotopic in $U-Y$.

The cellularity criterion and being cellular measure how wildly a subset is
embedded into a space. The next theorem compares cell-like and cellular sets
in a PL manifold. We omit its difficult proof here.

######  Theorem 3.10.

Let $C$ be a cell-like subset of a PL $n$-dimensional manifold, where $n\geq
4$. Then $C$ is cellular if and only if $C$ satisfies the cellularity
criterion.

In dimension $2$ we have a simpler statement:

######  Theorem 3.11.

Every cell-like subset in a $2$-dimensional manifold $X$ is cellular.

###### Proof.

At first suppose $X=\mathbb{R}^{2}$ and $C\subset\mathbb{R}^{2}$ is a cell-
like set. Let $U$ be a bounded nbhd of $C$ and let $V\subset U$ a nbhd of $C$
such that the inclusion $V\to U$ is homotopic to constant. Choose another nbhd
$W\subset V$ of $C$ as well such that $\mathrm{cl}\thinspace W\subset V$. Take
a compact smooth $2$-dimensional manifold $H\subset V$ such that
$C\subset\mathrm{int}H$, $\partial H\subset V-\mathrm{cl}\thinspace W$ and
$\mathrm{int}H$ is connected. Such an $H$ can be obtained by taking a Morse
function $f\colon V\to[0,1]$ which maps the nbhd $W$ of $C$ into $0$ and a
small nbhd of $\mathbb{R}^{2}-V$ into $1$. Then the preimage of a regular
value $r$ close to $1/2$ is a smooth $1$-dimensional submanifold of
$\mathbb{R}^{2}$ and the preimage of $(-\infty,r]$ is a compact subset
containing $W$ and $C$, denote this $f^{-1}((-\infty,r])$ by $H$. Then $H$ is
a compact smooth $2$-dimensional submanifold of $\mathbb{R}^{2}$, see Figure
5. Take its connected component (this is also a path-connected component
because $H$ is a manifold) which contains $C$ and denote this by $H$ as well.

$C$$U$$V$$W$${{\color[rgb]{0,0,1}H_{2}}}$${{\color[rgb]{0,0,1}H_{1}}}$

Figure 5. The compact manifold $H=H_{1}\cup H_{2}$. Its component $H_{2}$
contains $C$. Since $H_{2}-C$ is path-connected, there is a path (dashed in
the figure) in $H_{2}$ connecting two different components of the boundary of
$H_{2}$.

We show that $H-C$ is connected. For this consider the commutative diagram

$\begin{CD}H_{1}(H;\mathbb{Z}_{2})@>{}>{}>H_{1}(H,H-C;\mathbb{Z}_{2})@>{}>{}>H_{0}(H-C;\mathbb{Z}_{2})@>{}>{}>H_{0}(H)@>{}>{}>0\\\
@V{}V{}V@V{}V{}V@V{}V{}V@V{}V{}V\\\
H_{1}(\mathbb{R}^{2};\mathbb{Z}_{2})@>{}>{}>H_{1}(\mathbb{R}^{2},\mathbb{R}^{2}-C;\mathbb{Z}_{2})@>{}>{}>H_{0}(\mathbb{R}^{2}-C;\mathbb{Z}_{2})@>{i_{*}}>{}>H_{0}(\mathbb{R}^{2})@>{}>{}>0\end{CD}$

coming from the long exact sequences and the inclusion
$(H,H-C)\subset(\mathbb{R}^{2},\mathbb{R}^{2}-C)$. This is just the diagram

$\begin{CD}H_{1}(H;\mathbb{Z}_{2})@>{}>{}>H_{1}(H,H-C;\mathbb{Z}_{2})@>{}>{}>H_{0}(H-C;\mathbb{Z}_{2})@>{}>{}>\mathbb{Z}_{2}@>{}>{}>0\\\
@V{}V{}V@V{}V{\cong}V@V{}V{}V@V{}V{\cong}V\\\
0@>{}>{}>H_{1}(\mathbb{R}^{2},\mathbb{R}^{2}-C;\mathbb{Z}_{2})@>{}>{}>H_{0}(\mathbb{R}^{2}-C;\mathbb{Z}_{2})@>{i_{*}}>{}>\mathbb{Z}_{2}@>{}>{}>0\end{CD}$

If the group $H_{0}(\mathbb{R}^{2}-C;\mathbb{Z}_{2})$ is $\mathbb{Z}_{2}$,
i.e. the manifold $\mathbb{R}^{2}-C$ is connected, then exactness implies that
$H_{0}(H-C;\mathbb{Z}_{2})\cong\mathbb{Z}_{2}$ so $H-C$ is connected. To show
that $\mathbb{R}^{2}-C$ is connected, we apply [HW41, Theorem VI.5, page 86],
which implies that if $C$ is a closed subset of a space $D$ and $f,g$ are
homotopic maps of $C$ into $S^{1}$ such that $f$ extends to $D$, then $g$
extends to $D$ and the extensions are homotopic. Suppose the open set
$\mathbb{R}^{2}-C$ is not connected, then it is the disjoint union of two open
sets $A$ and $B$. At least one of these is bounded because for large enough
$s$ the set $\mathbb{R}^{2}-[-s,s]^{2}$ is disjoint from $C$ and it is
connected hence it is in $A$ or $B$ but then $[-s,s]^{2}$ contains $B$ or $A$,
respectively. Suppose $A$ is bounded, $p\in A$ and $q\in B$. For a subset
$S\subset\mathbb{R}^{2}$ and point $x\in\mathbb{R}^{2}$ denote by
$\pi_{S,x}\colon S-\\{x\\}\to S^{1}$ the radial projection of $S-\\{x\\}$ to
the circle $S^{1}$ of radius $1$ centered at $x$. Then $\pi_{C,q}$ extends to
$\mathbb{R}^{2}-\\{q\\}$ so also to $A\cup C$ but $\pi_{C,p}$ does not extend
to $A\cup C$ because such an extension would extend to a much larger disk $P$
centered at $p$ as well by radial projection and then a retraction of $P$ onto
its boundary (if we identify it with the target circle of $\pi_{C,p}$) would
exists. Consequently $\pi_{C,q}$ and $\pi_{C,p}$ are not homotopic and so at
least one of them is not homotopic to constant. This means if
$\mathbb{R}^{2}-C$ is not connected, then there is a map $C\to S^{1}$ which is
not homotopic to constant. But since the inclusion $V\subset U$ and then also
$C\subset U$ are homotopic to constant, we get that $\mathbb{R}^{2}-C$ is
connected.

Finally, we get that $H-C$ is a path-connected smooth $2$-dimensional manifold
with boundary. Hence if the number of components of $\partial H$ is larger
than one, then there exists a smooth curve transversal to $\partial H$,
disjoint from $C$ and connecting different components of $\partial H$. We can
cut $H$ along this curve and by repeating this process we end up with
$\partial H$ being a single circle. By the Jordan curve theorem $H$ is a
compact $2$-dimensional disk. In this way we get

$C\subset W\subset\mathrm{int}H\subset H\subset V\subset U.$

Since in $\mathbb{R}^{2}$ every compact set is a countable intersection of
open sets which form a decreasing sequence, we have
$C=\cap_{n=1}^{\infty}U_{n}$, where $U_{1}\supset U_{2}\supset\cdots\supset
U_{n}\supset\cdots$, where the sets $U_{n}$ are open. We can also assume that
for each $n$ we have $\mathrm{cl}\thinspace U_{n+1}\subset U_{n}$.

We obtain countably many compact $2$-dimensional disks $H_{1},H_{2},\ldots$ by
the previous construction, which satisfy

$C\subset U_{n+1}\subset\mathrm{int}H_{n}\subset H_{n}\subset V\subset U_{n}.$

Hence $C=\cap_{n=1}^{\infty}H_{n}$ so $C$ is cellular.

In the case of $X$ is an arbitrary $2$-dimensional manifold, since $C$ is
cell-like, there exists a nbhd of $C$ which is homotopic to constant so $C$ is
contained in a simply-connected $2$-dimensional manifold nbhd, which is
homeomorphic to $\mathbb{R}^{2}$. Hence a similar argument gives that $C$ is
cellular. ∎

######  Proposition 3.12.

If $C$ is cell-like in a smooth $n$-dimensional manifold $X$, where $n\geq 3$,
then $C\times\\{0\\}$ is cellular in $X\times\mathbb{R}^{3}$.

###### Proof.

It is enough to show that $C\times\\{0\\}$ satisfies the cellularity
criterion. It is easy to see that $C\times\\{0\\}$ is cell-like in
$X\times\mathbb{R}^{3}$. Let $U$ be a nbhd of $C\times\\{0\\}$ in
$X\times\mathbb{R}^{3}$. It is obvious that there is a nbhd $V\subset U$ of
$C\times\\{0\\}$ such that every loop $\gamma\colon[0,1]\to V$ is null-
homotopic in $U$. Let $\gamma$ be an arbitrary loop in $V-C\times\\{0\\}$, it
is homotopic to a smooth loop in $\tilde{\gamma}\colon V-C\times\\{0\\}$ by a
homotopy $H$. A homotopy of $\tilde{\gamma}$ to constant can be approximated
by a smooth map $\tilde{H}\colon D^{2}\to U$, where $\tilde{H}|_{\partial
D^{2}}=\tilde{\gamma}$. In the subspace $X\times\\{0\\}$ of
$X\times\mathbb{R}^{3}$ let $W$ be a nbhd of $C\times\\{0\\}$ which is
disjoint from the homotopy $\tilde{H}$. Perturb $\tilde{H}$ keeping
$\tilde{H}|_{\partial D^{2}}$ fixed to get a transversal map to the
$n$-dimensional manifold $W$ in $U$, hence we get that $\gamma$ is null-
homotopic in $U-C\times\\{0\\}$. So the cellularity criterion holds for
$C\times\\{0\\}$. ∎

###  3.2. Antoine’s necklace

Take the defining sequence where

  * •

$C_{1}$ is a solid torus,

  * •

$C_{2}$ is a finite number of solid tori embedded in $C_{1}$ in such a way
that each torus is unknotted and linked to its neighbour as in a usual chain,

  * •

$C_{3}$ is again a finite number of similarly linked solid tori,

…, etc., see Figure 6.

Figure 6. A sketch of the defining sequence of Antoine’s necklace. We can see
the solid torus $C_{1}$, the linked tori $C_{2}$ and some linked tori from the
collection $C_{3}$, etc. The number of components of $C_{n+1}$ in $C_{n}$ is
large enough to make the diameters of the tori converge to $0$.

We always consider at least three tori in each $C_{n}$. We require that the
maximal diameter of tori in $C_{n}$ converges to $0$. The set
$\cap_{n=1}^{\infty}C_{n}$ is called Antoine’s necklace and denoted by
$\mathcal{A}$. It is easy to see that each of its components is cell-like.
Unlike Whitehead continuum the components of $\mathcal{A}$ are cellular
because every component of $C_{n+1}$ is inside a ball in $C_{n}$.

Recall that the Cantor set is the topological space

$D_{1}\times D_{2}\times\cdots\times D_{n}\times\cdots,$

where every space $D_{n}$ is a finite discrete metric space with $|D_{n}|\geq
2$.

######  Proposition 3.13.

The space $\cap_{n=1}^{\infty}C_{n}$ is homeomorphic to the Cantor set.

###### Proof.

Denote the number of tori embedded in $C_{1}$ by $m_{1}$, these tori are

$C_{2,1},\ldots,C_{2,m_{1}}$

whose disjoint union is $C_{2}$. For $1\leq i_{1}\leq m_{1}$ take the
$i_{1}$-th torus $C_{2,i_{1}}$ and denote the number of tori embedded into it
by $m_{2,i_{1}}$, these tori are

$C_{3,i_{1},1},\ldots,C_{3,i_{1},m_{2,i_{1}}}$

whose disjoint union is $C_{3}$. Again for $1\leq i_{2}\leq m_{2,i_{1}}$ take
the $i_{2}$-th torus $C_{3,i_{1},i_{2}}$ and denote the number of tori
embedded into it by $m_{3,i_{1},i_{2}}$, these tori are

$C_{4,i_{1},i_{2},1},\ldots,C_{4,i_{1},i_{2},m_{3,i_{1},i_{2}}}$

whose disjoint union is $C_{4}$. In general in the $n$-th step for $1\leq
i_{n}\leq m_{n,i_{1},\ldots,i_{n-1}}$ take the $i_{n}$-th torus
$C_{n+1,i_{1},\ldots,i_{n}}$ and denote the number of tori embedded into it by
$m_{n+1,i_{1},\ldots,i_{n}}$, these tori are

$C_{n+2,i_{1},\ldots,i_{n},1},\ldots,C_{n+2,i_{1},\ldots,i_{n},m_{n+1,i_{1},\ldots,i_{n}}}$

whose disjoint union is $C_{n+2}$.

Now we construct a Cantor set $\mathcal{C}$ in the interval $[0,1]$. Divide
$[0,1]$ into $2m_{1}-1$ closed intervals

$I_{2,1},\ldots,I_{2,2m_{1}-1}\subset[0,1]$

of equal length and disjoint interiors. Then divide the $i_{1}$-th interval
$I_{2,i_{1}}$, where $i_{1}$ is odd, into $2m_{2,i_{1}}-1$ closed intervals

$I_{3,i_{1},1},\ldots,I_{3,i_{1},2m_{2,i_{1}}-1}$

of equal length. Then divide the $i_{2}$-th interval $I_{3,i_{1},i_{2}}$,
where $i_{2}$ is odd, into $2m_{3,i_{1},i_{2}}-1$ closed intervals

$I_{4,i_{1},i_{2},1},\ldots,I_{4,i_{1},i_{2},2m_{3,i_{1},i_{2}}-1}$

of equal length. In the $n$-th step divide the $i_{n}$-th interval
$I_{n+1,i_{1},\ldots,i_{n}}$, where $i_{n}$ is odd, into the closed intervals

$I_{n+2,i_{1},\ldots,i_{n},1},\ldots,I_{n+2,i_{1},\ldots,i_{n},2m_{n+1,i_{1},\ldots,i_{n}}-1}$

of equal length and so on. So all the intervals $I_{n+1,i_{1},\ldots,i_{n}}$
have length

$\frac{1}{(2m_{1}-1)\cdots(2m_{n,i_{1},\ldots,i_{n-1}}-1)}.$

Then let

$\mathcal{C}=\bigcap_{n=1}^{\infty}\bigcup_{{\begin{smallmatrix}1\leq
i_{1}\leq m_{1}\\\ 1\leq i_{2}\leq m_{2,i_{1}}\\\ \cdots\\\ 1\leq i_{n}\leq
m_{n,i_{1},\ldots,i_{n-1}}\end{smallmatrix}}}I_{n+1,2i_{1}-1,\ldots,2i_{n}-1}.$

Assign to a point $x\in\cap_{n=1}^{\infty}C_{n}$ the point

$\bigcap_{n=1}^{\infty}I_{n+1,2i_{1}(x)-1,\ldots,2i_{n}(x)-1},$

which is the intersection of the closed intervals containing $x$. This defines
a map

$f\colon\cap_{n=1}^{\infty}C_{n}\to\mathcal{C},$

which is clearly surjective. It is injective as well because if $x\neq
x^{\prime}$, then for large $n$ they are in different $C_{n}$ so they are
mapped into different intervals as well. The map $f$ is continuous because if
$x$ and $x^{\prime}$ are in the same $C_{n}$ until some large enough $n$, then
they are mapped to the same intervals until a large index so $f(x)$ and
$f(x^{\prime})$ are close enough. Then $f$ is a homeomorphism since its domain
is compact and it maps injectively into a $T_{2}$ space. ∎

Of course the components of $\mathcal{A}$ are points so the decomposition
space is obviously $\mathbb{R}^{3}$. An important property of $\mathcal{A}$ is
that it is _wild_ , i.e. there is no self-homeomorphism of $\mathbb{R}^{3}$
mapping $\mathcal{A}$ onto the Cantor set in a line segment. To prove this, we
study the local behaviour of the complement of $\mathcal{A}$.

######  Definition 3.14.

Let $k\geq 0$. A closed subset $A$ of a space $X$ is locally $k$-co-connected
($k$-LCC for short) if for every point $a\in A$ and for every nbhd $U$ of $a$
in $X$ there is a nbhd $V\subset U$ of $a$ in $X$ such that if
$\varphi\colon\partial D^{k+1}\to V-A$ is a map of the $k$-sphere, then
$\varphi$ extends to a map of $D^{k+1}$ into $U-A$.

######  Proposition 3.15.

The set $\mathcal{A}$ in $\mathbb{R}^{3}$ is not $1$-LCC.

###### Sketch of the proof.

At first we show that if $\alpha\colon S^{1}\to C_{1}$ is the meridian of the
torus $C_{1}$, then every smooth embedding $\tilde{\alpha}\colon
D^{2}\to\mathbb{R}^{3}$ extending $\alpha$ is such that
$\tilde{\alpha}(D^{2})$ intersects $\mathcal{A}$. If this was not true, then
$\tilde{\alpha}(D^{2})$ would intersect at most finitely many tori
$C_{1},\ldots,C_{n}$ and it would be possible to perturb $\tilde{\alpha}$ to
get a smooth embedding transversal to each $\partial C_{n}$. Then it is
possible to show that there is a disk $D_{1}\subset D^{2}$ such that
$\tilde{\alpha}(\partial D_{1})$ intersects some torus $\partial C_{2,i_{1}}$
in a meridian. Inductively, $\tilde{\alpha}(D^{2})$ has to intersect some
torus $\partial C_{m,i_{1},\ldots,i_{m-1}}$ for arbitrarily large $m>n$, which
is a contradiction. Suppose that $\mathcal{A}$ is $1$-LCC. Let $\beta\colon
D^{2}\to\mathbb{R}^{3}$ be a smooth embedding such that $\beta(\partial
D^{2})$ is a meridian of $C_{1}$. Cover $\beta(D^{2})\cap\mathcal{A}$ by open
sets $\\{U_{\gamma}\\}_{\gamma\in\Gamma}$ around each of its points, then
there is a covering $\\{V_{\gamma}\\}_{\gamma\in\Gamma}$ such that for all
$\gamma\in\Gamma$ we have $V_{\gamma}\subset U_{\gamma}$ and each map
$\partial D^{2}\to(\mathbb{R}^{3}-\mathcal{A})\cap V_{\gamma}$ can be extended
to a map $D^{2}\to(\mathbb{R}^{3}-\mathcal{A})\cap U_{\gamma}$. We can also
suppose that $\cup_{\gamma}U_{\gamma}$ is disjoint from $\beta(\partial
D^{2})$. By Lebesgue lemma there is a refinement of $D^{2}$ into finitely many
small disks with disjoint interiors such that each of their boundary circles
is mapped by $\beta$ into some $V_{\gamma}$. After a small perturbation we can
suppose that each of the $\beta$-images of these boundary circles is disjoint
from $C_{n}$ for some common large $n$ but it is still in some $V_{\gamma}$.
Now change $\beta$ on each of the small disks to get a map into
$(\mathbb{R}^{3}-\mathcal{A})\cap U_{\gamma}$. By Dehn’s lemma there are
embeddings as well of the small disks into $(\mathbb{R}^{3}-\mathcal{A})\cap
U_{\gamma}$. In this way we get an embedding of the original disk $D^{2}$
which is disjoint from $\mathcal{A}$. This contradicts to the fact that every
embedded disk $D^{2}\subset\mathbb{R}^{3}$ with boundary circle being a
meridian of $C_{1}$ intersects $\mathcal{A}$. ∎

The standard Cantor set
$C\subset\mathbb{R}\times\\{0\\}\times\\{0\\}\subset\mathbb{R}^{3}$ is
$1$-LCC, because having a small loop in its complement $\mathbb{R}^{3}-C$
yields by approximation a small smooth loop in $\mathbb{R}^{3}-C$ transversal
to and disjoint from $\mathbb{R}\times\\{0\\}\times\\{0\\}$. Then deform this
loop by compressing it in a direction parallel to
$\mathbb{R}\times\\{0\\}\times\\{0\\}$ until the loop sits in the plane
$\\{x\\}\times\mathbb{R}^{2}$ for some number $x\in\mathbb{R}^{3}-C$. After
these the loop can be squeezed easily inside this plane to a point in
$\mathbb{R}^{3}-C$. This implies that Antoine’s necklace is a wild Cantor set
in $\mathbb{R}^{3}$.

###  3.3. Bing decomposition

If in the construction of Antoine’s necklace there are always two tori
components of $C_{n+1}$ in each component of $C_{n}$, then we call the arising
decomposition Bing decomposition. Apriori there could be many different Bing
decompositions depending on how the solid tori are embedded into each other.
It is not obvious that we can arrange the components of $C_{n}$ embedded in
such a way that $\cap_{n=1}^{\infty}C_{n}$ is a Cantor set, which would follow
if the maximal diameter of the tori in $C_{n}$ converges to $0$. A random
defining sequence can be seen in Figure 7.

Figure 7. A sketch of a defining sequence of the Bing decomposition. We can
see the torus $C_{1}$, the two torus components of $C_{2}$ and the four torus
components of $C_{3}$. The maximal diameter of tori in $C_{n}$ does not
converge to $0$ necessarily.

Now we construct a defining sequence, where the maximal diameter of the tori
in $C_{n}$ converges to $0$. For this, consider the following way to define a
finite sequence of finite sequences of embeddings:

$D_{0},$ $D_{0}\supset D_{2,1},$ $D_{0}\supset D_{3,1}\supset D_{3,2},$
$D_{0}\supset D_{4,1}\supset\cdots\supset D_{4,3},$

$D_{1}$$D_{2,1}$$D_{3,2}$$D_{4,3}$$D_{5,4}$

Figure 8. A sketch of constructing the tori $D_{n+1,n}$. Instead of the solid
tori we just draw their center circles. We always take the previously obtained
linked tori $D_{n,n-1}$, squeeze them to become “flat” as the figure shows,
then curve them a little and link them with another copy at the two “endings”.
Hence we get $D_{n+1,n}$. The sequence of embeddings $D_{0}\supset
D_{n+1,1}\supset\cdots\supset D_{n+1,n}$ can be kept in sight by checking all
the smaller linkings.

…, etc., where $D_{n,0}=D_{0}$ is a solid torus, $D_{n,k}$ is a disjoint union
of $2^{k}$ copies of solid tori and the components of $D_{n,k}$ are pairwise
embedded in the components of $D_{n,k-1}$, moreover these pairs are linked
just like in the defining sequence of Bing decomposition, for further
subtleties see Figure 8.

Arriving to the tori $D_{n+1,n}$ and assuming that their meridional size is
small enough, we obtain a regular $2n$-gon-like arrangement of $2^{n}$ copies
of solid tori as Figure 8 shows. Two conditions are satisfied: the meridional
size of all the tori is small and an “edge” of this $2n$-gon is also small.
This means that if this $D_{n+1,n}$ is embedded into a torus (as the figure
suggests) whose meridional size is small, then the maximal diameter of the
torus components of $D_{n+1,n}$ is small if $n$ is large.

######  Proposition 3.16.

There is a defining sequence $C_{1},\ldots,C_{n},\ldots$ of the Bing
decomposition, where the maximal diameter of tori in $C_{n}$ converges to $0$.
Hence $\cap_{n=1}^{\infty}C_{n}$ is homeomorphic to the Cantor set.

###### Proof.

Let $\varepsilon_{n}>0$ be a sequence whose limit is $0$. Let $n_{1}$ be so
large that in $C_{n_{1}}$ in a defining sequence the meridional size of tori
is smaller than $\varepsilon_{1}$. Let $m_{1}$ be so large that we can embed
$D_{m_{1}+1,m_{1}}$ into the torus components of $C_{n_{1}}$ so that the
maximal diameter of tori in the obtained $C_{n_{1}+m_{1}}$ is smaller then
$\varepsilon_{1}$. Then let $n_{2}>n_{1}+m_{1}$ be so large that in a
continuation of the defining sequence in $C_{n_{2}}$ the meridional size of
tori is smaller than $\varepsilon_{2}$. Let $m_{2}$ be so large that we can
embed $D_{m_{2}+1,m_{2}}$ into the torus components of $C_{n_{2}}$ so that the
maximal diameter of tori in the obtained $C_{n_{2}+m_{2}}$ is smaller then
$\varepsilon_{2}$. And so on. It is easy to see that the maximal diameter of
tori converges to $0$. ∎

This implies that the decomposition space of this decomposition is
$\mathbb{R}^{3}$. For an arbitrary defining sequence the space
$\cap_{n=1}^{\infty}C_{n}$ may be not the Cantor set, however the
decomposition space could be still homeomorphic to the ambient space
$\mathbb{R}^{3}$. It is a very important observation that the embedding of the
tori in $D_{n+1,n}$ can be obtained by an isotopy of
$C_{1}\subset\cdots\subset C_{n+1}$ in any defining sequence, see [Bi52]. By
such an isotopy for a given defining sequence we can manage something similar
to the previous statement: if $n$ is large enough, then the meridional size of
the torus components in $C_{n}$ is smaller than a given $\varepsilon>0$. Then
apply the required isotopy for $C_{n+1},\ldots,C_{n+k}$ for some large $k$ to
make the maximal diameter of the torus components of $C_{n+k}$ smaller than
$\varepsilon$. Note that since $n$ is large enough and all the isotopy happens
inside $C_{n}$, all the isotopy happens inside an arbitrarily small nbhd of
$\cap_{n=1}^{\infty}C_{n}$. This means that for every $\varepsilon>0$ there is
a self-homeomorphism $h$ of $\mathbb{R}^{3}$ with support $C_{1}$ such that
$h(\mathcal{D})<\varepsilon$ for every decomposition element
$\mathcal{D}\subset\cap_{n=1}^{\infty}C_{n}$ and also $\pi\circ
h(\mathcal{D})$ stays in the $\varepsilon$-nbhd of $\pi(\mathcal{D})$ for some
metric on the decomposition space. This condition is called shrinkability
criterion and it implies that the decomposition space is homeomorphic to the
ambient space $\mathbb{R}^{3}$ as we will see in the next section.

##  4\. Shrinking

Let $X$ be a topological space and $\mathcal{D}$ a decomposition of $X$. An
open cover $\mathcal{U}$ of $X$ is called $\mathcal{D}$-saturated if every
$U\in\mathcal{U}$ is a union of decomposition elements.

######  Definition 4.1 (Bing shrinkability criterion).

Let $\mathcal{D}$ be a usc decomposition of the space $X$. We say
$\mathcal{D}$ is _shrinkable_ if for every open cover $\mathcal{V}$ and
$\mathcal{D}$-saturated open cover $\mathcal{U}$ there is a self-homeomorphism
$h$ of $X$ such that for every $D\in\mathcal{D}$ the set $h(D)$ is in some
$V\in\mathcal{V}$ and for every $x\in X$ there is a $U\in\mathcal{U}$ such
that $x,h(x)\in U$. In other words, $h$ shrinks the elements of $\mathcal{D}$
to arbitrarily small sets and $h$ is $\mathcal{U}$-close to the identity. We
say $\mathcal{D}$ is _strongly shrinkable_ if for every open set $W$
containing all the non-degenerate elements of $\mathcal{D}$ the decomposition
$\mathcal{D}$ is shrinkable so that the support of $h$ is in $W$.

In other words $\mathcal{D}$ is shrinkable if its elements can be made small
enough simultaneously so that this shrinking process does not move the points
of $X$ too far in the sense of measuring the distance in the decomposition
space. If $X$ has a shrinkable decomposition, then we expect that the local
structure of $X$ is similar to the structure of the nbhds of the decomposition
elements.

######  Proposition 4.2.

Let $X$ be a regular space and let $\mathcal{D}$ be a shrinkable usc
decomposition of $X$. If every $x\in X$ has arbitrarily small nbhds satisfying
a fixed topological property, then every $D\in\mathcal{D}$ has arbitrarily
small nbhds satisfying the same property.

###### Proof.

Let $W$ be an arbitrary nbhd of an element $D\in\mathcal{D}$. Then there is a
saturated nbhd $\tilde{U}_{1}$ of $D$ such that $\tilde{U}_{1}\subset W$. Let
$U_{1}$ denote $\pi(\tilde{U}_{1})$. Since $X_{\mathcal{D}}$ is regular, there
are open sets $U_{2}$ and $U_{3}$ such that

$\pi(D)\subset U_{3}\subset\mathrm{cl}\thinspace U_{3}\subset
U_{2}\subset\mathrm{cl}\thinspace U_{2}\subset U_{1}.$

Then take the sets

$\pi^{-1}(U_{3}),\mbox{\ }\pi^{-1}(U_{2})-D,\mbox{\
}\pi^{-1}(U_{1}-\mathrm{cl}\thinspace U_{3}),\mbox{\ and\
}X-\pi^{-1}(\mathrm{cl}\thinspace U_{2}),$

see Figure 9. These yield a $\mathcal{D}$-saturated open cover $\mathcal{U}$
of $X$. Let $\mathcal{V}$ be an open cover of $X$ which refines $\mathcal{U}$
and consists of open sets with our fixed property. Since $\mathcal{D}$ is
shrinkable, we have a homeomorphism $h\colon X\to X$ such that $h(D)\subset V$
for some $V\in\mathcal{V}$ and $h$ is $\mathcal{U}$-close to the identity.
Then $D\subset h^{-1}(V)$ so it is enough to show that

$h^{-1}(V)\subset W.$

Suppose there exists some $x\in h^{-1}(V)-W$, then $x\in
h^{-1}(V)-\pi^{-1}(U_{1})$ since $\pi^{-1}(U_{1})\subset W$. Hence among the
sets in $\mathcal{U}$ only $X-\pi^{-1}(\mathrm{cl}\thinspace U_{2})$ contains
$x$ so $h(x)\in V$ has to be in $X-\pi^{-1}(\mathrm{cl}\thinspace U_{2})$ as
well. This implies that $V\subset X-\pi^{-1}(\mathrm{cl}\thinspace U_{3})$
because $V$ is a subset of some sets in $\mathcal{U}$. Also we know that
$h(D)\subset\pi^{-1}(U_{3})$ since $D\subset\pi^{-1}(U_{3})$. This means that
$h(D)\subset V\subset X-\pi^{-1}(\mathrm{cl}\thinspace U_{3})$ cannot hold so
$h^{-1}(V)\subset W$. ∎

$X$$D$

Figure 9. The set $D$ and the $\pi$-preimages of the sets $U_{3}\subset
U_{2}\subset U_{1}$ in $X$. In the figure the sets $\pi^{-1}(U_{3})$ and
$\pi^{-1}(U_{2})-D$ are shaded.

For example every decomposition element of a shrinkable decomposition of a
manifold is cellular.

It is often not too difficult to check whether a decomposition of a space $X$
is shrinkable. A corollary of shrinkability is that the decomposition space is
homeomorphic to $X$. This is often applied when we want to construct embedded
manifolds and the construction uses mismatched pieces, which we eliminate by
taking them as the decomposition elements and then looking at the
decomposition space.

######  Definition 4.3 (Near-homeomorphism, approximating by homeomorphism).

Let $X$ and $Y$ be topological spaces. An $f\colon X\to Y$ surjective map is a
_near-homeomorphism_ if for every open covering $\mathcal{W}$ of $Y$ there is
a homeomorphism $h\colon X\to Y$ such that for every $x\in X$ the points
$f(x)$ and $h(x)$ are in some $W\in\mathcal{W}$, in other words $h$ is
_$\mathcal{W}$ -close_ to $f$.

If $(Y,\varrho)$ is a metric space, then $f$ being a near-homeomorphism
implies that $f$ can be approximated by homeomorphisms in the possibly
infinite-valued metric $d(f,g)=\sup\\{\varrho(f(x),g(x))\\}$. Notice that if
$f\colon X\to Y$ is a near-homeomorphism, then $X$ and $Y$ are actually
homeomorphic.

The main result is that a usc decomposition yields a homeomorphic
decomposition space if the decomposition is shrinkable. This is applied in
major $4$-dimensional results: in the disk embedding theorem and in the proof
of the $4$-dimensional topological Poincaré conjecture [Fr82, BKKPR]. It is
extensively applied in constructing approximations of manifold embeddings in
dimension $\geq 5$, see [AC79] and Edwards’s cell-like approximation theorem.

For an open cover $\mathcal{W}$ of a space $X$ and a subset $A\subset X$ let
$St(A,\mathcal{W})$ denote the subset

$\bigcup\\{W\in\mathcal{W}:W\cap A\neq\emptyset\\}.$

This is called the _star_ of $A$ and it is a nbhd of $A$. Of course if
$A\subset B$, then $St(A,\mathcal{W})\subset St(B,\mathcal{W})$. If
$\mathcal{W}^{\prime}$ is an open cover which is a refinement of the open
cover $\mathcal{W}$, then obviously $St(A,\mathcal{W}^{\prime})\subset
St(A,\mathcal{W})$. We will use often that if the covering
$\mathcal{W}^{\prime}$ is a _star-refinement_ of the covering $\mathcal{W}$,
that is the collection

$\left\\{St(W_{\alpha},\mathcal{W}^{\prime}):W_{\alpha}\in\mathcal{W}^{\prime}\right\\}$

of stars of elements of $\mathcal{W}^{\prime}$ is a refinement of
$\mathcal{W}$, then for every point $x\in X$ we have
$St(\\{x\\},\mathcal{W}^{\prime})\subset W$ for some $W\in\mathcal{W}$.

The following theorem requires a complete metric on the space $X$, for example
the statement holds for an arbitrary manifold.

######  Theorem 4.4.

Let $\mathcal{D}$ be a usc decomposition of a space $X$ admitting a complete
metric. Then the following are equivalent:

  1. (1)

the decomposition map $\pi\colon X\to X_{\mathcal{D}}$ is a near-
homeomorphism,

  2. (2)

$\mathcal{D}$ is shrinkable.

If additionally $X$ is also locally compact and separable, then shrinkability
is equivalent to

  1. (3)

if $C\subset X$ is an arbitrary compact set, $\varepsilon>0$ and $\mathcal{U}$
is a $\mathcal{D}$-saturated open cover of $X$, then there is a homeomorphism
$h\colon X\to X$ such that ${\mathrm{diam}}\thinspace h(D)<\varepsilon$ for
every $D\subset C$, $D\in\mathcal{D}$ and $h$ is $\mathcal{U}$-close to the
identity.

###### Proof.

Near-homeomorphism (1) implies shrinking (2) and (3). Of course (2) implies
(3) so we are going to prove only that (1) implies (2). At first, suppose that
the decomposition map $\pi\colon X\to X_{\mathcal{D}}$ is a near-
homeomorphism. We have to show that $\mathcal{D}$ is shrinkable by finding an
appropriate homeomorphism $h$. We know that since $X$ is metric, the
decomposition space $X_{\mathcal{D}}$ is metrizable hence it is paracompact.
(To show that $\mathcal{D}$ is shrinkable, we will use only that the space $X$
is paracompact and $T_{4}$.) Let $\mathcal{V}$ be an open cover and let
$\mathcal{U}$ be a $\mathcal{D}$-saturated open cover of $X$. Take the open
covering $\\{\pi(U):U\in\mathcal{U}\\}$ of $X_{\mathcal{D}}$. Since
$X_{\mathcal{D}}$ is paracompact, this covering has a star-refinement
$\mathcal{W}_{0}$, i.e. $\mathcal{W}_{0}$ is a covering and the collection of
stars of elements of $\mathcal{W}_{0}$, that is the collection

$\left\\{St(W_{\alpha},\mathcal{W}_{0}):W_{\alpha}\in\mathcal{W}_{0}\right\\}$

is a refinement of $\\{\pi(U):U\in\mathcal{U}\\}$, see [Du66, Section 8.3].
Similarly $\mathcal{W}_{0}$ has a star-refinement covering $\mathcal{W}_{1}$.
Then there is a homeomorphism

$h_{1}\colon X\to X_{\mathcal{D}}$

which is $\mathcal{W}_{1}$-close to $\pi$ because $\pi$ is a near-
homeomorphism. Take the open cover

$\mathcal{W}_{1}\bigcap h_{1}(\mathcal{V})=\\{W\cap
h_{1}(V):W\in\mathcal{W}_{1},V\in\mathcal{V}\\}$

and a star-refinement $\mathcal{W}_{2}$ of it. Of course $\mathcal{W}_{2}$ is
a star-refinement of $\mathcal{W}_{1}$ and $h_{1}(\mathcal{V})$ as well. There
is a homeomorphism

$h_{2}\colon X\to X_{\mathcal{D}}$

which is $\mathcal{W}_{2}$-close to $\pi$. Let $h\colon X\to X$ be the
composition

$h_{1}^{-1}\circ h_{2}.$

At first we show that $h$ shrinks every decomposition element
$D\in\mathcal{D}$ into some $V\in\mathcal{V}$. Let $D\in\mathcal{D}$. It is
enough to show that $h_{2}(D)\subset h_{1}(V)$ for some $V\in\mathcal{V}$. We
have that for every $x\in D$ the points $\pi(D)$ and $h_{2}(x)$ are in the
same $W_{x}\in\mathcal{W}_{2}$ so

$h_{2}(D)\subset St(\\{\pi(D)\\},\mathcal{W}_{2})\subset h_{1}(V)$

for some $V\in\mathcal{V}$ because $\mathcal{W}_{2}$ is a star-refinement of
$h_{1}(\mathcal{V})$. Now we show that $h$ is $\mathcal{U}$-close to the
identity. We have that for every $x\in D$ the points $\pi(D)$ and $h_{1}(x)$
are in the same $W_{x}\in\mathcal{W}_{1}$ because $h_{1}$ is
$\mathcal{W}_{1}$-close to $\pi$ so

$h_{1}(D)\subset St(\\{\pi(D)\\},\mathcal{W}_{1}).$

Since $\mathcal{W}_{2}$ is a refinement of $\mathcal{W}_{1}$, we have

$h_{2}(D)\subset St(\\{\pi(D)\\},\mathcal{W}_{2})\subset
St(\\{\pi(D)\\},\mathcal{W}_{1}).$

These imply that

$h_{1}(D)\cup h_{2}(D)\subset St(\\{\pi(D)\\},\mathcal{W}_{1})\subset W_{0}$

for some $W_{0}\in\mathcal{W}_{0}$ because $\mathcal{W}_{1}$ is a star-
refinement of $\mathcal{W}_{0}$. Hence for every $D\in\mathcal{D}$ we have

$D\cup h(D)=h_{1}^{-1}\circ h_{1}(D\cup h(D))=h_{1}^{-1}(h_{1}(D)\cup
h_{2}(D))\subset h_{1}^{-1}(W_{0})$

so if we show that

$h_{1}^{-1}(W_{0})\subset U$

for some $U\in\mathcal{U}$, then we prove the statement. Since $h_{1}$ and
$\pi$ are $\mathcal{W}_{1}$-close, they are $\mathcal{W}_{0}$-close as well.
This means that for every $x\in X$ the points $\pi(x)$ and $h_{1}(x)$ are in
the same $W_{x}\in\mathcal{W}_{0}$. So if $x\in h^{-1}_{1}(W_{0})$, then

$\pi(x)\in St(W_{0},\mathcal{W}_{0}),$

which gives that

$\pi(h^{-1}_{1}(W_{0}))\subset St(W_{0},\mathcal{W}_{0})\subset\pi(U)$

for some $U\in\mathcal{U}$ because $W_{0}$ is a star-refinement of
$\pi(\mathcal{U})$. Then the statement follows because

$h^{-1}_{1}(W_{0})\subset\pi^{-1}\circ\pi(h^{-1}_{1}(W_{0}))\subset\pi^{-1}\circ\pi(U)=U.$

Shrinking (2) or (3) implies near-homeomorphism (1). At first observe that in
the case of (3) if $X$ is locally compact and separable, then $X$ is
$\sigma$-compact so $X$ is the union $\cup_{n=1}^{\infty}C_{n}$ of countably
many compact sets

$C_{1}\subset C_{2}\subset\cdots\subset C_{n}\subset\cdots.$

We also suppose that every $C_{n}$ is $\mathcal{D}$-saturated and has non-
empty interior. Let $\mathcal{W}$ be an arbitrary open cover of
$X_{\mathcal{D}}$. We have to construct a homeomorphism $h\colon X\to
X_{\mathcal{D}}$ which is $\mathcal{W}$-close to $\pi$. At first, we construct
a sequence

$\mathcal{U}_{0},\mathcal{U}_{1},\ldots,\mathcal{U}_{n},\ldots$

of $\mathcal{D}$-saturated open covers of $X$ and a sequence

$h_{0},h_{1},\ldots,h_{n},\ldots$

of self-homeomorphisms of $X$ with some useful properties. Let
$\mathcal{U}_{0}$ be a $\mathcal{D}$-saturated open cover of $X$ such that the
collection of the closures of the elements of $\mathcal{U}_{0}$ refines the
open cover $\pi^{-1}(\mathcal{W})$. This obviously exists because
$X_{\mathcal{D}}$ is regular so around every point of $X_{\mathcal{D}}$ there
is a small closed nbhd contained in some element of $\mathcal{W}$. Let $h_{0}$
be the identity homeomorphism. Let $\varepsilon_{n}>0$ be a decreasing
sequence converging to $0$. Define $\varepsilon_{0}$ to be $\infty$. Denote
the metric on $X$ by $d$. Suppose inductively that we constructed already the
covers $\mathcal{U}_{0},\ldots,\mathcal{U}_{n}$ and the homeomorphisms
$h_{0},\ldots,h_{n}$ with the following properties:

  1. (1)

    1. (a)

$\mathcal{U}_{i+1}$ is a $\mathcal{D}$-saturated open cover, which refines
$\mathcal{U}_{i}$ for $0\leq i\leq n-1$,

    2. (b)

for all $0\leq i\leq n$ the set $\mathcal{U}_{i}$ refines the collection of
$\varepsilon_{i}$-nbhds of the elements of $\mathcal{D}$ and also refines the
collection $\\{\pi^{-1}(B_{\varepsilon_{i}}(y)):y\in X_{\mathcal{D}}\\}$,
where $B_{\varepsilon_{i}}(y)$ is the open ball of radius $\varepsilon_{i}$
around $y$,

  2. (2)

    1. (a)

for every $0\leq i\leq n-1$ every $D\in\mathcal{D}$ has a nbhd
$U\in\mathcal{U}_{i}$ such that for every $U^{\prime}\in\mathcal{U}_{i+1}$
which contains $D$ we have

$h_{i}(U^{\prime})\cup h_{i+1}(U^{\prime})\subset h_{i}(U),$

    2. (b)

for every $0\leq i\leq n$ the diameter of each $h_{i}(U)$,
$U\in\mathcal{U}_{i}$, is smaller than $\varepsilon_{i}$,

    3. (b′)

in the case of $X$ is $\sigma$-compact we require only that for every $0\leq
i\leq n$ and for every nbhd $U\in\mathcal{U}_{i}$ such that $U\cap
C_{i}\neq\emptyset$ the diameter of each $h_{i}(U)$ is smaller than
$\varepsilon_{i}$.

There will be some important corollaries of these constructions. Part (a) of
(2) implies that every $D\in\mathcal{D}$ has a nbhd $U\in\mathcal{U}_{i}$ such
that for every $k\geq 1$ and $U^{\prime}\in\mathcal{U}_{i+k}$ which contains
$D$ we have

(4.1) $h_{i+k}(U^{\prime})\subset h_{i}(U).$

For $k=1$ this is immediate from (2)(a) and for $k\geq 2$ this follows by a
simple induction. This means that once we have $U_{n}$ and $h_{n}$ for every
$n\in\mathbb{N}$ satisfying (1) and (2), the sequence $h_{n}$ is a Cauchy
sequence in the sense of local uniform convergence in the space of maps of $X$
into $X$. Indeed, if $x\in X$, then for some $D\in\mathcal{D}$ we have $x\in
D$ and then $D$ has a nbhd $U\in\mathcal{U}_{n}$ for every $n$ such that by
applying (4.1) for all $k\in\mathbb{N}$

(4.2) $h_{n+k}(D)\subset h_{n}(U),$

which means that $d(h_{n}(x),h_{n+k}(x))<\varepsilon_{n}$ for all $x\in X$ by
(2)(b). In the case of $X$ is $\sigma$-compact we have that for some
$m\in\mathbb{N}$ the intersection $D\cap C_{n}\neq\emptyset$ for $n\geq m$
hence by (2)(b′) we have $\mathrm{diam}\thinspace h_{n}(U)<\varepsilon_{n}$
for every $n\geq m$ and nbhd $U\in\mathcal{U}_{n}$ of $D$. This implies that
for all $n\geq m$ we get $d(h_{n}(x),h_{n+k}(x))<\varepsilon_{n}$ for all $k$
and $x\in D$, where $D\subset C_{n}$. Since $(X,d)$ is complete, the sequence
$h_{n}$ converges locally uniformly to a continuous map

$\chi\colon X\to X,$

which will be a good candidate for obtaining our desired near-homeomorphism.

Defining $\mathcal{U}_{n+1}$ and $h_{n+1}$. So let us return to the definition
of the covers $\mathcal{U}_{n}$ and homeomorphisms $h_{n}$. Suppose
inductively that we constructed already the covers
$\mathcal{U}_{0},\ldots,\mathcal{U}_{n}$ and the homeomorphisms
$h_{0},\ldots,h_{n}$ with the properties (1) and (2). We are going to define
$\mathcal{U}_{n+1}$ and $h_{n+1}$. The metrizable space $X_{\mathcal{D}}$ is
paracompact so the open cover $\pi(\mathcal{U}_{n})$ has a star-refinement
whose $\pi$-preimage $\mathcal{U}_{n}^{\prime}$ is a $\mathcal{D}$-saturated
open cover of $X$, which star-refines $\mathcal{U}_{n}$. Let $\mathcal{V}$ be
an open cover of $X$ such that the diameter of each of its elements is smaller
than $\varepsilon_{n+1}$. Then we have two possibilities.

  * •

If $\mathcal{D}$ is shrinkable, then there is a self-homeomorphism $H$ of $X$,
which is $h_{n}(\mathcal{U}_{n}^{\prime})$-close to the identity and shrinks
the elements of $h_{n}(\mathcal{D})$ into the sets of $\mathcal{V}$. Let

$h_{n+1}=H\circ h_{n}.$

Clearly the diameter of each $h_{n+1}(D)$, where $D\in\mathcal{D}$, is smaller
than $\varepsilon_{n+1}$.

  * •

If only those elements of $\mathcal{D}$ are shrinkable which are in a chosen
compact set as we suppose in (3) of the statement of Theorem 4.4, then there
is a homeomorphism $H_{0}\colon X\to X$ such that the elements of
$\mathcal{D}$ in the compact set $C_{n+1}$ are mapped by $H_{0}$ into some
element of $h_{n}^{-1}(\mathcal{V})$ and $H_{0}$ is
$\mathcal{U}_{n}^{\prime}$-close to the identity. This implies that
$h_{n}\circ H_{0}\circ h_{n}^{-1}$ is such a self-homeomorphism of $X$ that
maps the elements of $h_{n}(\mathcal{D})$ which are in $h_{n}(C_{n+1})$ into
the sets of $\mathcal{V}$ and it is $h_{n}(\mathcal{U}_{n}^{\prime})$-close to
the identity. Denote $h_{n}\circ H_{0}\circ h_{n}^{-1}$ by $H$. Then let

$h_{n+1}=H\circ h_{n}.$

So $h_{n+1}$ maps every $D\in\mathcal{D}$, $D\subset C_{n+1}$ into a set of
diameter smaller than $\varepsilon_{n+1}$.

The definition of $\mathcal{U}_{n+1}$ is a little more complicated. For every
$U_{n}^{\prime}\in\mathcal{U}_{n}^{\prime}$

$h_{n+1}(U_{n}^{\prime})\subset
h_{n}(St(U_{n}^{\prime},\mathcal{U}_{n}^{\prime}))$

because

$h_{n+1}(U_{n}^{\prime})=H\circ h_{n}(U_{n}^{\prime})\subset
St(h_{n}(U_{n}^{\prime}),h_{n}(\mathcal{U}_{n}^{\prime}))$

since $H$ is $h_{n}(\mathcal{U}_{n}^{\prime})$-close to the identity and also

$St(h_{n}(U_{n}^{\prime}),h_{n}(\mathcal{U}_{n}^{\prime}))=h_{n}(St(U_{n}^{\prime},\mathcal{U}_{n}^{\prime})).$

The covering $\mathcal{U}_{n}^{\prime}$ star-refines $\mathcal{U}_{n}$ so for
every $U_{n}^{\prime}\in\mathcal{U}_{n}^{\prime}$ there is an
$U_{n}\in\mathcal{U}_{n}$ such that

$h_{n}(St(U_{n}^{\prime},\mathcal{U}_{n}^{\prime}))\subset h_{n}(U_{n}),$

which obviously implies that for every
$U_{n}^{\prime}\in\mathcal{U}_{n}^{\prime}$ there is an
$U_{n}\in\mathcal{U}_{n}$ such that

$h_{n}(U_{n}^{\prime})\cup h_{n+1}(U_{n}^{\prime})\subset
h_{n}(St(U_{n}^{\prime},\mathcal{U}_{n}^{\prime}))\subset h_{n}(U_{n}).$

Let $\mathcal{S}$ be a $\mathcal{D}$-saturated open cover of $X$ with the
following properties:

  1. (i)

the elements of $\mathcal{S}$ are nbhds of the elements of $\mathcal{D}$ such
that the diameter of each $h_{n+1}(S)$, where $S\in\mathcal{S}$, is smaller
than $\varepsilon_{n+1}$ (in the case of $S\cap C_{n+1}\neq\emptyset$ if $X$
is $\sigma$-compact),

  2. (ii)

$\mathcal{S}$ refines the collection of $\varepsilon_{n+1}$-nbhds of the
elements of $\mathcal{D}$,

  3. (iii)

$\mathcal{S}$ also refines the $\mathcal{D}$-saturated coverings

    1. (a)

$\mathcal{U}_{n}^{\prime}$ and

    2. (b)

the collection $\\{\pi^{-1}(B_{\varepsilon_{n+1}}(y)):y\in
X_{\mathcal{D}}\\}$,

  4. (iv)

for every $S\in\mathcal{S}$ there is a $U_{n}\in\mathcal{U}_{n}$ such that

$h_{n}(S)\cup h_{n+1}(S)\subset h_{n}(U_{n}).$

Let $\mathcal{U}_{n+1}$ be the $\pi$-preimage of an open cover of
$X_{\mathcal{D}}$ which star-refines the open cover $\pi(\mathcal{S})$. It
follows that $\mathcal{U}_{n+1}$ star-refines $\mathcal{S}$. After we defined
$\mathcal{U}_{n+1}$ and $h_{n+1}$ let us check if
$\mathcal{U}_{0},\ldots,\mathcal{U}_{n+1}$ and $h_{0},\ldots,h_{n+1}$ satisfy
the conditions (1) and (2) on page 4. The cover $\mathcal{U}_{n+1}$ refines
the cover $\mathcal{U}_{n}$ because $\mathcal{U}_{n}^{\prime}$ refines
$\mathcal{U}_{n}$, $\mathcal{S}$ refines $\mathcal{U}_{n}^{\prime}$ by
(iii)(a) and $\mathcal{U}_{n+1}$ refines $\mathcal{S}$. So (1)(a) holds. Also
(1)(b) holds because of (ii) and (iii)(b). To prove (2)(a) observe that for
every $D\in\mathcal{D}$ the set $St(D,\mathcal{U}_{n+1})$ is a subset of
$St(U,\mathcal{U}_{n+1})$ for some $U\in\mathcal{U}_{n+1}$. Then
$St(D,\mathcal{U}_{n+1})\subset S$ for some $S\in\mathcal{S}$ since
$\mathcal{U}_{n+1}$ star-refines $\mathcal{S}$. By (iv) there exists a
$U\in\mathcal{U}_{n}$ such that

$h_{n}(S)\cup h_{n+1}(S)\subset h_{n}(U)$

so

$h_{n}(St(D,\mathcal{U}_{n+1}))\cup h_{n+1}(St(D,\mathcal{U}_{n+1}))\subset
h_{n}(U).$

But every $U^{\prime}\in\mathcal{U}_{n+1}$ which contains $D$ is in
$St(D,\mathcal{U}_{n+1})$ so we have

$h_{n}(U^{\prime})\cup h_{n+1}(U^{\prime})\subset h_{n}(U),$

which proves (2)(a). Finally, the diameter of each $h_{n+1}(U)$,
$U\in\mathcal{U}_{n+1}$, is smaller than $\varepsilon_{n+1}$ (if $U\cap
C_{n+1}\neq\emptyset$ in the case of $\sigma$-compact $X$), because
$\mathcal{U}_{n+1}$ refines $\mathcal{S}$ and we can apply (i).

Constructing the near-homeomorphism. After having these infinitely many
$\mathcal{D}$-saturated open coverings

$\mathcal{U}_{0},\mathcal{U}_{1},\ldots$

and homeomorphisms

$h_{0},h_{1},\ldots$

take the map

$\chi\colon X\to X$

that we obtained applying (4.2) and defined to be the pointwise limit of the
sequence $h_{n}$. At first, we show that $\chi$ is surjective. Let $x\in X$
and $x_{n}=h_{n+1}^{-1}(x)$. Let $D_{n}\in\mathcal{D}$ be such that $x_{n}\in
D_{n}$, then by (2)(a) we get a nbhd $U_{n}\in\mathcal{U}_{n}$ of $D_{n}$ such
that for every $U^{\prime}\in\mathcal{U}_{n+1}$ containing $D_{n}$ we have

$h_{n}(U^{\prime})\cup h_{n+1}(U^{\prime})\subset h_{n}(U_{n}).$

In this way we get a decreasing sequence

$U_{1}\supset U_{2}\supset\cdots\supset U_{n}\supset\cdots$

because of the following. It is enough to show that $D_{n}\subset U_{n+1}$ as
well, then by (2)(a) we obtain $h_{n}(U_{n+1})\subset h_{n}(U_{n})$ so
$U_{n+1}\subset U_{n}$. But $D_{n}\subset U_{n+1}$ because

$h_{n+2}(V)\subset h_{n+1}(U_{n+1})$

by (2)(a) for every $V\in\mathcal{U}_{n+2}$ containing $D_{n+1}$, so we also
have

$h_{n+2}(x_{n+1})\in h_{n+2}(D_{n+1})\subset h_{n+2}(V),$

which implies

$x\in h_{n+1}(U_{n+1})$

hence

$h_{n+1}(x_{n})\in h_{n+1}(U_{n+1})\mbox{\ \ \ \ and so\ \ \ \ }x_{n}\in
U_{n+1}$

but $U_{n+1}$ is $\mathcal{D}$-saturated hence also $D_{n}\subset U_{n+1}$.

The sequence $(x_{n})$ has a Cauchy (hence convergent) subsequence: since
$x_{n}\in U_{n}$, for all $k\geq 0$

$x_{n+k}\in U_{n}$

and for every $\varepsilon>0$ there is $\varepsilon_{n}<\varepsilon$ such that
$U_{n}$ is in the $\varepsilon$-nbhd of some $D\in\mathcal{D}$ by (1)(b).
Since the metric space $X$ is complete, there is an $x_{0}\in X$ such that a
subsequence $(x_{n_{k}})$ of $(x_{n})$ converges to $x_{0}$.

All of these imply that because of the definition of $\chi$ and the locally
uniform convergence of $h_{n}$ we have

$\chi(x_{0})=\lim_{k\to\infty}h_{n_{k}+1}(x_{n_{k}})=\lim_{k\to\infty}x=x.$

This means $\chi$ is surjective.

It will turn out that $\chi$ is not injective so it is not a homeomorphism.
However, the composition $\pi\circ\chi^{-1}$ of the relation $\chi^{-1}$ and
the decomposition map $\pi$ is a homeomorphism. To see this, we show that the
sets $\chi^{-1}(x)$, where $x\in X$, are exactly the decomposition elements of
$\mathcal{D}$. By (4.2) for every $n\in\mathbb{N}$ and $D\in\mathcal{D}$ there
is a nbhd $U\in\mathcal{U}_{n}$ of $D$ such that for every $k\geq 0$

$h_{n+k}(D)\subset h_{n}(U)$

hence

$\chi(D)=\lim_{k\to\infty}h_{n+k}(D)\subset\mathrm{cl}\thinspace h_{n}(U).$

It is a fact that ${\mathrm{diam}}\thinspace\mathrm{cl}\thinspace
A={\mathrm{diam}}\thinspace A$ for an arbitrary subset $A$ of a metric space
so by (2)(b) we obtain $\chi(D)<\varepsilon_{n}$ for each $n$ and by (2)(b′)
for some $C_{m}\supset D$ we obtain $\chi(D)<\varepsilon_{n}$ for each $n\geq
m$, which implies that $\chi(D)$ is a point. To show that the $\chi$-preimage
of a point is not bigger than a decomposition element, observe that for
different elements $D_{1}$ and $D_{2}$ and for large enough $n$ by (1)(b)
there are $U,V\in\mathcal{U}_{n}$ which lie in the small
$\varepsilon_{n}$-nbhds of $D_{1}$ and $D_{2}$, respectively, hence $U$ and
$V$ are disjoint. Then similarly to above,

$\chi(D_{1})\subset\mathrm{cl}\thinspace h_{n}(U)\mbox{\ \ \ \ and\ \ \ \
}\chi(D_{2})\subset\mathrm{cl}\thinspace h_{n}(V),$

which implies that $\chi(D_{1})$ and $\chi(D_{2})$ are different so the sets
$\chi^{-1}(x)$, where $x\in X$, are exactly the decomposition elements of
$\mathcal{D}$.

This means that $\pi\circ\chi^{-1}$ is a bijection. Its inverse is continuous
because $\chi$ is continuous and $\pi$ is a closed map since the decomposition
is usc. To prove that $\pi\circ\chi^{-1}$ is continuous it is enough to show
that $\chi$ is a closed map. Let $A\subset X$ be a closed set and observe that
a point $y\in X$ is in $X-\chi(A)$ if and only if
$\chi^{-1}(y)\cap\chi^{-1}(\chi(A))=\emptyset$, which holds exactly if
$\chi^{-1}(y)\cap\pi^{-1}(\pi(A))=\emptyset$. This means that in order to show
that $\chi(A)$ is closed it is enough to prove that for any decomposition
element $D$ such that $D\cap\pi^{-1}(\pi(A))=\emptyset$ the point $\chi(D)$ is
an inner point of $X-\chi(A)$. If $\varepsilon_{n}$ is small enough, then
since $D\cap\pi^{-1}(\pi(A))=\emptyset$, by (1)(b) for every
$U_{n}\in\mathcal{U}_{n}$ containing $D$ we have

$\mathrm{cl}\thinspace U_{n}\cap\mathrm{cl}\thinspace
St(A,\mathcal{U}_{n})=\emptyset.$

By (4.2) we have $\chi(D)\in h_{n}(\mathrm{cl}\thinspace U_{n})$ and obviously

$\chi(A)\subset h_{n}(\mathrm{cl}\thinspace
St(A,\mathcal{U}_{n}))=\mathrm{cl}\thinspace h_{n}(St(A,\mathcal{U}_{n}))$

so finally we get

$\chi(D)\in X-\mathrm{cl}\thinspace h_{n}(St(A,\mathcal{U}_{n}))\subset
X-\chi(A)$

implying that $\chi(D)$ is an inner point of $X-\chi(A)$. As a consequence the
map $\pi\circ\chi^{-1}$ is a homeomorphism. We have to prove that it is
$\mathcal{W}$-close to the identity. By (4.1) for every $D$ and for all $n$
there exist $U_{n}\in\mathcal{U}_{n}$ nbhds of $D$ such that

$U_{0}=h_{0}(U_{0})\supset h_{1}(U_{1})\supset\cdots\supset
h_{n}(U_{n})\supset\cdots.$

So $h_{n}(D)\in U_{0}$ for every $n$ and then $\chi(D)\in\mathrm{cl}\thinspace
U_{0}$. Since the collection of the closures of the elements in
$\mathcal{U}_{0}$ refines the cover $\pi^{-1}(\mathcal{W})$, both of $D$ and
$\chi(D)$ are in the same $\pi^{-1}(W)\in\mathcal{W}$. This implies that if we
denote $\chi(D)$ by $x$, then both of $\chi^{-1}(x)$ and $x$ are in
$\pi^{-1}(W)$. As a result $\chi(\chi^{-1}(x))=x$ and $\chi(x)$ are in
$\chi(\pi^{-1}(W))$ so by applying the map $\pi\circ\chi^{-1}$ we get that

$\pi\circ\chi^{-1}(x)\mbox{\ \ \ \ and\ \ \ \
}\pi\circ\chi^{-1}(\chi(x))=\pi(x)$

are in $W$. This shows that $\pi\circ\chi^{-1}$ is $\mathcal{W}$-close to
$\pi$. ∎

The goal of most of the applications of shrinking is to obtain some kind of
embedding of a manifold by the process of approximating a given map. Let
$\mathbb{R}^{d}_{+}$ denote the closed halfspace in $\mathbb{R}^{d}$.

######  Definition 4.5 (Flat subspace and locally flat embedding).

Let $A\subset X$ be a chosen subspace of a topological space $X$. We say that
the subspace $B\subset X$ homeomorphic to $A$ is _flat_ if there is a
homeomorphism $h\colon X\to X$ such that $h(B)=A$. Let $X$ be an
$n$-dimensional manifold. An embedding $e\colon B\to X$ of a $d$-dimensional
manifold $B$ is _locally flat_ if every point $e(b)$ has a nbhd $U$ in $X$
such that the pair

$(U,e(B)\cap U)\mbox{\ is homeomorphic to\
}\left\\{\begin{array}[]{ccc}(\mathbb{R}^{n},\mathbb{R}^{d})&\mbox{if $b$ is
an inner point of $B$}\\\ (\mathbb{R}^{n},\mathbb{R}^{d}_{+})&\mbox{if $b$ is
a boundary point of $B$}.\end{array}\right.$

######  Definition 4.6 (Collared and bicollared subspaces).

The subspace $A\subset X$ is _collared_ if there is an embedding $f\colon
A\times[0,1)\to X$ onto an open subspace of $X$ such that $f(a,0)=a$. The
subspace $A\subset X$ is _bicollared_ if there is an embedding $f\colon
A\times(-1,1)\to X$ such that $f(a,0)=a$. The subspace $A\subset X$ is
_locally collared_ (or _locally bicollared_) if every $a\in A$ has a nbhd $U$
in $X$ such that $A\cap U$ is collared (resp. bicollared).

A typical application of shrinking is the following.

######  Theorem 4.7.

Let $X$ be an $n$-dimensional manifold with boundary $\partial X$. Then
$\partial X$ is collared in $X$.

###### Proof.

Attach the manifold $\partial X\times[0,1]$ to $X$ along $\partial X\subset X$
by the identification

$\varphi\colon\partial X\times\\{0\\}\to X,$ $\varphi(x,0)=x.$

In this way we get a manifold $\tilde{X}$, which contains the attached
$\partial X\times[0,1]$ as a subset. The boundary of $\tilde{X}$ is $\partial
X\times\\{1\\}$ and so the boundary $\partial\tilde{X}$ is obviously collared.
Let $\mathcal{D}$ be the decomposition of $\tilde{X}$ into the intervals
$\\{\\{x\\}\times[0,1]:x\in\partial X\\}$ and the singletons in
$\tilde{X}-\partial X\times[0,1]$. Then $X$ and the quotient space
$\tilde{X}_{\mathcal{D}}$ are homeomorphic by the map

$\alpha\colon X\to\tilde{X}_{\mathcal{D}},$ $\alpha(x)=[x],$

where $[x]$ denotes the equivalence class of $x$. Indeed, $\alpha$ is a
bijection mapping $X-\partial X$ to the classes consisting of single points
and mapping the boundary points $x\in\partial X$ to the class $[x]$. It is
easy to see that $\alpha$ and also $\alpha^{-1}$ are continuous so $\alpha$ is
a homeomorphism. If we prove that $\tilde{X}$ is also homeomorphic to the
decomposition space $\tilde{X}_{\mathcal{D}}$ by a map $\beta$ as the diagram

$X$$\tilde{X}$$\tilde{X}_{\mathcal{D}}$$\alpha$$\beta$

shows, then we obtain that $X$ and $\tilde{X}$ are homeomorphic through the
map $\beta^{-1}\circ\alpha$, which finishes the proof. A homeomorphism $\beta$
exists if we prove that ${\mathcal{D}}$ is shrinkable because then
$\pi\colon\tilde{X}\to\tilde{X}_{\mathcal{D}}$ is a near-homeomorphism. Let
$\mathcal{V}$ be an arbitrary open cover of $\tilde{X}$ and let $\mathcal{U}$
be a $\mathcal{D}$-saturated open cover of $\tilde{X}$. Let $\mathcal{W}$ be a
refinement of $\mathcal{V}$ such that $\mathcal{W}$ contains all the small
nbhds of the form $U_{x}\times[1,1-\varepsilon_{x})$ for all
${(x,1)}\in\partial\tilde{X}$ and for some appropriate $\varepsilon_{x}>0$ and
relative nbhd $U_{x}\subset\partial X$. We also suppose that in $\mathcal{W}$
the nbhds of the inner points of $\tilde{X}$ are only these or such nbhds
which do not intersect $\partial\tilde{X}$. We will apply Theorem 4.4. Let
$C\subset\tilde{X}$ be a compact set and let $E\subset\tilde{X}$ be a compact
set containing the attached $\\{x\\}\times[0,1]$ for all
$(x,1)\in\partial\tilde{X}$ such that $\\{x\\}\times[0,1]$ intersects $C$.
Since $E$ is compact, there are finitely many nbhds in $\mathcal{W}$ and also
in $\mathcal{U}$ which cover $E$. Let us restrict ourselves to these finitely
many nbhds. Let $\varepsilon>0$ be such that $\varepsilon<\varepsilon_{x}$ for
all these finitely many points $(x,1)\in\partial\tilde{X}$. Let $U$ be the
union of the chosen finitely many nbhds in $\mathcal{U}$ and let $\delta>0$ be
such that for a metric on $\tilde{X}$ the $\delta$-nbhd of

$\bigcup_{(x,1)\in\partial\tilde{X}\cap E}\\{x\\}\times[0,1]$

is inside $U$. Then define a homeomorphism $h\colon\tilde{X}\to\tilde{X}$
which maps

$\bigcup_{(x,1)\in\partial\tilde{X}\cap E}\\{x\\}\times[0,1]$

into

$\bigcup_{(x,1)\in\partial\tilde{X}\cap E}\\{x\\}\times[1,1-\varepsilon)$

by mapping each arc $\\{x\\}\times[0,1]$, where $(x,1)\in\partial\tilde{X}$,
into itself. We suppose that the support of $h$ is inside the $\delta/2$-nbhd
of $\bigcup_{(x,1)\in\partial\tilde{X}\cap E}\\{x\\}\times[0,1]$. This $h$
satisfies (3) of Theorem 4.4 so $\pi$ is a near-homeomorphism which yields the
claimed homeomorphism $\beta$. ∎

##  5\. Shrinkable decompositions

The following notions are often used to describe types of decompositions which
turn out to be shrinkable.

######  Definition 5.1.

Let $\mathcal{D}$ be a usc decomposition of $\mathbb{R}^{n}$.

  * •

$\mathcal{D}$ is _cell-like_ if every decomposition element is cell-like,

  * •

$\mathcal{D}$ is _cellular_ if every decomposition element is cellular,

  * •

the decomposition elements are _flat arcs_ if for every $D\in\mathcal{D}$
there is a homeomorphism $h\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that
$h(D)$ is a straight line segment,

  * •

$\mathcal{D}$ is _starlike_ if every decomposition element $D$ is a starlike
set, that is, $D$ is a union of compact straight line segments with a common
endpoint $x_{0}\in\mathbb{R}^{n}$,

  * •

$\mathcal{D}$ is _starlike-equivalent_ if for every $D\in\mathcal{D}$ there is
a homeomorphism $h\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that $h(D)$ is
starlike,

  * •

$\mathcal{D}$ is _thin_ if for every $D\in\mathcal{D}$ and every nbhd $U$ of
$D$ there is an $n$-dimensional ball $B\subset\mathbb{R}^{n}$ such that
$D\subset B\subset U$ and $\partial B$ is disjoint from the non-degenerate
elements of $\mathcal{D}$,

  * •

$\mathcal{D}$ is _locally shrinkable_ if for each $D\in\mathcal{D}$ we have
that for every nbhd $U$ of $D$ and open cover $\mathcal{V}$ of
$\mathbb{R}^{n}$ there is a homeomorphism
$h\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ with support $U$ such that
$h(D)\subset V$ for some $V\in\mathcal{V}$,

  * •

$\mathcal{D}$ _inessentially spans_ the disjoint closed subsets
$A,B\subset\mathbb{R}^{n}$ if for every $\mathcal{D}$-saturated open cover
$\mathcal{U}$ of $\mathbb{R}^{n}$ there is a homeomorphism
$h\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ which is $\mathcal{U}$-close to the
identity and no element of $\mathcal{D}$ meets both of $h(A)$ and $h(B)$,

  * •

the decomposition element $D$ has _embedding dimension_ $k$ if for every
$(n-k-1)$-dimensional smooth submanifold $M$ of $\mathbb{R}^{n}$ and open
cover $\mathcal{V}$ of $\mathbb{R}^{n}$ there is a homeomorphism
$h\colon\mathbb{R}^{n}\to\mathbb{R}^{n}$ which is $\mathcal{V}$-close to the
identity, $h(M)\cap D=\emptyset$ and this is not true for $(n-k)$-dimensional
submanifolds.

Most of these notions have the corresponding verisons in arbitrary manifolds
or spaces. A condition that is obviously satisfied by at least $5$-dimensional
Euclidean spaces is the following.

######  Definition 5.2 (Disjoint disks property).

The metric space $X$ has the _disjoint disks property_ if for arbitrary maps
$f_{1}$ and $f_{2}$ from $D^{2}$ to $X$ and for every $\varepsilon>0$ there
are approximating maps $g_{i}$ from $D^{2}$ to $X$ $\varepsilon$-close to
$f_{i}$, $i=1,2$, such that $g_{1}(D^{2})$ and $g_{2}(D^{2})$ are disjoint.

The next theorem [Ed78] is one of the fundamental results of decomposition
theory, we omit its proof here.

######  Theorem 5.3.

Let $X$ be an at least $5$-dimensional manifold and let $\mathcal{D}$ be a
cell-like decomposition of $X$. Then $\mathcal{D}$ is shrinkable if and only
if $X_{\mathcal{D}}$ is finite dimensional and has the disjoint disks
property.

Recall that a separable metric space is finite dimensional if every point has
arbitrarily small nbhds having one less dimensional frontiers and dimension
$-1$ is by definition the dimension of the empty set. For example, a manifold
is finite dimensional.

In the following statement we enumerate several conditions which imply that a
(usc) decomposition is shrinkable.

######  Theorem 5.4.

The following decompositions are strongly shrinkable:

  1. (1)

cell-like usc decompositions of a $2$-dimensional manifold,

  2. (2)

countable usc decompositions of $\mathbb{R}^{n}$ if the decomposition elements
are flat arcs,

  3. (3)

countable and starlike usc decompositions of $\mathbb{R}^{n}$,

  4. (4)

countable and starlike-equivalent usc decompositions of $\mathbb{R}^{3}$,

  5. (5)

null and starlike-equivalent usc decompositions of $\mathbb{R}^{n}$,

  6. (6)

thin usc decompositions of $3$-manifolds,

  7. (7)

countable and thin usc decompositions of $n$-dimensional manifolds,

  8. (8)

countable and locally shrinkable usc decompositions of a complete metric space
if $\cup\mathcal{H}_{\mathcal{D}}$ is $G_{\delta}$,

  9. (9)

monotone usc decompositions of $n$-dimensional manifolds if $\mathcal{D}$
inessentially spans every pair of disjoint, bicollared $(n-1)$-dimensional
spheres,

  10. (10)

null and cell-like decompositions of smooth $n$-dimensional manifolds if the
embedding dimension of every $D\in\mathcal{D}$ is $\leq n-3$,

Before proving Theorem 5.4 let us make some observations and preparations. At
first, note that there are usc decompositions of $\mathbb{R}^{3}$ into
straight line segments which are not shrinkable: in the proof of Proposition
3.1 for any given compact metric space $Y$ we constructed a decomposition of
$\mathbb{R}^{3}$ into straight line segments and singletons such that $Y$ is a
subspace of the decomposition space. Since $\mathbb{R}^{3}$ is a complete
metric space and the decomposition is usc it is also shrinkable if and only if
$\pi$ is approximable by homeomorphisms. This means that if $Y$ cannot be
embedded into $\mathbb{R}^{3}$, then the decomposition space cannot be
homeomorphic to $\mathbb{R}^{3}$ and then this decomposition is not
shrinkable.

If the decomposition is countable, then we can shrink successively the
decomposition elements if there is a guaranty of not expanding an already
shrunken element while shrinking another one. The next proposition is a
technical tool for this process.

######  Proposition 5.5.

Let $\mathcal{D}$ be a countable usc decomposition of a locally compact metric
space $X$. Suppose for every $D\in\mathcal{D}$, for every $\varepsilon>0$ and
for every homeomorphism $f\colon X\to X$ there exists a homeomorphism $h\colon
X\to X$ such that

  1. (1)

outside of the $\varepsilon$-nbhd of $D$ the homeomorphism $h$ is the same as
$f$,

  2. (2)

$\mathrm{diam}\thinspace h(D)<\varepsilon$ and

  3. (3)

for every $D^{\prime}\in\mathcal{D}$ we have $\mathrm{diam}\thinspace
h(D^{\prime})<\varepsilon+\mathrm{diam}\thinspace f(D^{\prime})$.

Then $\mathcal{D}$ is strongly shrinkable.

###### Sketch of the proof.

Let $\varepsilon>0$ and let $\mathcal{U}$ be a $\mathcal{D}$-saturated open
cover of $X$. We enumerate the non-degenerate elements of $\mathcal{D}$ which
have diameter at least $\varepsilon/2$ as $D_{1},D_{2},\ldots$. We can find
$\mathcal{D}$-saturated open sets $U_{1},U_{2},\ldots$ such that for all $n$
we have $D_{n}\subset U_{n}$ and all sets $U_{n}$ are pairwise disjoint or
coincide. These $U_{n}$ are subsets of sets in $\mathcal{U}$ and they will
ensure $\mathcal{U}$-closeness. We produce a sequence
${\mathrm{id}}=h_{0},h_{1},\ldots$ of self-homeomorphisms of $X$ and a
sequence $C_{1},C_{2},\ldots$ of $\mathcal{D}$-saturated closed nbhds of
$D_{1},D_{2},\ldots$, respectively, such that a couple of conditions are
satisfied for every $n\geq 1$:

  1. (a)

$h_{n}|_{X-U_{n}}=h_{n-1}|_{X-U_{n}}$,

  2. (b)

${\mathrm{diam}}\thinspace h_{n}(D_{n})<\varepsilon$,

  3. (c)

for every $D\in\mathcal{D}$ we have ${\mathrm{diam}}\thinspace
h_{n}(D)<(1-\frac{1}{2^{n}})\frac{\varepsilon}{2}+{\mathrm{diam}}\thinspace
D$,

  4. (d)

$h_{n+1}|_{C_{1}\cup\cdots\cup C_{n}}=h_{n}|_{C_{1}\cup\cdots\cup C_{n}}$,

  5. (e)

if some $D\in\mathcal{D}$ is in $C_{n}$, then ${\mathrm{diam}}\thinspace
h_{n}(D)<\varepsilon$ and

  6. (f)

$h_{n}=h_{n-1}$ if ${\mathrm{diam}}\thinspace h_{n-1}(D_{n})<\varepsilon$.

The sets $C_{n}$ serve as protective buffers in which no further motion will
occur. For $n=1$ by the conditions (1), (2) and (3) in the statement of
Proposition 5.5 with the choice $f={\mathrm{id}}$ we can find a homeomorphism
$h_{1}\colon X\to X$ satisfying (a), (b) and (c) and also an appropriate
$C_{1}$ such that (d) and (e) are satisfied as well. If $h_{k}$ and $C_{k}$
are defined already for $1\leq k\leq n$, then we find $h_{n+1}$ and $C_{n+1}$
as follows. If ${\mathrm{diam}}\thinspace h_{n}(D_{n+1})<\varepsilon$, then
let $h_{n+1}=h_{n}$. If the diameter of $h_{n}(D_{n+1})$ is at least
$\varepsilon$, then by the conditions (1), (2) and (3) with the choice
$f=h_{n}$ we can find a homeomorphism $h_{n+1}\colon X\to X$ satisfying

  1. (i)

$h_{n+1}|_{X-U_{n+1}}=h_{n}|_{X-U_{n+1}}$

  2. (ii)

${\mathrm{diam}}\thinspace h_{n+1}(D_{n+1})<\varepsilon/2^{n+2}$,

  3. (iii)

for every $D\in\mathcal{D}$ we have ${\mathrm{diam}}\thinspace
h_{n+1}(D)<\varepsilon/{2^{n+2}}+{\mathrm{diam}}\thinspace h_{n}(D)$

furthermore (iii) and (c) imply that for every $D\in\mathcal{D}$ we have

${\mathrm{diam}}\thinspace
h_{n+1}(D)<\varepsilon/{2^{n+2}}+\left(1-\frac{1}{2^{n}}\right)\frac{\varepsilon}{2}+{\mathrm{diam}}\thinspace
D=\left(1-\frac{1}{2^{n+1}}\right)\frac{\varepsilon}{2}+{\mathrm{diam}}\thinspace
D$

so (a), (b) and (c) are satisfied. It is not too difficult to get (d) and (e)
with some $C_{n+1}$ as well. After having all $h_{1},h_{2},\ldots$ and
$C_{1},C_{2},\ldots$ with properties (a)-(f) it is easy to see by (d), (e) and
(f) that every $D\in\mathcal{D}$ which is in $C_{1}\cup\cdots\cup C_{n}$ is
shrunk by $h_{n}$ to size smaller than $\varepsilon$ and other $h_{n+i}$ does
not modify this. If some $D\in\mathcal{D}$ had diameter smaller than
$\varepsilon/2$ originally, then (c) implies that its diameter is smaller than
$\varepsilon$ during all the process. These imply that the sequence
$h_{1},h_{2},\ldots$ is locally stationary and it converges to a shrinking
homeomorphism $h$. ∎

We are going to give a sketch of the proof of Theorem 5.4. For the detailed
proof of (1) see [Mo25], for the proofs of (2) and (3) see [Bi57], for the
proof of (4) see [DS83], for (5) see [Be67], for (6) see [Wo77] and for (7),
(8), (9) and (10) see [Pr66], [Bi57], [Ca78] and [Ca79, Ed16], respectively.

###### Sketch of the proof of Theorem 5.4.

(1) follows from the fact that in a $2$-dimensional manifold $X$ a cell-like
decomposition is thin. The reason of this is that an arbitrarily small
$2$-dimensional disk nbhd $B$ with the property $\partial
B\cap(\cup\mathcal{H}_{\mathcal{D}})=\emptyset$ can be obtained by finding the
circle $\partial B$ in $X$ as a limit of a sequence of maps $f_{n}\colon
S^{1}\to X$ avoiding smaller and smaller decomposition elements. A thin usc
decomposition of a $2$-dimensional manifold is shrinkable if the points of
$\pi(\bigcup\mathcal{H}_{\mathcal{D}})$ do not converge to each other in a too
complicated way. Since the quotient space $X_{\mathcal{D}}$ can be filtered in
a way which implies this, the decomposition map $\pi$ can be successively
approximated by maps which are homeomorphisms on the induced filtration in
$X$.

(2)-(5) follows from Proposition 5.5: the flat arcs, starlike sets and
starlike-equivalent sets can be shrunk successively because of geometric
reasons.

To prove (6) and (7) we also use Proposition 5.5. Let $D\in\mathcal{D}$ be a
non-degenerate decomposition element, $U$ a nbhd of $D$ and let $B$ be a ball
such that $D\subset B\subset U$ and $\partial B$ is disjoint from the non-
degenerate elements of $\mathcal{D}$. After applying a self-homeomorphism of
$X$, we can suppose that $B$ is the unit ball. Let $k$ be some large enough
integer and let $1>\delta_{0}>\delta_{1}>\cdots>\delta_{k-1}>0$ be such that
if $D^{\prime}\in\mathcal{D}$ intersects the $\delta_{n+1}$-nbhd of $\partial
B$, then $D^{\prime}$ is inside the $\delta_{n}$-nbhd of $\partial B$. Define
a homeomorphism $f\colon B\to B$ which is the identity on $\partial B$, keeps
the center of $B$ fixed and on each radius $R$ the point at distance
$\delta_{n}$ from $\partial B$, where $1\leq n\leq k-1$, is mapped to the
point at distance $n/k$ from the center. We require that the homeomorphism $f$
is linear between these points. After applying this homeomorphism, every
$D^{\prime}\in\mathcal{D}$ in $B$ is shrunk to size small enough.

In the proof of (8) we enumerate the non-degenerate decomposition elements and
we construct a sequence of homeomorphisms of the ambient space which shrink
the decomposition elements successively using the locally shrinkable property.

To prove (9) for a given $\varepsilon>0$ we cover the manifold by two
collections $\\{B_{\alpha}\\}_{\alpha\in A}$ and
$\\{B_{\alpha}^{\prime}\\}_{\alpha\in A}$ of $n$-dimensional balls such that
$B_{\alpha}\subset\mathrm{int}\thinspace B_{\alpha}^{\prime}$ and
$\mathrm{diam}\thinspace B_{\alpha}^{\prime}<\varepsilon$. Then the closed
sets $\pi(\partial B_{\alpha})$ and $\pi(\partial B_{\alpha}^{\prime})$ are
made disjoint by applying homeomorphisms $h_{\alpha}$ successively. This
implies that the homeomorphism $h$ obtained by composing all the
homeomorphisms $h_{\alpha}$ is such that for every $D\in\mathcal{D}$ the set
$h(D)$ is fully contained in some ball $B_{\alpha}^{\prime}$ so its diameter
is smaller than $\varepsilon$.

In the proof of (10) at first we obtain that every decomposition element $D$
is cellular because of the following. By assumption $D$ is cell-like and it
behaves like an at most $(n-3)$-dimensional submanifold so the $2$-skeleton of
the ambient manifold is disjoint from $D$. This means that $D$ satisfies the
cellularity criterion since the $2$-skeleton carries the fundamental group.
Hence $D$ is cellular, which implies that it is contained in an $n$-dimesional
ball and also in a starlike-equivalent set $C$ of embedding dimension $\leq
n-2$. Now it is possible to use an argument similar to the proof of
Proposition 5.5: we can shrink $C$ to become smaller than an $\varepsilon>0$
by successively compressing $C$ and in each iteration carefully controlling
and avoiding other decomposition elements close to $C$ which would become too
large during the compression procedure. ∎

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