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 second approach is to identify 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).


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.


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.”


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.


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).
  1. If s<t, then \beta([s,t])\subset f_m(A_2)\cap f_m(A_4).
  2. If t<s, then \beta([t,s])\subseteq f_m(A_1)\cap f_m(A_3).
  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.


[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.

This entry was posted in Free groups, Fundamental group, Hawaiian earring, Inverse Limit, Tree. Bookmark the permalink.

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

  1. Min Ro says:

    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?


    • 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.


      • Min Ro says:

        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).


  2. Pingback: The Hawaiian Earring Group | Wild Topology

  3. Pingback: Shape Injectivity of the Hawaiian Earring Part II | Wild Topology

  4. Pingback: Homotopically Reduced Paths (Part III) | Wild Topology

  5. Gabin Kolly says:

    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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