Shape injectivity of the earring space (Part II)

This is the sequel to Shape injectivity of the earring space (Part I)

We’re on our way to proving the canonical homomorphism $\phi:\pi_1(\mathbb{E},b_0)\to\varprojlim_{n}F_n$ from the earring group to the inverse limit of free groups is injective. Part I was mostly dedicated to proving that an inverse limit of trees contains no simple closed curve. I’d like to reiterate here that the proof I’m detailing is almost completely self-contained; we’ll need to review two classical results from Continuum Theory; both are proved in Sam Nadler’s very readable book [2].

As a bonus, I hope you’ll find, in this two-part post, another excellent example of why general mathematical theory is worth developing. Our proof uses several existence/structure theorems that, by themselves, don’t tell you how to do anything practical. However, they can be used together to prove the Shape Injectivity Theorem, which does provide something concrete: a practical way to study and do calculations in the most fundamental class of groups with non-commutative infinite products.

Peano Continua and Dendrites

Definition: A Peano continuum is a connected, locally path-connected, compact metrizable space.

The following theorem is an important characterization of Peano continua (See [Nadler, 8.18]) that every topologist should keep in their back pocket.

Hahn-Mazurkiewicz Theorem: A space $X$ is a Peano continuum if and only if it is Hausdorff and there is a continuous surjection $[0,1]\to X$ .

Definition: A dendrite is a Peano continuum containing no simple closed curve.

Based on an early lemma from Part I, we could define a dendrite to be a uniquely arc-wise connected Peano continuum. Intuitively, a dendrite is a one-dimensional Peano continuum without any holes.

First, consider the “arc hedgehog” space $ah(\omega)$ which is a one-point union of a shrinking sequence of arcs of length $1/2^n$ . This space came up in an early post about the category of locally path-connected spaces. It’s easy to see that $ah(\omega)$ is uniquely arc-wise connected and is therefore a dendrite.

The arc hedgehog dendrite

How complicated can a dendrite be? Start with $ah(\omega)$ , which has a single branch point, meaning that if we delete it, the subspace left has at least 3 components. At the midpoint $m$ of a segment of length $1/2^n$ , attach a copy of $ah(\omega)$ scaled to have diameter $1/2^{n+1}$ . Continue the process inductively in a dense pattern to construct the following dendrite called Wazewski’s Universal Dendrite.

Wazewski’s Universal Dendrite

Notice there are no open sets in the Universal Dendrite homeomorphic to an open interval. In fact, this dendrite contains a homeomorphic copy of every dendrite as a retract. Hence, as far as dendrites go, this is as complicated as they get.

The second result we’ll need from continuum theory provides us with a convenient way of writing down a dendrite as an inverse limit.

Dendrite Structure Theorem [Nadler, 10.27]: Every dendrite is homeomorphic to the inverse limit of a sequence of tree’s $T_n$ where $T_1=[0,1]$ , $\overline{T_{n+1}\backslash T_n}$ is an arc, and the bonding retractions $r_{n+1,n}:T_{n+1}\to T_n$ collapse the arc $\overline{T_{n+1}\backslash T_n}$ to the attachment point $\overline{T_{n+1}\backslash T_n}\cap T_n=\{p_n\}$ .

This structure theorem just tells you that some inverse system exists, it doesn’t help you find one. It’s a good exercise to figure out how you’d actually realize some dendrites as inverse limits of trees in the specified fashion. I recommend working out the arc-hedgehog space and Wazewski’s Universal Dendrite as examples. Hint: create an inverse system by enumerating the arcs by size.

We’ll use the Dendrite Structure Theorem to prove the last technical ingredient we’ll need. It appears as Exercise 10.51 in [2] and, I believe, is usually attributed to Borsuk.

Theorem: Dendrites are contractible.

Tempting but incomplete attempt. It’s tempting to try to quickly prove dendrites are contractible in the following way: pick point $v$ in dendrite $D$ . For each point $d\in D$ , let $A_d$ be the unique arc from $v$ to $d$ (which degenerates to a point when $v=d$ ). Now construct a homotopy $H:D\times [0,1]\to D$ mapping $\{d\}\times [0,1]$ bijectively to $A_d$ as a path from $v$ to $d$ and we’re done! Ok…so nothing is wrong yet (and indeed the correct contraction has this form) but we’re simply not done. A homotopy is, by definition, a parameterized path of paths. For each $d$ you actually need to choose a parameterization of $A_d$ and these all have to match up perfectly so that $H$ is continuous. It can be done by analyzing or altering the metric on $D$ ; however, my point here is that a rigorous proof requires some care.

Correct Proof of Dendrite Contraction Theorem. Suppose $D$ is a dendrite and that we have realized $D=\varprojlim_{n}(T_n,r_{n+1,n})$ as described in the Dendrite Structure Theorem, where $[0,1]=T_1\subseteq T_2\subseteq T_3\subseteq \dots$ are trees each with a single added edge. Since the bonding maps $r_{n+1,n}:T_{n+1}\to T_n$ are all retractions, we may take $v_0=0$ to be the basepoint for all of the trees. We will use the point $x_0=(v_0,v_0,v_0,\dots)$ in $D$ as the vertex of our contraction. Let $H_1:T_1\times [0,1]\to T_1$ be the canonical contraction $H_1(x,t)=xt$ . Recursively, suppose we have constructed contraction $H_{n-1}:T_{n-1}\times [0,1]\to T_{n-1}$ satisfying $H_{n-1}(x,0)=v_0$ and $H_{n-1}(x,1)=x$ . Define $H_n:T_n\times [0,1]\to T_n$ as follows: let $A=\overline{T_{n}\backslash T_{n-1}}$ be the added arc where $a_0\in T_{n-1}$ is the attaching point, Let $G_n:T_n\times [0,1]\to (T_{n-1}\times [0,1])\cup (A\times \{1\})$ be a deformation retraction (the identity on the codomain) satisfying $G_n(A\times [0,1])\subseteq \{a_0\}\times [0,1]\cup A\times \{1\}$ and $G_n(A\times \{0\})=(a_0,0)$ . Now let $K_n:(T_{n-1}\times [0,1])\cup (A\times \{1\})\to T_n$ be the map defined as $H_{n-1}$ on $T_{n-1}\times [0,1]$ and $K_n(a,1)=a$ for $a\in A$ . The composition $H_n=K_n\circ G_n$ is the desired contraction of $T_n$ .

contr3

We construct the contractions $H_n$ in this way so that they agree with the bonding maps on the nose, i.e. satisfy $H_{n}\circ (r_{n+1,n}\times id_{[0,1]})=r_{n+1,n}\circ H_{n+1}$ . We end up with the following inverse system:

contr

Taking the limit gives a map $\varprojlim_{n}H_n:\varprojlim_{n}(T_n\times [0,1])\to D$ . But since the second component of the top inverse system is always the identity of $[0,1]$ , there is a canonical homeomorphism $D\times [0,1]=\varprojlim_{n}T_n\times \varprojlim_{n}[0,1]\cong \varprojlim_{n}(T_n\times [0,1])$ . The composition provides our desired contraction $H:D\times [0,1]\to D$ . Some routine formula checks will show that $H(d,0)=x_0$ and $H(d,1)=d$ . $\square$

Proof of the Shape Injectivity Theorem

Finally, we get to the point. Ok, remember the homomorphism $\phi:\pi_1(\mathbb{E},b_0)\to \varprojlim_{n}F_n$ , $\phi([\alpha])=([r_1\circ\alpha],[r_2\circ\alpha],\dots)$ from the first post? To show it’s injective, we’ll show $\ker(\phi)$ is trivial. Suppose that $\alpha:[0,1]\to \mathbb{E}$ is a loop based at $b_0$ such that $r_n\circ\alpha:[0,1]\to X_n$ is null-homotopic for every $n\in\mathbb{N}$ . We must show that $\alpha$ is null-homotopic in $\mathbb{E}$ .

Each space $X_n$ is a wedge of circles and therefore has a universal covering space $\widetilde{X}_n$ , which is an infinite tree. In particular, $\widetilde{X}_n$ is the Caley graph of $F_n$ . Let $p_n:\widetilde{X}_n\to X_n$ be the universal covering map.

After making a choice of vertex basepoints $\tilde{x}_n\in (p_{n})^{-1}(b_0)$ , notice that the map $r_{n+1,n}\circ p_{n+1}:\widetilde{X}_{n+1}\to X_n$ from the simply connected covering space has a unique lift $s_{n+1,n}:(\widetilde{X}_{n+1},\tilde{x}_{n+1})\to (\widetilde{X}_{n},\tilde{x}_n)$ such that $p_n\circ s_{n+1,n}=r_{n+1,n}\circ p_{n+1}$ .

This gives us an inverse system of based covering maps.

Consider the inverse limit $\varprojlim_{n}p_n:\varprojlim_{n}\widetilde{X}_n\to \varprojlim_{n}X_n=\mathbb{E}$ . There are two things to be wary of: 1. the inverse limit of path-connected spaces is not always path-connected and 2. and an inverse limit of covering maps is not usually a covering map. But these general failures are not a deal-breaker in our situation.

First, pick a path component of $\varprojlim_{n}\widetilde{X}_n$ . Specifically, take $\widetilde{\mathbb{E}}$ to be the path component containing the basepoint $\tilde{b}_0=(\tilde{x}_1,\tilde{x}_2,\tilde{x}_3,\dots)\in \varprojlim_{n}\widetilde{X}_n$ . Let $p:\widetilde{\mathbb{E}}\to \mathbb{E}$ be the restriction of $\varprojlim_{n}p_n$ .

up1

While the map $p$ is not a traditional covering map, it is surjective and it enjoys all of the usual lifting properties of a covering map. It is a kind of “generalized covering map.” Also, the space $\widetilde{\mathbb{E}}$ is not a dendrite. In fact, with the subspace topology inherited from the inverse limit, it’s not even locally path connected. However, if you apply the locally path-connected coreflection to it, the result is a true generalized universal covering in the sense of [1]. For our purposes, we only need to recognize that $\widetilde{\mathbb{E}}$ is sitting inside an inverse limit of trees.

Lemma: $\widetilde{\mathbb{E}}$ is uniquely arc-wise connected. In particular, it contains no simple closed curves.

Proof. The main theorem proved in Part I is that the limit of an inverse system of trees contains no simple closed curves. Since each covering space $\widetilde{X}_n$ is a tree, this theorem applies and we conclude that $\varprojlim_{n}\widetilde{X}_n$ contains no simple closed curve. In particular, the path component $\widetilde{\mathbb{E}}$ contains no simple closed curve. $\square$

Recall that $\alpha:[0,1]\to \mathbb{E}$ is a loop based at $b_0$ such that, for all $n\in\mathbb{N}$ , the projection loop $\alpha_n=r_n\circ\alpha:[0,1]\to X_n$ onto the wedge of n-circles is null-homotopic in $X_n$ . We are looking to build a null-homotopy of $\alpha$ from a choice of null-homotopies of the approximating loops $\alpha_n$ despite the fact that these null-homotopies might seem completely unrelated.

up2

Let $\widetilde{\alpha}_n:([0,1],0)\to (\widetilde{X}_n,\widetilde{x}_n)$ be the unique lift of $\widetilde{\alpha}_n$ starting at $\widetilde{x}_n$ , i.e. so that $p_n\circ\widetilde{\alpha}_n=\alpha_n$ . Since $\alpha_n$ is null-homotopic in $X_n$ , it must be the case that each $\widetilde{\alpha}_n$ is actually a loop based at $\tilde{x}_n$ . The diagram below shows the following equalities hold: is8

is7

Since $s_{n+1,n}$ preserves the basepoints (by construction) and $\widetilde{\alpha}_n$ is the unique lift of $\alpha_n$ starting at $\widetilde{x}_n$ , the equality above tells us that $s_{n+1,n}\circ\widetilde{\alpha}_{n+1}=\widetilde{\alpha}_{n}$ . This means the lifted loops $\widetilde{\alpha}_{n}$ agree with the bonding maps of the covering space inverse system. The universal property of the top inverse system hands us a unique loop $\widetilde{\alpha}:[0,1]\to \varprojlim_{n}\widetilde{X}_n$ based at $\tilde{b}_0$ satisfying $s_n\circ \widetilde{\alpha}=\widetilde{\alpha}_n$ .

up3

Now $D=\widetilde{\alpha}([0,1])$ is the continuous image of $[0,1]$ in a Hausdorff space so, by the Hahn-Mazurkiewicz Theorem, $D$ is a Peano continuum. Moreover, since $D$ is path connected and contains $\tilde{b}_0$ , we have $D\subseteq \widetilde{\mathbb{E}}$ . Since $\widetilde{\mathbb{E}}$ contains no simple closed curves, neither does $D.$ Therefore, $D$ is a dendrite. Finally, we apply the theorem (from earlier in this post) all dendrites are contractible. Since $\widetilde{\alpha}$ factors through a contractible space, it is null-homotopic in $\widetilde{\mathbb{E}}$ . We conclude that $\alpha=p\circ\widetilde{\alpha}$ is null-homotopic in $\mathbb{E}$ . This completes the proof that $\ker(\phi)$ is trivial. $\square$

Concluding Thoughts

Where did the null-homotopy of $\alpha$ come from? It would have required a super-technical effort to build an explicit homotopy so we passed the hard work off to Borsuk’s Theorem that dendrites are contractible. Using Part I and the Hahn-Mazurkiewicz Theorem, we could say that the space $D=\widetilde{\alpha}([0,1])$ is a dendrite in the first place. Then follow any proof of the contractibility of dendrites to finish the job.

Notice that we basically didn’t use anything specific about the earring space except that the universal covers of the approximating spaces $X_n$ are trees! So we could replace the earring with any inverse limit of graphs and the same proof would go through. So actually, we proved:

One-Dimensional Shape Injectivity Theorem: If $(X,x_0)=\varprojlim_{n}(G_n,x_n)$ is an inverse limit of based graphs $G_n$ , then the canonical induced homomorphism $\phi:\pi_1(X,x_0)\to\varprojlim_{n}\pi_1(G_n,x_n)$ to the inverse limit of free groups is injective.

For example, any one-dimensional Peano continuum (including the Menger Curve and Sierpinski Carpet) can be written as the inverse limit of finite graphs and falls within the scope of this useful theorem.

If you feel comfortable with the end of the proof, then you can also prove as a quick exercise that every inverse limit of graphs is aspherical, i.e. has trivial higher homotopy groups!

References.

[1] H. Fischer and A. Zastrow, Generalized universal covering spaces and the shape group, Fund. Math. 197 (2007) 167-196.

[2] S. Nadler, Continuum Theory: An Introduction, Chapman & Hall/CRC Pure and Applied Mathematics. 1992.

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