# Non-metrizable manifolds and contractibility

Mathieu Baillif

###### Abstract

We investigate whether non-metrizable manifolds in various classes can be
homotopy equivalent to a CW-complex (in short: heCWc), and in particular
contractible. We show that a non-metrizable manifold cannot be heCWc if it has
one of the following properties: it contains a countably compact non-compact
subspace; it contains a copy of an $\omega_{1}$-compact subset of an
$\omega_{1}$-tree; it contains a non-Lindelöf closed subspace functionally
narrow in it. (These results hold for more general spaces than just
manifolds.) We also show that the positive part of the tangent bundle of
$\mathbb{L}_{+}$ is not heCWc (for any smoothing). These theorems follow from
stabilization properties of real valued maps. On a more geometric side, we
also show that the Prüfer surface, which has been shown to be contractible
long ago, has an open submanifold which is not heCWc. On the other end of the
spectrum, we show that there is a non-metrizable contractible Type I surface.

##  1 Introduction

By space we mean ‘topological Hausdorff space’, in particular ‘regular’ and
’normal’ imply Hausdorff. By manifold, or $n$-manifold if we want to empasize
the dimension, is meant a connected space each of whose points has a
neighborhood homeomorphic to the Euclidean space $\mathbb{R}^{n}$ for some
finite $n\geq 1$. A surface is a $2$-manifold. Every function is assumed
continuous otherwise stated. We use brackets $\langle\,,\,\rangle$ for ordered
pairs reserving parenthesis $(\,,\,)$ for open intervals. The letters
$\alpha,\beta,\gamma$ are used exclusively for ordinals, and an ordinal is
assumed to be the set of its predecessors. Undefined terminology used in this
introduction is explained either at its end or in later sections.

This article is concerned with homotopy questions for non-metrizable manifolds
and similar spaces. This is a somewhat exotic subject for which there exists a
small but growing literature; let us cite the book by D. Gauld [18, Chapter
5.2] and the articles by D.Gauld, S. Deo, A. Gabard and the author [17, 2, 7,
11], for instance. It is worth noting that the methods and results in these
papers (and the present one) belong mainly to general topology (with a small
set theoretic flavor) and have nothing in common with homotopy theory per se.

Acknowledging its relative unpopularity, let us explain why we think homotopy
in non-metrizable manifolds is an interesting subject.111The true and honest
reason is that we happen to like to think about these problems. It might
unfortunately be seen as slightly insuficient. Milnor showed in [26] that a
metrizable manifold is homotopy equivalent to a CW-complex (in short: heCWc),
hence metrizable manifolds that are contractible coincide with those having
vanishing homotopy groups by Whitehead Theorem (see, e.g., [22, p. 346]). If
one releases the metric assumption, this is not true anymore as there are
simple non-metrizable non-contractible manifolds with vanishing homotopy
groups (e.g. the longray described below). Of course, any space with a non-
trivial homotopy group is non-contractible. It was at first not clear to us
whether contractible non-metrizable manifolds do exist at all. We then noticed
that examples were found a long time ago by Calabi and Rosenlicht: the Prüfer
surface and some of its variants are contractible. (The original source is
[10], see also [17].)

While manifolds can be widly different one from another, they share the local
properties of the Euclidean space. In particular, any manifold is locally
compact and thus Tychonoff, locally (path) connected and locally second
countable. In addition, metric manifolds are normal (even perfectly normal and
monotonically normal, see e.g. [18, p. 23] for definitions), second countable
and thus Lindelöf and paracompact. Lindelöfness and paracompactness are
actually two of the many properties equivalent to metrizability for manifolds,
see [18, Chapter 2]. In metric spaces, Lindelöfness is equivalent to being
$\omega_{1}$-compact, and compactness to countable compactness. (Recall that a
space is $\omega_{1}$-compact or has countable extent iff its closed discrete
subsets are at most countable. A countably compact space is
$\omega_{1}$-compact.) Non-metrizable manifolds, on the contrary, can exhibit
some but not all of these features, and the mere existence of non-metrizable
manifolds with such-and-such properties sometimes depends on the axioms of set
theory. For example, perfectly normal manifolds do exist under CH and do not
under MA + $\neg$CH (due to M.E. Rudin and P. Zenor, see e.g. [18, Chapter
6]), and monotonically normal non-metrizable manifolds of dimension $>1$ do
not exist (Z. Balogh and M.E. Rudin, see [9, Cor. 2.3 (e)]).

Given this variety, we thought it might be interesting to see whether
contractible non-metrizable manifolds exist in such-and-such class, or on the
other hand if such-and-such topological property prevents contractibility
regardless of the homotopy groups. Some results are already available, for
instance the following theorem was proved by S. Deo and D. Gauld (and by the
author for manifolds):

######  Theorem 1.1 ([2, Prop. 7.1], [11, Thm 3.4]).

A locally metrizable space (in particular, a manifold) containing a copy of
$\omega_{1}$ is non-contractible.

Notice that the proper forcing axiom PFA implies that a countably compact non-
compact subset of a manifold contains a copy of $\omega_{1}$ (due to Balogh,
see [8]), an immediate corollary is that under PFA manifolds with such subsets
are non-contractible. We show below that PFA is actually not needed.

On the other side of the spectrum, the contractible Prüfer surface is neither
normal nor $\omega_{1}$-compact. (A version of it is realcompact, though, as
shown by A. Mardani [25].) It contains an open (Lindelöf) Euclidean set with a
non-metrizable (equivalently, non-Lindelöf) closure. In a terminology dating
to P. Nyikos, it is in the broad class of Type II manifolds (see just below
for a precise definition). In a loose sense, what makes it non-metrizable lies
just at the boundary of some perfectly nice Euclidean open contractible set.
We happen to be able to ‘push’ this non-metrizable stuff inside this Euclidean
open set, yielding the contractibility.

The other broad class of non-metrizable manifolds are the Type I manifolds
[27, Def. 2.10]. Denote as usual by $\overline{A}$ the closure of $A$ (in some
topological space clear from the context). A space $X$ is of Type I if and
only if $X=\cup_{\alpha\in\omega_{1}}X_{\alpha}$, where $X_{\alpha}$ is open,
$\overline{X_{\alpha}}$ Lindelöf for each $\alpha$, and
$\overline{X_{\alpha}}\subset X_{\beta}$ whenever $\alpha<\beta<\omega_{1}$
(and of Type II otherwise).

In case of manifolds, each $X_{\alpha}$ is an open metrizable submanifold.
Loosely speaking again, these manifolds ‘grow slowly’ instead of jumping at
once into non-metrizability. In a Type I space, Lindelöf subsets have Lindelöf
closure.

Let us give a summary of our results. The first one (obtained with Peter
Nyikos) shows that Type I contractible manifolds exist.

######  Theorem 1.2.

There is a contractible non-metrizable Type I surface.

The mapping class group of such a manifold, on the other hand, can be quite
large, see Section 7.2 for definitions and details. This surface is obtained
by “fattening” an $\mathbb{R}$-special $\omega_{1}$-tree and is non-normal.
(Trees in this article are to be understood in a set theoretic sense.
Definitions of the relevant notions are recalled in Section 6.) Our other
results are on the negative side, we show that spaces in various classes are
not contractible, actually not heCWc (which we recall is short for homotopy
equivalent to a CW-complex). We first obtained a generalization of Theorem
1.1. (Theorems 1.3–1.5 below are stated for manifolds in this introduction,
but we obtained results in larger classes.)

######  Theorem 1.3.

A manifold containing a countably compact non-compact subspace is not heCWc.

Since $\omega_{1}$-compactess is a weakening of countable compactness, it
seemed natural to see whether the same result holds in this class. What we
obtained is something quite weaker and involves subsets of trees. The order on
the tree is essential to our proof.

######  Theorem 1.4.

A manifold containing a copy of an $\omega_{1}$-compact subset of an
$\omega_{1}$-tree (in particular, a Suslin tree) is not heCWc.

Recall that the closed longray $\mathbb{L}_{\geq 0}$ is the product
$\omega_{1}\times[0,1)$ with lexicographic order topology (not the product
topology !). When dealing with it, we often see $\omega_{1}$ as a subset of
$\mathbb{L}_{\geq 0}$ by identifying $\alpha$ with $\langle\alpha,0\rangle$.
$\mathbb{L}_{+}$ is $\mathbb{L}_{\geq 0}$ with the $0$ point removed.

Our next result involves the concept of functionally narrow subspaces in Type
I spaces, which whose definition is recalled in Section 3. Loosely speaking, a
closed subspace in a Type I space $X$ is functionally narrow in $X$ if it
cannot be separated by a function $X\to\mathbb{L}_{\geq 0}$ into two non-
Lindelöf pieces, one with a bounded image and the other with an unbounded
image. Functional narrowness is relative to the ambient space, however a
closed copy of $\omega_{1}$ in any Type I space is functionally narrow in it.
There are spaces with discrete subspaces functionally narrow in them as well,
see [6, Example 5.12] (and the rest of this paper for more on the subject).

######  Theorem 1.5.

A Type I (non-metrizable) manifold containing a closed non-Lindelöf subspace
functionally narrow in it is not heCWc.

We also show that surfaces in a very interesting class studied extensively by
P. Nyikos in [28] are non-contractible: the (positive part of the) tangent
bundle of the longray $\mathbb{L}_{+}$. (We write the tangent bundle, but
actually there are a lot of non-homeomorphic ones, with distinct topological
properties.)

######  Theorem 1.6.

The positive part of the tangent bundle of $\mathbb{L}_{+}$ for a given
smoothing is not contractible (and thus not heCWc).

(See the preprint [3, Section 6] for related results.) The reason for the non-
homotopy equivalence in Theorems 1.3–1.6 relies on stabilization properties of
real valued maps on relevant subspaces. This is basically the same for Theorem
1.1, whose proof is based on the fact that any function
$\omega_{1}\to\mathbb{R}$ is eventually constant. However, a non-metrizable
manifold can be non-contractible for more ‘geometrical’ reasons: there is a
submanifold (with trivial homotopy groups) of the contractible Prüfer surface
which is non-contractible. To understand better the statement, let us briefly
recall that (the separable version of) Prüfer surface is obtained by adjoining
to the open Euclidean plane uncountably many boundary components
$\mathbb{R}_{c}$, each homeomorphic to $\mathbb{R}$. (This version of Prüfer
surface is actually a surface with boundary, see the end of this
introduction.) A precise definition is recalled in Section 5.

######  Theorem 1.7.

The Prüfer surface minus the $0$-points in each $\mathbb{R}_{c}$ is a non-
contractible (and hence not heCWc) non-metrizable submanifold of a
contractible manifold.

This last example gives us the impression that being heCWc is a somewhat rare
occurence in the world of non-metrizable manifolds: some kind of perfect
alignment of uncountably many pieces seems to be necessary. We are thus
inclined to believe that the problems below have negative answers, but were
not able to find any evidence corroborating our feeling.

######  Problem 1.8.

Is there a (consistent) example of a contractible $\omega_{1}$-compact
manifold ? More generally, is there a contractible locally metrizable space
containing a non-metrizable $\omega_{1}$-compact subspace ?

Theorem 1.4 is far from providing an answer to this problem. For instance,
with the help of $\diamondsuit$ one can build an $\omega_{1}$-compact
$2$-manifold which mimicks any given $\omega_{1}$-tree, in particular an
$\mathbb{R}$-special one, see [19, 20]. (Recall that the surface in Theorem
1.2 is build with an $\mathbb{R}$-special tree.) It is not clear to us whether
these manifolds are contractible. Under PFA, a contractible
$\omega_{1}$-compact manifold cannot be of Type I (see Theorem 2.3 below), but
this does not settle the general case. Another problem, for which we have no
information so far:

######  Problem 1.9.

Is there a normal contractible non-metrizable manifold ? A perfectly normal
one, consistently ?

Notice that by Jones Lemma (see e.g. [14, 1.7.12 (c) p.60]), a separable
normal manifold is $\omega_{1}$-compact if $2^{\omega}<2^{\omega_{1}}$, hence
Problems 1.8–1.9 might be related.

As said above, Theorems 1.1 and 1.3–1.5 hold for more general spaces than just
manifolds. It might be instructing to note that some amount of non-
compactness, first countability and local Lindelöfness has to be involved to
ensure non-contractibility, as the next simple examples show.

######  Example 1.10.

There are contractible non-metrizable spaces in the following classes:
(i) compact and first countable,
(ii) compact and containing a (non-closed) copy of $\omega_{1}$,
(iii) first countable, countably compact and containing a closed copy of
$\omega_{1}$.

###### Details.

For (i), Helly’s space [32, Ex. 107] is compact, first countable and
contractible. For (ii)–(iii), given a space $X$ we let
$CX=X\times[0,1]/\langle x,1\rangle\sim\langle y,1\rangle$ be the cone over
$X$. $CX$ is contractible for any space $X$. For (iii) take $C\omega_{1}$,
which is first countable (a neighborhood basis of the vertex of the cone is
given by the vertex union $\omega_{1}\times(1-1/n,1)$ for $n\in\omega$, as
well known). Notice that $C\omega_{1}$ is not locally Lindelöf. For (ii) take
$C(\omega_{1}+1)$. It is not first countable (or even countably tight, for
that matter) due to extra point in $\omega_{1}+1$. ∎

This paper is organized in somewhat independant sections. In Sections 2 & 3,
we prove Theorems 1.3 and 1.5 involving countable compactness and functionally
narrow subspaces; Section 4 deals with Theorem 1.6 (about tangent bundles of
$\mathbb{L}_{+}$) and Section 5 with Theorem 1.7 (about subspaces of Prüfer
surface). Then, Section 6 is a bit different as it studies $\omega_{1}$-trees
and the homotopic properties of their road spaces, generalizing Theorem 1.1 in
various ways. This enables us in particular to prove Theorem 1.4. We also
prove that the road space of an $\mathbb{R}$-special tree is contractible,
which gives half the proof of Theorem 1.2. The other half was given to us by
Peter Nyikos who found a way to “fatten” this road space to obtain a surface
homotopy equivalent to it. This part is given in Section 7. This Section
contains further results on the mapping class group of this contractible
surface and the Prüfer surface. (We note in passing that the results of
Section 6 appeared in the preprint [5].)

Let us finish this introduction with some definitions for further reference.
By cover of a space, we always mean a cover by open sets. A manifold with
boundary $M$ is a connected space each of whose points has a neighborhood
homeomorphic to $\mathbb{R}^{n-1}\times\mathbb{R}_{\geq 0}$ for some finite
$n>0$.222Whether a manifold possesses a boundary or not is mostly irrelevant
for the content of this paper, but we sometimes allow boundaries to simplify
some definitions. Its boundary $\partial M$ contains the points which do not
have a neighborhood homeomorphic to $\mathbb{R}^{n}$, it is an $n-1$-manifold
(without boundary) and a closed subset of $M$. The term club is a shorthand
for closed and unbouded. We use very often the notation $h_{t}(x)$ for
$h(x,t)$ and $h_{t}$ for $h(\cdot,t)$ when $h$ is a function with domain
$X\times[0,1]$. An homotopy between two maps $f,g:X\to Y$ is a function
$h:X\times[0,1]\to Y$ such that $h(\cdot,0)=f$ and $h(\cdot,1)=g$ (in short:
$h_{0}=f$, $h_{1}=g$). Two spaces $X,Y$ are homotopy equivalent if there are
$f:X\to Y$ and $g:Y\to X$ such that $f\circ g$ and $g\circ f$ are homotopic to
$id_{Y}$ and $id_{X}$, respectively. A contraction is an homotopy between the
identity map and a constant map, and a space is contractible if it admits a
contraction (or equivalently if it is homotopy equivalent to a one point
space).

We recall briefly the definition of a CW-complex. (Caution: to avoid confusion
with the notation $X^{n}$ usually reserved for the $n$-th power of the space
$X$ in point set topology, we will denote the $n$-skeleton by $X^{(n)}$.) A
CW-complex is a space $X$ constructed by starting with a discrete space
$X^{(0)}$ of $0$-cells (which are points). We then inductively form the
$n$-skeleton $X^{(n)}$ by attaching $n$-cells (which are copies of the open
$n$-disk) $e^{n}_{i}$ ($i$ in some index set) via continuous maps
$\varphi_{i}:S^{n-1}\to X^{(n-1)}$; that is, we identify $x\sim\varphi_{i}(x)$
for $x\in\partial D^{n}_{i}$, where $D^{n}_{i}$ is a copy of the closed
$n$-disk and $\partial D^{n}_{i}\simeq S^{n-1}$ its boundary. We endow
$X^{(n)}$ with the quotient topology. The $n$-cell $e^{n}_{i}$ is then the
homeomorphic image of $D^{n}_{i}-\partial D^{n}_{i}$.
$X=\cup_{n\in\omega}X^{(n)}$ is given the weak topology: $A\subset X$ is open
(or closed) if and only if $A\cap X^{(n)}$ is open (or closed) for all $n$. In
particular, $X^{(n)}$ is closed in $X$. Then, $e^{n}_{i}$ is the image under a
characteristic map $\Phi_{i}:D^{n}_{i}\to X$ (the composition of $\varphi_{i}$
and the inclusion) of $D^{n}_{i}-\partial D^{n}_{i}$. For more details, see
for instance Appendix A in [22], from which we borrowed the terminology. Any
CW-complex is hereditarily paracompact (see for instance [16, Thm 1.3.5 & Ex.
1 p. 33]) and hence hereditarily collectionwise normal (see [14, Theorem
5.1.18 p. 305] for the definition and proof).

##  2 Countably compact non-metrizable manifolds are non-contractible

A space is C-closed if any countably compact subspace is closed. The class of
C-closed spaces contains in particular the hereditarily paracompact spaces,
the sequential spaces and the regular spaces with $G_{\delta}$ points, see
e.g. [23]. In particular, manifolds and CW-complexes are C-closed. A cover of
a space is a chain cover if it is linearly ordered by the inclusion relation.
We may (and do) assume that chain covers are indexed by regular cardinals and
that a member with a strictly smaller index than another member is strictly
contained in it. We will prove the following.

######  Theorem 2.1.

Let $X$ be countably compact and $Y$ be C-closed such that each point in $Y$
has a Lindelöf (non-necessarily open) neighborhood. Let $h:X\times[0,1]\to Y$
be such that $h_{t_{0}}(X)$ is compact for some $t_{0}\in[0,1]$. Then
$h_{t}(X)$ is compact for each $t\in[0,1]$.

Recall that compactness is equivalent to linear compactness, that is, a space
$S$ is compact if and only if any chain cover of it contains $S$ as a member.
(To our knowledge, this dates back to Alexandroff and Urysohn [1].) Hence, if
$S$ is a non-compact space, there is a chain cover
$\mathcal{U}=\\{U_{\alpha}\,:\,\alpha\in\kappa\\}$ of $S$, with $\kappa$ a
regular cardinal, such that $S\not\subset U_{\alpha}$ for each
$\alpha\in\kappa$. Given such a chain cover $\mathcal{U}$, we say that
$E\subset S$ is $\mathcal{U}$-unbounded iff $E\not\subset U_{\alpha}$ for each
$\alpha\in\kappa$.

######  Lemma 2.2.

Let $X$ be countably compact and non-compact, and let
$\mathcal{U}=\\{U_{\alpha}\,:\,\alpha\in\kappa\\}$ be a chain cover such that
$X\not\subset U_{\alpha}$ for each $\alpha$. Then the following hold.
(a) Given $f:X\to Y$ such that $f(X)$ is compact and $Y$ is C-closed, there is
$c\in Y$ such that $f^{-1}(\\{c\\})$ is $\mathcal{U}$-unbounded.
(b) Let $E_{n}\subset X$ ($n\in\omega$) be closed and $\mathcal{U}$-unbounded
such that $E_{n+1}\subset E_{n}$. Then $E=\cap_{n\in\omega}E_{n}$ is closed
and $\mathcal{U}$-unbounded.

###### Proof.

(a) For each $\alpha\in\kappa$ choose $x_{\alpha}\in X-U_{\alpha}$ and set
$E_{\alpha}=\\{x_{\beta}\,:\,\beta\geq\alpha\\}$. Then
$\overline{E_{\alpha}}\cap U_{\alpha}=\varnothing$ and each $E_{\alpha}$ is
$\mathcal{U}$-unbounded. Since $\overline{E_{\alpha}}$ is countably compact,
$f(X)$ compact and $Y$ C-closed, then $f(\overline{E_{\alpha}})\subset f(X)$
is compact. Any finite intersection of $E_{\alpha}$ is nonempty, hence by
compactness $\cap_{\alpha\in\kappa}f(\overline{E_{\alpha}})\not=\varnothing$.
Take $c$ is this intersection, then
$f^{-1}(\\{c\\})\cap\overline{E_{\alpha}}\not=\varnothing$ for each $\alpha$,
which shows that $f^{-1}(\\{c\\})$ is $\mathcal{U}$-unbounded.
(b) For each $\alpha$, $\cap_{n\in\omega}(E_{n}-U_{\alpha})\not=\varnothing$
by countable compactness. ∎

###### Proof of Theorem 2.1.

If $X$ is compact, there is nothing to do. We assume that $X$ is non-compact.
Set $A=\\{t\in[0,1]\,:\,h_{t}(X)\text{ is compact}\\}$. We will show that $A$
is open and closed in $[0,1]$. Since $t_{0}\in A$, it is non-empty and thus
equal to $[0,1]$.
$A$ is open. Let $t\in A$. Assume that there is no open interval around $t$ in
$A$, hence there is a sequence $t_{n}$ converging to $t$ such that
$h_{t_{n}}(X)$ is non-compact for each $n\in\omega$. By local Lindelöfness of
$Y$ and compactness of $h_{t}(X)$, there is a Lindelöf $V\subset Y$ whose
interior contains $h_{t}(X)$. Since $Y$ is C-closed and $h_{t_{n}}(X)$ is
countably compact, the latter is closed for each $n$. Hence
$h_{t_{n}}(X)\not\subset V$ (otherwise it would be compact), choose $x_{n}\in
X$ such that $h_{t_{n}}(x_{n})\not\in V$. Take an accumulation point $x$ of
the $x_{n}$’s, then $h_{t}(x)$ is not in the interior of $V$, which contains
$h_{t}(X)$, a contradiction.
$A$ is closed. Let $t_{n}$ be a sequence converging to $t$ and assume that
$h_{t_{n}}(X)$ is compact for each $n$. We show that $h_{t}(X)$ is compact as
well. Assume it is not the case, hence there is a chain cover
$\mathcal{V}=\\{V_{\alpha}\,:\,\alpha\in\kappa\\}$ of $h_{t}(X)$ such that
$h_{t}(X)\not\subset V_{\alpha}$ for each $\alpha$. Let $\mathcal{U}$ be the
chain cover of $X$ given by the inverse images of $V_{\alpha}$ under $h_{t}$.
Set $E_{0}=X$. We assume by induction that $E_{n}$ is closed and
$\mathcal{U}$-unbounded, in particular $E_{n}$ is countably compact non-
compact. Since $h_{t_{n}}(E_{n})$ is compact, by Lemma 2.2 (a) there is
$c_{n}\in Y$ such that $E_{n+1}=h_{t_{n}}^{-1}(\\{c_{n}\\})\cap E_{n}$ is
$\mathcal{U}$-unbounded. By Lemma 2.2 (b), $E=\cap_{n\in\omega}E_{n}$ is
$\mathcal{U}$-unbounded. But $h_{t_{n}}$ is constant on $E$ for each $n$,
hence by continuity $h_{t}$ is also constant on $E$, taking value $c\in f(X)$.
Let $\alpha$ be such that $c\in V_{\alpha}$, then $E\subset
U_{\alpha}=h_{t}^{-1}(V_{\alpha})$, a contradiction. ∎

Notice that our only use of local Lindelöfness is to ensure that given any
compact subspace of $X$, there is a (non-necessarily open) neighborhood $V$ of
it such that any countably compact subset of $V$ is compact. Notice also that
the proof goes through if one replaces $[0,1]$ by any connected sequential
space. Theorem 1.3 is an immediate consequence of Theorem 2.1.

###### Proof of Theorem 1.3.

Manifolds are C-closed and each point has a Lindelöf (open) neighborhood. Let
thus $M$ be a manifold containing a countably compact non-compact subspace
$N$. Let $W$ be a CW-complex, and $f:M\to W$, $g:W\to M$ be given. Since
$f(N)$ is countably compact and CW-complexes are paracompact and C-closed,
$f(N)$ is compact, hence so is $g\circ f(N)$. It follows that $g\circ f$
cannot be homotopic to $id_{M}$ by Theorem 2.1. ∎

We close this section with a consequence of Theorem 2.1 under the locally
compact trichotomy axiom LCT due to Eisworth and Nyikos [13]. Recall that a
space is $\omega$-bounded iff any countable set has compact closure. LCT
states that a locally compact space is either a countable union of
$\omega$-bounded spaces, or contains an uncountable closed discrete set, or
has a countable subset with non-Lindelöf closure. This axiom holds for
instance under PFA or in models of the form PFA(S)[S], see [30].

######  Theorem 2.3 (LCT).

Let $X$ be locally compact, C-closed, $\omega_{1}$-compact and non-Lindelöf.
If $X$ is heCWc, then $X$ does not contain a closed Type I subspace.

###### Proof.

Recall that we assume our Type I spaces to be non-Lindelöf. It is well known
that in a locally compact space, each point has an open Lindelöf neighborhood.
If $Y\subset X$ is closed and of Type I, the closure (in $Y$, hence in $X$) of
any Lindelöf subset of $Y$ is Lindelöf. Hence by LCT $Y$ is a countable union
of countably compact subsets $Y_{n}$, which are thus closed in $X$. Since one
of the $Y_{n}$ must be non-compact, Theorem 2.1 shows that $X$ is not heCWc. ∎

##  3 A Type I space containing a closed non-Lindelöf subspace functionally
narrow in it is non-contractible

Let $X$ be of Type I, hence $X=\cup_{\alpha\in\omega_{1}}X_{\alpha}$ with
$X_{\alpha}$ open, $\overline{X_{\alpha}}$ Lindelöf and
$\overline{X_{\alpha}}\subset X_{\beta}$ whenever $\alpha<\beta$. The cover by
the $X_{\alpha}$’s is said canonical if
$X_{\alpha}=\cup_{\beta<\alpha}X_{\beta}$ for each limit $\alpha$. Each chain
cover indexed by ordinals $<\omega_{1}$ can be made canonical by adding
members whenever needed. (Note: our terminology differs slightly from that of
Nyikos in [27] who uses the terms canonical sequence instead of canonical
cover.) Two canonical covers agree on a club set of $\alpha$’s, as easily
seen. A subset of $X$ is bounded if it is contained in some $X_{\alpha}$ and
unbounded otherwise. A map between Type I spaces is bounded if its range is
bounded and unbounded otherwise. A slicer of (the canonical cover of) $X$ is a
map $\sigma:X\to\mathbb{L}_{\geq 0}$ such that
$\sigma(\overline{X_{\alpha+1}}-X_{\alpha})\subset[\alpha,\alpha+1]\subset\mathbb{L}_{\geq
0}$. If $X$ is regular, then $X$ is Tychonoff, each $\overline{X_{\alpha}}$ is
normal and there is a slicer of $X$ (see [6, Lemmas 2.2–2.3]). The following
is proved in [6, Theorem 3.8].

######  Lemma 3.1.

Let $X=\cup_{\alpha<\omega_{1}}X_{\alpha}$ be Type I and $D\subset X$ be club.
Then the following are equivalent.
(a) For each $f:X\to\mathbb{L}_{\geq 0}$ with $f(D)$ unbounded, the set
$S\subset\omega_{1}$ of $\alpha$ such that
$f(D-X_{\alpha})\subset[\alpha,\omega_{1})$ is club,
(b) For each $f:X\to\mathbb{L}_{\geq 0}$ with $f(D)$ unbounded,
$f^{-1}(\\{0\\})\cap D$ is bounded in $X$.

A closed subspace $D$ of $X$ is functionally narrow in $X$ if it satisfies (a)
and (b) in the above lemma. Whether a Type I space $X$ contains a non-Lindelöf
subset functionally narrow in it can be a tricky question: any closed
unbounded copy of $\omega_{1}$ is functionally narrow in $X$, but there are
spaces containing discrete subspaces functionally narrow in them as well, and
an Aronszajn tree (see the definition in Section 6) does not contain any
uncountable subspace functionally narrow in it. What we prove is the
following.

######  Theorem 3.2.

A regular Type I space containing a closed non-Lindelöf subset functionally
narrow in it is not heCWc.

For the proof, we need a fact about CW-complexes.

######  Lemma 3.3.

Let $W$ be a CW-complex, and $Y=\\{y_{\alpha}\,:\,\alpha\in\omega_{1}\\}$ be
an uncountable subset of $W$ (all $y_{\alpha}$ are distinct). Then, either
there is a cell $e^{n}_{i}$ such that $\overline{e^{n}_{i}}\cap Y$ is
uncountable, or there is an uncountable $E\subset\omega_{1}$ and for each
$\alpha\in E$, an open $V_{\alpha}\ni y_{\alpha}$ such that
$\\{V_{\alpha}\,:\,\alpha\in E\\}$ is a discrete family.

Recall that a family $\mathcal{F}$ of subsets of $W$ is discrete iff any point
in $W$ has a neighborhood which intersects at most one member of
$\mathcal{F}$. This implies in particular that for any
$\mathcal{G}\subset\mathcal{F}$, $\cup\\{\overline{F}\,:\,F\in\mathcal{G}\\}$
is closed.

###### Proof.

We can assume that $Y\subset W^{(n)}$ for the smallest $n$ such that
$W^{(n)}\cap Y$ is uncountable. Then, $Y\cap W^{(n-1)}$ is at most countable,
so we can assume that $Y\subset W^{(n)}-W^{(n-1)}$. So each $y_{\alpha}$ lies
in an $n$-cell. If $e^{n}_{i}\cap Y$ is at most countable for all $n$-cells
$e^{n}_{i}$, take an uncountable subset $E\subset\omega_{1}$ such that each
$e^{n}_{i}$ intersects $Y_{0}=\\{y_{\alpha}\,:\,\alpha\in E\\}$ in at most one
point. It thus follows that $Y_{0}$ is closed discrete. Since any CW-complex
is collectionwise normal, it implies in particular that $Y_{0}$ can be
expanded into a discrete collection of open sets, which gives us the required
$V_{\alpha}$. ∎

###### Proof of Theorem 3.2..

Let $X=\cup_{\alpha\in\omega_{1}}X_{\alpha}$ be a Type I regular space and
$D\subset X$ be closed, non-Lindelöf and functionally narrow in $X$. Let
$\sigma:X\to\mathbb{L}_{\geq 0}$ be a slicer and $h:X\times[0,1]\to X$ be such
that $h_{1}=id$. We show that $h_{t}(D)$ must be unbounded for each $t$. Set

$\tau=\inf\\{t\,:\,h_{s}(D)\subset X\text{ is unbounded }\forall s>t\\}.$

We first show that $h_{\tau}$ must be unbounded on $D$. This is obvious if
$\tau=1$. If $\tau<1$, let $\\{t_{n}\\}_{n\in\omega}$ be a dense subset of
$(\tau,1]$. By Lemma 3.1 (a) applied to $\sigma\circ h_{t_{n}}$, there is a
club $S_{n}\subset\omega_{1}$ such that $h_{t_{n}}(D-X_{\alpha})\subset
X-X_{\alpha}$ for all $\alpha\in S_{n}$. By continuity, this holds also for
$h_{\tau}$ and $\alpha\in S=\cap_{n\in\omega}S_{n}$. Since $S$ is also club,
$h_{\tau}(D)$ is unbounded. If $\tau>0$, take a sequence $t_{n}\nearrow\tau$
with $h_{t_{n}}(D)$ bounded. By continuity, $h_{\tau}(D)$ must also be
bounded. This shows that $\tau=0$, hence $h_{t}$ is unbounded for each $t$.
Now, for each $\alpha\in\omega_{1}$ choose $x_{\alpha}\in D-X_{\alpha}$. Let
$W$ be a CW-complex, and $f:X\to W$, $g:W\to X$ be homotopy equivalences. Set
$Y=\\{y_{\alpha}=f(x_{\alpha})\,:\,\alpha\in\omega_{1}\\}$. If $Y$ is
countable or there is an uncountable $Y^{\prime}\subset Y$ contained in a
countable (hence Lindelöf) subcomplex, then an unbounded subset of $D$ is sent
by $g\circ f$ to a bounded subset of $X$. Since $D$ is functionally narrow in
$X$, $\sigma\circ g\circ f(D)$ must be bounded by Lemma 3.1 (recall that
$\sigma$ is a slicer), thus so is $g\circ f(D)$. It follows that $g\circ f$
cannot be homotopic to the identity.
Thus, we can assume that no uncountable subset of $Y$ is contained in a
countable subcomplex of $W$. Let $E\subset\omega_{1}$ and the $V_{\alpha}\ni
y_{\alpha}$ be given by Lemma 3.3, and partition $E$ into two disjoint
uncountable subsets $E_{0},E_{1}$. Letting $U_{\alpha}=f^{-1}(V_{\alpha})$, by
discreteness the $\overline{U_{\alpha}}$ are $2$-by-$2$ disjoint and
$\overline{\cup_{\alpha\in E_{0}}U_{\alpha}}=\cup_{\alpha\in
E_{0}}\overline{U_{\alpha}}$. We want to define a continuous
$\chi:X\to\mathbb{L}_{\geq 0}$ such that $\chi(x_{\alpha})=\alpha$ when
$\alpha\in E_{0}$ and with value $0$ on the complement of
$\overline{\cup_{\alpha\in E_{0}}U_{\alpha}}$. This can be done since we
assumed $X$ to be regular and hence (since of Type I), Tychonoff: set $\chi$
to be $0$ on $X-\overline{\cup_{\alpha\in E_{0}}U_{\alpha}}$, and inside
$\overline{U_{\alpha}}$ take any function into
$[0,\alpha]\subset\mathbb{L}_{\geq 0}$ that maps $x_{\alpha}$ to $\alpha$ and
$\overline{U_{\alpha}}-U_{\alpha}$ to $0$. The inverse image of an interval
$(a,b)\subset\mathbb{L}_{\geq 0}$ (resp. $[0,b)\subset\mathbb{L}_{\geq 0}$) is
then an union of open subsets of the $U_{\alpha}$ (resp. together with
$X-\cup_{\alpha\in E_{0}}\overline{U_{\alpha}}$), and is then open. By
definition, $\chi(D)$ is unbounded (because of $E_{0}$), and
$\chi^{-1}(\\{0\\})\cap D$ is unbounded as well (because of $E_{1}$). By Lemma
3.1, $D$ is not functionally narrow in $X$, a contradiction. ∎

##  4 Tangent bundles of $\mathbb{L}_{+}$ are non-contractible

We will use the terminology for Type I spaces defined at the beginning of the
previous section. The following theorem shows that a topologically very
interesting class of surfaces with trivial homotopy groups does not contain
contractible ones. Given a smoothing of the longray $\mathbb{L}_{+}$, write
$TL^{+}$ for the resulting tangent bundle. P. Nyikos has extensively studied
these spaces in [28] and showed that they have very different topological
properties depending on the smoothing. We refer to this paper for precise
definitions. Since we will just use properties shared by all tangent bundles
of $\mathbb{L}_{+}$, we only need to recall that there is a bundle projection
$\pi:TL^{+}\to\mathbb{L}_{+}$ such that
$\\{\pi^{-1}((0,\alpha))\,:\,\alpha\in\omega_{1}\\}$ is a canonical cover for
$TL^{+}$ and $\pi^{-1}((0,\alpha))$ is homeomorphic to $\mathbb{R}^{2}$ for
each $\alpha$, hence $TL^{+}$ is a Type I surface. (Here and in the remainder
of this section, we view $\omega_{1}$ as a subset of $\mathbb{L}_{+}$ whenever
convenient.) $TL^{+}$ contains a copy of $\mathbb{L}_{+}$ (the $0$-section).
Removing it disconnects the surface in two homeomorphic subsurfaces
$T^{+},T^{-}$, none of which contain a copy of $\mathbb{L}_{+}$. Depending on
the smoothing, they may contain a copy of $\omega_{1}$. It is important to
note that $\pi^{-1}((0,\alpha))\cap T^{+}$ is also homeomorphic to
$\mathbb{R}^{2}$, thus we slightly abuse the notation by denoting the
restriction of $\pi$ to $T^{+}$ also by $\pi$. We say that a function $f$ with
domain $T^{+}$ is almost constant if there is some club $C$ of $\omega_{1}$
such that $f$ is constant on $\pi^{-1}(C)\subset T^{+}$.

######  Theorem 4.1.

If $h:T^{+}\times[0,1]\to\mathbb{L}_{+}$ is given such that $h_{t_{0}}(T^{+})$
is Lindelöf for some $t_{0}\in[0,1]$, then $h_{t}$ is almost constant for each
$t\in[0,1]$.

An immediate corollary is:

######  Corollary 4.2.

Neither $TL^{+}$ nor $T^{+}$ are heCWc.

###### Proof.

The homotopy groups of $T^{+}$ are all trivial since any compact subset is
contained in $\pi^{-1}((0,\alpha))$ (which is contractible) for some $\alpha$.
If $T^{+}$ was heCWc it would then be contractible, which would yield an
homotopy between $\pi:T^{+}\to\mathbb{L}_{+}$ and a constant map. Theorem 4.1
impedes it, proving the result for $T^{+}$. The result for $TL^{+}$ follows
from Theorem 1.1 since it contains a copy of $\omega_{1}$. ∎

Following [28], call a subset of $T^{+}$ large if it intersects all the fibers
above some stationary subset of $\omega_{1}$. We are going to use the
following two results of P. Nyikos.

######  Lemma 4.3 ([28, Thm 4.10 & Cor. 4.15]).

(i) If $U$ is a large open set of $T^{+}$, then $\overline{U}$ contains
$\pi^{-1}(C)$ for a club subset $C$ of $\omega_{1}$.
(ii) If $U\subset T^{+}$ is open and contains $\pi^{-1}(C)$ for some club $C$
of $\omega_{1}$, then given $f:U\to\mathbb{R}$, there is a club $D$ of
$\omega_{1}$ such that $f$ is constant on $\pi^{-1}(D)\cap U$.

Actually, Nyikos proves part (ii) only for $U=T^{+}$, but his proof works as
well in our case.

###### Proof of Theorem 4.1.

Let $h$ be as in the statement. Write
$T^{+}=\cup_{\alpha\in\omega_{1}}T^{+}_{\alpha}$ with
$T^{+}_{\alpha}=\pi^{-1}((0,\alpha))$ homeomorphic to $\mathbb{R}^{2}$. Set

$A=\\{t\in[0,1]\,:\,h_{t}\text{ is almost constant}\\}.$

We show that $A$ is open and closed in $[0,1]$. Since $h_{t_{0}}(T^{+})$ is
Lindelöf, it is contained in some $(0,\alpha)\subset\mathbb{L}_{+}$ which is
homeomorphic to $\mathbb{R}$, hence $t_{0}\in A$ by Lemma 4.3 (ii). It follows
that $A=[0,1]$.
$A$ is closed. Let $t_{n}\in A$ ($n\in\omega$) be a sequence converging to
$t\in[0,1]$ and let $C_{n}\subset\omega_{1}$ be club such that $h_{t_{n}}$ is
constant on $\pi^{-1}(C_{n})$. Then $C=\cap_{n\in\omega}C_{n}$ is club and by
continuity $h_{t}$ is constant on $\pi^{-1}(C)$.
$A$ is open. Let $t\in A$ and suppose in the way of contradiction that there
is a sequence $t_{n}\not\in A$ ($n\in\omega$) converging to $t$. Let $\gamma$
be such that $h_{t}(\pi^{-1}(C))$ is contained in
$(0,\gamma)\subset\mathbb{L}_{+}$ for some club $C$. If
$\pi\bigl{(}h_{t_{n}}^{-1}((0,\gamma))\bigr{)}$ is stationary, by Lemma 4.3
(i) $h_{t_{n}}^{-1}((0,\gamma+1))$ contains $\pi^{-1}(D_{n})$ for a club
$D_{n}$, hence by (ii) $t_{n}\in A$. There is thus a club
$C_{n}\subset\omega_{1}$ such that $h_{t_{n}}^{-1}((0,\gamma))$ misses the
fibers above $C_{n}$, that is,
$h_{t_{n}}(\pi^{-1}(C_{n}))\subset[\gamma,\omega_{1})$. By continuity,
$h_{t}(\pi^{-1}(B))\subset[\gamma,\omega_{1})$ for
$B=C\cap\cap_{n\in\omega}C_{n}$, a contradiction.
∎

We believe that the following problem has a negative answer, but cannot back
our claim with any evidence.

######  Problem 4.4.

Is it possible to have a locally trivial fiber bundle projection
$p:E\to\mathbb{L}_{+}$ with a contractible total space $E$ ?

##  5 A non-contractible sub-surface of the Prüfer surface

We first recall the definition of the Prüfer surface. We just sketch the
construction and refer to [18, Ex. I.25–28] for more details. Notice however
that our notation is slightly different. Let $H$ be the half plane $\\{\langle
x,y\rangle\in\mathbb{R}^{2}\,:\,x\geq 0\\}$, and $H_{0}$ contain the points of
$H$ with horizontal coordinate $>0$. By Prüferizing $H$ at height
$c\in\mathbb{R}$, we mean deleting $\langle 0,c\rangle\in H$ and replacing it
by a copy $\mathbb{R}_{c}$ of $\mathbb{R}$, with the following topology. Given
$E\subset\mathbb{R}$, we write $E_{c}$ for the copy of $E$ in
$\mathbb{R}_{c}$. The neighborhoods of the non-deleted points in $H$ are
unchanged, and a neighborhood basis of $u\in\mathbb{R}_{c}$ is given by all

$T_{a,b,\epsilon}^{c}=(a,b)_{c}\cup\\{\langle x,y\rangle\in
H_{0}\,:\,ax+c<y<bx+c\,,\,x<\epsilon\\},$

for $a,b\in\mathbb{R}_{c}$, $a<u<b$ and $\epsilon>0$. In words,
$T_{a,b,\epsilon}^{c}\cap H_{0}$ is the triangular region between the lines of
slopes $a,b$ passing through $\langle 0,c\rangle$ and the vertical line
$x=\epsilon$. If $A\subset\mathbb{R}$, by Prüferizing at $A$ we mean
Prüferizing at each point of $A$. If one Prüferizes at $A$ and deletes
$\\{0\\}\times(\mathbb{R}-A)$ from $H$, the resulting space $P_{A}$ is a
surface with boundary $\cup_{c\in A}\mathbb{R}_{c}$. A finite $A$ yields a
surface homeomorphic to $H$ minus finitely many points in the vertical axis.
The surface (with boundary) $P_{\mathbb{R}}$ obtained by Prüferizing at each
height is usually called the (separable) Prüfer surface and will be denoted by
$\mathsf{P}$. It is known for a long time that $P_{A}$ is contractible for any
$A\subset\mathbb{R}$, in particular $\mathsf{P}$ and its Moorized version (see
below) are both contractible, see [10]. We will show that deleting some points
in the boundary of $\mathsf{P}$ yields a non-contractible submanifold. Let
$0_{c}$ denote the $0$ point in $\mathbb{R}_{c}\subset\mathsf{P}$ and set
$Z=\cup_{c\in\mathbb{R}}\\{0_{c}\\}$. Then $Z$ is a closed discrete subset of
$\mathsf{P}$.

######  Example 5.1.

The separable surface with boundary $\mathsf{P}-Z$ has trivial homotopy groups
but is non-contractible (and hence not heCWc).

We first show the following variation, for which we have a simpler proof of
the non-contractibility.

######  Example 5.2.

Let $d>0$ and $N_{d}=\cup_{c\in\mathbb{R}}[-d,d]_{c}$. Then the separable
surface with boundary $\mathsf{P}-N_{d}$ has trivial homotopy groups but is
non-contractible (and hence not heCWc).

###### Details.

Recall that $H_{0}$ is the open half plane $\\{\langle
x,y\rangle\in\mathbb{R}^{2}\,:\,x>0\\}$. Firstly, it is clear that the
homotopy groups of $\mathsf{P}-N_{d}$ are trivial. Indeed, any compact subset
intersects finitely many $\mathbb{R}_{c}$, and is thus contained in a
submanifold homeomorphic to $H$ minus finitely many points in its boundary,
which is contractible. Secondly, if $d,d^{\prime}\not=0$, the map
$\mathsf{P}-N_{d}\to\mathsf{P}-N_{d^{\prime}}$ defined by $\langle
x,y\rangle\mapsto\langle\frac{d^{\prime}}{d}x,y\rangle$ for points in $H_{0}$
and $u\mapsto\frac{d^{\prime}}{d}u$ for $u\in\mathbb{R}_{c}$ is a
homeomorphism. We may thus assume that $d=\frac{1}{2}$ and prove that
$\mathsf{P}-N_{1/2}$ is not contractible.
By way of contradiction, assume
$h:(\mathsf{P}-N_{1/2})\times[0,1]\to\mathsf{P}-N_{1/2}$ to be such that
$h_{1}=id$ and $h_{0}\equiv*$. We may assume that $*\in H_{0}$. Write
$1_{c},-1_{c}$ for the points
$1,-1\in\mathbb{R}_{c}\subset\mathsf{P}-N_{1/2}$. We are going to use the
subsets of $H_{0}$ depicted in Figure 1. (Strictly speaking, $\langle
0,c\rangle$ belongs to $H$ and not to $H_{0}$, but it should cause no
confusion since it also does not belong to any of the subsets we consider.) We
allow $\delta_{1}$ to be $0$ and $\delta_{2}$ to be equal to $+\infty$.
$I^{\pm}_{c,\delta}$ are closed segments in $H_{0}$, while
$A^{\pm}_{c,\delta_{1},\delta_{2}},B^{\pm}_{c,\delta_{1},\delta_{2}}$ are open
(hence dashed lines do not belong to them).

Figure 1: Subsets of $H_{0}$.

Notice that in $\mathsf{P}-N_{1/2}$, for any $\delta>0$ the following hold:

  1. 1.

$1_{c}$ is the only accumulation point of $I^{+}_{c,\delta}$ not in
$I^{+}_{c,\delta}$ and $-1_{c}$ is the only accumulation point of
$I^{-}_{c,\delta}$ not in $I^{-}_{c,\delta}$,

  2. 2.

For any $\delta>0$, $A^{+}_{c,0,\delta}\cup(\frac{1}{2},+\infty)_{c}$ is a
neighborhood of $(\frac{1}{2},+\infty)_{c}$ and
$A^{-}_{c,0,\delta}\cup(-\infty,-\frac{1}{2})_{c}$ is a neighborhood of
$(-\infty,-\frac{1}{2})_{c}$.

######  Claim 1.

Assume that there are $\delta>0$, $u,v\in\mathbb{R}\cup\\{\pm\infty\\}$ and
$t_{0}\in[0,1]$ such that both $h_{t_{0}}(1_{c})$ and $h_{t_{0}}(-1_{c})$ are
in $H_{0}$ and have horizontal coordinate $>\delta$ for each $c\in(u,v)$.
Then:
(i) $\exists n\geq 1,\,u^{+},v^{+},a^{+},b^{+}\in\mathbb{Q}\cap[0,1]$, with
$t_{0}<a^{+}<b^{+}$ and $u\leq u^{+}<v^{+}\leq v$ such that both
$h_{t}(1_{c})$ and $h_{t}(-1_{c})$ are in the closure of
$B^{+}_{c,\delta/(n+1),\delta/n}$ for each $c\in[u^{+},v^{+}]$ and
$t\in[a^{+},b^{+}]$,
(ii) $\exists m\geq 1,\,u^{-},v^{-},a^{-},b^{-}\in\mathbb{Q}\cap[0,1]$, with
$t_{0}<a^{-}<b^{-}$ and $u\leq u^{-}<v^{-}\leq v$ such that both
$h_{t}(1_{c})$ and $h_{t}(-1_{c})$ are in the closure of
$B^{-}_{c,\delta/(m+1),\delta/m}$ for each $c\in[u^{-},v^{-}]$ and
$t\in[a^{-},b^{-}]$.

Figure 2: Claim 1.

This claim is summarized in Figure 2, where it is to be understood that the
left part implies the right part. Let us show why it implies the non-
contractibility of $\mathsf{P}-N_{1/2}$. Since $h_{0}$ sends everything to
$*\in H_{0}$, set $\delta_{0}$ to be half of $*$’s horizontal coordinate. We
now use induction on $k$ and the claim to find nested intervals
$[u_{k},v_{k}]\subset[u_{k-1},v_{k-1}]$, $t_{k}\geq t_{k-1}$ and
$\delta_{k}=\delta_{k-1}/(n(k)+1)$, $\delta^{\prime}_{k}=\delta_{k-1}/(n(k))$
for some $n(k)\geq 1$ such that the following holds:
(i) $h_{t_{k}}(1_{c})$ and $h_{t_{k}}(-1_{c})$ are both in
$\overline{B^{+}_{c,\delta_{k},\delta^{\prime}_{k}}}$ for $c\in[u_{k},v_{k}]$
when $k$ is even,
(ii) $h_{t_{k}}(1_{c})$ and $h_{t_{k}}(-1_{c})$ are both in
$\overline{B^{-}_{c,\delta_{k},\delta^{\prime}_{k}}}$ if $c\in[u_{k},v_{k}]$
when $k$ is odd.
(These two points correspond exactly to (i) and (ii) in the claim.) Let
$t=\sup_{k\in\omega}t_{k}$ and $c\in\cap_{k\in\omega}[u_{k},v_{k}]$. Then
$h_{t_{k}}(1_{c})$ and $h_{t_{k}}(-1_{c})$ should converge to $h_{t}(1_{c})$
and $h_{t}(-1_{c})$, but are divergent in $\mathsf{P}-N_{1/2}$ since they
zigzag around the horizontal at height $c$ while going to the boundary. (This
horizontal line converges to $0_{c}$ in $\mathsf{P}$, which is not present in
$\mathsf{P}-N_{1/2}$.) ∎

###### Proof of Claim 1.

We prove (i), as (ii) is entirely similar. Let $\pi:\mathsf{P}\to\mathbb{R}$
be the projection on the first coordinate on $H_{0}$ and take constant value
$0$ on the boundary components $\mathbb{R}_{c}$. Then $\pi$ is easily seen to
be continuous. For $n,k\geq 1$ and $p,q\in\mathbb{Q}\cap[0,1]$, Let

$C^{+}_{n,k,p,q}=\\{c\in(u,v)\,:\,h_{t}(I^{+}_{c,1/k})\subset
A^{+}_{c,\delta/(n+1),\delta/n}\,\forall t\in(p,q)\\}.$

We show that each $c\in(u,v)$ belongs to some $C^{+}_{n,k,p,q}$. By assumption
$\pi\circ h_{t_{0}}(1_{c})$ and $\pi\circ h_{t_{0}}(-1_{c})$ are $>\delta$.
Notice that $(\frac{1}{2},+\infty)_{c}$ is a connected component of
$(\mathsf{P}-N_{1/2})-H_{0}$, hence $1_{c}$ must first enter $H_{0}$ before
potentially entering any other boundary component, and $1_{c}$ does leave
$(\frac{1}{2},+\infty)_{c}$ since $\pi\circ h_{t_{0}}(1_{c})=\delta$. Since
$A^{+}_{c,0,1}\cup(\frac{1}{2},+\infty)_{c}$ is a neighborhood of
$(\frac{1}{2},+\infty)_{c}$, $1_{c}$ enters $A^{+}_{c,0,1}$ after leaving
$(\frac{1}{2},+\infty)_{c}$. There is thus an interval $(p,q)$ (with
$p>t_{0}$) and $n\geq 1$ such that $h_{t}(1_{c})\in
A^{+}_{c,\delta/(n+1),\delta/n}$ for each $t\in(p,q)$. Now, $\\{1_{c}\\}\cup
I^{+}_{c,\eta}$ being compact for each $\eta>0$, if we shrink the interval
$(p,q)$ and take $\eta$ small enough, then $h_{t}(I^{+}_{c,\eta})\subset
A^{+}_{c,\delta/(n+1),\delta/n}$ for each $t\in(p,q)$. Taking $k$ such that
$\frac{1}{k}<\eta$, we obtain that $c\in C^{+}_{n,k,p,q}$.
Since any interval of $\mathbb{R}$ is a Baire space and there are countably
many $C^{+}_{n,k,p,q}$, one of them must be non-meagre. In particular, for
some $n,k,p,q$, $C^{+}_{n,k,p,q}$ is dense in an interval
$[u^{\prime},v^{\prime}]\subset(u,v)$. If $w=\langle x,y\rangle\in H_{0}$, set
$c(w)=y-x$. By continuity of $h$, if $w$ is in the parrallelogram in $H_{0}$
defined by the lines of slope $1$ emerging from $\langle 0,u^{\prime}\rangle$
and $\langle 0,v^{\prime}\rangle$, the vertical line at $\frac{1}{k}$ and the
boundary, then $w$ is sent to the closure of
$A^{+}_{c(w),\delta/(n+1),\delta/n}$ by $h_{t}$ for $t\in(p,q)$, as shown on
the lefthandside of Figure 3.

Figure 3: Proof of Claim 1.

By looking at righthandside of the same figure, we see that if we take
$c\in(u^{\prime},v^{\prime})$ and $\eta$ small enough, then $I^{-}_{c,\eta}$
is sent in the closure of $B^{+}_{c,\delta_{n+1},\delta/n}$ by $h_{t}$ for
$t\in(p,q)$. (For those keen on deciphering formulas, take $\eta$ such that
$u^{\prime}-2\eta>v$ and $\eta<\min(\frac{\delta}{4(n+1)},\frac{1}{k})$.)
Since $-1_{c}$ is in the closure of $I^{-}_{c,\eta}$, this holds also for
$-1_{c}$. To finish the proof, take a closed subinterval
$[u^{+},v^{+}]\subset(u^{\prime},v^{\prime})$. ∎

The proof that $\mathsf{P}-Z$ is also non-contractible (Example 5.1) is very
similar, but with extra steps. A shortcut would be to answer the following
question in the affirmative, but we could not decide whether it is the case or
not.

######  Questions 5.3.

Given $d>0$, are $\mathsf{P}-N_{d}$ and $\mathsf{P}-Z$ homeomorphic ?

Let us summarize the argument of $\mathsf{P}-Z$’s non-contractibility. Assume
as before that $h:(\mathsf{P}-Z)\times[0,1]\to\mathsf{P}-Z$ is a contraction
with $h_{1}=id$ and $h_{0}\equiv*$. We also consider $1_{c},-1_{c}$. Set
$t^{+}(c)=\sup\\{t\in[0,1]\,:\,h_{t}(1_{c})\in(0,+\infty)_{c}\\}$. Then
$0<t^{+}(c)$ and $h_{t^{+}(c)}(1_{c})$ is a member of $(0,+\infty)_{c}$ that
we denote by $s(c)$. Define $A^{\pm,\ell}_{c,\delta_{1},\delta_{2}}$ as
$A^{\pm}_{c,\delta_{1},\delta_{2}}$, but with slope $1/\ell$ instead of $1/2$.
Take $\ell$ such that $\frac{1}{\ell}<\frac{s(c)}{2}$. Then there are some
$p,q\in\mathbb{Q}$, $p<q<1$, and $n\in\omega$ such that $h_{t}(1_{c})$ is
contained in $A^{+,\ell}_{c,0,+\infty}$ if $p<t$ and in
$A^{+,\ell}_{c,1/(n+1),1/n}$ if $p<t<q$. Up to shrinking $(p,q)$, the same
holds for $I^{+}_{c,\delta}$ if $\delta$ is small enough. The same Baire
argument as before yields an interval $[u,v]$ and $\ell,n,p,q$ such that for
each $c\in[u,v]$, $h_{t}(1_{c})\in A^{+,\ell}_{c,1/(n+1),1/n}$ if $p<t<q$.
Then $h_{t}(I^{-}_{c,\eta})\subset A^{+,\ell}_{c,1/(n+1),1/n}$ as well for the
same $t$’s and $\eta$ small enough (depending on $\ell$). This shows in
particular that $h_{t}(-1_{c})$ is in the closure of $B^{+}_{c,1/(n+1),1/n}$
for the same $t$’s. Now proceed with the $-1_{c}$, etc, to find the same
zigzaging diverging sequence.

Our examples so far all have a boundary, but we may get rid of it by
Mooreizing the boundary components (see again [18, Ex. I.27] for details). In
its classical version, we take a boundary component $\mathbb{R}_{c}$ obtained
by Prüferization and identify $x$ with $-x$ in it. This has the effect of
folding $\mathbb{R}_{c}$ which is thus ‘swallowed’ by the manifold and is not
a boundary component anymore. We can perform this identification in all the
boundary components or only in some of them. If one takes the Prüfer surface
and Moorize each boundary component, we obtain a contractible boundaryless
non-metrizable surface. We can also fold around another point than $0_{c}$, or
use more complicated functions for the identification. For instance, if as
above $0_{c}$ has been removed from the boundary component $\mathbb{R}_{c}$,
we may fold $(0,+\infty)_{c}$ around $1_{c}$ by using any homeomorphism
between $(0,1]_{c}$ and $[1,+\infty)_{c}$ for the identification, and
similarly fold $(-\infty,0)_{c}$ around $-1_{c}$.

######  Example 5.4.

The separable surface obtained by Mooreizing the boundary components of
$\mathsf{P}-Z$, by folding $(0,+\infty)_{c}$ around $1_{c}$ and
$(-\infty,0)_{c}$ around $-1_{c}$ is non-contractible.

The proof of non-contractibility in Example 5.1 works verbatim. Another way to
get rid of boundary components is to glue collars to them, that is, copies of
$\mathbb{R}\times[0,1)$. Doing it to $\mathsf{P}$ yields a contractible non-
separable surface, while doing it to $\mathsf{P}-Z$ yields a non-contractible
subsurface of the former. For more on these variations of the Prüfer surface,
see [4].

##  6 $\omega_{1}$-trees and their road spaces

The contents of this section appeared in the preprint [5]. The objects we
study are set theoretic trees and their road spaces, which are obtained by
joining consecutive points by a line segment with a topology that makes it
locally embeddable in $\mathbb{R}^{2}$ (the details are given below). These
spaces are good toy models for Type I non-metrizable manifolds.

Recall that a tree $T$ is a partially ordered set such that each point has a
well ordered set of predecessors. We define the height of $x\in T$ and of $T$,
the $\alpha$-th-level $\text{\rm Lev}_{\alpha}(T)$, the chains and antichains
in $T$ as usual, see for instance [24, Section II.5] or [29]. A subset $E$ of
$T$ is order-dense iff for any $y\in T$ there is $y\in E$ with $x<y$. All
trees in this note are endowed with the order (also called interval) topology.
A tree is rooted iff it has a unique minimal element called the root and
binary iff each member has exactly $2$ immediate successors. A subtree $S$ is
a subset of $T$ with order restricted to $S$. Notice that the induced topology
on a subtree $S$ is finer (sometimes strictly) than the one given by the order
restricted to $S$. Both topologies agree if $S$ is closed in $T$. We assume
that our trees are Hausdorff, that is, if $x,y\in T$ are at a limit level and
have the same predecessors, then $x=y$. (This could be false for a subtree.)
An $\omega_{1}$-tree has countable levels and height $\omega_{1}$. A tree is
Aronszajn (resp. Suslin) if it has height $\omega_{1}$ and its chains (resp.
chains and antichains) are at most countable. When $x\in T$ and $\alpha$ is an
ordinal, write $T_{\geq x}=\\{y\in T\,:\,y\geq x\\}$, $x\upharpoonright\alpha$
for the unique predecessor of $x$ at level $\alpha$ (if $x$ is below the
$\alpha$-th level, $x\upharpoonright\alpha=x$) and $T_{\leq\alpha}$ for the
subset of elements at level $\leq\alpha$. If $E\subset T$, set
$E^{\downarrow}=\\{x\in T\,:\,\exists y\in E\text{ with }x\leq y\\}$ to be its
downward closure. We say that the tree $T$ is $\mathbb{R}$-special iff there
is a strictly increasing (not necessarily continuous) function
$T\to\mathbb{R}$. Recall that $\mathbb{R}$-special binary $\omega_{1}$-trees
exist in ZFC.

The road space $R_{T}$ of a tree $T$ is obtained by joining consecutive points
by an interval $[0,1]$, with $0$ glued to the lowest point and $1$ to the
highest. We extend the order in the obvious way. When convenient we consider
$T$ as a subset of $R_{T}$. The topology on the interior of the added
intervals is that of $(0,1)$. For $x\in T\subset R_{T}$, in order for $R_{T}$
to be (locally) connected any open set containing $x$ must contain a small
portion of each interval emanating from $x$. In order for the space to be
locally metrizable (and hence first countable), we take these portions
uniformly as follows. Denote by $[0,1]_{x}^{y}\subset R_{T}$ the interval
between the two consecutive points $x,y\in T$. If $A\subset[0,1]$, then
$A_{x}^{y}$ is understood as the corresponding subset of $[0,1]_{x}^{y}$. For
singletons we usually write $a_{x}^{y}$ instead of $\\{a\\}_{x}^{y}$. If $x\in
T$, denote by $s(x)$ the set containing its immediate successors and set
$W_{x,n}=\displaystyle\bigcup_{y\in s(x)}[0,1/n)_{x}^{y}$. If $x\in s(z)$ with
$z\in T$, a basis for the neighborhoods of $x$ in $R_{T}$ is given by
$\\{W_{x,n}\cup(1-1/n,1]_{z}^{x}\,:\,n\in\omega\\}$ (see Figure 4, left). A
basic neighborhood of $x\in T\subset\mathbb{R}_{T}$ at a limit level is
obtained by choosing some $z\in T$, $z<x$ and $n\in\omega$ and taking the
segment $\\{y\in R_{T}\,:\,z<y<x\\}$ union each $W_{w,n}$ for those $w\in T$
with $z<w\leq x$ (see Figure 4, right). This makes $R_{T}$ locally embeddable
in $\mathbb{R}^{2}$ (as seen by induction, or by extrapolating from Figure 4).
The induced topology on $T\subset R_{T}$ is that of $T$, and $R_{T}$ is arc
connected if and only if $T$ is rooted.

Figure 4: Neighboroods in $R_{T}$.

The first result of this section is the following.

######  Theorem 6.1.

The road space $R_{T}$ of a rooted $\mathbb{R}$-special tree is contractible.

This gives half of the proof of Theorem 1.2, and is proved in Subsection 6.1.
The other two theorems of this section are generalizations of Theorem 1.1. We
depart somewhat from the manifold theme of this paper and investigate quite
precise properties of subspaces, trying to see whether they prevent global
contractibility. The first version is to weaken the assumption in Theorem 1.1
that the subspace is a copy of $\omega_{1}$.

######  Theorem 6.2.

Let $T$ be a tree, $S\subset T$ be uncountable and $\omega_{1}$-compact in the
subspace topology, and let $X$ be a locally metrizable space. Then there is no
continuous $h\colon S\times[0,1]\to X$ such that $h_{1}\colon S\to X$ has
uncountable image and $h_{0}$ is constant. In particular, if $X$ contains a
copy of $S$ then $X$ is not contractible.

Our proof of Theorem 6.2 is entirely topological, for one using forcing see
[5]. The first step is to show that one can assume without loss of generality
that $T$ is an Aronszajn tree. Then, the bulk of our argument consists of
showing that many well known properties of Suslin trees also hold for
$\omega_{1}$-compact subsets of trees of height $\omega_{1}$. These
generalizations are probably part of the folklore and no more than standard
exercices, but we found convenient to gather the proofs of most of them in
Subsection 6.2. Then the ‘real’ proof of this theorem (and the fact that it
implies Theorem 1.4) is contained in Subsection 6.3. The property that is
central to our argument is that any continuous map from $S$ to a locally
metrizable space must be constant on the points above $x\in S$ if $x$ is high
enough in the tree (see Lemma 6.7 (i) below). This is a slight generalization
of a result of Steprāns [33].
Notice that an uncountable $\omega_{1}$-compact subset of a tree cannot be
metrizable, since $\omega_{1}$-compactness is equivalent to Lindelöfness in
metrizable spaces (see, e.g., [14, Thm 4.1.15]), and a Lindelöf subset of a
tree with countable levels is countable.
A quick corollary, which is another exhibit of the similarities between
$\omega_{1}$ and Suslin trees, is the following. Its proof is immediate since
the road space of a tree satisfying the assumptions is locally metrizable and
a Suslin tree is $\omega_{1}$-compact, as well known (see Lemma 6.6 below).

######  Corollary 6.3.

Let $T$ be a tree of height $\omega_{1}$. If $T$ contains an uncountable
subset $S$ which is $\omega_{1}$-compact in the subspace topology (in
particular, if $T$ is a Suslin tree), then its road space $R_{T}$ is non-
contractible.

Our second generalization of Theorem 1.1 is to weaken the local metrizability
of the target space $X$. First countability alone is not enough, as shown by
Example 1.10. What we were able to prove is the following.

######  Theorem 6.4.

Let $X$ be a regular space with $G_{\delta}$ points such that each point has a
closed $[\aleph_{0},\aleph_{1}]$-compact neighborhood. Let $S$ be a stationary
subset of $\omega_{1}$ endowed with the subspace topology. Then there is no
continuous $h\colon S\times[0,1]\to X$ such that $h_{1}\colon S\to X$ has an
uncountable image and $h_{0}$ is constant. In particular, if $X$ contains a
copy of $S$ then $X$ is not contractible.

Recall that a space is $[\aleph_{0},\aleph_{1}]$-compact if and only if any
cover of $X$ of size $\leq\aleph_{1}$ has a countable subcover. This is
equivalent to the property that given any uncountable subspace $E$ of $X$
there is $x\in X$ which is a condensation point of $E$, that is, any
neighborhood of $x$ contains uncountably many points of $E$. We do not know
whether this theorem holds if $S$ is a Suslin tree but note in passing that
our main tool (any continuous map from $S$ to $X$ must be constant on the
points above $x\in S$ if $x$ is high enough, see Lemma 6.11 below) does not.
(This should appear in a later paper.) Theorem 6.4 holds if $X$ is locally
metrizable. Indeed, a stationary subset of $\omega_{1}$ is
$\omega_{1}$-compact, $\omega_{1}$-compactness is preserved by continuous
functions and is equivalent to Lindelöfness in metrizable spaces; hence the
image of $S\times[0,1]$ in $X$ is (hereditarily) Lindelöf.

Corollary 6.3 inspires the following question.

######  Questions 6.5.

Let $T$ be a tree and $R_{T}$ be its road space.
(a) Does the contractibility of $R_{T}$ implies that $T$ is
$\mathbb{R}$-special ?
(b) Does the non-contractibility of $R_{T}$ imply that $T$ contains a Suslin
subtree ?

Notice that it is not possible for (a) and (b) to both have a positive answer
since there are models of set theory with a non-$\mathbb{R}$-special tree $T$
that does not contain a Suslin subtree, see [31]. We do not know whether these
trees have contractible road spaces.

###  6.1 Proof of Theorem 6.1

###### Proof.

Let $T$ be a rooted $\mathbb{R}$-special tree with root $r$ and let
$f:T\to\mathbb{R}$ be a strictly increasing function. By replacing $f(x)$ by
$\sup_{y<x}f(y)$ when $x$ is at limit levels, we may assume that $f$ is
continuous. By composing with a strictly increasing function, we may assume
that the range of $f$ is contained in $[0,1]$. We first start to define the
contraction as a map $h:T\times[0,1]\to R_{T}$. First, set $h_{1}(x)=x$ and
then define $h$ such that $x$ travels downwards in $R_{T}$, starting to move
exactly at time $t=f(x)$ and reaching $y<x$ ($y\in T$) exactly at time
$t=f(y)$. Since $f$ is strictly increasing, there is time available to cross
the interval between consecutive points. For those who like (less readable)
formulas, let $x$ at level $\alpha$ be given. If $f(x)\leq t$, we set
$h_{t}(x)=x$. If $\beta<\alpha$ and $f(x\upharpoonright\beta)\leq t\leq
f(x\upharpoonright\beta+1)$, we set $h_{t}(x)$ to be
$u_{x\upharpoonright\beta}^{x\upharpoonright\beta+1}$ where

$u=\frac{t-f(x\upharpoonright\beta)}{f(x\upharpoonright\beta+1)-f(x\upharpoonright\beta)}.$

Finally, if $t\leq f(r)$, we set $h_{t}(x)=r$. It should be clear that $h$ is
continuous (on $T\times[0,1]$) and that $h_{0}(x)=r$ for all $x\in T$. Notice
that $h$ has the property that at time $f(x)$, all of $T_{\geq x}$ is squished
onto $x$. Actually, each point on the tree starts to move exactly when all the
points above it reach it, all at the same time. This enables to extend easily
the map to all of $R_{T}$, a point in the segment $[0,1]_{x}^{y}$ does not
move until $y$ reaches it, and then it follows it until the end. This gives
the required contraction. ∎

Notice that if one sets $j_{t}(x)=h_{1-t}(x)$ for $t\in[0,1]$, $j_{t}(x)=x$
when $t<0$ and $j_{t}(x)=r$ when $t>1$, then $j$ is actually a flow, that is
$j_{t}\circ j_{s}=j_{t+s}$.

###  6.2 A collection of facts on uncountable $\omega_{1}$-compact subsets of
trees

Notations: Given a tree $T$, if $A,B\subset T$ and $x\in T$, $A<B$ means that
each member of $A$ is $<$ each member of $B$. We denote $\\{x\\}<A$ by $x<A$
and $T_{\geq x}\cap A$ by $A_{\geq x}$. Notice that
$\text{Lev}_{\alpha}(A^{\downarrow})=\text{Lev}_{\alpha}(T)\cap
A^{\downarrow}$.

######  Lemma 6.6.

Let $T$ be a tree of height $\omega_{1}$ and $S\subset T$ be uncountable and
endowed with the subspace topology. Then the following hold.
(a) The subspace topology on $S^{\downarrow}$ always agrees with the topology
given by the induced order. If $S$ is closed in $T$, then the subspace
topology agrees with the topology given by the induced order on $S$.
(b) If $S$ is $\omega_{1}$-compact, then it intersects a stationary subset of
levels of $S^{\downarrow}$ (and of $T$).
(c) An antichain is closed discrete in $T$, and a closed discrete subset of
$T$ is an at most countable union of antichains.
(d) If $S$ is $\omega_{1}$-compact then so is $S^{\downarrow}$.
(e) If $S$ is $\omega_{1}$-compact and does not contain an uncountable chain,
then $S^{\downarrow}$ is Suslin.
(f) If $A\subset S\subset T$, then $A$ is a maximal antichain in $S$ if and
only if it is a maximal antichain in $S^{\downarrow}$.

###### Proof.

(a) and (f) are straightforward. For (b), if $S$ avoids a club set of levels
of $S^{\downarrow}$ then taking one point between consecutive avoided levels
(when available) yields an uncountable discrete subset which is closed in $S$.
Item (c) is proved e.g. in [29, Thm 4.11], and (d)–(e) follow immediately from
it. ∎

######  Lemma 6.7.

Let $T$ be a tree of height $\omega_{1}$ and $S\subset T$ be uncountable and
$\omega_{1}$-compact in the subspace topology. Then the following hold.
(a) $S^{\downarrow}$ is the disjoint union of an at most countable set, a
Suslin tree and at most countably many copies of $\omega_{1}$. In particular,
$S^{\downarrow}$ is an $\omega_{1}$-tree.
(b) There is $\alpha\in\omega_{1}$ such that $|S_{\geq
x}|=|(S^{\downarrow})_{\geq x}|=\aleph_{1}$ when $x$ is above level $\alpha$.
(c) If $C\subset S^{\downarrow}$ is uncountable, there is $x\in S$ such that
$(S^{\downarrow})_{\geq x}\subset C^{\downarrow}$ and $|S_{\geq
x}|=|(S^{\downarrow})_{\geq x}|=\aleph_{1}$.
(d) $C^{\downarrow}$ is $\omega_{1}$-compact for any $C\subset
S^{\downarrow}$.
(e) If $C\subset S^{\downarrow}$ is club in $S^{\downarrow}$, then
$\\{\gamma\,:\,\text{Lev}_{\gamma}(C^{\downarrow})\subset C\\}$ is club in
$\omega_{1}$.
(f) If $C\subset S^{\downarrow}$ is club in $S^{\downarrow}$, then $S\cap C$
intersects a stationary subset of levels.
(g) If $E_{n}\subset S^{\downarrow}$ are closed and order-dense in
$S^{\downarrow}$ for $n\in\omega$, then $\cap_{n\in\omega}E_{n}$ is also
closed and order-dense.
(h) Let $E,F\subset S\subset T$ be closed in $S$. If $|E\cap
F|\leq\aleph_{0}$, then $|E^{\downarrow}\cap F^{\downarrow}|\leq\aleph_{0}$.
(i) Let $f:S\to Y$ be continuous where $Y$ is a metrizable space. Then there
is $\beta\in\omega_{1}$ such that $f(S_{\geq x})$ is a singleton whenever $x$
is above level $\beta$ in $T$. In particular, $f$ has a countable image.

###### Proof.

We will use several times (without acknowledging it explicitly) the fact that
an at most countable intersection of club subsets of $\omega_{1}$ is club.

(a) If $S$ contains an uncountable branch $B$, then there is some minimal
$x_{B}\in B$ such that $S_{\geq x_{B}}$ is linearly ordered. Indeed, otherwise
$S$ contains an uncountable antichain (take points branching away from $B$
above each height) and hence an uncountable closed discrete subset by Lemma
6.6 (c). Since the minimal elements of $\\{x\,:\,S_{\geq x}\subset S\text{ is
an uncountable branch}\\}$ is an antichain, $S$ contains at most countably
many disjoint uncountable branches. Notice that a maximal branch in
$S^{\downarrow}$ contains an unbounded branch of $S$. Removing the branches
above each minimal $x_{B}$ in $S^{\downarrow}$, what remains is either
countable or a Suslin tree by Lemma 6.6 (e).

(b) follows immediately from (a) and the equivalent statement for Suslin
trees.

(c) If $C\cap B$ is uncountable for some branch $B\subset S^{\downarrow}$, we
are over. If not, by (a) and (b) we can assume that $S^{\downarrow}$ is a
Suslin tree such that $|S_{\geq x}|=\aleph_{1}$ for each $x\in
S^{\downarrow}$. Then the result is well known (see, for instance, the claim
in Theorem 2.1 in [12]).

(d) is immediate by (a) since an uncountable (downward closed) subset of a
Suslin tree is a Suslin tree.

(e) By (d), $C^{\downarrow}$ is $\omega_{1}$-compact. Fix $\alpha$ given by
(b) and some $\beta>\alpha$. Since each uncountable maximal branch $B$ of
$C^{\downarrow}$ is a copy of $\omega_{1}$, $C$ contains a club set of levels
of $B$. By (a) we may assume that $C^{\downarrow}$ is Suslin. By Lemma 6.6
(f), for each $n\in\omega$ we may find a countable antichain $A_{n}\subset C$
above level $\beta$ which is maximal in $C^{\downarrow}$ and such that
$A_{n+1}>A_{n}$. Let $\gamma=\sup_{n\in\omega}\sup\\{\text{height}(x)\,:\,x\in
A_{n}\\}$. By construction $\text{Lev}_{\gamma}(C^{\downarrow})$ is the set of
limit points of $\cup_{n\in\omega}A_{n}$, hence since $C$ is closed
$\text{Lev}_{\gamma}(C^{\downarrow})\subset C$. This shows that
$\\{\gamma\,:\,\text{Lev}_{\gamma}(C^{\downarrow})\subset C\\}$ is unbounded
in $\omega_{1}$, and closedness is obvious.

(f) If $C\cap B$ is unbounded for some maximal uncountable branch $B\subset
S^{\downarrow}$, then it is homeomorphic to a club subset of $\omega_{1}$. By
Lemma 6.6 (b) $S\cap C\cap B$ is thus stationary. If $C\cap B$ is bounded for
each uncountable branch of $S^{\downarrow}$, we may assume by (a) that
$S^{\downarrow}$ is a Suslin tree in which $C$ is unbounded. By (c) it follows
that $S$ is unbounded in $C^{\downarrow}$ as well. By (e) and Lemma 6.6 (b),
$S$ intersects $C$ on a stationary set of levels.

(g) Closedness is immediate, and order-density follows immediately by (a) and
the fact that the result holds for Suslin trees and $\omega_{1}$.

(h) Let $\overline{E},\overline{F}$ be the closures in $S^{\downarrow}$ of
$E,F$. If $\overline{E}\cap\overline{F}$ is unbounded in $S^{\downarrow}$, by
(f)
$S\cap\overline{E}\cap\overline{F}=(S\cap\overline{E})\cap(S\cap\overline{F})=E\cap
F$ is unbounded, a contradiction. Hence $\overline{E}\cap\overline{F}$ is
bounded and thus $\overline{E}$ and $\overline{F}$ are disjoint above some
level $\alpha$. It follows that $E$ and $F$ cannot be both unbounded in the
same uncountable branch. It is well known (see e.g. [34, Thm 6.18] or [12, Thm
2.1]) that if $A,B$ are disjoint closed sets in a Suslin tree, then
$A^{\downarrow}\cap B^{\downarrow}$ is at most countable. Together with (a),
this shows that $|E^{\downarrow}\cap F^{\downarrow}|\leq\aleph_{0}$.

(i) Our proof is a slight adaptation of Steprāns topological proof in [33]
that a real valued map with domain a Suslin tree has a countable image. Denote
the distance in $Y$ by $\text{dist}(\cdot,\cdot)$. By (b) me may assume that
$S_{\geq x}$ is uncountable for each $x$. Set $D(\epsilon)=\\{x\in
S\,:\,\text{diam}(f(S_{\geq x}))\leq\epsilon\\}$, where diam stands for
diameter, that is, the supremum of the distances between two points in a set.
Assume for now that $D(\epsilon)$ is order-dense in $S$ (and thus in
$S^{\downarrow}$) when $\epsilon>0$. Let $\overline{D(\epsilon)}$ denote the
closure of $D(\epsilon)$ in $S^{\downarrow}$. By (g),
$D=\cap_{n\in\omega,n>0}\overline{D(1/n)}$ is closed and order dense in
$S^{\downarrow}$, hence by (f) $S\cap D$ intersects stationary many levels
above each $x\in S$ (since $|S_{\geq x}|=\aleph_{1}$). Moreover, $D$ is upward
closed in $S$. Denote by $A$ the minimal elements of $D\cap S$. Then $A$ is an
antichain of $S^{\downarrow}$, let $\beta$ be the supremum of the heights of
its members. For each $x\in S$ above level $\beta$, the diameter of
$f\bigl{(}S_{\geq x}\bigr{)}$ is $0$, hence $f\bigl{(}S_{\geq
x}\bigr{)}=f(\\{x\\})$ and the lemma is proved.
To finish, we now prove that $D(\epsilon)$ is order-dense in $S$. Suppose that
it is not the case and let $x\in S$ be such that $\text{diam}\bigl{(}f(S_{\geq
y})\bigr{)}>\epsilon$ for each $y>x$, $y\in S$. We build inductively
antichains $A^{\alpha}_{n}$ ($n\in\omega,\alpha\in\omega_{1}$) such that the
following hold.
$\bullet$ $A_{n}^{\alpha}\subset S$ is maximal above $x$, that is, in
$(S^{\downarrow})_{\geq x}$,
$\bullet$ $A_{n+1}^{\alpha}>A_{n}^{\alpha}>A_{m}^{\beta}$ for each
$n,m\in\omega$ and $\alpha>\beta$,
$\bullet$ If $u\in A_{n}^{\alpha}$, $v\in A_{n+1}^{\alpha}$, then
$\text{dist}(f(u),f(v))\geq\epsilon/4$.
Assume that $A_{n}^{\alpha}$ is defined. Set

$E=\\{z\in S\,:\,z>A_{n}^{\alpha}\text{ and
dist}(f(z),f(u))\geq\epsilon/4\text{, where $u$ is the member of
$A_{n}^{\alpha}$ below $z$}\\}.$

It is enough to see that $E$ is order-dense in $(S^{\downarrow})_{\geq x}$,
since then we may put its minimal elements in $A_{n+1}^{\alpha}$. Let thus
$w\in(S^{\downarrow})_{\geq x}$, $w>u\in A_{n}^{\alpha}$. Up to going further
up, we may assume that $w\in S$. If $\text{dist}(f(w),f(u))\geq\epsilon/4$,
then $w\in E$. If not, then $\text{dist}(f(w),f(u))<\epsilon/4$. Choose $v\in
S_{\geq w}$ such that $\text{dist}(f(w),f(v))>\epsilon/2$ (which exists since
we assumed $\text{diam}\bigl{(}f(S_{\geq y})\bigr{)}>\epsilon$ for each
$y>x$). Then

$\text{dist}(f(v),f(u))\geq\text{dist}(f(v),f(w))-\text{dist}(f(w),f(u))>\epsilon/4,$

and $v\in E$. If $A_{n}^{\gamma}$ is chosen for each $n\in\omega$ and each
$\gamma<\alpha$, set $A_{0}^{\alpha}$ to be an antichain in $S$, maximal in
$(S^{\downarrow})_{\geq x}$, whose members are all
$>\cup_{n\in\omega,\gamma<\alpha}A_{n}^{\gamma}$. This defines
$A_{n}^{\alpha}$ for each $n\in\omega,\alpha\in\omega_{1}$.
Set $\beta(\alpha)$ to be
$\sup\\{\text{height}(y)\,:\,y\in\cup_{n\in\omega}A_{n}^{\alpha}\\}$, let $C$
be the closure in $\omega_{1}$ of
$\\{\beta(\alpha)\,:\,\alpha\in\omega_{1}\\}$, and $C^{\prime}$ be its derived
set (that is its limit points). By construction, if there is $y\in S_{\geq x}$
whose height in $S^{\downarrow}$ is in $C^{\prime}$, then $f$ is not
continuous at $y$ as there is a sequence of points in $S$ converging to $y$
whose images are $\geq\epsilon/4$ apart. But by Lemma 6.6 (b) (and the fact
that $S_{\geq x}$ is $\omega_{1}$-compact), there must be such an $y$, a
contradiction. This shows that $D(\epsilon)$ is order-dense and concludes the
proof. ∎

###  6.3 Proof of Theorem 6.2

.

Our proof relies on simple consequences of the properties given in Lemma 6.7,
especially (i). When $S$ is a subset of a tree $T$, the height of a point of
$S$ is to be understood as its height in $T$. Recall that $S_{\geq x}=T_{\geq
x}\cap S$.

######  Lemma 6.8.

Let $X$ be a space containing a closed metrizable $G_{\delta}$ subset
$B\subset X$, $S$ be an uncountable $\omega_{1}$-compact subset of a tree of
height $\omega_{1}$ and $f:S\to X$ be continuous. Then there is
$\alpha\in\omega_{1}$ such that either $f(S_{\geq x})\cap B=\varnothing$ or
$f(S_{\geq x})$ is a singleton whenever $x$ is at height $\geq\alpha$.

###### Proof.

By Lemma 6.7 (b), above some level each $S_{\geq x}$ is uncountable, we assume
for simplicity that this holds for each $x\in S$. Let $U_{n}$ be open sets
such that $\cap_{n\in\omega}U_{n}=B$. By Lemma 6.7 (h), there is
$\beta\in\omega_{1}$ such that for each $n$ we have

$\left(\,\left(f^{-1}(B))\right)^{\downarrow}\cap\left(f^{-1}(X-U_{n})\right)^{\downarrow}\right)\,-T_{\leq\beta}=\varnothing.$

It follows that if $x\in S$ is at level above $\beta$, either $f(S_{\geq
x})\subset B$ or $f(S_{\geq x})\cap B=\varnothing$. Let $E=\\{x\in
S\,:\,\text{\rm Lev}(x)\geq\beta\text{ and }f(S_{\geq x})\subset B\\}$. Then
$E$ is an upward closed subspace of $S$, in particular it is an
$\omega_{1}$-compact subspace of $T$. By Lemma 6.7 (i), there is
$\alpha\geq\beta$ such that $f$ is constant on $S_{\geq x}$ whenever $x\in E$
is at level $\geq\alpha$. This proves the lemma. ∎

######  Corollary 6.9.

Let $X$ be a space and $U\subset X$ be open such that $\overline{U}$ is
contained in a metrizable open $V\subset X$. Let $S$ be an uncountable
$\omega_{1}$-compact subset of a tree of height $\omega_{1}$. Let
$h:S\times[0,1]\to X$ be continuous. Then there is $\alpha\in\omega_{1}$ such
that for each $t\in[0,1]$ and each $x\in S$ above level $\alpha$, either
$h_{t}^{-1}(U)\cap S_{\geq x}=\varnothing$, or $h_{t}$ is constant on $S_{\geq
x}$.

###### Proof.

Let $\\{t_{n}\,:\,n\in\omega\\}$ be a countable dense subset of $[0,1]$. Set
$B=\overline{U}$. Since $B$ is closed in the metrizable subset $V$, it is a
$G_{\delta}$. The previous lemma shows that there is some $\alpha$ such that
when $x$ is above level $\alpha$, either $h_{t_{n}}^{-1}(B)\cap S_{\geq
x}=\varnothing$ or $h_{t_{n}}$ is constant on $S_{\geq x}$ for each
$n\in\omega$. The result follows by continuity. ∎

###### Proof of Theorem 6.2.

We first show that we may assume that $T$ is an Aronszajn tree. Indeed,
consider $S^{\downarrow}\subset T$; if some level of $S^{\downarrow}$ is
uncountable, then $S$ is not $\omega_{1}$-compact (by Lemma 6.6 (c) & (d)). If
$S^{\downarrow}$ contains an uncountable branch $E$, then $E$ is
$\omega_{1}$-compact in the subspace topology and thus homeomorphic to a
stationary subset of $\omega_{1}$, and we may apply Theorem 6.4 (whose proof
given later is independent from this one).
Let $h:S\times[0,1]\to X$ be continuous such that $h_{0}(x)=u_{0}\in X$ and
$h_{1}:S\to X$ has uncountable image. The set of $x\in S$ such that $S_{\geq
x}$ has uncountable image under $h_{1}$ is uncountable and downward closed,
hence by Lemma 6.7 (c) there is $x$ such that the image of $S_{\geq z}$ under
$h_{1}$ is uncountable for each $z\geq x$. Up to replacing $S^{\downarrow}$ by
$(S^{\downarrow})_{\geq x}$ we assume that this holds for all $z\in
S^{\downarrow}$. For $x\in S$, set

$\tau(x)=\sup\\{t\,:\,h_{t}\text{ is constant on }S_{\geq x}\\}.$

Then $\tau$ is an increasing map $S\to\mathbb{R}$, however $\tau$ is a priori
neither continuous nor strictly increasing. We will show that there is a
closed unbounded $C\subset\omega_{1}$ such that $\tau$ is strictly increasing
on the subspace of members of $S$ at levels belonging to $C$. Such a subspace
is uncountable (and thus unbounded) in $S$ by Lemma 6.7 (f). Hence as a
partially ordered space $S$ contains an $\mathbb{R}$-special tree and thus (at
least) an uncountable antichain (see for instance [29, Thm 4.29]), a
contradiction with the fact that $S$ is $\omega_{1}$-compact.
It is enough to show that for each $x\in S$, there is $\alpha$ such that
$\tau(y)>\tau(x)$ whenever $y>x$ is at level $\geq\alpha$. Indeed, since the
levels of $S^{\downarrow}$ are countable by Lemma 6.7 (a) and $\tau$ is
increasing, a simple induction provides $C$. Let $x\in S$ be fixed. By
continuity, $h_{\tau(x)}$ is constant on $S_{\geq x}$ with value
$u=h_{\tau(x)}(x)$ and thus $\tau(x)<1$ since $h_{1}(S_{\geq x})$ is
uncountable. (While it is not needed, notice that $h$ restricted to the
subspace $S_{\geq x}\times[\tau(x),1]$ ’contracts’ all of $S_{\geq x}$ to the
point $u$.) Since $X$ is locally metrizable, we may choose an open $U\ni u$
such that $B=\overline{U}$ is contained in an open metrizable set. Let
$\alpha$ be given by Corollary 6.9. We may assume that
$\alpha>\text{height}(x)$. Assume that there is $y$ at level above $\alpha$
such that for each $t>\tau(x)$, $h_{t}$ is not constant on $S_{\geq y}$. By
definition of $\alpha$, this implies that $h_{t}(S_{\geq y})\cap
U=\varnothing$ and in particular that $h_{t}(y)\not\in U$ for each
$t>\tau(x)$. But this contradicts the continuity of $h$ since
$h_{\tau(x)}(y)=u\in U$. Hence, $h_{t}$ is constant on $S_{\geq y}$ for at
least one $t>\tau(x)$ and thus $\tau(y)>\tau(x)$. This finishes the proof. ∎

Theorem 1.4 follows almost immediately.

###### Proof of Theorem 1.4.

Suppose that $W$ is a CW-complex and $f:M\to W$, $g:W\to M$ are such that
$g\circ f$ is homotopic to $id_{M}$. If $f(S)$ meets uncountably many open
cells, then $S$ contains an uncountable closed discrete subset (see for
instance [16, Theorem 1.5.19]) which contradicts its $\omega_{1}$-compactness.
Hence $f(S)$ meets the interior of at most countably many cells and is
therefore Lindelöf. It follows that $g\circ f(S)$ is contained in a Lindelöf
hence metrizable submanifold of $M$. By Lemma 6.7 (i), $g\circ f(S)$ is
constant on $S_{\geq x}$ if $x$ is above some fixed level $\alpha$ in $T$.
Since $S_{\geq x}$ is uncountable for some $x$, Theorem 6.2 yields a
contradiction. ∎

###  6.4 Proof of Theorem 6.4

The proof is almost exactly the same as that of Theorem 6.2 once we have an
equivalent of Corollary 6.9 in this context. This is given by Corollary 6.12
below. Lemma 6.11 below plays the role of Lemma 6.7 (i). We first state the
following easy fact.

######  Lemma 6.10.

Let $S\subset\omega_{1}$ be stationary, endowed with the subspace topology.
Then an at most countable family of club subsets of $S$ has a club
intersection.

###### Proof.

A direct proof is straightforward, but notice that since $S$ is an
$\omega_{1}$-compact subspace of the tree $\omega_{1}$, the result is also a
consequence of Lemma 6.7 (f)–(g). ∎

######  Lemma 6.11.

Let $S$ be a stationary subset of $\omega_{1}$ endowed with the subspace
topology. If $Y$ is regular, $[\aleph_{0},\aleph_{1}]$-compact and has
$G_{\delta}$ points, then any continuous $f:S\to Y$ is eventually constant,
that is, there is $\alpha\in\omega_{1}$ such that $f(\beta)=f(\alpha)$ for
each $\beta\geq\alpha$, $\beta\in S$.

###### Proof.

We start by showing that there is some $c\in Y$ such that $f^{-1}(\\{c\\})$ is
club in $S$. If $f(S)$ is countable, then this is immediate. We thus assume
that $f(S)$ is uncountable. Since $Y$ is $[\aleph_{0},\aleph_{1}]$-compact,
$f(S)$ has a condensation point $c\in Y$. Since $Y$ is regular and has
$G_{\delta}$ points, we may choose open sets $U_{n}\ni c$, $n\in\omega$, such
that $\cap_{n\in\omega}U_{n}=\cap_{n\in\omega}\overline{U_{n}}=\\{c\\}$. Since
$c$ is a condensation point, $f^{-1}(\overline{U_{n}})$ is club in $S$ for
each $n$, hence
$f^{-1}(\\{c\\})=f^{-1}(\cap_{n\in\omega}\overline{U_{n}})=\cap_{n\in\omega}f^{-1}(\overline{U_{n}})$
is club in $S$ by Lemma 6.10.
Now, since $f^{-1}(Y-U_{n})$ is closed, it must be bounded, otherwise it
intersects $f^{-1}(\\{c\\})$. It follows that
$f^{-1}(Y-\\{c\\})=\cup_{n\in\omega}f^{-1}(Y-U_{n})$ is bounded in $S$, say by
$\alpha$. Hence $f$ is constant on $S$ above $\alpha$. ∎

######  Corollary 6.12.

Let $S$ be a stationary subset of $\omega_{1}$ endowed with the subspace
topology. Let $Y$ be a regular space with $G_{\delta}$ points. Let $U\subset
Y$ be open such that $\overline{U}$ is $[\aleph_{0},\aleph_{1}]$-compact. Let
$h:S\times[0,1]\to Y$ be continuous. Then there is $\alpha\in\omega_{1}$ such
that for each $t\in[0,1]$ either $h_{t}^{-1}(U)\cap[\alpha,\omega_{1})\cap
S=\varnothing$, or $h_{t}$ is constant on $[\alpha,\omega_{1})\cap S$.

###### Proof.

Fix a countable dense subset $\\{t_{n}\,:\,n\in\omega\\}\subset[0,1]$. By
Lemmas 6.10–6.11 we may fix $\alpha$ such that for each $n$ either
$[\alpha,\omega_{1})\cap S\cap h_{t_{n}}^{-1}(\overline{U})=\varnothing$ or
$h_{t_{n}}$ is constant above $\alpha$. The result follows by continuity. ∎

The proof of Theorem 6.4 can now be done exactly along the same lines as that
of Theorem 6.2, we thus only give a sketch. Set $\tau=\sup\\{t\,:\,h_{t}\text{
is eventually constant}\\}$. Then $h_{\tau}$ is eventually constant and
$\tau<1$ since $h_{1}$ has uncountable image. Fix $\alpha$ minimal in $S$ such
that $h_{\tau}$ is constant above $\alpha$, take an open $U$ containing
$h_{\tau}(\alpha)$ such that $\overline{U}$ is
$[\aleph_{0},\aleph_{1}]$-compact. By Corollary 6.12 this contradicts the
continuity of $h$.

##  7 Surfaces from trees

The goal of this section is to define a surface $M_{T}$ homotopy equivalent to
the road space $R_{T}$ of a given rooted binary $\omega_{1}$-tree, and to
check some additional properties involving the mapping class group. We recall
that we assume that our trees are Hausdorff. (Everything in this section could
be done starting from any rooted $\omega_{1}$-tree, but for simplicity we
shall consider only binary ones.)

###  7.1 A surface homotopy equivalent to the road space of a tree

This construction was found and explained to us by Peter Nyikos in 2006. We
fix a binary $\omega_{1}$-tree $T$. Let us define the manifold $M_{T}$. As a
set, $M_{T}=\cup_{x\in T}M_{x}$, where each piece $M_{x}$ is the union of a
right isoceles triangle with sides $1,1,\sqrt{2}$ and two rectangles with
sides $1,\frac{1}{2}$ glued together as in Figure 5 (dashed boundaries do not
belong to $M_{x}$).

Figure 5: The ‘pieces’ $M_{x}$.

The topology in the interior of $M_{x}$ is the usual topology. We shall need
an arbitrary total order $\vartriangleleft$ on $T$. If $y,y^{\prime}$ are the
two immediate $\leq$-successors of $x$ in $T$ and $y\vartriangleleft
y^{\prime}$, we glue $M_{y}$ and $M_{y^{\prime}}$ on the top of $M_{x}$ (after
multiplying them by $1/\sqrt{2}$) as in Figure 6 below (we put the
$\vartriangleleft$-bigger on the right), which gives a topology for points in
the bottom boundary of $M_{x}$ for $x$ at successor levels.

Figure 6: Gluing successor levels

If now $p\in M$ belongs to the bottom boundary of $M_{x}$ for $x$ at a limit
level $\alpha$ of $T$, a neighborhood of $p$ is given by a half ball in
$M_{x}$ of diameter $\epsilon$ centered at $p$, together with a ‘zigzag’
running between some level $\gamma<\alpha$ and $\alpha$ which follows $p$ in
$M_{T}$, that is, the union of open strips of width $\epsilon$ in $M_{y}$ for
all $y<x$ in levels $\geq\gamma$ (see Figure 7), choosing the strip that
points eastwards or westwards according to $\vartriangleleft$. Then, to make
it open, remove the part that lies in the bottom boundary of $M_{y}$, for
$y\in\text{\rm Lev}_{\gamma}(T)$.

Figure 7: Neighborhoods at limit levels

This gives $M_{T}$ a structure of Type I surface without boundary with
canonical cover given by $M_{\alpha}=\cup_{\beta<\alpha}\cup_{x\in\text{\em
Lev}_{\beta}(T)}M_{x}$. (Strictly speaking, there is a remaining boundary in
$M_{0}$, $0$ the root of $T$, but we remove it by adding an extra open strip
below it.) Since a countable increasing union of $2$-discs is homeomorphic to
a $2$-disc (the famous M. Brown theorem, here in dimension $2$, see for
instance [18, Section 3.1]), by induction $M_{\alpha}$ is homeomorphic to a
$2$-disc, and $M_{T}$ is a surface. $T$ and $R_{T}$ embed in $M_{T}$ as seen
in Figure 8 (where $R_{x}=R\cap M_{x}$). It is clear that all homotopy groups
of $M_{T}$ are trivial, because the image of a compact set by a continuous map
will be contained in some $M_{\alpha}$. Notice that $T$ is closed in $M_{T}$
but $R_{T}$ is not: if $x$ is at a limit level in $T$, then at least one of
the left or right half part of the bottom boundary of $M_{x}$ is in the
closure of the segments in $R_{T}$ emanating from the points of $T$ below $x$.

Figure 8: $T$ and $R_{T}$ in $M_{T}$.

######  Theorem 7.1.

(P. Nyikos) The inclusion $R_{T}\subset M_{T}$ is a homotopy equivalence.

This theorem is a direct consequence of the following lemma (see Exercise 4 p.
18 in [22]):

######  Lemma 7.2.

There is a deformation retraction in the weak sense from $M_{T}$ to $R_{T}$,
i.e. there is a homotopy $h:M_{T}\times[0,1]\to M_{T}$ with $h_{0}=\text{\rm
id}$, $h_{1}(M_{T})=R_{T}$ and $h_{t}(R_{T})=R_{T}$ for all $t\in[0,1]$.

The geometric features of $M_{T}$ will be used in the proof. It is interesting
to note that it is not possible to obtain a (real) deformation retraction
(i.e. $h_{t}\upharpoonright R_{T}=\text{\rm id}_{R_{T}}$ for each $t$), in
fact there is even no retraction $h:M_{T}\to R_{T}$ (i.e. with
$h\upharpoonright R_{T}=\text{\rm id}_{R_{T}}$), since $R_{T}$ is not closed
in $M_{T}$.

###### Proof of Lemma 7.2..

The homotopy we shall describe is the same in every piece $M_{x}$ of $M_{T}$,
and acts symmetrically on $M_{x}$, we thus only need to explain what happens
in its left part (see Figure 9). Let us first describe how the homotopy works
globally. The points in $R_{x}$ will stay in it, those on the segment between
$x$ and $x+1/2$ (see Figure 9, left, for the definition) will be all
contracted to the point $x$ while the segment between $x+1/2$ and the next
level will eventually cover all of $R_{x}$. Meanwhile, the south-west and
north-east ‘sides’ will shrink at unit speed to $R_{x}$, and the triangle will
be completely contracted to $x$ (see Figure 9, on the right).

Figure 9: The homotopy globally seen.

Let us now be more specific and describe the trajectories of the points under
the homotopy. First, the triangle. The points at height $x+1/2$ start moving
at once, while the points in the triangle start moving when the points at
height $x+1/2$ ‘reach them’. Then, they go at speed $\frac{1}{2}$ parallel to
$R_{x}$ until they hit the bottom or the vertical line above $x$, and then
they head directly toward $x$ at speed $\frac{\sqrt{2}}{2}$ (see Figure 10, on
the left). Hence, $h_{1}(p)=x$ for all points in the triangle. Points in the
rectangle have one velocity component parallel to $R_{x}$ which is
proportional to the distance to the next level (that is, between $0$ and
$\frac{1}{2}$), the other component is at speed $\frac{1}{2}$ and goes
perpendicular to $R_{x}$. The first component starts immediately, but the
second one only when the south-west or the north-east side reaches it (see
Figure 10, on the right).

Figure 10: Trajectories under $h_{t}$.

The homotopy $h_{t}$ described above is continuous with $h_{0}=id$ and
$h_{1}(M_{T})=R_{T}$. ∎

Theorem 1.2 follows immediately:

###### Proof of Theorem 1.2.

Let $T$ be a binary $\mathbb{R}$-special $\omega_{1}$-tree, and set $M=M_{T}$.
Then, by Theorems 7.1 and 6.1, $M$ is a Type I contractible surface. ∎

Figure 11: Lemma 7.3.

The tree $T$ embeds in $M_{T}$ as a closed subset. If $T$ is
$\mathbb{Q}$-special (that is, there is a strictly increasing function
$T\to\mathbb{Q}$), then $T$ is normal in some models of set theory, for
instance under MA + $\neg$CH, see e.g. [29]. We might hope that normality
‘lifts up’ to $M_{T}$, but this is never the case if $T$ is Aronszajn.

######  Lemma 7.3.

If $T$ is an Aronszajn tree, then $M_{T}$ is not normal.

The proof is reminiscent of Lemma 4.9 in [28].

###### Proof.

For each $x\in T$, let $x^{-},x^{+}\in M_{T}$ be defined as on the
lefthandside of Figure 11. Define $E^{\pm}=\\{x^{\pm}\,:\,x\in T\\}$, then
$E^{+},E^{-}$ are closed subsets of $M_{T}$ and have empty intersection. Let
$U^{+}\supset E^{+},U^{-}\supset E^{-}$ be open. For each $x\in T$, choose
$U^{+}_{x},U^{-}_{x}$ to be basic connected open subsets of $U^{+},U^{-}$
containing respectively $x^{+}$ and $x^{-}$. If $x$ is at limit level
$\alpha$, we may assume that $U^{+}_{x},U^{-}_{x}$ are union of strips as in
Figure 7, hence there is $\gamma(x)<\alpha$ such that $y^{+}\in U^{+}_{x}$ and
$y^{-}\in U^{-}_{x}$, where $y=x\upharpoonright\gamma(x)$. By the pressing
down lemma for $\omega_{1}$-trees (see e.g. [21, p. 154]), there is
$\gamma\in\omega_{1}$ and $S\subset T$ meeting a stationary set of levels such
that $\gamma(x)=\gamma$ for each $x\in S$. Since $\text{\rm Lev}_{\gamma}(T)$
is countable, there is some $y\in T$ at level $\gamma$ such that $y^{+}\in
U^{+}_{x}$ and $y^{-}\in U^{-}_{x}$ for $x$ in an uncountable set
$S^{\prime}\subset T$. It follows that both $V^{+}=\cup_{x\in
S^{\prime}}U^{+}_{x}$ and $V^{-}=\cup_{x\in S^{\prime}}U^{-}_{x}$ are
unbounded connected subsets of $U^{+}$ and $U^{-}$, respectively. Since $T$ is
Aronzsajn and $S^{\prime}$ is uncountable, there are $x,y,z$ in $S^{\prime}$
with $x<y,x<z$. Since $T$ is Hausdorff there is $w$ at maximal level such that
$w<y,w<z$. By connectedness, either $V^{+}$ intersects $V^{-}$ below the level
of $w$, or they intersect in $M_{w}$, as shown on the righthandside of Figure
11. This shows that $U^{+}\cap U^{-}\not=\varnothing$, hence $M_{T}$ is not
normal. ∎

###  7.2 Isotopies and the mapping class group

Our goal is to prove the following theorem.

######  Theorem 7.4.

(1) The mapping class group of the Prüfer surface $\mathsf{P}$ (which is
contractible) has cardinality $\geq\mathfrak{c}$.
(2) There is a contractible Type I surface with mapping class group of
cardinality $\geq\mathfrak{c}$.

Recall that two homeomorphisms are isotopic if there is a homotopy $h_{t}$
between them which is an homeomorphism for each $t\in[0,1]$. The mapping class
group $\text{\rm MG}(M)$ of a topological space $X$ is the group of
homeomorphisms of $X$ quotiented by the isotopy classes. Theorem 7.4 shows a
sharp contrast between metrizable and non-metrizable surfaces: two
homeomorphisms of a metrizable surface are homotopic if and only if they are
isotopic (except for some trivial exceptions). To our knowledge this is due to
D.B.A Epstein [15].

We shall prove (2) in detail, and then explain why (1) holds as well. We first
recall some definitions from [27]. Given a non-metrizable Type I manifold $M$
with canonical cover $\langle M_{\alpha}\,:\,\alpha\in\omega_{1}\rangle$, the
tree of non-metrizable-component boundaries $\Upsilon(M)$ is defined as the
set of topological boundaries $\partial C$ such that $C$ is a non-metrizable
connected component of $M-M_{\alpha}$ for some $\alpha\in\omega_{1}$, with the
following order: $\sigma\geq\tau$ if $\sigma$ is a subset of a component whose
boundary is $\tau$. (Of course, $\Upsilon(M)$ depends on the choice of the
canonical cover, but this is not inconsequential for our considerations.)
Recall that an $\omega_{1}$-tree $T$ is normalized if $T_{\geq x}$ is
uncountable for each $x\in T$. Since we only take non-metrizable components,
$\Upsilon(M)$ is normalized, and if one starts with a normalized $T$, then
$\Upsilon(M_{T})=T$ for the surface $M_{T}$ constructed in Section 7.1.

Given a tree $T$, we say that a map $T\to T$ is an automorphism of $T$ if it
is an order preserving bijection. A map $f:M\to M$ of a Type I manifold $M$
(with some given canonical cover) is a $\Upsilon$-automorphism if it is a
homeomorphism of $M$ such that for all $\sigma\in\Upsilon(M)$,
$f(\sigma)=\tau$ for some $\tau\in\Upsilon(M)$, and if the induced map
$\widehat{f}:\Upsilon(X)\to\Upsilon(X)$ is an automorphism. Being a
$\Upsilon$-automorphism is a strong assumption, however it is not difficult to
see that any homeomorphism of a Type I manifold induces an
$\Upsilon$-automorphism of $\Upsilon(M)$ for a well choosen canonical cover.

######  Lemma 7.5.

Let $M=\cup_{\alpha\in\omega_{1}}M_{\alpha}$ be a Type I manifold. If
$f,g:M\to M$ are $\Upsilon$-automorphisms such that
$\widehat{f}\not=\widehat{g}$, then $f$ and $g$ are non isotopic.

We shall not give the shortest proof of this lemma but rather an argument
which can be easily adapted for the Prüfer surface.

###### Proof.

Let $h_{t}$ be a homotopy such that $h_{0}=f$, $h_{1}=g$. Let
$\sigma\in\Upsilon(M)$ be such that
$\widehat{f}(\sigma)\not=\widehat{g}(\sigma)$. We will show that $h_{t}$
cannot be an isotopy. Since $\widehat{f},\widehat{g}$ are automorphisms,
$\widehat{f}(\tau)\not=\widehat{g}(\tau)$ for all $\tau\geq\sigma$ in
$\Upsilon(M)$, and both $\widehat{f}(\sigma)$ and $\widehat{g}(\sigma)$ belong
to the same level $\gamma$ of $\Upsilon(M)$. By definition of $\Upsilon(M)$,
there is some $\alpha$ such that $\widehat{f}(\tau)$ and $\widehat{g}(\tau)$
are in different connected components of $M-M_{\alpha}$ for all $\tau>\sigma$.
Thus, for $\tau\geq\sigma$ and all $y\in f(\tau)\subset M$, there is an open
(connected) interval $I_{y}\subset[0,1]$ such that $h_{t}(y)\in M_{\alpha}$
for $t\in I_{y}$. Now, since $\Upsilon(M)$ is normalized, for all
$\beta\in\omega_{1}$ with $\beta>\gamma$, we may choose $\tau_{\beta}$ in the
$\beta$-level of $\Upsilon(M)$ and $y_{\beta}$ in $f(\tau_{\beta})$. Since
$[0,1]$ is second countable, there is an interval $J$ contained in
$I_{y_{\beta}}$ for uncoutably many $\beta$. Thus, for $t\in J$, $h_{t}$ sends
an unbounded and therefore non-metrizable subset of $M$ inside $M_{\alpha}$,
which is metrizable. Hence, $h_{t}$ cannot be a homeomorphism, and $h_{t}$ is
not an isotopy. ∎

###### Proof of Theorem 7.4..

We first prove (2). Let $T$ be any rooted binary $\omega_{1}$-tree and
$M_{T}\supset T$ be as in Section 7.1. Take now copies $T_{i}\subset M_{i}$ of
$T,M_{T}$ for each $i\in\mathbb{Z}$. Build the manifold $M^{\omega}$ by gluing
all these $M_{i}$ on a open disc as in Figure 12 below. Let $T^{\omega}$ be
$\cup_{i\in\mathbb{Z}}T_{i}$ together with a new root $x$ as shown in Figure
12. For $m\in\mathbb{Z}$ let $\varphi_{m}:M^{\omega}\to M^{\omega}$ be the
homeomorphism acting by ‘rotating’ $M_{i}$ to $M_{i+m}$. It is clear that
$\varphi_{m}$ is a $\Upsilon$-automorphism of $M^{\omega}$.

Figure 12: $M^{\omega}$.

We now define $M=M^{\omega,\omega}$ with copies $M_{j}^{\omega}$ of
$M^{\omega}$ for $j\in\mathbb{Z}$ in the same way. Again,
$\Upsilon(M)=T^{\omega,\omega}$, where $T^{\omega,\omega}$ is the
$\omega_{1}$-tree obtained from the $T^{\omega}_{i}$ by adding a common root.
For $s\in\mathbb{Z}^{\mathbb{Z}}$, let $\phi_{s}:M\to M$ be given by
$\phi_{s}\upharpoonright M^{\omega}_{j}=\varphi_{s(j)}:M^{\omega}_{j}\to
M^{\omega}_{j}$ and $\phi_{s}(u)=u$ if
$u\not\in\cup_{j\in\mathbb{Z}}M^{\omega}_{j}$. It is clear that $\phi_{s}$ is
an $\Upsilon$-automorphism. By Lemma 7.5, $s\mapsto[\phi_{s}]\in\text{\rm
MG}(M)$ is one–to–one, hence $\text{\rm MG}(M)$ has cardinality at least
$\mathfrak{c}$. If $T$ is a rooted binary $\mathbb{R}$-special tree, then $M$
is contractible.
We now briefly explain how to prove (1). We use the notations of Section 5 for
everything related to the Prüfer surface $\mathsf{P}$. Given $a\in\mathbb{R}$,
let $f_{a}:\mathsf{P}\to\mathsf{P}$ send all of $\mathbb{R}_{c}$ to
$\mathbb{R}_{c+a}$ pointwise, and be defined by $\langle
x,y\rangle\mapsto\langle x,y+a\rangle$ in $H_{0}$. Then each $f_{a}$ is a
homeomorphism. For each $c\in\mathbb{R}$ take the $0$-point
$0_{c}\in\mathbb{R}_{c}$. The proof of Lemma 7.5 can be easily adapted to show
that if $h$ is an homotopy between $f_{a}$ and $f_{b}$ (with $a\not=b$), then
there is some $t$ such that $h_{t}$ sends $0_{c}$ inside $H_{0}$ for an
uncountable subset $A\subset\mathbb{R}$ of $c$. Recall that $\\{0_{c}\,:\,c\in
A\\}$ is closed discrete for any $A$. Since $H_{0}$ is hereditarily separable,
the image of these points cannot be closed discrete, so $h_{t}$ is not an
homeomorphism and $h$ is not an isotopy. This shows that the map
$\mathbb{R}\to\text{\rm MG}(\mathsf{P})$ defined by $a\mapsto[f_{a}]$ is one-
to-one. ∎

The construction described in the proof of Point (2) but using the first
octant in the square of the longray instead of $M_{T}$ yields an
$\omega$-bounded surface with mapping class group of cardinality
$\geq\mathfrak{c}$, see [18, Thm 5.31].

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