3.1 Step 1: Fragmentation

We first fragment an element of $G({\Omega})$ into two maps that have fixed points in the orbit of x.

Note that ${\Omega}$ is a stable neighborhood with respect to any point in the $G({\Omega})$ -orbit of x. Thus, the previous lemma allows us to reduce our problem to only considering maps in $\mathrm{Homeo}({\Omega},x)$ . Next we fragment again, this time into bricks.

Proof. Choose indices $n_k$ and $m_k$ , where $m_k \in [n_{k-1}+1, n_k-1]$ inductively as follows. Let $n_0~=~0$ and $m_1=1$ . Then, assuming $n_1, \dots, n_{k-1}$ and $m_1, \dots, m_k$ are given, choose $n_k \gt m_k$ large enough so that

\[f^{\pm 1}(Y_{m_k}) \subset \bigsqcup_{i=n_{k-1}+1}^{n_k-1} Y_i \]

and then choose $m_{k+1}$ large enough so that

\[f^{\pm 1}(Y_{m_{k+1}}) \cap \Big(\bigsqcup_{i=1}^{n_k} Y_i \Big) = \emptyset. \]

Define

\[\mathcal {A}_m = \bigsqcup_{k=1}^\infty Y_{m_k}\quad \text{and}\quad \mathcal {A}_n = \bigsqcup_{k=1}^\infty Y_{n_k}. \]

Let $\mathcal {A}_m'$ and $\mathcal {A}_n'$ be complementary bricks to $\mathcal {A}_m$ and $\mathcal {A}_n$ , respectively. Note that $f(\mathcal {A}_m)$ is disjoint from $\mathcal {A}_n$ . Therefore, there is a homeomorphism $g:X \to X$ with support in $\mathcal {A}_n'$ such that $f|_{\mathcal {A}_m} = g|_{\mathcal {A}_m}$ . In particular, $h=g^{-1}f$ preserves the brick $\mathcal {A}_m$ . But g preserves the brick $\mathcal {A}_n$ and $f = gh$ . This finishes the proof.

Thus, combining these two lemmas, we see that any map $h \in G({\Omega})$ can be fragmented as $h = h_{1}h_{2}h_{3}h_{4}$ , where each $h_{i}$ is in $G({\Omega})$ and is supported on a brick (although not necessarily centered around the same point). Our new goal is to show that any map supported on a brick satisfies a Steinhaus condition.

3.2 Step 2: Finding commutators

3.3 Step 3: Diagonalization

Next we will find a ‘good’ subbrick. Namely, any map supported on this subbrick will be in $W^{8}$ . First, we see that we can find a subbrick on which we can approximate maps. A similar step is carried out in [Reference Rosendal and SoleckiRS07, Reference RosendalRos08, Reference MannMan16, Reference MannMan24].

1. (i) there is some $i\geq 1$ such that any homeomorphism of $\mathcal {A}_i$ extends to a homeomorphism of X supported on $\mathcal {A}$ that belongs to $g_iW$ ; and

2. (ii) moreover, every homeomorphism of $\mathcal {A}_i$ extends to a homeomorphism of X supported on $\mathcal {A}$ that belongs to $W^2$ .

Next we upgrade this approximation.

The support of u is contained in $\mathcal {C}$ , and so we can apply Lemma 3.13 to find a $\bar{u}$ such that $\bar{u} \in W^{2}$ , $\bar{u}$ is supported on $\mathcal {B}$ , and $\bar{u}\vert_{\mathcal {C}} = u$ . Similarly, the support of v is contained in $\mathcal {C} \subset \mathcal {B}$ . Lemma 3.13 implies that there exists $\bar{v}$ such that $\bar{v} \in W^{2}$ , $\bar{v}$ is supported on $\mathcal {A}$ , and $\bar{v}\vert_{\mathcal {B}}= v$ . In particular, $\bar{v}$ is the identity in $\mathcal {B} - \mathcal {C}$ .

Since $\bar{u}$ is the identity in $X - \mathcal {B}$ and $\bar{v}$ is the identity in $\mathcal {B} - \mathcal {C}$ , we have $[\bar{u},\bar{v}]$ is the identity in $X-\mathcal {C}$ . Thus, we have $[\bar{u},\bar{v}]= [u,v] = h$ and hence $h \in W^{8}$ .

3.4 Step 4: Pigeonhole

In the previous step we found a ‘good’ subbrick. Next we will see that we can conjugate any map supported on any given brick by maps in W in order to have support on this good brick (up to a finite amount of error). This will allow us to prove that any map supported on a brick is in $W^{24}$ .

To prove the claim, first note that the support of F is contained in $\bigcup_{i=1}^\infty f_\beta^{-1}f_\alpha(A_i)$ . Recall that $\mathcal {A}$ is disjoint from $\mathcal {Z}$ and hence from $\mathcal {Z}_\alpha$ , so $f_\alpha(A_i)$ is contained in $\mathcal {Z}_\alpha$ for all i. Furthermore, if i is sufficiently large, then $f_\alpha(A_i)$ will be disjoint from $\mathcal {Z}_\beta$ , since $\mathcal {Z}_\alpha\cap \mathcal {Z}_\beta$ is disjoint from a neighborhood of x. It follows that, for all sufficiently large i, $f_\beta^{-1}f_\alpha(A_i)\subset Z_\beta\subset Z$ , which proves the claim.

Thus, we can write $F = F_{1}F_{2}$ , where $F_{1}$ is supported on finitely many of the $Y_{k}$ , $\bigsqcup_{k=1}^{n} Y_{k}$ , and $F_{2} \in G(\mathcal {Z})$ . By Lemma 3.14 we have that $F_{2} \in W^{8}$ . It only remains to check that $F_{1}$ is in $W^{12}$ . By the property (iii) of Proposition 3.2, there is a homeomorphism $g \in G({\Omega})$ that maps $\bigsqcup_{i=1}^{n} Y_{i}$ into $\mathcal {Z}$ . Since $W^{2}$ is dense in $G({\Omega})$ , we take $\tilde{g} \in W^{2}$ with this same property (this $\tilde{g}$ may not fix x and may not belong to $G({\Omega})$ ). Thus, we have that $gF_{1}g^{-1} \in G(\mathcal {Z}) \subset W^{8}$ and therefore $F_{1} \in W^{12}$ .

Therefore, we have $F = (f_\beta^{-1}f_\alpha)h(f_\beta^{-1}f_\alpha)^{-1} \in W^{20}$ and so $h \in W^{24}$ .

3.5 Step 5: Wrapping up

We will first verify a version of the Steinhaus condition for stable neighborhoods.

• – G is covered by countably many sets $g_iW$ , $g_i\in G$ , and

• – $W^2$ is dense in $G({\Omega})$ .

Then $G({\Omega})\subseteq W^{96}$ .

Finally, we can upgrade this to the case that X is stable. We first need a short lemma on stable spaces.

We will need a slight extension of Lemmas 3.13–3.15 and Theorem 3.16.

Suppose we have a finite collection of distinct points $x_i\in X$ , $i=1,\ldots,N$ , and pairwise disjoint clopen subsets $\Omega_i$ , $i=1,\ldots,N$ , so that $\Omega_i$ is a stable neighborhood of $x_i$ for all i. We will abbreviate this by $\vec {\Omega}$ and $\vec x\in \vec {\Omega}$ . A multibrick $\vec{\mathcal {B}}$ is a collection of bricks $\mathcal {B}_i$ in each ${\Omega}_i$ centered at $x_i$ . The group $\mathrm{Homeo}(\vec{\Omega})$ is the group of homeomorphisms of $\sqcup {\Omega}_i$ preserving each ${\Omega}_i$ , and $\mathrm{Homeo}(\vec {\Omega},\vec x)$ is the subgroup that in addition preserves each $x_i$ .

We are finally ready to prove the main theorem of this section.

Proof of Theorem 3.1. We verify the Steinhaus property. Let W be a symmetric set in $G=\mathrm{Homeo}(X)$ so that G is the union of countably many translates of W. By Lemma 2.3, there is an open neighborhood $\mathcal {U}$ of the identity in G so that $W^2$ is dense in it. There is a finite clopen partition $A_1\sqcup\cdots\sqcup A_m$ of X so that any $g\in G$ that preserves each $A_i$ belongs to $\mathcal {U}$ . Now, applying Lemma 3.17 to each $A_i$ produces a finer clopen partition ${\Omega}_1\sqcup\cdots\sqcup {\Omega}_N$ of X, where each partition element is a stable neighborhood. This finer partition determines an open set $\mathcal {U}' \subset \mathcal {U}$ . Every $g \in \mathcal {U}'$ thus restricts on each subset in the partition to a map on a stable neighborhood. We can apply the Proposition 3.18 version of Theorem 3.16 to see that $g \in W^{96}$ .

4.1 Background

We recall some terminology and structural lemmas on end spaces of surfaces introduced in [Reference Mann and RafiMR23]. For a surface ${\Sigma}$ we denote by $E({\Sigma})$ the space of ends of ${\Sigma}$ . This space is defined as

\begin{align*} E({\Sigma}) = \varprojlim_{K} \pi_{0}({\Sigma} \setminus K),\end{align*}

where the inverse limit is taken over all connected, compact subsurfaces of ${\Sigma}$ . We denote by

\[E_{g}({\Sigma}) \subset E({\Sigma}) \]

the subset of $E({\Sigma})$ consisting of non-planar ends. The end space is equipped with the inverse limit topology. This makes it into a second-countable Stone space and $E_{g}({\Sigma})$ into a closed subset. When the underlying surface is fixed, we will often shorten the pair $(E({\Sigma}),E_{g}({\Sigma}))$ to just $E({\Sigma})$ .

Every homeomorphism of $\Sigma$ induces a homeomorphism of $E({\Sigma})$ that preserves $E_{g}({\Sigma})$ . Hence, when we talk about a homeomorphism between subsets of x, we always assume that our maps preserve the set $E_{g}({\Sigma})$ . That is, we think of points in $E_{g}({\Sigma})$ as being colored following the convention of § 3. Furthermore, this pair $(E({\Sigma}),E_{g}({\Sigma}))$ effectively classifies infinite-type surfaces.

Given a clopen subset $U \subseteq (E({\Sigma}),E_{g}({\Sigma}))$ , we will let ${\Sigma}_{U}$ denote a connected subsurface of ${\Sigma}$ , closed in ${\Sigma}$ , with either infinite or zero genus so that $\partial {\Sigma}_{U}$ is a single simple closed curve and the space of ends of ${\Sigma}_U$ , $E({\Sigma}_{U})$ , is U. Given a neighborhood U of an end $x \in E$ , we will say that ${\Sigma}_{U}$ is a neighborhood of x in the surface ${\Sigma}$ . Furthermore, if we have a pair $U \subset V$ , we will assume that $\Sigma_{U}$ and $\Sigma_{V}$ are chosen so that $\Sigma_{U} \subset \Sigma_{V}$ . Note that $\Sigma_U$ has infinite genus if and only if $U \cap E_{g}({\Sigma}) \not = \emptyset$ .

In the other direction, as in § 3, any homeomorphism $\phi : U \to V$ is assumed to send the set $U \cap E_{g}({\Sigma})$ to $V \cap E_{g}({\Sigma})$ . In particular, ${\Sigma}_U$ has infinite genus (or genus zero) if and only if ${\Sigma}_V$ does. Since they both have one boundary component, the same genus, and homeomorphic end spaces, they are homeomorphic by the classification of surfaces.

Abusing notation, we denote the partial order on the set of equivalence classes induced by $\preceq$ again by $\preceq$ .

We say that an end y is of Cantor type if there exists some neighborhood U of y such that $E(y) \cap U$ is homeomorphic to a Cantor set. Note that E(y) may not be a Cantor set if y is not maximal. For example, it is possible that only a finite number of points in $\overline E(y)$ are non-planar. However, note the following.

Proof. Let V be a stable neighborhood of y. For every $z \in V$ , E(z) intersects every stable neighborhood of y, hence $z \preceq y$ . This implies that either $z \sim y$ and $z \in E(y)$ or $y \not \preceq z$ and $z \not \in \overline{E(y)}$ . That is,

\[\overline{E(y)} \cap V = E(y) \cap V \]

and hence $E(y) \cap V$ is closed.

To see the second assertion, it is immediate from the definition that if y is of Cantor type, then E(y) is uncountable. In the other direction, since E(y) is uncountable and $E({\Sigma})$ is compact, there exists at least one point in E(y) that is an accumulation point of E(y). Moreover, because all stable neighborhoods of points in E(y) are homeomorphic to one another, every point in $E(y) \cap V$ is an accumulation point of E(y). This implies that $E(y) \cap V$ is a perfect set. Given that $E({\Sigma})$ is totally disconnected, $ E(y) \cap V$ forms a Cantor set, which means that y is of Cantor type.

Note that, for a maximal end x, saying x is of Cantor type is the same as saying E(x) is a Cantor set. This is because, for every point $y \in \overline{E(x)}$ , we have $x \preceq y$ and, since x is maximal, $y \preceq x$ .

Armed with these, we construct our example as follows. Starting with a sphere minus a Cantor set, we picture $S^2$ as a one-point compactification of $\mathbb{R}^2$ and the Cantor set C as the standard $1/3$ -Cantor set in $\mathbb{R}^2$ . That is, C is obtained from [0,1] by removing $2^n$ open intervals of size ${1}/{3^n}$ . We refer to these as the missing intervals.

We now decorate the Cantor set as follows. For every missing interval of size $1/{3^n}$ , place $\mathbb{Z}$ copies of disks $D_n$ . That is, remove a $\mathbb{Z}$ -family of disks from $\mathbb{R}^2$ that accumulate to the end points of the missing interval and replace them with homeomorphic copies of $D_n$ . Do this for every $n \geq 1$ and denote the resulting surface $\Sigma$ .

Now, for every n, $\Sigma$ has exactly $2^n$ ends that are accumulation points of $E(z_n)$ . We denote this set by $B_n$ . The union $B=\bigcup_n B_n$ is a countable and dense subset of the Cantor set C. We claim that all other points in the Cantor set are of the same type. That is, for $y \in C-B$ , we have $E(y) = C-B$ . To see this, we note that all intervals of type

\[\left[\frac {k}{3^n}, \frac{3k+1}{3^{n+1}} \right] \]

that are not contained in a missing interval are homeomorphic to each other by a horizontal translation and they are homeomorphic to intervals of type

\[\left[\frac {3k-1}{3^{n+1}}, \frac{k}{3^n} \right]\]

by a flip and a horizontal translation. We leave the details to the reader.

Now, for every $y \in C-B$ , $E(y)= C-B$ is uncountable but it is not locally closed. Note that, for every n, we can choose a small enough neighborhood of y that does not intersect $E(z_n)$ . Hence, the homeomorphism type of a small neighborhood of y does not stabilize and so $\Sigma$ is not stable.

This proposition arises from the following. The version we have written here is not exactly the same as was written in [Reference Mann and RafiMR23], but follows directly from their statements.

Proof of Proposition 4.9. Let ${\Sigma}$ be a stable surface. Using compactness of the end space of ${\Sigma}$ , we can cover $E({\Sigma})$ by a finite collection of stable neighborhoods $U_{1},\ldots,U_{n}$ . Now, by the previous lemma, each $U_{i}$ has only a single type of maximal end. Furthermore, each maximal end in $E({\Sigma})$ must appear as a maximal end of some $U_{i}$ . Therefore, we see that n is an upper bound on the number of maximal equivalence classes of ends.

Next, we recall the following.

We restate this lemma slightly.

Proof. For every $y \in E\setminus V$ , let $x \in V$ be an accumulation point of E(y). Let $V_x \subset V$ and $U_y \subset E\setminus V$ be as in Lemma 4.11. Then

\[V \cup U_y = (V\setminus V_x) \cup (V_x \cup U_y) \simeq (V\setminus V_x) \cup V_x \simeq V. \]

Since $E\setminus V$ is compact, it can be covered with finitely many such sets $U_{y_1}, \dots, U_{y_k}$ . Making these sets smaller, we can assume they are disjoint. Then, from the above argument, we have

\[E = (V \sqcup U_{y_1}) \sqcup U_{y_2} \sqcup \dots \sqcup U_{y_k} \simeq(V \sqcup U_{y_2}) \sqcup U_{y_3} \sqcup \dots \sqcup U_{y_k} \simeq\dots \simeq (V \sqcup U_{y_k} ) \simeq V. \]

This finishes the proof.

4.2 Telescoping and drawtubes

From now on, we will always be working under the assumption that ${\Sigma}$ is a stable surface.

1. (i) an isolated puncture;

2. (ii) of Cantor type;

3. (iii) not isolated in $E_{g}({\Sigma})$ and is a successor with all predecessors of Cantor type.

Let U be a stable neighborhood of x, let A be a tube in the interior of ${\Sigma}_U$ around x, and let A’ be the component of ${\Sigma}_U-A$ that does not contain x. We say A is a drawtube Footnote ¹ for the neighborhood U of x if both A and A’ contain all types of ends in U other than possibly x itself. If x is an isolated puncture, then ${\Sigma}_U$ is a punctured disk and every tube in ${\Sigma}_U$ around x is a drawtube.

Now we check some basic properties of stable neighborhoods and drawtubes relative to a telescoping end, as well as prove an alternate characterization of telescoping.

Let $V \subset U$ be a small neighborhood of x that is disjoint from E(A) and E(B). The end space E(A’) is a subset of the end space of $E(U-V)$ . Every $y \in E(U-V)$ is either a predecessor of x or accumulates to a predecessor of x. Either way, E(y) has an accumulation point in E(A’). Corollary 4.12 implies that E(A’) is homeomorphic to $E(U-V)$ . Also, since x is a stable end, ${\Sigma}_V$ is homeomorphic to $A \cup A''$ . Therefore, there is a homeomorphism $\phi_A$ of ${\Sigma}_U$ fixing x that sends $A \cup A''$ to ${\Sigma}_V$ and A’ to ${\Sigma}_{U}-{\Sigma}_V$ . Similarly, there is a homeomorphism $\phi_B$ of ${\Sigma}_U$ fixing x that sends $B \cup B''$ to ${\Sigma}_V$ and B’ to ${\Sigma}_{U}-{\Sigma}_V$ . Then $\phi_B^{-1}\phi_A$ sends A’ to B’.

Hence, we can assume $A' = B'$ . Now, applying the same argument, we can find a homeomorphism $\psi_A$ of $A \cup A''$ fixing x that sends A” to ${\Sigma}_V$ and a homeomorphism $\psi_B$ of $B \cup B''$ fixing x that sends B” to ${\Sigma}_V$ . Then $\phi_B^{-1}\phi_A$ sends A to B and we are done.

Lemma 4.16. Let x be an end and U a stable neighborhood of x. Then x is a telescoping end of ${\Sigma}$ if and only if there exists a subdivision

\[{\Sigma}_U = \bigcup_{i=0}^{\infty} Y_{i} \]

of ${\Sigma}_U$ into tubes $Y_i$ around x so that

1. (i) $Y_{i} \cap Y_{j} = \emptyset$ if $|i-j| \gt 1$ ,

2. (ii) $Y_{i}$ and $Y_{i+1}$ intersect only along one of their boundary components, and

3. (iii) for $0 \leq i \leq j$ , we have $\bigcup_{k=i}^j Y_k$ is homeomorphic to $Y_0$ .

In particular, each $E(Y_i)$ contains all types of ends in U other than possibly x itself, and for $i \geq 1$ , $Y_i$ is a drawtube.

Proof. Assume x is a telescoping end and let U be a stable neighborhood of x. Let

\[U=U_0 \supset U_1 \supset U_2 \supset\cdots \]

be a family of nested subsets of U that descends to x, $U_k \searrow x$ . We immediately note that if x is an isolated puncture, then we can take the collection $\{Y_{i}\}$ to be nested tubes that descend to x. From now on, we assume that x is not an isolated puncture. First we show that, for $k_1$ large enough, $U - U_{k_1}$ contains all types of ends in U other than possibly x itself.

There are two cases to consider. If x is of Cantor type then, for every end $y \in U$ , $E(y) \cap U$ accumulates to x and hence it accumulates to every point in $E(x) \cap U$ . So, we only need to take $k_1$ large enough so that $(U - U_{k_1}) \cap E(x)$ is not empty and this ensures that $(U - U_{k_1})$ contains all types of ends. If x is not of Cantor type, then it has finitely many predecessors $y_1, \dots ,y_n$ and they are all of Cantor type. Then, for every end $z \in U$ , E(z) accumulates to some $y_i$ and hence to all points in $E(y_i) \cap U$ . Hence, if $k_1$ is large enough so that $(U - U_k) \cap E(y_i)$ is not empty for $i=1, \dots, n$ , then $(U - U_k)$ contains all ends except possibly x itself.

We proceed in this way and choose indices $k_1, k_2, \dots$ such that $U_{k_i} - U_{k_{i+1}}$ contains all types of ends in U other than possibly x itself. Let ${\Sigma}_U$ be a stable neighborhood of x in ${\Sigma}$ and denote the boundary of ${\Sigma}_U$ by $\alpha_0$ . Choose a sequence of curves $\alpha_i$ in ${\Sigma}_U$ that exit the end x such that the component of ${\Sigma}_U - \alpha_i$ that contains x has the end space $U_{k_i}$ . Let $Y_i $ be the tube bounded by $\alpha_i$ and $\alpha_{i+1}$ . Then ${\Sigma}_U = \bigcup_{i=0}^{\infty} Y_{i}$ and the first two assumptions of the lemma hold by construction.

To see that the final assumption holds, it is sufficient to show that $Y_0 \cup Y_1$ is homeomorphic to both $Y_0$ and $Y_1$ . If ${\Sigma}_U$ has infinite genus, then x is a non-planar end. Since x is not isolated in $E_G$ , some ends in some $Y_i$ (and hence every $Y_i$ ) have to be non-planar. Therefore, every $Y_i$ has infinite genus. Otherwise, ${\Sigma}_U$ has genus zero and hence every $Y_i$ has genus zero. Hence, $Y_0 \cup Y_1$ , $Y_0$ and $Y_1$ have the same genus. Also, they all have two boundary components. We only need to check that their end spaces are homeomorphic. But this follows from Corollary 4.12 applied to $E(Y_0) \subset E(Y_0 \cup Y_1)$ or to $E(Y_1) \subset E(Y_0 \cup Y_1)$ . This finishes the proof of the one direction.

To see the other direction, let ${\Sigma}_U$ be a stable neighborhood of x in ${\Sigma}$ and let ${\Sigma}_U = \bigcup_{i=0}^{\infty} Y_{i}$ be a decomposition of ${\Sigma}_U$ into tubes as described in the statement. We show that x is a telescoping end by checking the conditions of Definition 4.13.

If x is the only non-planar end in ${\Sigma}_U$ , then there is a tube $Y_k$ that has a finite non-zero genus. The $Y_0$ also has finite non-zero genus and hence $Y_0$ cannot be homeomorphic to $Y_0 \cup \dots \cup Y_k$ , which contradicts assumption (iii). Therefore, either x is a planar end, or it is not isolated in $E_G({\Sigma})$ .

If x is an isolated puncture or is of Cantor type, then we are done by definition. Otherwise, we need to show that x is a successor and the predecessors are of Cantor type. By Lemma 4.9, $E(Y_0)$ has finitely many different maximal types, say $y_1, \dots, y_n$ . Then every $z \in U$ , $E(z)\cap E(Y_0)$ is non-empty and hence E(z) accumulates to some $y_i$ . This, by Definition 4.5, implies that x is a successor.

The assumption (iii) in particular implies that $E(Y_0)$ is homeomorphic to $E(Y_k)$ for every k, and in particular $E(y_i) \cap E(Y_k)$ is non-zero for every k. If $E(y_i) \cap E(Y_0)$ is finite, then $E(y_0) \cap (Y_0 \cup Y_1)$ is larger than $E(y_0) \cap Y_0$ and hence $Y_0$ cannot be homeomorphic to $Y_0 \cup Y_1$ , which contradicts assumption (iii). If $E(y_i) \cap E(Y_0)$ is infinite then $E(y_i)$ has an accumulation point in $E(Y_0)$ . But $y_i$ is maximal in $E(Y_0)$ , hence the accumulation point has to be in $E(y_i)$ . Therefore, every point in $E(y_i) \cap E(Y_0)$ is an accumulation point of $E(y_i)$ , and $E(y_i)$ is uncountable. Lemma 4.7 implies that $y_i$ is of Cantor type. This holds for every $i=1, \dots, n$ , therefore x is a telescoping end.

We call a decomposition as in this previous lemma a telescoping decomposition of the neighborhood $\Sigma_{U}$ . We also give yet another characterization of telescoping in Lemma 4.35. We will not make use of this alternate characterization until § 4.9 and so do not include it here, but the curious reader may want to jump ahead.

We have now also seen that the definition of telescoping, Definition 4.13, fits into the same framework as Proposition 3.2 and Definition 3.3 from the end-space case. Thus, we can use this to define a brick in the surface setting.

For example, fixing a tubular decomposition ${\Sigma}_U=\bigcup_{i=0} Y_i$ , the union $\cup Y_{n_i}$ is a brick for any infinite and co-infinite subset $\{n_i\}\subset\{1,2,\ldots\}$ (it is important not to include $Y_0$ since $Y_0$ is not a drawtube). From Lemma 4.15 we see that for any two bricks there is a homeomorphism of ${\Sigma}_U$ taking one to the other. In particular, any brick can be enlarged by inserting drawtubes into the complementary components, including the complementary component that contains $\partial{\Sigma}_U$ , and therefore we could choose a homeomorphism between two bricks to fix a neighborhood of $\partial{\Sigma}_U$ that contains a drawtube.

Standing Assumptions. As in the end-space case, we now set up some notation and standing assumptions. Unless explicitly stated, these will hold until the end of § 4.7.

1. (1) If A is a tube or a finite-type subsurface of ${\Sigma}$ , we will use G(A) to denote the subgroup of $G=\mathrm{Homeo}({\Sigma})$ consisting of maps supported on the interior of A. If U is a stable neighborhood of a telescoping end x, then $G({\Sigma}_U)$ is the set of homeomorphisms supported on some ${\Sigma}_{U'}$ so that $U\smallsetminus U'$ contains all types of ends in U except possibly x.

2. (2) Fix a telescoping end x and ${\Sigma}_U =\bigcup_{i=0}^{\infty} Y_{i}$ a subdivision as in Lemma 4.16.

3. (3) Assume that $W\subset G$ is a symmetric set so that G is covered by countably many sets $g_{i}W$ with $g_{i}\in G$ and $W^{2}$ is dense in $G({\Sigma}_{U})$ (i.e. $\overline{W^{2}} \supset G({\Sigma}_{U})$ ).

4. (4) For any brick, $\mathcal {B}$ , we write $G(\mathcal {B})$ for the closed subgroup of homeomorphisms of ${\Sigma}_{U}$ or ${\Sigma}$ supported on $\mathcal {B}$ .

4.3 Step 1: Fragmentation

Again, note that if U is a stable neighborhood of an end x, then U is also a stable neighborhood of $g \cdot x$ for all $g \in G({\Sigma}_U)$ . Keeping this in mind, the statements that follow will apply when one replaces the telescoping end x with an end of the form $g \cdot x$ for $g \in G({\Sigma}_{U})$ . We will slightly abuse notation and not keep track of this distinction.

Just as in the end-space case, these two lemmas allow us to write any $h \in G({\Sigma}_U)$ as $h=h_{1}h_{2}h_{3}h_{4}$ with each $h_{i}\in G({\Sigma}_U)$ supported on a brick limiting to a fixed point in the orbit of x.

4.4 Step 2: Finding commutators

For surfaces, this step will be drastically different than in the end-space case. For end spaces, since we were dealing with totally disconnected spaces, we could use maps that shift an entire brick and restrict to the identity elsewhere. Now, the topology of the surface obstructs using such a shift. Instead, we will fragment our map again using an Eilenberg–Mazur swindle [Reference BassBas63, Reference MazurMaz59] into two maps, still supported on bricks, that are themselves commutators.

Proof. Let h be as in the definition of alternating support. Let $f_{1}$ denote the homeomorphism that agrees with f on $A_{1}$ and is the identity outside of $A_{1}$ . Then we can write

\begin{align*} f = f_{1} h (f_{1})^{-1} h^{-1} = [f_{1},h]. \end{align*}

If $\mathcal {A}$ is a brick, then we can simply apply this to each tube of $\mathcal {A}$ and collect terms since the tubes are all pairwise disjoint

Proof. The brick $\mathcal {A}$ is the disjoint union of drawtubes ordered linearly, which we write as

$$A_1\sqcup A_2\sqcup A_3\sqcup\cdots,$$

with $A_{i+1}$ separating $A_i$ from x for $i\gt 0$ . Denote by $f_j=f|A_j$ the restriction of f to $A_j$ . We will also use $f_j$ to denote the extension by the identity to the rest of ${\Sigma}_U$ . Thus, we can write f as the infinite composition $f_1\circ f_2 \circ \cdots$ . We will denote by $B_0$ the closure of the complementary component of $\mathcal {A}$ in ${\Sigma}_U$ that contains the boundary component, and we will denote by $B_i$ the drawtube that shares a boundary component with both $A_i$ and $A_{i+1}$ for $i\gt 0$ . Thus,

$${\Sigma}_U=B_0\cup A_1\cup B_1\cup A_2\cup B_2\cup \cdots,$$

respecting the linear order.

Next, we subdivide each $B_i$ , $i\gt 0$ , into $4i+1$ drawtubes, writing in linear order

$$B_i=C_{i,1}\cup C_{i,2}\cup \cdots\cup C_{i,2i}\cup D_i\cup E_{i,1}\cup E_{i,2}\cup \cdots\cup E_{i,2i}.$$

We will also fix a homeomorphism between $B_0$ and every drawtube A we named so far, so that the boundary curve of ${\Sigma}_U$ is mapped to the boundary component of A not separated from x by the other boundary component of A. In particular, we have preferred homeomorphisms between any two tubes above by composing the homeomorphisms with $B_0$ .

Now define a new homeomorphism $h_1:{\Sigma}_U\to {\Sigma}_U$ as follows:

• – $h_1|A_i=f_i$ ;

• – $h_1|B_0=\mathrm{id}$ ;

• – $h_1|D_i=\mathrm{id}$ for $i\gt 0$ ;

• – $h_1|E_{i,2j-1}=\phi_{i,2j-1}f_j\phi_{i,2j-1}^{-1}$ for $j=1,\ldots,i$ , where $\phi_{i,2j-1}:A_j\to E_{i,2j-1}$ is the preferred homeomorphism;

• – $h_1|C_{i,2j}=\psi_{i,2j}f_j^{-1}\psi_{i,2j}^{-1}$ for $j=1,\ldots,i$ , where $\psi_{i,2j}:A_j\to E_{i,2j}$ is the preferred homeomorphism;

• – $h_1|C_{i,2j-1}=\mathrm{id}$ , $h_1|E_{i,2j}=\mathrm{id}$ , for $j=1,\ldots,i$ .

We also define $h_2:{\Sigma}_U\to {\Sigma}_U$ in the same way, except that $h_2|A_i=\mathrm{id}$ for $i\gt 0$ . Thus, $f=h_1h_2^{-1}$ and it remains to argue that $h_1,h_2$ have alternating support on suitable bricks.

We start with $h_2$ . We consider the supporting brick $B_1\sqcup B_2\sqcup\cdots$ and show that $h_2|B_i$ has alternating support for every i. Indeed, write $B_i$ as a subdivision into five drawtubes:

$$B_i=C_{i,1}\cup (C_{i,2}\cup C_{i,3}\cup \cdots \cup C_{i,2i})\cup D_i\cup (E_{i,1}\cup E_{i,2}\cup \cdots \cup E_{i,2i-1})\cup E_{i,2i}.$$

Then $h_2|B_i$ is the identity on the first, third and fifth of these subtubes. Let $g_i$ be a homeomorphism of $B_i$ that restricts to the preferred homeomorphisms $C_{i,2j}\to E_{i,2j-1}$ , $j=1,2,\ldots,i$ . This homeomorphism exists by the classification of surfaces, as in the proof of Lemma 4.15, and it conjugates $h_2$ restricted to the second tube to the inverse of the restriction on the fourth tube, as required.

Now consider $h_1$ . It will be convenient to subdivide each $D_i$ into three drawtubes as $D_i=D_i^1\cup D_i^2\cup D_i^3$ . We consider the supporting brick, which is the complement of $\bigcup_{i=1}^\infty D_i^2$ and a subtube of $B_0$ containing $\partial {\Sigma}_U$ . Thus, a tube of this brick has the form

$$D_i^3\cup (E_{i,1}\cup E_{i,2}\cup \cdots\cup E_{i,2i}\cup A_{i+1})\cup C_{i+1,1}\cup (C_{i+1,2}\cup \cdots\cup C_{i+1,2i+2})\cup D_{i+1}^1$$

and is written as the union of five tubes (the very first tube in the brick is slightly different and its consideration is left to the reader). Note that $h_1$ is the identity on the first, third and fifth tubes. As in the case of $h_2$ , we can construct a homeomorphism $k_i$ of this tube that restricts to preferred homeomorphisms $E_{i,2j-1}\to C_{i+1,2j}$ for $j=1,2,\ldots,i$ , and $A_{i+1}\to C_{i+1,2i+2}$ . Then $k_i$ will conjugate the restriction of $h_1$ to the second tube to the inverse of its restriction on the fourth tube.

The following schematic summarizes the construction. The given map f can be represented as

$$\mathbf{1}\quad \mathbf{2} \quad \mathbf{3}\quad \mathbf{4} \quad \cdots,$$

where we only keep indices from various $f_i$ . The map $h_1$ can then be represented as

$$\mathbf{1}\mathit{\overline 1 1}\mathbf{2}\mathit{\overline 1 \overline 2 1 2}\mathbf{3} \mathit{\overline 1 \overline 2 \overline 3 123}\mathbf{4}\mathit{\overline 1 \overline 2 \overline 3 \overline 4\cdots},$$

where italicized numbers refer to the restriction to subdivision tubes, and number i means that the map is conjugate to $f_i$ , while the bar denotes the inverse. Between any two drawtubes with consecutive numbers, there is a drawtube on which the map is the identity, giving a buffer. Each block of italicized numbers between two bold numbers defines a map on a single drawtube with alternating supports:

$$\mathbf{1}\overbrace{\mathit{\overline 1}\mathit{1}}\mathbf{2}\overbrace{\mathit{\overline 1 \overline 2}\mathit {1 2}}\mathbf{3} \overbrace{\mathit{\overline 1 \overline 2 \overline 3}\mathit{123}}\mathbf{4}\overbrace{\mathit{\overline 1 \overline 2 \overline 3 \overline 4}\mathit{1234}}\cdots.$$

Therefore, $h_2$ has alternating supports.

But so does $h_1$ :

$$\underbrace{\mathbf{1}\mathit{\overline 1}}\underbrace{\mathit{1}\mathbf{2}\mathit{\overline 1 \overline 2}}\underbrace{\mathit {1 2}\mathbf{3} \mathit{\overline 1 \overline 2 \overline 3}} \underbrace{\mathit{123}\mathbf{4} \mathit{\overline 1 \overline 2 \overline 3 \overline 4}}\cdots.$$

Here we get alternating supports corresponding to the groups of numbers indicated after adding a drawtube on each side of the group (and using the definition of a brick that guarantees that there is a drawtube to the left of the first group).

The culmination of this step is to combine these two lemmas in order to write a map supported on a brick as a product of two commutators, each of which are also supported on the brick.

4.5 Step 3: Diagonalization

Again, we next want to find a ‘good’ brick. The first lemma is identical to the end-space case.

1. (i) there is some $i\geq 1$ such that any homeomorphism of $\mathcal {A}_i$ extends to a homeomorphism of ${\Sigma}$ supported on $\mathcal {A}$ that belongs to $g_iW$ , and

2. (ii) moreover, every homeomorphism of $\mathcal {A}_i$ extends to a homeomorphism of ${\Sigma}$ supported on $\mathcal {A}$ that belongs to $W^2$ .

Next we upgrade this approximation. Notably, here we will only be doing it for maps with alternating support.

4.6 Step 4: Pigeonhole

Next we see that, except for a finite error, we can push any brick into a specific subbrick.

Since the collection $\{\Lambda_{\alpha}\}_{\alpha \in \mathbb{R}}$ is uncountable, there exist some pair $\alpha \neq \beta$ so that $f_{\alpha}$ and $f_{\beta}$ are in the same left translate of W. See Figure 3 for a schematic of the arrangement of these tubes and the maps $f_{\alpha}$ and $f_{\beta}$ . Thus, we have that $f_{\alpha}^{-1} f_{\beta}$ and $f_{\beta}^{-1}f_{\alpha}$ are both in $W^{2}$ . Set $a_{1} =f_{\beta}^{-1} f_{\alpha}$ and $a_{2} = f_{\alpha}^{-1}f_{\beta}$ . Let $\mathcal {A}_{0} = \mathcal {B}_{\alpha} \cap \mathcal {B}_{\beta}$ , note that this set is finite, let $\mathcal {A}_{1} = \mathcal {A} \setminus \mathcal {B}_{\alpha}$ and $\mathcal {A}_{2} = \mathcal {A} \setminus \mathcal {B}_{\beta}$ . Note that $\mathcal {A} = \mathcal {A}_{0} \cup \mathcal {A}_{1} \cup \mathcal {A}_{2}$ . Now, given any tube A in $\mathcal {A}$ that is not in the finite exceptional set $\mathcal {A}_{0}$ , we have that either $A \in \mathcal {A}_{1}$ or $A \in \mathcal {A}_{2}$ . If $A \in \mathcal {A}_{1}$ , then $a_{1}A \in \mathcal {B}_{\beta} \subset \mathcal {B}$ and if $A \in \mathcal {A}_{2}$ , then $a_{2}A \in \mathcal {B}_{\alpha} \subset \mathcal {B}$ .

Figure 3.

A schematic of the pigeonhole argument. The initial two tubes represent the finitely many tubes in $\mathcal {B}_{\alpha}\cap \mathcal {B}_{\beta}$ . The wavy-lined and straight-lined tubes are the tubes of $\mathcal {B}_{\alpha}$ and $\mathcal {B}_{\beta}$ , respectively.

Finally, we can put all of these pieces together to see that any map supported on a brick is in $W^{72}$ .

Apply Lemma 4.26 to $\mathcal {A}$ to find $\mathcal {Z}$ so that any map in $G(\mathcal {Z})$ with alternating support is in $W^{8}$ . Next, apply Lemma 4.27 to $\mathcal {Z}$ as a subbrick of $\mathcal {A}$ to obtain maps $a_{1},a_{2} \in W^{2}$ . Write $\mathcal {A} = F \sqcup X \sqcup Y$ , where F is the finite collection of tubes for which Lemma 4.27 fails, X is the set of tubes for which $a_{1}A \in \mathcal {Z}$ for all $A \in X$ , and Y is the complement of these two sets. We can write $h = h_{F}h_{X}h_{Y}$ for the restrictions of h to each of these sets. Note that $a_{1}h_{X}a_{1}^{-1}$ and $a_{2}h_{Y}a_{2}^{-1}$ are two maps with alternating support in $\mathcal {Z}$ and therefore $h_{X},h_{Y} \in W^{12}$ .

It remains to deal with $h_{F}$ . Let $g \in G({\Sigma}_{U})$ be a map that sends the set F of tubes into $\mathcal {Z}$ . By assumption, $W^{2}$ is dense in $G({\Sigma}_{U'})$ and so we may approximate g by $g'\in W^2$ (which may not belong to $G({\Sigma}_U)$ or even be an identity outside of ${\Sigma}_U$ ). Thus, we see that $g'h_{F}g'^{-1}$ is again a map with alternating support in $\mathcal {Z}$ and hence $h_{F} \in W^{12}$ . We conclude that $h \in W^{36}$ and thus our original map is contained in $W^{72}$ .

4.7 Step 5: Wrapping up

We again first verify a version of the Steinhaus condition for a stable neighborhood of a telescoping end.

• – $\mathrm{Homeo}({\Sigma})$ is covered by countably many sets $g_{i}W$ with $g_{i} \in \mathrm{Homeo}({\Sigma})$ , and

• – $W^{2}$ is dense in $G({\Sigma}_U)$ .