Testing the limits of Eda’s Theorem

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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