Component labels and annular depth

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

A continuum without cut points and its graph describes a compact connected space $L$ without cut points, a group $H$, and a graph $K$ whose vertices accumulate on $L$. Paths of $K$ stay near a point of $L$ when their ends are near it. In the next lesson such paths are pushed towards $L$, and for this some vertices of the graph have to be removed. This lesson prepares two tools.

Component labels. When a set of vertices is removed, the remaining graph falls into components. Near a point of $L$ at which the removed vertices do not accumulate, all remaining vertices lie in one component, and this component depends on the point in a locally constant way (Theorem 2.1). At a removed vertex that lies in $L$ this uses the vanishing of the first Čech cohomology of $L$.
Annular depth. There is a family of pairs of disjoint closed sets, called annuli, that is invariant under $H$, in which only finitely many members have two large sides, and which nests arbitrarily often around every point of $L$ (Theorem 3.2). Counting nested annuli between a set and a point gives an invariant depth, which is finite between group elements and large near $L$ (Theorem 4.1).

The lesson follows Section 8 of [OpenAI-S]; systems of annuli of this kind were introduced by Bowditch [Bowditch 3]. It uses the lesson named above and the extension of locally constant maps across a point from First Čech cohomology of a compact metric space.

1. The setting

We use the following data and properties. They hold in the two cases of the lesson on a continuum without cut points and its graph (Theorem 2.1 there); from here on only this list is used.

(S1) $(Y,\rho)$ is a compact metric space, the disjoint union of a group $H$ and a set $L$. The points of $H$ are isolated in $Y$ and dense in $Y$. The group $H$ acts on $Y$ by homeomorphisms that extend left multiplication on $H$ and map $L$ to itself. The space $L$ is compact and connected, has more than one point and no cut point.
(S2) If $(h_n)$ is a sequence in $H$ with $h_n\to a\in L$ and $h_n^{-1}\to b\in L$, then for every closed set $C\subseteq Y\setminus\{b\}$ and every open set $U\subseteq Y$ containing $a$, $h_n(C)\subseteq U$ for all but finitely many $n$.
(S3) $K$ is a connected graph with vertex set $I\subseteq Y$; $H$ maps $I$ to itself and edges to edges, with finitely many orbits of vertices and of edges. Either $I=H$ or $I\subseteq L$. Every point of $L$ is the limit of a sequence of pairwise distinct points of $I$.
(S4) For every $w\in I$: $\rho(hw,h)\to0$ as $h\to\infty$ in $H$.
(S5) For every $x\in L$ and every neighbourhood $W$ of $x$ in $Y$ there is a neighbourhood $V$ of $x$ in $Y$ such that any two points of $I\cap V$ are joined by a path in $K$ all of whose points lie in $W$.
(S6) If $I\subseteq L$, then for every $p\in I$ there is $h\in H$ of infinite order with $hp=p$ and $h^jz\to p$ as $j\to\infty$ for every $z\in L$.
(S7) $\check H^1(L;k)=0$, where $k$ is the group with two elements $0,1$.

By (S1), a sequence of pairwise distinct elements of $H$ has a subsequence $(h_n)$ such that $h_n$ and $h_n^{-1}$ converge to points of $L$: limits in the compact space $Y$ exist, and a limit of distinct points is not an isolated point. Since $L$ is connected with more than one point, no point of $L$ is isolated in $L$. Balls $B(y,r)$ are open balls of $Y$.

2. Component labels

For a set $D\subseteq Y$ let $D'$ be the set of limits of sequences of pairwise distinct points of $D$. It is closed.

Let $T\subseteq I$ be a set of vertices, the allowed vertices; the vertices in $I\setminus T$ are deleted. Let $K[T]$ be the graph on $T$ whose edges are the edges of $K$ with both ends in $T$; its components are the classes of the relation "joined by a path in $K[T]$". Let $\Omega_T=L\setminus(I\setminus T)' .$ This is the open subset of $L$ of the points at which deleted vertices do not accumulate.

Theorem 2.1. Let $T\subseteq I$ and $x\in\Omega_T$.

$x$ is the limit of a sequence of pairwise distinct allowed vertices.
There is exactly one component $Q_T(x)$ of $K[T]$ such that all allowed vertices in some neighbourhood of $x$ in $Y$ belong to $Q_T(x)$.
The map $x\mapsto Q_T(x)$ is locally constant on $\Omega_T$.
If $x$ is itself an allowed vertex, then $x\in Q_T(x)$.

Proof. There is a neighbourhood $W$ of $x$ in $Y$ that contains no deleted vertex other than possibly $x$: otherwise there would be deleted vertices different from $x$ at distance less than $1/n$ from $x$ for every $n$, and $x$ would lie in $(I\setminus T)'$. By (S3), $x$ is the limit of a sequence of distinct points of $I$; those in $W$ other than $x$ are allowed. This proves (1). It also shows that at most one component can have the property in (2).

Case 1: $x$ is not a deleted vertex. Then $W$ contains no deleted vertex. By (S5) there is an open neighbourhood $V\subseteq W$ of $x$ such that any two points of $I\cap V$ are joined by a path of $K$ inside $W$. Such a path consists of allowed vertices, so it is a path of $K[T]$. Hence all allowed vertices in $V$ lie in one component, which is $Q_T(x)$; if $x$ is an allowed vertex, it is one of them. For $y\in V\cap L$, the point $y$ lies in $\Omega_T$ and is not a deleted vertex, and the allowed vertices near $y$ lie in $V$; so $Q_T(y)=Q_T(x)$.

Case 2: $x$ is a deleted vertex. Then $I\subseteq L$, by (S3). Let $V=W\cap L$, which we may take open in $L$. Every $y\in V\setminus\{x\}$ lies in $\Omega_T$ and is not a deleted vertex, so by Case 1 the map $q$ from $V\setminus\{x\}$ to the set of components of $K[T]$, $q(y)=Q_T(y),$ is defined and locally constant. We show that $q$ is constant on $V_0\setminus\{x\}$ for some neighbourhood $V_0\subseteq V$ of $x$ in $L$.

For every map $c$ from the set of components to $k$, the map $c\circ q$ is locally constant on $V\setminus\{x\}$. Since $L\setminus\{x\}$ is connected and $\check H^1(L;k)=0$, it extends to a locally constant map on $V$ (Corollary 4.5 of the lesson on Čech cohomology). So:

(E) for every $c$, the map $c\circ q$ is constant on $V_c\setminus\{x\}$ for some neighbourhood $V_c$ of $x$.

Call a component recurrent if it is a value of $q$ in every neighbourhood of $x$.

If two different components are recurrent, let $c$ be $0$ on one and $1$ on the other; this contradicts (E).

If no component is recurrent, we choose points $y_1,y_2,\dots$ of $V\setminus\{x\}$ converging to $x$ with pairwise different values of $q$: when $y_1,\dots,y_n$ are chosen, each of their values is absent from some neighbourhood of $x$, and $y_{n+1}$ is taken in the intersection of these neighbourhoods with the ball of radius $1/n$ around $x$, which contains points of $L$ other than $x$. Let $c$ be $0$ on $q(y_n)$ for even $n$ and $1$ on all other components; this contradicts (E).

So exactly one component $A$ is recurrent. If $q$ took values other than $A$ in every neighbourhood of $x$, the map $c$ that is $0$ on $A$ and $1$ elsewhere would contradict (E). Hence $q=A$ on $V_0\setminus\{x\}$ for some neighbourhood $V_0$ of $x$ in $L$.

Every allowed vertex $t\in V_0$ is a point of $L$ other than $x$, and by Case 1 it lies in $Q_T(t)=q(t)=A$. All allowed vertices lie in $L$, so the allowed vertices in a neighbourhood of $x$ in $Y$ whose trace on $L$ is $V_0$ belong to $A$. So $Q_T(x)=A$, and $Q_T=A$ on $V_0$. $\blacksquare$

3. A system of annuli

An annulus is an ordered pair $\sigma=(\sigma^-,\sigma^+)$ of disjoint closed subsets of $Y$, its sides, such that each side meets $L$ and $\sigma^-\cup\sigma^+\ne Y$. For annuli $\sigma,\tau$ write $\sigma<\tau\quad\Longleftrightarrow\quad\sigma^+\cup\tau^-=Y .$

Lemma 3.1.

If $\sigma<\tau$, then $\sigma^-\subseteq\tau^-$ and $\tau^+\subseteq\sigma^+$.
The relation $<$ is transitive, and $\sigma<\sigma$ never holds. So the annuli of a chain $\sigma_1<\sigma_2<\dots<\sigma_n$ are pairwise different.
The image $g\sigma=(g\sigma^-,g\sigma^+)$ of an annulus under $g\in H$ and its reversal $(\sigma^+,\sigma^-)$ are annuli, and $\sigma<\tau$ implies $g\sigma<g\tau$.

Proof. (1) $\sigma^-$ is disjoint from $\sigma^+$, so it lies in $\tau^-$; likewise $\tau^+$, disjoint from $\tau^-$, lies in $\sigma^+$. (2) If $\sigma<\tau<\upsilon$, then $\sigma^+\cup\upsilon^-\supseteq\sigma^+\cup\tau^-=Y$ by (1). And $\sigma^+\cup\sigma^-\ne Y$. (3) $g$ is a homeomorphism of $Y$ that maps $L$ to itself. $\blacksquare$

A chain $\sigma_1<\dots<\sigma_n$ goes from $F$ to $E$, for subsets $F,E$ of $Y$, if $F\subseteq\sigma_1^-$ and $E\subseteq\sigma_n^+$. Its length is $n$.

Theorem 3.2. There is a set $\mathcal A$ of annuli, mapped to itself by $H$ and by reversal, such that:

for every $\delta>0$, only finitely many annuli of $\mathcal A$ have both sides of diameter at least $\delta$;
for every closed set $F\subseteq Y$, every $x\in L\setminus F$ and every $n\ge1$, there is a chain of length $n$ in $\mathcal A$ from $F$ to a neighbourhood of $x$ in $Y$.

Proof. Step 1: a consequence of (S2). For all $r,\varepsilon>0$, all but finitely many $g\in H$ have a point $b_g\in L$ such that $\operatorname{diam}g\bigl(Y\setminus B(b_g,r)\bigr)<\varepsilon .$ Otherwise there are pairwise distinct $g_n$ without such a point. Pass to a subsequence with $g_n\to a\in L$ and $g_n^{-1}\to b\in L$. By (S2), with the closed set $Y\setminus B(b,r)$ and the ball of radius $\varepsilon/3$ around $a$, the point $b$ serves as $b_{g_n}$ for large $n$.

Step 2: the annuli. Choose numbers $r_1>r_2>\dots$ tending to $0$, all smaller than a third of the diameter of $L$, and let $\varepsilon_i=2^{-i}$. Let $E_i\subseteq H$ be the finite set of exceptions in Step 1 for $r=r_i/4$ and $\varepsilon=\varepsilon_i$. The finitely many maps in $E_i$ are uniformly continuous on $Y$, so there is $s_i$ with $0<s_i<r_i/4$ such that the image under each $g\in E_i$ of each closed ball of radius $s_i$ has diameter less than $\varepsilon_i$. Choose finitely many points $y_{ij}\in L$ such that the balls $B(y_{ij},s_i/2)$ cover $L$, and let $\sigma_{ij}$ be the pair $\begin{gathered} \sigma_{ij}^-=Y\setminus B(y_{ij},r_i),\\ \sigma_{ij}^+=\{y\mid\rho(y,y_{ij})\le s_i\}. \end{gathered}$ The two sides are closed and disjoint, and any point of one has distance more than $3r_i/4$ from any point of the other. The side $\sigma_{ij}^+$ contains $y_{ij}\in L$. The side $\sigma_{ij}^-$ meets $L$: if $u,v\in L$ have the largest possible distance, one of them has distance at least half the diameter of $L$ from $y_{ij}$, which is more than $r_i$. The two sides do not cover $Y$: otherwise their traces on $L$ would be two disjoint nonempty closed sets covering the connected space $L$. So $\sigma_{ij}$ is an annulus.

Let $\mathcal A$ be the set of all annuli $g\sigma_{ij}$, $g\in H$, and their reversals. It is mapped to itself by $H$ and by reversal.

Step 3: one side is small. Every annulus $g\sigma_{ij}$ has a side of diameter less than $\varepsilon_i$. For $g\in E_i$ it is $g\sigma_{ij}^+$, by the choice of $s_i$. For $g\notin E_i$: the ball $B(b_g,r_i/4)$ has diameter at most $r_i/2$, so it cannot meet both sides of $\sigma_{ij}$; the side that it does not meet lies in $Y\setminus B(b_g,r_i/4)$, and its image under $g$ has diameter less than $\varepsilon_i$.

Step 4: proof of (1). Let $\delta>0$. By Step 3 an annulus $g\sigma_{ij}$ or its reversal with both sides of diameter at least $\delta$ has $\varepsilon_i\ge\delta$, which leaves finitely many pairs $(i,j)$. Fix one of them and write $\sigma=\sigma_{ij}$. Suppose there are infinitely many different annuli $g\sigma$ with both sides of diameter at least $\delta$. Choose pairwise distinct $g_n$ giving such annuli, and a subsequence with $g_n\to a\in L$ and $g_n^{-1}\to b\in L$. One of the two disjoint closed sides of $\sigma$ does not contain $b$, and by (S2) the diameter of its images under $g_n$ tends to $0$. This is a contradiction.

Step 5: proof of (2). Let $F$ be closed and $x\in L\setminus F$. Choose $i$ so large that $r_i+s_i/2$ is smaller than the distance from $x$ to $F$, if $F$ is not empty, and $j$ with $\rho(x,y_{ij})<s_i/2$. Then $F\subseteq\sigma_{ij}^-$, and $\sigma_{ij}^+$ contains the ball $B(x,s_i/2)$. This is a chain of length one.

Suppose a chain from $F$ is given whose last annulus $\sigma$ has $B(x,\eta)\subseteq\sigma^+$. Choose $i'$ with $r_{i'}+s_{i'}/2<\eta$ and $j'$ with $\rho(x,y_{i'j'})<s_{i'}/2$, and let $\tau=\sigma_{i'j'}$. Then $B(y_{i'j'},r_{i'})\subseteq B(x,\eta)\subseteq\sigma^+$, so $\sigma^+\cup\tau^-=Y$: the chain is prolonged by $\tau$, and $\tau^+$ contains a ball around $x$. Repeating this gives chains of every length. $\blacksquare$

4. Annular depth

Fix a set $\mathcal A$ as in Theorem 3.2. For $F\subseteq Y$ and $v\in Y$ let $D(F,v)$ be the largest $n$ such that there is a chain of length $n$ in $\mathcal A$ from $F$ to $\{v\}$; it is $0$ if there is no such chain, and $\infty$ if there are chains of every length. For a point $u$ we write $D(u,v)$ for $D(\{u\},v)$.

Theorem 4.1.

$D(gF,gv)=D(F,v)$ for all $g\in H$.
If $F$ contains a point of $H$ and $v\in H$, then $D(F,v)$ is finite.
If $F$ is closed, $x\in L\setminus F$ and $N\ge1$, there is a neighbourhood $V_N$ of $x$ in $Y$ with $D(F,v)\ge N$ for all $v\in V_N$.
If $F\subseteq F'$, then $D(F',v)\le D(F,v)$.

Proof. (1) $g$ maps chains from $F$ to $\{v\}$ to chains from $gF$ to $\{gv\}$, and $g^{-1}$ maps them back (Lemma 3.1(3)).

(2) Let $h\in F\cap H$. The points $h$ and $v$ are not in the closed set $L$; let $\delta>0$ be smaller than their distances from $L$. Let $\sigma_1<\dots<\sigma_n$ be a chain from $F$ to $\{v\}$. By Lemma 3.1(1), $h\in\sigma_1^-\subseteq\sigma_i^-$ and $v\in\sigma_n^+\subseteq\sigma_i^+$ for every $i$. Each side also meets $L$, so both sides of every $\sigma_i$ have diameter at least $\delta$. By Theorem 3.2(1) there are finitely many such annuli, and the annuli of a chain are pairwise different. So $n$ is bounded.

(3) By Theorem 3.2(2) there is a chain of length $N$ from $F$ to a neighbourhood $V_N$ of $x$; it goes from $F$ to $\{v\}$ for every $v\in V_N$.

(4) A chain from $F'$ to $\{v\}$ goes from $F$ to $\{v\}$. $\blacksquare$

5. Exercises

5.1. Let $L$ be a circle and $x\in L$. Give a locally constant map on $V\setminus\{x\}$, for a small open arc $V$ around $x$, that does not extend to a locally constant map on $V$. Which assumption of this lesson fails for the circle?

5.2. Show that in Theorem 2.1 the set $\Omega_T$ contains every point of $L$ if $I\setminus T$ is finite.

5.3. Show that $D(F,v)\ge1$ implies $v\notin F$, and that $D(u,v)=D(v,u)$ for points $u,v\in Y$. Which property of the set $\mathcal A$ is used for the second statement?

5.4. Let $u\in H$. Show that $D(u,v)\to\infty$ as $v\in H$ tends to infinity.

6. Solutions

5.1. The set $V\setminus\{x\}$ consists of two open arcs; the map that is $0$ on one and $1$ on the other is locally constant and has no locally constant extension to $V$. The circle is connected without cut points and has dimension one, but its first Čech cohomology with coefficients in $k$ is not zero (Corollary 4.3 of the lesson on Čech cohomology), so (S7) fails.

5.2. A finite set has no sequence of pairwise distinct points, so $(I\setminus T)'$ is empty.

5.3. If $F\subseteq\sigma_1^-$ and $v\in\sigma_n^+\subseteq\sigma_1^+$, then $v\notin F$, because the sides of $\sigma_1$ are disjoint. For the symmetry, let $\sigma_1<\dots<\sigma_n$ be a chain from $\{u\}$ to $\{v\}$, and let $\bar\sigma_i$ be the reversal of $\sigma_i$. Then $\bar\sigma_n<\dots<\bar\sigma_1$: indeed $\bar\sigma_{i+1}^+\cup\bar\sigma_i^-=\sigma_{i+1}^-\cup\sigma_i^+=Y$. This chain goes from $\{v\}$ to $\{u\}$, and it lies in $\mathcal A$ because $\mathcal A$ is mapped to itself by reversal; this is the property used. So $D(v,u)\ge D(u,v)$, and equality follows by symmetry.

5.4. Suppose not: there are $N$ and pairwise distinct $v_n\in H$ with $D(u,v_n)<N$. A subsequence converges to a point $x\in L$, and $x\notin\{u\}$. By Theorem 4.1(3), $D(u,v)\ge N$ for all $v$ near $x$, a contradiction.

References

[OpenAI-S] OpenAI, The Singer conjecture in dimension four, OpenAI Math Release preprint, 25 September 2026, Section 8. https://github.com/openai/math/tree/main/preprints/The-Singer-conjecture-in-dimension-four-September-25-2026
[Bowditch 3] B. H. Bowditch, A topological characterisation of hyperbolic groups, Journal of the American Mathematical Society 11 (1998), Sections 6 and 7. Author's version: http://homepages.warwick.ac.uk/~masgak/papers/bhb-topchar.pdf