Homotopically Reduced Paths (Part II)

Posted on December 23, 2020 by Jeremy Brazas

Understanding this post requires reading Part I where I explain what reduced paths are and how we might go about proving they exist. In this second part, I’ll go into detail about how to use Zorn’s lemma to identifying a maximal cancellation and therefore ensure every path is path-homotopic to a reduced path. The argument we’ll use is a generalization of Cannon and Conner’s [3, Theorem 3.9], which is based on Curtis and Fort’s original argument in [4, Lemma 3.1]. Both of those papers restrict only to one-dimensional spaces and here we’re going to include lots of other spaces using a definition introduced in [2]. While writing this post, I’ve been surprised at how far the argument can be generalized.

We’ll use the same notation as in Part I: X is a space, \alpha:[0,1]\to X is a non-constant path, and \mathcal{C}(\alpha) is the partially ordered set of cancellations of \alpha.

When can we be sure that pinching off null homotopic loops on a maximal cancellation results in a reduced loop?

At the start of Part I, I gave some examples of familiar spaces where some or all homotopy classes of loops failed to have any reduced representative. This means in order to continue this discussion, we must identify some hypothesis on X that will allow us to move forward. The following definition has shown up in my work a good bit in the past two years. Curiously, it’s popping up again in a natural way. To distinguish notation, note that \pi_1 denotes the fundamental group and \Pi_1 denotes the fundamental groupoid.

Definition 8: A space X has well-defined transfinite \Pi_1-products if for any closed set A \subseteq [0,1] containing \{0,1\} and paths \alpha,\beta:[0,1]\to X such that \alpha|_{A}=\beta|_{A} and \alpha|_{[a,b]}\simeq \beta|_{[a,b]} for every component (a,b) of [0,1]\backslash A, we have \alpha\simeq\beta.

This definition says that all infinite product operations \{[\alpha_j]\}_{j\in L}\mapsto \prod_{j\in L}[\alpha_j] (indexed by any countable linear order L) in the fundamental groupoid \Pi_1(X) are well-defined on homotopy classes: homotopic factors result in homotopic products. For instance, in the case that A=\{0,1/2,2/3,3/4,4/5,...\}, having well-defined transfinite products means that if we have path-homotopies \alpha_1\simeq\beta_1, \alpha_2\simeq \beta_2, \alpha_3\simeq \beta_3,… and we can form the infinite concatenations , then we have \prod_{n=1}^{\infty}\alpha_n\simeq \prod_{n=1}^{\infty}\beta_n (see the figure below). Why this is a hypothesis (and doesn’t just always happen for free) is because you may have no control over the size of the homotopies \alpha_n\simeq \beta_n and you need them to shrink in order to form a continuous homotopy on the product.

If the space has well-defined infinite products (a special case of transfinite products) then if the top (blue) paths can be deformed into the bottom (red) paths individually, then the entire top path can be homotoped to the entire bottom path.

The set A in the definition could be the Cantor set, which means that we also need to consider the scenario in which the components of [0,1]\backslash A are densely ordered. That’s the worst case scenario, which is still not so bad, since there can only be countably many components. Spaces with well-defined transfinite \Pi_1-products abound, including one-dimensional spaces, planar sets, and lots of other spaces where “small null-homotopic loops are bounded by small disks.”

One remarkable thing about Definition 8 is that, for metrizable X, this slightly-more-algebraic-flavored property is equivalent to X admitting a generalized universal covering space . I haven’t had a chance to write about those yet, but they’re expected to be the primary tool for understanding wild higher homotopy groups.

Lemma 9 (Construction of reduced paths): Suppose X has transfinite \Pi_1-products, \alpha:[0,1]\to X is a path, \mathscr{M} is a maximal cancellation for a \alpha and A=[0,1]\backslash\bigcup\mathscr{M}. If \beta:[0,1]\to X is the path induced by a collapse function r:[0,1]\to[0,1] for \bigcup\mathscr{M}, i.e. \beta\circ r|_{A}=\alpha|_{A}, then \beta is reduced and \alpha\simeq\beta.

Proof. Recall that since \mathscr{M} is maximal any two open intervals, which are elements of \mathscr{M} must have disjoint closures. Let A=[0,1]\backslash \bigcup\mathscr{M} so that \mathscr{M} is the set of connected components of [0,1]\backslash A. Let \gamma:[0,1]\to X be the path, defined as \alpha on A and constant on each element of \mathscr{M} (just like in the Pinching-off Lemma in Part I). If (a,b)\in \mathscr{M}, then \alpha|_{[a,b]} and \gamma|_{[a,b]} are both null-homotopic. So of course, \alpha|_{[a,b]}\simeq\gamma|_{[a,b]}. Therefore, since X is assumed to have well-defined transfinite \Pi_1-products, we have \alpha\simeq\gamma. Now by Lemma 2 in Part I, we may find a collapse function r for \bigcup\mathscr{M} and a nowhere constant path \beta:[0,1]\to X such that \beta\circ r=\gamma. As pointed out in the comment after Lemma 2, we have \beta\simeq\gamma. Therefore, \alpha\simeq\beta.

Finally, we need to make sure that \beta is reduced. Suppose to the contrary that there are 0\leq c<d\leq 1 such that \beta|_{[c,d]} is a null-homotopic loop. Find c',d'\in A=[0,1]\backslash \bigcup\mathscr{M} such that 0\leq c'<d'\leq 1 and r(c')=c and r(d')=d. Since \gamma|_{[c',d']}=\beta|_{[c,d]}\circ r|_{[c',d']}, it must be that \gamma|_{[c',d']} is a null-homotopic loop. But remember that \alpha|_{[c',d']} agrees with \gamma|_{[c',d']} on [c',d']\backslash\bigcup\mathscr{M} and on any I\in \mathscr{M} that happens to lie in the interior of [c',d'], \alpha is null-homotopic and \gamma is constant. We once again apply the assumption that X has well-defined transfinite \Pi_1-products to see that \alpha|_{[c',d']}\simeq\gamma|_{[c',d']}. Hence, \alpha|_{[c',d']} is a null-homotopic loop. This allows us to define \mathscr{M}'=\{(I\in\mathscr{M}\mid I\cap [c',d']=\emptyset\}\cup \{(c',d')\}. Let’s see why this gives us a contradiction.

If there was no I\in \mathscr{M} contained in (c',d') or if there is some I\in \mathscr{M} such that I is a proper subset of (c',d'), then \mathscr{M}' is a strictly larger cancellation than \mathscr{M}. This would violate the maximality of \mathscr{M}. Because c',d'\notin\bigcup\mathscr{M}, the only other possibility is that (c',d')\in\mathscr{M}. However this would mean that r([c',d']) is a point, contradicting the assumption that c=r(c')<r(d')=d. Either way, we run into a problem. So it must be that \beta is reduced. \square

Comment about understanding Lemma 9: Lemma 9 is packed with terminology from Part I so here’s how you can think about it. As long as X has this well-defined products property, then given any maximal cancellation \mathscr{M}\in\mathcal{C}(\alpha), we can delete the subloops of \alpha defined on the components of \mathscr{M} (each of which is null-homotopic). I wrote the statement of Lemma 9 to avoid the case where \alpha itself is a null-homotopic loop. In that case, \beta and r won’t exist and you can always represent with the constant loop anyway.

When do maximal cancellations exist?

Ok, so this is the last piece of the puzzle and this is the one that requires Zorn’s Lemma. What we’re trying to do is start with a path \alpha and show that the partially ordered set \mathcal{C}(\alpha) of all cancellations of \alpha has a maximal element. So this is ripe for an application of Zorn’s Lemma, which is equivalent to the Axiom of Choice. You shouldn’t really expect this to work without Zorn’s lemma because maximal cancellations are not unique and very much are like choosing a contraction of a tree.

One again, the well-definedness property from Definition 9 is the natural hypothesis to complete the proof.

Lemma 10 (Existence of Maximal Cancellations): Suppose X has well-defined transfinite \Pi_1-products and \alpha:[0,1]\to X is a non-constant path. Then there exists a maximal cancellation \mathscr{M} for \alpha.

Proof. Let \mathcal{L} be a linearly ordered subset of the partially ordered set \mathcal{C}(\alpha). The result will follow from Zorn’s Lemma, if we can show that \mathcal{L} is bounded above in \mathcal{C}(\alpha). Let U=\bigcup_{\mathscr{U}\in\mathcal{L}}\left(\bigcup\mathscr{U}\right). In otherwords, just union all of the open intervals together. Take \mathscr{M} be the set of connected components of U. The main thing we’ll need to do is prove that \mathscr{M} is a cancellation. However, once this is done it’s not too hard to see that \mathscr{U}\leq\mathscr{M} in \mathcal{C}(\alpha) for all \mathscr{U}\in\mathcal{L}.

The trickier part is showing that \mathscr{M} is a cancellation, i.e. that \alpha|_{[a,b]} is a null-homotopic loop for all (a,b)\in \mathscr{M}. Let’s fix such a connected component (a,b) of U. Notice that a,b\notin U. We need to pause and proof the following claim.

Claim: For every closed interval [c,d]\subseteq (a,b), there exists a \mathscr{U}\in\mathscr{L} and (x,y)\in \mathscr{U} such that [c,d]\subseteq (x,y)\subseteq (a,b).

Proof of Claim. Fix [c,d]\subseteq (a,b). For each t\in [c,d], we have t\in U and so there exists \mathscr{U}_t\in\mathscr{L} and I_t\in\mathscr{U}_t such that t\in I_t\subseteq (a,b). Now \{I_t\mid t\in [c,d]\} is an open cover of the compact space [c,d] by open sets in (a,b) and so we may find a finite subcover \{I_{t_1},I_{t_2},\dots ,I_{t_m}\}. Since \mathscr{L} is a linear order, \mathscr{V}=\max\{\mathscr{U}_{t_i}\mid 1\leq i\leq m\} exists. For s\in [c,d], we have s\in I_{t_i} for some i and since \mathscr{U}_{t_i}\leq \mathscr{V}, we have I_{t_i}\subseteq (x,y) for some (x,y)\in\mathscr{V}. We must have (x,y)\subseteq (a,b) since a,b\notin U.

Equipped with this now-proven Claim, we’re really going places. Start with some [c_1,d_1]\subseteq (a,b). Find

a<\cdots <c_3<c_2<c_1<d_1<d_2<d_3<\cdots b

such that \{c_n\}\to a and \{d_n\}\to b. Using the Claim, for every n\in\mathbb{N}, we may find \mathscr{U}_n\in\mathcal{L} and (x_n,y_n)\in\mathscr{U}_n such that [c_n,d_n]\subseteq (x_n,y_n)\subseteq (a,b). This means that:

  • (x_1,y_1)\subseteq (x_2,y_2)\subseteq (x_3,y_3)\subseteq \cdots\subseteq (a,b)
  • \displaystyle\bigcup_{n\in\mathbb{N}}(x_n,y_n)=(a,b)
  • Since each \mathscr{U}_n is a cancellation, \alpha|_{[x_n,y_n]} is a null-homotopic loop!

Important early observation: Since the sequence \{\alpha(x_n)\}=\{\alpha(y_n)\} converges to both \alpha(a) and \alpha(b) and we’ve assumed from the start that X is Hausdorff, it must be the case that \alpha(a)=\alpha(b). All of the basepoints involved might be different but that’s ok… we’ll still work it out.

Define

\gamma_m=\begin{cases} \alpha|_{[y_{m},y_{m+1}]} , &\text{ if }m\geq 1 \\ \alpha|_{[x_1,y_1]} , &\text{ if }m=0 \\ \alpha|_{[x_{-m-1},x_{-m}]}, &\text{ if }m\leq -1 \end{cases}

Warning: It is possible that x_n=x_{n+1} or y_{n+1}=y_n. In fact, we could have x_n and/or y_n=b for sufficiently large n. To make sure that this proof doesn’t require a bunch of tedious cases, we allow \alpha|_{[p,q]} to mean the constant loop at \alpha(p) when p=q.

Now, \alpha|_{[x_n,y_n]} is homotopic to the \mathbb{Z}-indexed concatenation \cdots \gamma_{-3}\cdot\gamma_{-2}\cdot\gamma_{-1}\cdot\gamma_{0}\cdot \gamma_{1}\cdot\gamma_{2}\cdot\gamma_{3}\cdots. At worst, we’re inserting a sequence of constant subpaths and reparameterizing, which never changes the path-homotopy class.

Now we go back and use that third bullet point. First, \gamma_0=\alpha|_{[x_1,y_1]} is null-homotopic. Also, for n\geq 1, we have [\alpha|_{[x_{n+1},y_{n+1}]}]= [\gamma_{-n}][\alpha|_{[x_n,y_n]}][\gamma_{n}] where \alpha|_{[x_n,y_n]} and \alpha|_{[x_{n+1},y_{n+1}]} are null-homotopic loops. Hence \gamma_{n}^{-}\simeq \gamma_{-n} for all n\geq 1.

Define \beta:[a,b]\to X to be the loop, which is defined as \gamma_0 on [x_1,y_1], \gamma_n on [y_n,y_{n+1}] and \gamma_{n}^{-} on [x_{n+1},x_{n}].

Notice that A=\{a,b\}\cup \{x_n\mid n\geq 1\}\cup\{y_n\mid n\geq 1\} is closed and and for every component latex I of [a,b]\backslash A, we have \alpha|_{\overline{I}}\simeq \beta|_{\overline{I}}. Therefore, since X is assumed to have well-defined transfinite \Pi_1-products, we have \alpha|_{[a,b]}\simeq\beta. However, \beta is a reparemeterization of the path conjugate \left(\prod_{n=1}^{\infty}\gamma_n\right)^{-}\cdot \gamma_0\cdot \left(\prod_{n=1}^{\infty}\gamma_n\right) of the null-homotopic loop \gamma_0. Since \beta is a null-homotopic loop, it follows that \alpha|_{[a,b]} is null-homotopic.

Whew! Let’s put a little square here and be done with it. \square

An observation for those familiar with the Homotopically Hausdorff property: The well-defined transfinite products property is required when you run into densely ordered products. We really only needed products indexed by the naturals in the proof of Lemma 10. The subtleties of well-defined infinite path/loop products are studied in detail in [1] and are shown to be equivalent to the homotopically Hausdorff property when X is first countable. So Lemma 10 could be strengthened as follows: Suppose X is first countable and homotopically Hausdorff and \alpha:[0,1]\to X is a non-constant path. Then there exists a maximal cancellation \mathscr{M} for \alpha.

Putting it all together

Finally, combining Lemmas 9 and 10, we finish off the existence of reduced paths.

Theorem 11: If X has well-defined transfinite \Pi_1-products, then every path in X is path-homotopic to a reduced path induced by deleting subloops on a maximal cancellation.

Proof. If \alpha:[0,1]\to X is a null-homotopic loop, then the constant loop is the desired reduced representative. Otherwise, \alpha is a non-constant path. Lemma 10 ensures that \mathcal{C}(\alpha) contains a maximal cancellation \mathscr{M}. Lemma 9 ensures that if A=[0,1]\backslash\bigcup\mathscr{M} and \beta:[0,1]\to X is the path induced by a collapse function r:[0,1]\to[0,1] for \bigcup\mathscr{M} (these details were laid out in Part I), then \beta is reduced and \alpha\simeq\beta. \square

What else is there to wonder about?

To what extent are reduced representatives of paths unique? In general, there is definitely no uniqueness within homotopy classes – recall the cylinder example from Part I. Even relative to the subloop deletion construction, the result won’t be unique. For suppose that \gamma_1,\gamma_2:[0,1]\to X are reduced paths that are path-homotopic to each other. Then \alpha=\gamma_{1}\cdot\gamma_{1}^{-}\cdot\gamma_{2}, reduces to \gamma_1 using the cancellation \{(1/3,1)\} or \gamma_2 using the cancellation \{(0,2/3)\}. There is choice involved! That choice will typically result in one of many possible reduced path results. This shouldn’t be too surprising though. It’s a slightly more topological version of the fact that in a group that is not free, shortest representation of an element using finite products of generators may not be unique.

However, I hinted in Part I that for one-dimensional spaces that every path-homotopy class has a unique (up to reparameterization) reduced path – I love this result. Maybe a Part III?

References:

[1] J. Brazas, Scattered products in fundamental groupoids. Proc. Amer. Math. Soc. 148 (2020), no 6, 2655-2670. arXiv

[2] J. Brazas, H. Fischer, Test map characterizations of local properties of fundamental groups. Journal of Topology and Analysis. 12 (2020) 37-85. arXiv

[3] J.W. Cannon, G.R. Conner, On the fundamental groups of one-dimensional spaces, Topology Appl. 153 (2006) 2648–2672.

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

Posted in Fundamental groupoid, reduced paths | Tagged , , , , , , , | 1 Comment

Homotopically Reduced Paths (Part I)

Posted on December 18, 2020 by Jeremy Brazas

The topic of this post focuses on a general concept that is heavily used in the wild topology world. Writing this post has been a fun exploration I’ve been meaning to spend time on for a while now. I will assume that all spaces involved are Hausdorff.

Definition 1: A path \alpha:[0,1]\to X is (homotopically) reduced if either \alpha is constant or if there is no interval [a,b]\subseteq [0,1] such that \alpha|_{[a,b]} is a null-homotopic loop, i.e. if \alpha has no null-homotopic subloops.

I’ll usually just refer to a homotopically reduced path as a reduced path.

There are some other, more geometric, notions of “homotopically reduced” that are good for other purposes but this one is particularly relevant for one-dimensional spaces. Soon, we’ll combine new stuff with an old post to see that every path in a one-dimensional Hausdorff space is path-homotopic to a unique reduced path (unique up to reparameterization). Just as the uniqueness of reduced words in free groups is central to their theory and applications, the same kind of uniqueness for homotopy classes is important for proving things about fundamental groups of one-dimensional spaces.

History: The idea behind reduced paths in the one-dimensional case is based off of Curtis and Fort’s work in [2] from the 1950’s. I should note that Curtis-Fort use the notion of a “normal loop” which permits constant subpaths and is not exactly the same as in Definition 1. The modern definition of “reduced path,” which is far more effective for applications, does not permit constant subpaths. This modern version appears to appear first in Section 2 of Eda’s 2002 paper [3]. There is also a nice proof in Cannon and Conner’s outstanding 2006 paper [1].

Fun question to ponder: What do you think a homotopically reduced map [0,1]^n\to X should be for n\geq 2?

Let’s start with some basic observations.

  1. A homotopically reduced path \alpha:[0,1]\to X that is not a constant path must be nowhere constant, i.e. there is no open set U\subseteq [0,1] on which \alpha is constant. For then we could find [a,b]\subseteq U such that \alpha|_{[a,b]} is the constant loop, which is null-homotopic.
  2. In general, it’s possible for a path-homotopy class to be represented by many different reduced paths. For instance let X=S^1\times [0,1] be the cylinder and for z\in[0,1], consider the paths  \ell_{z}(t)=\left(\cos(2 \pi t),\sin(2 \pi t),z\right) and \mu_z(t)=(1,0,tz). Now the family of paths \mu_{z}\cdot \ell_{z}\cdot \mu_{z}^{-}, z\in [0,1] are all reduced paths that represent the same path-homotopy class. See the gif below.

    A family of reduced loops all representing the same homotopy class.

    It’s worth pointing out that the cylinder S^1\times [0,1] is a 2-dimensional space (with whatever notion of topological dimension is your favorite).

  3. If \mathbb{G} is the Griffiths twin cone with basepoint b_0, then every non-trivial homotopy class 1\neq [\alpha]\in \pi_1(\mathbb{G},b_0) has no reduced representative. This is true since every loop \gamma:[0,1]\to \mathbb{G} satisfying \alpha^{-1}(b_0)=\{0,1\} is null-homotopic in \mathbb{G}. You can start with any loop and begin pinching off null-homotopic subloops and you’ll never stop or arrive at a reduced loop. In fact it’s worse than that. No loop based at b_0 except for the constant loop will be reduced. Interestingly, only some elements of the fundamental group of the harmonic archipelago fail to have reduced representatives. Can you find one?

What I’d like to do in this post is discuss the existence of reduced paths in homotopy classes. When can we be sure that a path \alpha:[0,1]\to X is path-homotopic to some reduced path?

Say we start with a non-constant path \alpha. We want to “pinch off” or delete subloops which are null-homotopic loops. If you want to pinch things off one-by-one you’ll end up in an infinite deletion procedure, which could get pretty messy. So I think we should try to delete infinitely many subloops at a time. For example, if you have an infinite concatenation \displaystyle\prod_{n=1}^{\infty}(\alpha_n\cdot\alpha_{n}^{-}) of inverse pairs you can delete the inverse pairs one-by-one or you could just delete them all at the same time. One difficulty we could face is that if \prod_{n=1}^{\infty}\alpha_n=\alpha_1\cdot\alpha_2\cdot\alpha_3\cdots is an infinite concatenation of null-homotopic loops, this product itself might not be null-homotopic. This phenomenon occurs precisely when the space in question fails to be homotopically Hausdorff, a concept very relevant to this post.

A Lemma about deleting constant subloops

Lemma 2: For every path \alpha:[0,1]\to X, there exists a non-decreasing continuous function r:[0,1]\to [0,1] and a nowhere constant path \beta:[0,1]\to X such that \beta\circ r=\alpha.

Proof. Let U be the set of t\in[0,1] such that \alpha is locally constant at t, i.e. there is an open neighborhood of t on which \alpha is constant. Let C(U) be the connected components of U and notice that if I,J\in C(U), then I and J must have disjoint closures. We give C(U) the natural linear ordering inherited from [0,1]. For each I\in C(U), pick a rational number q_I\in I. We define r as follows:

  • for each I\in C(U), we set r(\overline{I})=q_I.
  • if I=(a,b)<J=(c,d) in C(U) are consecutive (no element of C(U) is between them) then for t\in (b,c), we use the linear function \displaystyle r(t)=\frac{q_J-q_I}{c-b}(t-b)+q_I. Doing so adds the line segment connecting (b,q_I) and (c,q_J) to the graph of r.
  • If we haven’t defined r(t) yet, then t is the limit point of a monotone sequence of intervals I_n in C(U), in the sense that \{q_{I_n}\}\to t. Therefore, we define r(t)=t.

A little real analysis will finish the proof that r:[0,1]\to[0,1] is a well-defined, continuous, non-decreasing function. As a specific example, if U is the complement of the middle-third Cantor set, and we choose the q_I to be the dyadic rational that is the midpoint of I. In this case, r will be a modified version of the Cantor function.

The Cantor Function

In general, r is always a kind of step function like the Cantor function that is constant on the components of U.

Let’s finish this argument. Since r is constant precisely on the components of U. Since r is a quotient map, and \alpha is constant on the fibers of r, there exists a unique map \beta:[0,1]\to X such that \beta\circ r=\alpha.

How do we know \beta is nowhere constant? Suppose otherwise that there exists a<b such that \beta((a,b))=x\in X. Since r is non-decreasing, continuous, and onto there exists c<d such that r(c)=a, r(d)=b, and r((c,d))=(a,b). Now \alpha((c,d))=\beta(r((c,d))=\beta((a,b))=x and so we must have (c,d)\subseteq U. By the definition of r, this means that r((c,d)) is a single point in [0,1], which contradicts a<c. Thus \beta cannot be constant on any open subset of [0,1]. \square

Here’s why that lemma is a useful and necessary starting place: Since \alpha=\beta\circ r, we have that \alpha and \beta are path-homotopic by the homotopy H(s,t)=\beta(r(s)t+s(1-t)). So, if we’re given a path \alpha, we can go ahead and delete a maximal family of constant subloops in one single step without changing the path-homotopy class. Hence, toward our goal for this post, we may assume from the start that \alpha is nowhere constant.

But….to proceed we should dissect the proof to formalize the idea of “pinching off subloops.”

Definition 3: The function r doesn’t need \alpha to be constructed. In fact, given any open set U in [0,1] and the choice of a point in each connected component of U, we can construct the function r. We’ll refer to such a function as a collapse function for U.

Notice that if r is a collapse function for U and A=[0,1]\backslash U, then r(A)=[0,1].

Lemma 4 (Pinch-off Lemma): Suppose \alpha:[0,1]\to X is a path and A\subseteq [0,1] is a closed set such that for each connected component I\in C([0,1]\backslash A), \alpha|_{\overline{I}} is a loop, that is, \alpha maps the two points of \partial I to a single point. If r is a collapse function for [0,1]\backslash A, then there exists path \beta:[0,1]\to X such that \beta\circ r|_{A}=\alpha|_{A}.

Proof. Suppose \alpha(\partial I)=x_I for I\in C(U). Define an intermediate path \gamma:[0,1]\to X so that

\gamma(t)=\begin{cases} \alpha(t), & \text{ if }t\in A\\ x_I, & \text{ if }t\in I, I\in C([0,1]\backslash A)\end{cases}

In other words, \gamma is the same as \alpha except that you force it to be constant on each component I of U.

Altering a path to be constant on a given (possibly infinite) family of subloops.

Because \alpha is continuous, \gamma is continuous. Moreover, \gamma is now constant on each component of U. Since r is a quotient map and \gamma is constant on the fibers of r, there exists a unique path \beta:[0,1]\to X such that \beta\circ r=\gamma. Thus \beta\circ r|_{A}=\gamma|_{A}=\alpha|_{A}

Now this new pinch-off lemma is, in a way, much stronger than our first lemma. But we must be responsible with all this power. It just says that we can always pinch of subloops to obtain some continuous path. It doesn’t say that the result will be homotopic to the original….The resulting path will be “obviously” path-homotopic to the original only if we’re pinching off just finitely many null-homotopic subloops.

Maximal Cancellations

Definition 5: A cancellation of a path \alpha is a collection \mathscr{U} of disjoint, connected open sets in [0,1] such that for each I\in \mathscr{U}, the subpath \alpha|_{\overline{I}} is a null-homotopic loop.

Let \mathcal{C}(\alpha) be the set of all cancellations of \alpha. Notice that \mathcal{C}(\alpha) has a natural partial order: given \mathscr{U},\mathscr{V}\in\mathcal{C}(\alpha), we say \mathscr{U}\leq \mathscr{V} if for every I\in \mathscr{U}, there exists a J\in \mathscr{V} such that I\subseteq J.

Definition 6: A cancellation \mathscr{M}\in \mathcal{C}(\alpha) is maximal if it is a maximal element of the partially ordered set \mathcal{C}(\alpha),\leq).

Observation 7: If \mathscr{M} is a maximal cancellation of \alpha, then the elements of \mathscr{M} have disjoint closures. For if we have (a,b),(b,c)\in \mathscr{M}, then \alpha_{[a,c]}\simeq \alpha_{[a,b]}\cdot \alpha_{[b,c]} would be the concatenation of two null-homotopic loops and would therefore be null-homotopic. This would mean we could define \mathscr{M}'\in\mathcal{C}(\alpha) by replacing (a,b) and (b,c) with (a,c). This would give \mathscr{M}'>\mathscr{M}, violating the maximality of \mathscr{M}.

So, we want to know two things:

  1. When does a path \alpha admit a maximal cancellation?
  2. If \alpha does admit a maximal cancellation \mathscr{M}, and we form \beta using a collapsing function r for \bigcup\mathscr{M}, must \beta be path-homotopic to \alpha?

We’ll work to answer these questions in Part II.

References:

[1] J.W. Cannon, G.R. Conner, On the fundamental groups of one-dimensional spaces, Topology Appl. 153 (2006) 2648–2672.

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

[3] K. Eda, The fundamental groups of one-dimensional spaces and spatial homomorphisms, Topology Appl. 123 (2002) no. 3, 479-505.

Posted in Fundamental group, Homotopy theory | Tagged , | 2 Comments

Testing the limits of Eda’s Theorem

Posted on October 18, 2020 by Jeremy Brazas

With everything going on, it’s been a bit tough to find time to post about the things I’d like to. I’ve promised posts about topological homotopy groups for a while but I want to do those right, so it may take longer to find the time to write them how I’d really like to. This post is about a cute counterexample I was thinking about the other day.

What is Eda’s Theorem? Here, I’m referring to Katusya Eda’s “homotopy classification of one-dimensional Peano continua” [1] which states that if X,Y are one-dimensional Peano continua (connected, locally path-connected, compact metric spaces) and \pi_1(X,x_0)\cong \pi_1(Y,y_0), then X and Y are homotopy equivalent. Why is this amazing? Because it includes spaces like the Hawaiian earring, double Hawaiian earring, double Hawaiian earring, dyadic arc space, wild circle, Sierpinski triangle and square, Menger Cube, etc. Even though these fundamental groups are very large, their algebraic structure is so rigid that it completely determines the homotopy type of the space. 

Why do I care so much? Well… fundamental group(oid)s of one-dimensional spaces are basically “free” objects among group(oid)s with natural infinite product operations. Everything about them – from the fact that each homotopy class has a reduced representative (just like in a graph) to the fact that they have \mathbb{R}-tree “generalized” coverings (just like graphs have trees as universal covers) tells you that this is the case.

A key part of the proof of Eda’s Theorem is that any homomorphism f:\pi_1(X,x_0)\to \pi_1(Y,y_0) is induced up to basepoint-change, that is, there exists a map g:X\to Y and a path \beta:[0,1]\to Y from g(x_0) to y_0 such that if \varphi_{\beta}:\pi_1(Y,g(y_0))\to \pi_1(Y,y_0), \varphi_{\beta}([\alpha])=[\beta^{-}\cdot\alpha\cdot\beta] is the basepoint-change isomorphism, then \varphi_{\beta}\circ g_{\#}=f.

Can we extend Eda’s classification theorem to non-compact or non-metrizable spaces? It’s not so obviously false since the continuous image of a circle or disk in any Hausdorff space is, in fact, a Peano continuum! Not only does this kind of logical exploration often result in some fun examples, it helps clarify what is really going on. By this, I mean it may help answer the question: which parts of being a “Peano continuum” are actually crucial to the argument? 

The thing is, giving adjunction spaces (like CW-complexes) the weak topology usually pays off big-time in algebraic topology. However, in this post, we’ll see that the weak topology works against us. Metrizability is crucial because one needs spaces to be “uniform” in a sense to ensure that algebra determines topological structure.

Lately, I’ve become a fan of using spaces a invariants of homotopy type. In particular, the topological 1-wild set \mathbf{w}(X)=\{x\in X\mid X\text{ is not semilocally simply connected at }x\} with the subspace topology is a homotopy invariant of X, i.e. if X\simeq Y, then \mathbf{w}(X)\simeq\mathbf{w}(Y). There is also a more algebraic version of this space, which I call the algebraic 1-wild set \mathbf{aw}(X), which I contrast with \mathbf{w}(X) in this post.

Here’s a useful statement that works for arbitrary spaces.

Lemma: If a map f:X\to Y is \pi_1-injective, i.e. the induced homomorphism f_{\#}:\pi_1(X,x)\to \pi_1(Y,f(x)) is injective for all x\in X, then f(\mathbf{w}(X))\subseteq\mathbf{w}(Y).

Proof. Suppose f is \pi_1-injective, x\in \mathbf{w}(X), and, to obtain a contradiction, that f(x)\notin \mathbf{w}(Y). Then Y is semilocally simply connected at f(x) and so there exists a neighborhood U of f(x) such that every loop in U based at f(x) is null-homotopic in X. Find an open neighborhood latex V of x such that f(V)\subseteq U, Given a loop \alpha in V based at x, f\circ \alpha is a loop in U and is therefore null-homotopic in X. However, since f is \pi_1-injective, \alpha itself must be null-homotopic in X. Since we’ve shown that every loop in V based at x contracts in X, we conclude that X is semilocally simply connected at x; a contradiction. \square

From this lemma, you might be able to start working out how the homotopy type of \mathbf{w}(X) is a homotopy invariant. If f:X\to Y and g:Y\to X are homotopy inverses, then f(\mathbf{w}(X))\subseteq\mathbf{w}(Y) and g(\mathbf{w}(Y))\subseteq\mathbf{w}(X). Restricting the homotopies g\circ f\simeq id_{X} and f\circ g\simeq id_{Y} to \mathbf{w}(X)\times[0,1] and \mathbf{w}(Y)\times [0,1] respectively will show that the restrictions f|_{\mathbf{w}(X)}:\mathbf{w}(X)\to\mathbf{w}(Y) and g|_{\mathbf{w}(Y)}:\mathbf{w}(Y)\to\mathbf{w}(X) are homotopy inverses.

Theorem: If f:X\to Y and g:Y\to X are homotopy inverses, then so are the restrictions f|_{\mathbf{w}(X)}:\mathbf{w}(X)\to\mathbf{w}(Y) and g|_{\mathbf{w}(Y)}:\mathbf{w}(Y)\to\mathbf{w}(X).

What is particularly special about being 1-dimensional is that such spaces have a property called the “discrete monodromy property,” which implies that the homeomorphism type of \mathbf{w}(X) is a homotopy invariant of X. This means that if X,Y are 1-dimensional Hausdorff spaces and X\simeq Y, then \mathbf{w}(X)\cong \mathbf{w}(Y). This is an exceptionally useful rigidity result – if you want to deform the one-dimensional space in your hand, the wild points can’t go anywhere. Proving this requires knowing some special things about how loops can contract in one-dimensional spaces but the good news is that we can do everything we need to in this post without it.

Example 1: A weak wild circle

Here’s the first space. Consider the circle S^1 (with distinguished point t_0) and for every t\in S^1, let \mathbb{H}_t be a homeomorphic copy of the Hawaiian earring with basepoint b_t. Now let X be obtained by attaching each \mathbb{H}_t to S^1 by t\sim b_t. We give X the weak topology with respect to the collection of subset \{S^1\}\cup \{\mathbb{H}_t\mid t\in S^1\} so that any compact subset of X must lie in at most finitely many of the attached earrings. You can think of it as a weak topology version of the metric wild circle:

The wild circle

Things to notice:

  1. X is a one-dimensional, locally path-connected, Hausdorff space but it’s not metrizable since its not first countable at any point of S^1.
  2. Since we’ve attached an earring at each point of S^1, and \mathbb{H}_t is locally contractible at each point in \mathbb{H}_t\backslash\{b_t\}, we have \mathbf{w}(X)=S^1 with its usual topology.

If F\subseteq S^1 is a finite set, define X_F=S^1\cup \bigcup\{\mathbb{H}_t\mid t\in F\}. In other words, X_F is the circle with finitely many earrings attached. The usual van Kampen Theorem shows that \pi_1(X_F,t_0) is the free product \mathbb{Z}\ast \left(\ast_{t\in F}\pi_1(\mathbb{H}_t,b_t)\right) where the \mathbb{Z} comes from S^1 and the other free factors come from the finitely many attached earrings in X_F. Since every path and homotopy of paths in X lies in X_F for some F (this is the whole point of using the weak topology), it follows that

\pi_1(X,t_0)\cong \varinjlim_{F}\pi_1(X_F,t_0)\cong \mathbb{Z}\ast \left(\ast_{t\in S^1}\pi_1(\mathbb{H}_t,b_t)\right)

where the limit is directed by the set of finite subsets of S^1.

Example 2: A one-point union

Here’s the second example. At first, it might feel like S^1\vee\bigvee_{t\in S^1}\mathbb{H}_t is the thing we want. But actually, this won’t do because even when you take a one-point union (A,a)\vee (B,b) of two first countable spaces where a\in\mathbf{w}(A) and b\in\mathbf{w}(B), the natural homomorphism \pi_1(A)\ast\pi_1(B)\to \pi_1(A\vee B) is not an isomorphism because it’s not surjective. So we need to be a bit more careful not to attach wild points to each other.

Let \mathbb{H}_{t}^{+}=\mathbb{H}_{t}\cup [0,1]/\mathord{\sim} , b_t\sim 1 be the space obtained by attaching a “whisker” to the earring \mathbb{H}_t at the wild point. We take the image of 0 (the end of the whisker), denoted b_{t}^{+}, to be the basepoint of \mathbb{H}_{t}^{+}.

Notice that (\mathbb{H}_{t}^{+},b_{t}^+) is now locally contractible at its basepoint so it’s now a good idea to consider the one-point union:

Y=S^1\vee \bigvee_{t\in S^1}\mathbb{H}_{t}^{+}.

The Space Y: an uncountable one-point union

Things to notice:

  1. Y is a one-dimensional, locally path-connected, Hausdorff space but it’s not metrizable since its not first countable at the wedge point.
  2. The wild points b_t still exist at the connected of each whisker and \mathbb{H}_t and so \mathbf{w}(Y) is homeomorphic to S^1 with the discrete topology.
  3. If you really wanted to, you could allow S^1 to become one of the circles in one of the earrings, i.e. Y\cong \bigvee_{t\in S^1}\mathbb{H}_{t}^{+} but I feel like doing this would just add confusion to the whole point of this.

So, now the van Kampen theorem does apply to Y and we have:

\pi_1(Y,y_0)\cong\mathbb{Z}\ast \left(\ast_{t\in S^1}\pi_1(\mathbb{H}_t,b_t)\right).

The Takeaway

So what we have is a pair of one-dimensional, locally path-connected Hausdorff spaces X and Y with isomorphic fundamental groups \pi_1(X,t_0) and \pi_1(Y,t_0).

Proposition: Even though X and Y have isomorphic fundamental groups, X and Y are not homotopy equivalent.

Proof. We proceed by contradiction. Suppose f:X\to Y is a homotopy equivalence. Then f is \pi_1-injective and the theorem from earlier in the post implies that the restriction f|_{\mathbf{w}(X)}:\mathbf{w}(X)\to\mathbf{w}(Y) is a well-defined homotopy equivalence. However, \mathbf{w}(X)\cong S^1 with it’s usual topology and \mathbf{w}(Y) is homeomorphic to S^1 with the discrete topology. These spaces cannot be homotopy equivalent (e.g. one is path connected and the other is not), which gives us our contradiction. \square

It is possible to construct a map Y\to X that induces an isomorphism on fundamental groups. This makes sense since it’s possible to map the discrete circle continuously onto the usual circle. However, the above argument shows that no injective homomorphism \pi_1(X,t_0)\to \pi_1(Y,t_0) can be induced by a map X\to Y up to basepoint-change. A modification of these spaces could definitely be built so that an isomorphism is not induced in either direction.

Admission: Ok, so there’s nothing special about using S^1 except that you get a non-simply connected wild set, which is interesting to me right now. You could replace it with [0,1] and the analogous examples would get the same idea across.

Can we do better?…What are some other limits to Eda’s Theorem?

Eda’s Theorem is absolutely incredible where it works. But where else does it fail to hold? There are some not-so-hard non-locally path-connected cases where Eda’s Theorem doesn’t work. But what about non-compact metric spaces? This leaves me with the following questions, which at the moment I don’t know the answer to. 

Question: Are there examples of connected, locally path-connected, one-dimensional metric spaces X,Y with isomorphic fundamental groups but which are not homotopy equivalent? What about separable examples? It would be really interesting if there was a pair where one of them was a compact and the other was not. 

My guess is that that there are non-compact counterexamples.

[1] K. Eda, Homotopy types of one-dimensional Peano continua, Fund. Math. 209 (2010) 27-42.

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