Shape injectivity of the earring space (Part I)

One of my posts where I did some substantial hand-waving is my original post on the fundamental group of the earring space. I wrote about how to understand and work with this group, but I never gave a proof of the key fact that the earring group naturally injects into an inverse limit of free groups $\varprojlim_{n}F_n$ . This is one of the two primary viewpoints that researchers take to study and apply the beautiful algebra of this group (and more generally fundamental groups of one-dimensional spaces). Seriously, I’m using this machinery like 1.) it’s going out of style and 2.) I understand fashion. The other approach avoids inverse limits by identifying the earring group as a group of reduced countable, linear words over a countable alphabet. They’re logically equivalent, but sometimes one is more convenient than the other.

To be honest, I hesitated about writing this post. I say with confidence that there is no completely elementary proof. While I’ve read and understood many different proofs of shape injectivity, most of them are either super technical or they gloss over details by applying continuum and dimension theory. Some inquisitive and kind readers have given me the motivation to do it.

When trying to write this post, I dug deep into the literature trying to weasel out an almost entirely self-contained proof that a grad student would believe. After some reading, I worked things out and these posts are the results of the effort. This first post will mostly be used to set up the technical tools about arcs in inverse limits that we’ll need to prove shape injectivity.

Let $C_n=\{(x,y)\in\mathbb{R}^2\mid (x-1/n)^2+y^2=(1/n)^2\}$ be the circle of radius $1/n$ centered at $(1/n,0)$ . Then $\mathbb{E}=\bigcup_{n\in\mathbb{N}}C_n$ is the earring with basepoint $b_0=(0,0)$ . We need to set up a little more notation:

Let $X_n=\bigcup_{k=1}^{n}C_k$ be the bouquet of the first n-circles with free fundamental group $F_n=\pi_1(X_n,b_0)$ .
Let $r_{n+1,n}:X_{n+1}\to X_n$ be the retraction, which collapses $C_{n+1}$ to $b_0$ and is the identity elsewhere.
Let $r_n:\mathbb{E}\to X_n$ be the retraction, which collapses the smaller copy of the earring $\mathbb{E}_{\geq n+1}=\bigcup_{k=n+1}^{\infty}C_k$ to $b_0$ and is the identity elsewhere.

This gives an inverse system of retracts:

The closed mapping theorem should help convince you that the earring is homeomorphic to the inverse limit of this system of bouquets. Now apply $\pi_1$ to this inverse system: the maps $r_n:\mathbb{E}\to X_n$ induce homomorphisms $(r_n)_{\#}:\pi_1(\mathbb{E},b_0)\to F_n$ , which together induce a canonical homomorphism $\phi:\pi_1(\mathbb{E},b_0)\to\varprojlim_{n}(F_n,(r_{n+1,n})_{\#})$ defined by $\phi([\alpha])=([r_1\circ\alpha],[r_2\circ\alpha],[r_3\circ\alpha],\dots)$ .

up4

The inverse limit of free groups, which we can abbreviate as $\varprojlim_{n}F_n$ is precisely the first shape (or Cech) homotopy group $\check{\pi}_1(\mathbb{E},b_0)$ .

First, let’s notice that $\mathbb{E}$ is homeomorphic to $\varprojlim_{n}X_n$ . Basically, we just need to believe that $\mathbb{E}$ is precisely the infinite wedge $\bigvee_{n}S^1$ viewed as a subspace of the infinite torus $\prod_{n}S^1$ with the product topology. It’s then very tempting to think that $\phi$ is an isomorphism. However, $\pi_1$ doesn’t always preserve inverse limits!

Nevertheless, we can still understand and work with $\pi_1(\mathbb{E},b_0)$ if we identify it as a subgroup of $\varprojlim_{n}F_n$ . Hence, the motivation for showing that $\phi$ is injective.

Shape Injectivity Theorem: $\phi:\pi_1(\mathbb{E},b_0)\to\varprojlim_{n}(F_n,(r_{n+1,n})_{\#})$ is injective.

Basically, this theorem says that a loop $\alpha$ in $\mathbb{E}$ is null-homotopic if and only if every projection $r_n\circ\alpha$ is null-homotopic in the wedge of circles $X_n$ . The contrapositive says that $\alpha$ is not null-homotopic in $\mathbb{E}$ iff there exists some $n\geq 1$ such that $r_n\circ\alpha$ represents a non-trivial word in $F_n$ .

Why this is not so obvious

Let $\ell_n$ be the loop going once around $C_n$ counterclockwise and let $\ell_{n}^{-}$ be its reverse loop. Given a loop $\alpha:[0,1]\to\mathbb{E}$ , we may assume that for each component $(a,b)$ of $[0,1]\backslash\alpha^{-1}(b_0)$ , the restriction of $\alpha$ to $[a,b]$ is one of the paths $\ell_n$ or $\ell_{n}^{-}$ . Obviously, for each $n\geq 1$ , the loops $\ell_{n}$ and $\ell_{n}^{-}$ can show up as subloops at most finitely many times or we would violate the uniform continuity of $\alpha$ .

Suppose we know $r_n\circ\alpha$ is null-homotopic in $X_n$ . The primary difficulty is that the null-homotopies $H_n:[0,1]^2\to X_n$ for $r_n\circ\alpha$ might have nothing to do with each other. We just know one exists for each approximation level. There is no guarantee that we can “fix em up right” so that they agree with the bonding maps, i.e. satisfy $r_{n+1,n}\circ H_{n+1}=H_n$ and thus induce a null-homotopy of $\alpha$ in $X$ .

Here is a more algebraic way to look at it. It doesn’t hurt to think each projection loop $r_n\circ\alpha$ as an unreduced finite word in the letters $\{\ell_{k}^{\pm}\mid 1\leq k\leq n\}$ . Then $\ell_{n}^{m}$ means a concatenation $\ell_n\cdot\ell_n\cdots \ell_n$ of length $m$ and $\ell_{n}^{-m}$ means a concatenation $\ell_{n}^{-}\cdot\ell_{n}^{-}\cdots \ell_{n}^{-}$ of length $m$ . For instance, suppose $\alpha$ is the loop described by the following projections.

$r_1\circ\alpha\equiv \ell_{1}^{1}\ell_{1}^{-1}$
$r_2\circ\alpha\equiv\ell_{2}^{1}\ell_{2}^{-1}\ell_{1}^{1}\ell_{2}^{2}\ell_{2}^{-2}\ell_{1}^{-1}\ell_{2}^{3}\ell_{2}^{-3}$
$r_3\circ\alpha\equiv\ell_{3}^{1}\ell_{3}^{-1}\ell_{2}^{1}\ell_{3}^{2}\ell_{3}^{-2}\ell_{2}^{-1}\ell_{3}^{3}\ell_{3}^{-3}\ell_{1}^{1}\ell_{3}^{4}\ell_{3}^{-4}\ell_{2}^{2}\ell_{3}^{5}\ell_{3}^{-5}\ell_{2}^{-2}\ell_{3}^{6}\ell_{3}^{-6}\ell_{1}^{-1}\ell_{3}^{7}\ell_{3}^{-7}\ell_{2}^{3}\ell_{3}^{8}\ell_{3}^{-8}\ell_{2}^{-3}\ell_{3}^{9}\ell_{3}^{-9}$
and so on where between any two letters in the previous projection you insert an inverse pair of the form $\ell_{n}^{k}\ell_{n}^{-k}$ .

Notice that deleting the $\ell_2$ ‘s from $r_2\circ\alpha$ gives $r_1\circ\alpha$ , deleting the $\ell_3$ ‘s from $r_3\circ\alpha$ gives $r_2\circ\alpha$ , and so on. Moreover, each letter $\ell_n$ is only used finitely many times. We conclude that this does indeed give a loop in $\mathbb{E}$ . Notice that even though these finite projection words are getting pretty long, the homotopy class $[r_n\circ\alpha]$ will cancel to the trivial word in the free group $F_n$ . After all, we just inserted inverse pairs that cancel!

But the infinite limit loop $\alpha$ in $\mathbb{E}$ is a “transfinite word” of dense order. It is null-homotopic (due to the main result of the post) but it’s much harder to come up with an explicit contraction because there is no finite reduction scheme that can do the job. Between any two of the inverse pairs that you wanted to cancel in the n-th projection, you actually had letters $\ell_{n+1}^{\pm}$ up in the next level. You can’t just cancel in the $n$ -th level, forget about what you just did, and then move on up to the $n+1$ -st level.

The trouble is that the process of cancellation requires choice. Ok, there’s not much choice in cancelling the first word. But look at the second one.

$r_2\circ\alpha\equiv\ell_{2}^{1}\ell_{2}^{-1}\ell_{1}^{1}\ell_{2}^{2}\ell_{2}^{-2}\ell_{1}^{-1}\ell_{2}^{3}\ell_{2}^{-3}$

You could cancel all the $\ell_2$ ‘s first and then the remaining $\ell_1$ ‘s. Or you could cancel the middle $\ell_2$ ‘s first, then the $\ell_1$ ‘s, and then the remaining $\ell_2$ ‘s. It may not seem like a big deal but these are different homotopies! The only thing we have going for us it that we have some contracting reduction for each $[r_n\circ\alpha]$ . This leaves us with an infinite sequence of reductions for the projection words – one for each level. We have no idea if these reductions match up or can be chosen so that the projection of a reduction of $[r_n\circ\alpha]$ to the next level down is exactly the reduction for $[r_{n-1}\circ\alpha]$ . Sure, once I choose a reduction/null-homotopy for $[r_n\circ\alpha]$ , I can project it down to make sure the reductions on lower levels match up, but then you have to start over and worry about $[r_{n+1}\circ\alpha]$ . If you project down and fix all the lower levels and continue this process all the way up, you’re going to end up with a rearranging the reduction choice at each level infinitely many times. There is no guarantee this can be done “continuously.”

History

The first attempt to identify $\pi_1(\mathbb{E},b_0)$ was by H.B. Griffiths [3]. However, there was a critical error in Griffiths’ proof of the injectivity. The error was observed and a correct proof finally given (30 years later!) by Morgan and Morrison [4]. Many years back when I read the original proof for the first time, I was a bit unsatisfied with how specific and technical it all was. Later on, I read the proof given by Eda and Kawamura in [2], which felt more intuitive because all I had to do was understand inverse limits and believe a little continuum theory. Bonus: It applies to all spaces with Lebesgue covering dimension 1, not just $\mathbb{E}$ . The key idea is originally due to the work in [1] by Curtis and Fort from the 1950’s.

Trees and Inverse Limits

An important theme in wild topology is the idea of a space being “uniquely arc-wise connected.” Here an “arc” in a space $X$ refers to a subspace of $X$ homeomorphic to $[0,1]$ . The image of $0$ and $1$ in $X$ are the endpoints of the arc. A “simple closed curve” in $X$ is a homeomorphic copy of the unit circle $S^1$ .

Definition: A space $X$ is uniquely arc-wise connected if for all distinct points $x,y\in X$ , there is a unique arc in $X$ whose endpoints are $x$ and $y$ .

The next proposition gives another useful way to describe uniquely arc-wise connected spaces.

Proposition: If $X$ is uniquely arc-wise connected, then $X$ is path connected and contains no simple closed curves. The converse holds if $X$ is weakly Hausdorff.

Proof. Since $S^1$ is not uniquely arc-wise connected, one direction is obvious. Now suppose $X$ is weakly Hausdorff and not uniquely arc-wise connected. Then there are distinct arcs $A,B\subseteq X$ sharing the same endpoints. Since $A\neq B$ , without loss of generality, we may suppose there is a point $a\in A\backslash B$ . Note that since $X$ is weakly Hausdorff, $A$ and $B$ are closed. It follows that $A\cap B$ is non-empty and closed in $A$ and thus $A\backslash (A\cap B)$ is open in $A$ . Choosing a homeomorphism $h: [0,1]\to A$ , let $(c,d)$ be the component of $h^{-1}(A\backslash B\cap A)$ containing $h^{-1}(a)$ . Now $A_1=h([c,d])$ is a subarc of $A$ with endpoints $\{h(c), h(d)\}\subseteq A\cap B$ . If $B_1$ is the subarc of $B$ with endpoints $h(c)$ and $h(d)$ , then we have $A_1\cap B_1=\{h(c),h(d)\}$ . Now it’s clear that $A_1\cup B_1$ is a homeomorphic image of a circle, i.e. a simple closed curve. $\square$

The uniquely arc-wise connected spaces you’re most likely to already be familiar with are trees.

Definition: A simplicial tree is a one-dimensional simplicial complex without any cycles. A (topological) tree is a space, which is the geometric realization of a simplicial tree.

Basic algebraic topology tells us that trees are contractible and uniquely arc-wise connected. Since a tree $T$ is simply connected, between any two points $x,y\in T$ there is a single homotopy (rel. endpoints) class of paths from $x$ to $y$ . This means $\beta:[0,1]\to T$ of the unique arc from $x$ to $y$ is a reduced representative of the single homotopy class of paths from $x$ to $y$ in the sense that it has no null-homotopic subloops. This reduced representative is unique up to reparameterization. A non-reduced path in a tree would have some null-homotopic zig-zags that we could “delete” by a homotopy to obtain a reduced representative. Of course, there could be infinitely many zig-zags but since trees are semilocally simply connected, this is not much of an obstacle to overcome.

Now what about an inverse limit $\varprojlim_{n}(T_n,f_{n+1,n})$ of trees $T_n$ ? Informally, such an inverse limit “glues” together the trees $T_n$ according to their bonding maps. The result should be one-dimensional and if $f_{n+1,n}$ maps $x,y\in T_{n+1}$ to the same point of $T_n$ , then $f_{n+1,n}$ will send the unique arc connecting $x$ and $y$ to a finite topological subtree of $T_n$ . So there should be no way for a simple closed curve to magically appear in the gluing process. We’ll prove exactly this using the simplest proof I could come up with.

Recall that an inverse limit $\varprojlim_{n}(X_n,f_{n+1,n})=\{(x_n)\mid x_n\in X_n\text{ and }f_{n+1,n}(x_{n+1})=x_n\}$ is topologized as a subspace of $\prod_{n\in\mathbb{N}}X_n$ . If $f_n:X\to X_n$ are the projection maps, then a point $x\in X$ is represented by the sequence $x=(f_1(x),f_2(x),f_3(x),\dots)$ . A basic open neighborhood latex of $x$ is of the form $X\cap \prod_{n\in\mathbb{N}}U_n$ where $U_n$ is an open neighborhood of $f_n(x)$ and there is an $M$ such that $U_n=X_n$ for $n>M$ . Since the functions $f_{n+1,n}$ are continuous and $f_{n+1,n}(f_{n+1}(x))=f_n(x)$ , we may replace $U_2$ with $U_2\cap W_2$ where $f_{2,1}(W_2)\subseteq U_1$ . Terminating at $n=M$ , we can inductively replace $U_n$ with $U_n\cap W_n$ where $f_{n,n-1}(W_n)\subseteq U_{n-1}$ . In this way, we may take a basic open neighborhood $X\cap \prod_{n\in\mathbb{N}}U_n$ of $x$ to satisfy $f_{n+1,n}(U_{n+1})\subseteq U_n$ for $1\leq n\leq M-1$ and $U_n=X_n$ for $n>M$ .

Lemma: Suppose $X=\varprojlim_{n}(X_n,f_{n+1,n})$ is an inverse limit of Hausdorff spaces and $f_n:X\to X_n$ are the projection maps. If $A,B$ are disjoint compact subsets of $X$ , then there exists an $n\geq 1$ such that $f_n(A)\cap f_n(B)=\emptyset$ .

Proof. Suppose, to obtain a contradiction, that for every $n\geq 1$ there exists $a_n\in A$ and $b_n\in B$ such that $f_n(a_n)=f_n(b_n)$ . Notice that since the coordinates of each $a_n$ and $b_n$ must agree with the bonding maps $f_{n+1,n}$ , this means $f_k(a_n)=f_k(b_n)$ for all $1\leq k\leq n$ . Since $A$ and $B$ are compact, we may find a subsequences $\{a_{n_j}\}$ and $\{b_{n_j}\}$ that converge to $a\in A$ and $b\in B$ respectively. We’re going to prove that $\{b_{n_j}\}$ also converges to $a$ . Consider a basic open neighborhood $U=\prod_{n}U_n$ of $a$ . Let $N$ be the minimal $n$ such that $U_n\neq X_n$ . Since $\{a_{n_j}\}\to a$ , there exists a $J\geq 1$ such that $a_{n_j}\in \prod_{n}U_n$ for all $j\geq J$ . We can choose $J$ large enough so that $n_J>N$ . Now pick any $j\geq J$ . Since $f_{n_j}(a_{n_j})=f_{n_j}(b_{n_j})$ , we have $f_{k}(b_{n_j})=f_{k}(a_{n_J})\in U_k$ for all $1\leq k\leq n_j$ . Since $N<n_J\leq n_j$ , this implies that $f_{k}(b_{n_j})\in U_k$ for $1\leq k\leq N$ . Hence $b_{n_j}\in U$ (for $j\geq J$ ) and we conclude that $\{b_{n_j}\}\to a$ . However, this means the $\{b_{n_j}\}$ converges to both of the points $a$ and $b$ ; an impossibility in a Hausdorff space. $\square$

Remark: Notice that if $f_m(A)\cap f_m(B)=\emptyset$ , then it must also be the case that $f_k(A)\cap f_k(B)=\emptyset$ for $k\geq m$ . So for given $A$ and $B$ , we can choose $m$ to be as large as we want.

Theorem: An inverse limit $X=\varprojlim_{n}(T_n,f_{n+1,n})$ of trees contains no simple closed curves.

Proof. Since topological trees are always Hausdorff, $X$ is Hausdorff. Suppose, to obtain a contradiction, that $f:S^1\to X$ is an embedding. For $i=1,2,3,4$ , let $S^{1}_{i}$ be the intersection of $S^1$ and $i$ -th quadrant of the plane (include the bounding axes). Now $A_i=f(S^{1}_{i})$ are four (compact) arcs in $X$ the meet at endpoints to form the simple closed curve $f(S^1)$ . Let $x=f(1,0)$ and $y=f(-1,0)$ . Notice that $A_1\cap A_3=\emptyset$ and $A_2\cap A_4=\emptyset$ . Let $\gamma_i$ be the paths that trace the arcs $A_i$ with the orientation shown below so that $\gamma_1\cdot\gamma_2$ and $\gamma_4\cdot\gamma_3$ are injective paths from $x$ to $y$ .

ilim2

According to the previous Lemma (and the following Remark), if we denote the projection maps by $f_n:X\to T_n$ , then we can find an $m$ such that $f_m(A_1)\cap f_m(A_3)=\emptyset$ and $f_m(A_2)\cap f_m(A_4)=\emptyset$ . Note that $f_m(x)$ and $f_m(y)$ are distinct points in the tree $T_m$ (as $f_m(A_1)$ and $f_m(A_3)$ are disjoint) and are thus connected in $T_m$ by a unique arc. Let $\beta:[0,1]\to X$ trace out this arc from $f_m(x)$ to $f_m(y)$ . Now $\beta$ is the reduced representative of both $f_m\circ(\gamma_1\cdot\gamma_2)=(f_m\circ\gamma_1)\cdot(f_m\circ\gamma_2)$ and $f_m\circ(\gamma_4\cdot\gamma_3)=(f_m\circ\gamma_4)\cdot(f_m\circ\gamma_3)$ .

Considering the reduction of the path $(f_m\circ\gamma_1)\cdot(f_m\circ\gamma_2)$ to $\beta$ , we see that an initial segment $\beta|_{[0,s]}$ has image in $f_m(A_1)$ and the terminal segment $\beta|_{[s,1]}$ has image in $f_m(A_2)$ .
Considering the reduction of the path $(f_m\circ\gamma_4)\cdot(f_m\circ\gamma_3)$ to $\beta$ , we see that an initial segment $\beta|_{[0,t]}$ has image in $f_m(A_4)$ and the terminal segment $\beta|_{[t,1]}$ has image in $f_m(A_3)$ .

If $s<t$ , then $\beta([s,t])\subset f_m(A_2)\cap f_m(A_4)$ .
If $t<s$ , then $\beta([t,s])\subseteq f_m(A_1)\cap f_m(A_3)$ .
If $s=t$ , then $\beta(s)$ lies in every $f_m(A_i)$ .

An illustration of the two cancellations to $\beta$ that leads to Case 1. Note that a middle portion may cancel in the reduction to $\beta$ (indicated by the dashed lines).

In any of these cases, even the degenerate ones where one of $s$ or $t$ is $0$ or $1$ , we arrive at a contradiction. $\square$

Corollary: Every path-component of the limit of an inverse system of trees is uniquely arc-wise connected.

In Part II, the fact that an inverse limit of trees contains no simple closed curves will be a critical part of proving the Shape Injectivity Theorem.

References.

[1] M.L. Curtis and M.K. Fort, The fundamental group of one-dimensional spaces, Proc. Amer. Math. Soc. 10 (1959) 140-148.

[2] K. Eda and K. Kawamura, The fundamental groups of one-dimensional spaces, Topology and its Applications 87 (1998) 163-172.

[3] H.B. Griffiths, Infinite products of semigroups and local connectivity, Proc. London Math. Soc. (3), 6 (1956), 455-485.

[4] J.W. Morgan and I. Morrison, A van Kampen theorem for weak joins, Proceedings of the London Mathematical Society 53 (1986) 562-576.

9 Responses to Shape injectivity of the earring space (Part I)

Min Ro says:

August 4, 2019 at 8:41 pm

I should quit being lazy and do this myself, especially provided this excellent post, but if you know off-hand, can you compute the complex K-theory for the Hawaiian earring? And in particular is it a free abelian group?

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- Jeremy Brazas says:
  
  August 5, 2019 at 1:01 pm
  
  I’ve never seen this done but I think this would probably not be so bad. I’m pretty sure it’s free abelian because when you consider a vector bundle over it you have local triviality at the wild point. This means you can only have interesting bundle structure over finitely many of the circles.
  
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  - Min Ro says:
    
    August 5, 2019 at 5:07 pm
    
    Ah, sorry, I’m in the bad habit of talking about all the K-groups when using the term K-theory. You’re certainly right about the K^0 group. I was curious about the K^{-1} group. If I’m not mistaken, it should be isomorphic to the first cohomotopy group of classes of maps to the circle (with multiplication defined by the unit circle in the complex plane).
    
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Gabin Kolly says:

July 3, 2021 at 8:19 pm

Nice article!
I think that in the proposition about uniquely arc-wise connected space, the converse is not exactly correct: a path-connected weakly hausdorff space without simple closed curve is not arc-wise connected in general, so we need to suppose arc-wise connectedness.
Also, I’m a bit confused with the proof of the proposition about inverse limit of Hausdorff spaces: you deduce that a subset is sequentially compact from the fact that it is compact, something which is not true in general. Why can we conclude that here? In the theorem, if the trees are locally finite (each vertex has finitely many edges incident to it), they are metrizable spaces, hence the inverse limit is also a metrizable space, and therefore a subset is compact if and only if it is sequentially compact, and the proof would be correct. But I don’t see why it would be true for arbitrary trees, or for arbitrary Hausdorff spaces.

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- Gabin Kolly says:
  
  July 6, 2021 at 8:08 am
  
  I just discovered that for a Hausdorff space, path-connected is equivalent to arc-connected, so the first proposition is correct, sorry.
  
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