Topologized Fundamental Groups: The Quotient Topology Part 3 (Why isn’t it always a topological group?)

In Part 1, I mentioned that one of the surprising things about the natural quotient topology on the fundamental group is that the resulting group with topology \pi_{1}^{qtop}(X,x_0) often fails to be a topological group. In fact, I’d say it’s usually not a topological group when \pi_{1}^{qtop}(X,x_0) is not discrete. This surprises a lot of folks because the concatenation function c:\Omega(X,x_0)\times \Omega(X,x_0)\to \Omega(X,x_0), (\alpha,\beta)\mapsto \alpha\cdot\beta IS continuous and it descends to the multiplication operation \mu:\pi_{1}^{qtop}(X,x_0)\times \pi_{1}^{qtop}(X,x_0)\to \pi_{1}^{qtop}(X,x_0), \mu([\alpha],[\beta])=[\alpha\cdot\beta]. In this post, I’ll detail out an important example showing why \mu isn’t always continuous. 

The reason why the “obvious” proof doesn’t work is actually not to hard to see: If q:\Omega(X,x_0)\to \pi_{1}^{qtop}(X,x_0) is the natural quotient map and the product q\times q is a quotient map, then because \mu \circ (q\times q)=q\circ c, the universal property of quotient maps ensures that \mu is continuous. However, in genreal topology the product of two quotient maps is NOT always a quotient map!

Ok, so the obvious proof doesn’t work….but this doesn’t really constitute a counterexample, does it? In fact, it’s not really known if, for given space X, q\times q being quotient is equivalent to \mu being continuous. So let’s talk through a counterexample. This one is a slightly modified version of the example given by Paul Fabel (MS State) in the 2011 paper [1].

We don’t have to go too far in our search for examples. Consider the usual infinite earring space \mathbb{E}=\bigcup_{n\in\mathbb{N}}C_n where C_n\subseteq \mathbb{R}^2 is the circle of radius \frac{1}{n} centered at (\frac{1}{n},0). We take the basepoint x_0 to be the origin. For brevity, let’s let L=\Omega(\mathbb{E}_n,x_0) denote the loop space and G=\pi_{1}^{qtop}(\mathbb{E},x_0) be the earring group with the natural quotient topology.

the infinite earring space

The infinite earring space

Let \ell_n:[0,1]\to C_n be the standard paramterization of C_n going once around counterclockwise and x_n=[\ell_n]\in G. The group G is locally free and although we can form infinite words in the letters x_n in G, we only need to use finite words to work through this example. If a and b are elements of L or G, we’ll use [a,b] to denote the commutator aba^{-1}b^{-1} and w^n will be the n-fold product www\cdots w of a word w with itself. We’ll write e for both the constant loop at x_0 and it’s homotopy class, which is the identity element of G.

We’re going to define two doubly indexed sequences and make some observations about them. Some of these observations are going to be something of a warm-up for the real argument later.

First doubly indexed sequence: Let w_{m,n}=[x_m,x_n]^{m} for m,n\in\mathbb{N}.

  1. Note that w_{n,n}=e and the word length otherwise is |w_{m,n}|=4m.
  2. Fix n and suppose n<m_1<m_2<m_3<\cdots. Then w_{m,n_k}=[x_{m_k},x_{n}]^{m_k} is a sequence of reduced words of unbounded length. In fact, the number of appearrances of the letter x_n is unbounded as k\to\infty. Hence, if we considering representing loops \alpha_k\in w_{m_k,n}, then, for each k, the loop \alpha_k may not be reduced (in the sense of not having any null-homotopic subloops) but because w_{m_k,n} is a reduced word, the loop \alpha_k must go around the circle C_n (in one direction or the other) at least 2m_k-many distinct times. Since the topology of L agrees with the topology of uniform convergence, there is no way for the sequence \{\alpha_k\} with to converge to any loop in L.
  3. Fix m. Then [\ell_m,\ell_n]^{m} has fixed “word length” as n\to \infty and converges to the null-homotopic loop [\ell_m,e]^m. Therefore, \{w_{m,n}\}_{n}\to e when m is fixed, showing that e is a limit point of the set W=\{w_{m,n}\mid m,n>1, m\neq n\}. In fact, we show something stronger next.
  4. Let U be an open neighborhood of e in G. Then q^{-1}(U) is an open neighborhood of the constant loop at x_0 in L. Since L has the compact-open topology, we can find an open neighborhood O of x_0 in \mathbb{E} such that all loops ([0,1],\{0,1\})\to (O,x_0) lie in q^{-1}(U). Find N such that C_N\subseteq O. Let m,n\geq N with m\neq n. Since \ell_m and \ell_{n} have image in O, so does [\ell_{m},\ell_{n}]^{m}. Thus [\ell_{m},\ell_{n}]^{m}\in q^{-1}(U) which gives w_{m,n}=q([\ell_{m},\ell_{n}]^{m})\in U for all m,n\geq N.

Second doubly indexed sequence: Let v_{m,n}=[x_1,x_m]^{n} for m,n\in\mathbb{N}.

  1. Note that v_{1,n}=e is trivial and the word length otherwise is |v_{m,n}|=4n.
  2. Fix m>1 and suppose n_1<n_2<n_3<\cdots. Then v_{m,n_k}=[x_1,x_{m}]^{n_k} is a sequence of reduced words of unbounded length such that the number of appearrances of x_1 is unbounded as k\to\infty. Hence, if we considering representing loops \alpha_k\in v_{m,n_k}, then \alpha_k must have at least n_k-many copies of the loop \ell_{1}^{\pm} as a subloop. Thus there is no way for the sequence \{\alpha_k\} to a loop in L.
  3. Fixing n, note that since [\ell_1,\ell_{m}]^{n} has fixed length, it converges uniformly to [\ell_1,e]^{n}, which is null-homotopic. The continuity of q, then ensures that \{v_{m,n}\}_{m>1}\to e in G.
  4. It follows from 3. that the identity element e \in G is a limit point of V=\{w_{m,n}\mid m>1, m\neq n\}.

Now, we use the sets W=\{w_{m,n}\mid m,n>1,m\neq n\} and V=\{w_{m,n}\mid m,n>1,m\neq n\} to define the set we really want to focus on. In particular, let

C=WV=\{w_{m,n}v_{m,n}\mid m,n>1,m\neq n\}

By definition C consists of the doubly indexed sequences:


We add the restrictions m,n>1 and m\neq n just to ensure that these words are, in fact, reduced words. In particular, e\notin C. Let’s show that C is a desired counterexample, that is, it’s a closed subset C of G such that \mu^{-1}(G) is not closed in G\times G. First the easy direction.

Proposition 1: \mu^{-1}(C) is not closed in G\times G.

Proof. Since e\notin C and \mu(e,e)=e, the pair (e,e) cannot be an element of \mu^{-1}(C). However, by construction of C, we have W\times V\subseteq \mu^{-1}(C). The fourth observation for each doubly-indexed sequence above tells us that e is both a limit point of W and V in G. Thus (e,e) is a limit point of W\times V in G\times G. Since W\times V\subseteq \mu^{-1}(C), (e,e) is a limit point of \mu^{-1}(C). Since \mu^{-1}(C) is missing at least one limit point in G, it is not closed in G\times G. \square

For the other direction, we’ll need some observations. The first one is from genreal topology.

Definition 2: A subset C of a topological space X is sequentially closed if C is closed under limits of convergent sequences, that is, if whenever \{x_n\}\to x in X and x_n\in C for all n\in\mathbb{N}, we have x\in C.

Definition 3: A topological space X is a sequential space if a set C\subseteq X is closed if and only if C is sequentially closed.

All metrizable spaces are sequential spaces and all quotient spaces of sequential spaces are also sequential spaces. In particular, since the loop space L is metrizable and G is a quotient of L, both L and G are sequential spaces.

The world would be a different place if whenever we have a quotient map q:X\to Y and a convergent sequence \{y_n\}\to y in Y, we could lift it to a convergent sequence in X that maps to \{y_n\}\to y. “Biquotient maps” have this lifting property but ordinary quotient maps do not. However, even when lifting all convergent sequences is not at our disposal, we can often at least find a subsequence that lifts.

Lemma 4: If X is a Hausdorff sequential space and q:X\to Y is a quotient map where Y is Hausdorff, then for every convergent sequence \{y_n\}\to y in Y, there exists n_1<n_2<n_3<\cdots and a convergent sequence \{x_{n_k}\}\to x in X such that q(x_{n_k})=y_{n_k} and q(x)=y.

Proof. If there is a subsequence of y_n that is constant at y, then the conclusion is easy to acheive. Suppose y\notin C=\{y_n\mid n\in\mathbb{N}\}. Then C is not closed in Y. Since q is quotient, q^{-1}(C) is not closed in X. Since X is assumed to be sequential, q^{-1}(C) is not sequentially closed in X. Thus there exists a convergent sequence \{x_k\}\to x in X such that x_k\in q^{-1}(C) for all k\geq 1 and x\notin q^{-1}(C). We have q(x_k)=y_{n_k} for some n_k\in\mathbb{N}. If the sequence of integers \{n_k\} is bounded by M, then \{x_k\} has image in the finite set F=\{x_1,x_2,\dots,x_M\}. But since X is Hausdorff, F is closed and so we must have x\in F\subseteq q^{-1}(C); a contradiction. Thus \{n_k\} must be unbounded. In particular, we may replace \{n_k\} with an increasing subsequence. Doing so gives n_1<n_2<n_3<\cdots where q(x_{n_k})=y_{n_k}. The continuity of q ensures \{y_{n_k}\}\to q(x) and since Y is Hausdorff and \{y_n\}\to y, we must have q(x)=y.

We’re going to apply this lemma to the quotient map q:L\to G. Since L is metrizable, it is sequential and Hausdorff. The only hypothesis that may be a little unclear is that G is Hausdorff.

Theorem 5: G is Hausdorff.

Proof. Let X_n=\bigcup_{k\leq n}C_k be the finite wedge of circles so that \pi_1(X_n,x_0) is the free group F_n on the letters x_1,x_2,\dots,x_n. From Part 2, we know that F_n is discrete. Also, we have a canonical retraction r_n:\mathbb{E}\to X_n that collapse \bigcup_{k>n}C_k to x_0 and, which induce continuous homomorphisms R_n:G\to F_n. We also have bonding retractions r_{n+1,n}:X_{n+1}\to X_n, which collapse C_{n+1} to x_0. These induce maps R_{n+1,n}:F_{n+1}\to F_n on the discrete free groups. The inverse limit \varprojlim_{n}(F_n,R_{n+1,n}) is topologized as a subspace of the product \prod_{n}F_n. Since this product is Hausdorff, so is \varprojlim_{n}(F_n,R_{n+1,n}). Finally, note that since R_{n+1,n}\circ R_{n+1}=R_n, we have an induced continuous homomorphism \phi:G\to \varprojlim_{n}(F_n,R_{n+1,n}) given by \phi(g)=(R_1(g),R_2(g),R_3(g),\dots). We proved in a previous post that \phi is injective (warning! \phi is not a topological embedding). Therefore, since G continuously injects into a Hausdorff group, G must be Hausdorff. \square

Finally, we have the ingredients to finish our argument. I love this next part of the argument. It highlights why doubly indexed sequences are the thing to use.

Lemma 6: The set C=\{w_{m,n}v_{m,n}\mid m,n>1,m\neq n\} is closed in G.

Proof. Since G is the quotient of the sequential space L, G is sequential. Therefore, it suffices to prove that C is sequentially closed. Let g_{k}\to g be a convergent sequence in G such that g_k\in C for all k\geq 1. In particular, we have


for integers m_k,n_k>1 with m_k\neq n_k. We must check that g\in C. Note that by Lemma 4, there exists integers k_1<k_2<k_3<\cdots and a convergent sequence of loops \{\alpha_j\}\to \alpha in L such that [\alpha_j]=g_{k_j} and [\alpha]=g.

Case I: Suppose both sequence of integers \{m_k\} and \{n_k\} are bounded. Then F=\{g_k\mid k\geq 1\} is a finite subset of C. Since G is Hausdorff by Theorem 5, F is closed and it follows that g\in F\subseteq C.

Case II: Suppose \{n_k\} is unbounded. Then \{n_{k_j}\}_{j\geq 1} is unbounded and the (possibly unreduced) loop \alpha_j must go around the circle C_1 in one-direction or the other at least 2n_{k_j}-many times (this is contributed by the subword v_{m_{k_j},n_{k_j}} of g_k). But this makes it impossible for the sequence \{\alpha_j\} to converge uniformly to any loop; a contradiction. Therefore, this case is impossible.

Case III: Suppose \{m_k\} is unbounded and \{n_k\} is bounded. Then \{m_{k_j}\}_{j\geq 1} is unbounded and \{n_{k_j}\}_{j\geq 1} is bounded. By the Pigeon-Hole Principle, there must exist some N such that n_{k_j}=N for infinitely many j. By replacing k_j with the subsequence of k_j for, which n_{k_j}=N, we may assume that n_{k_j}=N for all j\geq 1. Recall that (the possibly unreduced loop) \alpha_{k_j} represents [x_{m_{k_j}},x_{N}]^{m_{k_j}}[x_1,x_{m_{k_j}}]^{N} and so \alpha_j must go around the circle C_N in one-direction or the other at least (2m_{k_j})-many times (this is contributed by the subword w_{m_{k_j},N} of g_k). But this makes it impossible for the sequence \{\alpha_j\} to converge uniformly to any loop in L; a contradiction. Therefore, this case is impossible.

Thus only Case I is possible and in this case, we verified that g\in C. \square

Theorem [Fabel]: The infinite earring group G equipped with the natural quotient topology is not a topological group.

Proof. By Lemma 6, C=\{w_{m,n}v_{m,n}\mid m,n>1,m\neq n\} is closed in G. However, by Proposition 1, \mu^{-1}(C) is not closed in G\times G where mu:G\times G\to G is the group operation. Thus \mu is not continuous and G is not a topological group.

Despite not being a topological group, we discussed in Part 1 that \pi_{1}^{qtop}(X,x_0) is always a quasitopological group. Thus G is a nice example of a quasitopological group that is “close” to being a topological group but doesn’t quite make the cut.

Deeper discussion: Even if you make it through the logic of the proof I presented here, you might feel that you still don’t really have a good feel for what’s going on. This is a common feeling about this continuity fiasco, so I’m going to offer another perspective that is a little informal but which I think could be made formal. The reason that G fails to be a topological group is because L and G fail to be locally compact in a certain way.

Our initial observations about the words w_{m,n} tells us that we can topologically identify W\cup \{e\} with the planar subset B=\{(0,0)\}\cup \{(\frac{1}{n},\frac{1}{mn})\mid m,n\geq 1\}. Here I’m fudging the fact that we needed to get rid of the diagonal and the first row and column in V but that’s superficial.

The planar set $latex B$ consisting of points clustering toward the origin.

The planar set B consisting of the origin and points (\frac{1}{n},\frac{1}{mn}).

This space B is important because every non-locally compact metric space contains a homeomorphic copy of B as a subspace. In the same manner, we’ve found a copy of B within G.

The elements of V=\{v_{n,m}\mid m,n>1, m\neq n\} work a little differently. If we lay out these points on the integer grid in the same way (so v_{m,n} is at (m,n)), then none of the vertical sequences converge. In fact, any sequence which meets infinitely many rows cannot converge. However, along the fixed row at height n, \{v_{m,n}\}_{m} converges to the identity e=[x_1,e]^{n}. This essentially means that V\cup\{e\} is a copy of the Frechet Uryshon Fan, which I’ll denote by A=\{a_0\}\cup\{a_{m,n}\mid m,n\in\mathbb{N}\}. The Frechet Urysohn Fan can be constructed by infinitely many disjoint convergent sequences S_m=\{a_0\}\cup \{a_{m,n}\mid n\geq 1\} with their limits points a_0 identified. We then give A the weak topology with respect to the subspaces S_m.

The Frechet-Uryshon Fan, a fan of converging sequences

The Frechet-Uryshon Fan

The set V\cup\{e\} is a copy of A inside of G. We’ve managed to play these embeddings of A and B off of each other to win the game. In particular, we’ve embedded these non-locally compact subspaces in G in a way that (1) doesn’t allow for them to algebraically intermingle and (2) lifts to similar behavior in L. In my view, that is what really is causing discontinuity. Admittedly, this is a little informal but all of the discontinuity proofs I’ve seen, even for higher homotopy groups [2] and the \pi_1-action [3] all kind of look like this one. So, I suspect there is a deeper thread here to unravel. 


[1] Paul Fabel, Multiplication is discontinuous in the Hawaiian earring group (with the Quotient Topology). Bulletin of the Polish Academy of Sciences 59 (2011), no. 1, 77-83.

[2] Paul Fabel, Compactly generated quasitopological homotopy groups with discontinuous multiplication, Topology Proc. 40 (2012) 303–309.

[3] J. Brazas, On the discontinuity of the pi_1-action, Topology Appl. 247 (2018) 29-40.

This entry was posted in compact-open topology, earring group, Free groups, Fundamental group, quotient topology, topological fundamental group, Topological groups, Uncategorized and tagged , , , , , , , , , , . Bookmark the permalink.

1 Response to Topologized Fundamental Groups: The Quotient Topology Part 3 (Why isn’t it always a topological group?)

  1. Pingback: How to “topologize” the fundamental group: a primer | Wild Topology

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s