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 often fails to be a topological group. In fact, I’d say it’s usually not a topological group when is not discrete. This surprises a lot of folks because the concatenation function , IS continuous and it descends to the multiplication operation , . In this post, I’ll detail out an important example showing why isn’t always continuous.

The reason why the “obvious” proof doesn’t work is actually not to hard to see: If is the natural quotient map and the product is a quotient map, then because , the universal property of quotient maps ensures that 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 , being quotient is equivalent to 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 * where is the circle of radius centered at . We take the basepoint to be the origin. For brevity, let’s let denote the loop space and be the earring group with the natural quotient topology.

Let be the standard paramterization of going once around counterclockwise and . The group is locally free and although we can form infinite words in the letters in , we only need to use finite words to work through this example. If and are elements of or , we’ll use to denote the commutator and will be the -fold product of a word with itself. We’ll write for both the constant loop at and it’s homotopy class, which is the identity element of .

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 for .

- Note that and the word length otherwise is .
- Fix and suppose . Then is a sequence of reduced words of unbounded length. In fact, the number of appearrances of the letter is unbounded as . Hence, if we considering representing loops , then, for each , the loop may not be reduced (in the sense of not having any null-homotopic subloops) but because is a reduced word, the loop must go around the circle (in one direction or the other) at least -many distinct times. Since the topology of agrees with the topology of uniform convergence, there is no way for the sequence with to converge to
*any*loop in . - Fix . Then has fixed “word length” as and converges to the null-homotopic loop . Therefore, when is fixed, showing that is a limit point of the set . In fact, we show something stronger next.
- Let be an open neighborhood of in . Then is an open neighborhood of the constant loop at in . Since has the compact-open topology, we can find an open neighborhood of in such that all loops lie in . Find such that . Let with . Since and have image in , so does . Thus which gives for all .

**Second doubly indexed sequence:** Let for .

- Note that is trivial and the word length otherwise is .
- Fix and suppose . Then is a sequence of reduced words of unbounded length such that the number of appearrances of is unbounded as . Hence, if we considering representing loops , then must have at least -many copies of the loop as a subloop. Thus there is no way for the sequence to a loop in .
- Fixing , note that since has fixed length, it converges uniformly to , which is null-homotopic. The continuity of , then ensures that in .
- It follows from 3. that the identity element is a limit point of .

Now, we use the sets and to define the set we really want to focus on. In particular, let

By definition consists of the doubly indexed sequences:

.

We add the restrictions and just to ensure that these words are, in fact, reduced words. In particular, . Let’s show that is a desired counterexample, that is, it’s a closed subset of such that is not closed in . First the easy direction.

**Proposition 1:** is not closed in .

*Proof*. Since and , the pair cannot be an element of . However, by construction of , we have . The fourth observation for each doubly-indexed sequence above tells us that is both a limit point of and in . Thus is a limit point of in . Since , is a limit point of . Since is missing at least one limit point in , it is not closed in .

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

**Definition 2:** A subset of a topological space is *sequentially closed* if is closed under limits of convergent sequences, that is, if whenever in and for all , we have .

**Definition 3:** A topological space is a *sequential space* if a set is closed if and only if 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 is metrizable and is a quotient of , both and are sequential spaces.

The world would be a different place if whenever we have a quotient map and a convergent sequence in , we could lift it to a convergent sequence in that maps to . “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 is a Hausdorff sequential space and is a quotient map where is Hausdorff, then for every convergent sequence in , there exists and a convergent sequence in such that and .

*Proof*. If there is a subsequence of that is constant at , then the conclusion is easy to acheive. Suppose . Then is not closed in . Since is quotient, is not closed in . Since is assumed to be sequential, is not sequentially closed in . Thus there exists a convergent sequence in such that for all and . We have for some . If the sequence of integers is bounded by , then has image in the finite set . But since is Hausdorff, is closed and so we must have ; a contradiction. Thus must be unbounded. In particular, we may replace with an increasing subsequence. Doing so gives where . The continuity of ensures and since is Hausdorff and , we must have .

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

**Theorem 5:** is Hausdorff.

Proof. Let be the finite wedge of circles so that is the free group on the letters . From Part 2, we know that is discrete. Also, we have a canonical retraction that collapse to and, which induce continuous homomorphisms . We also have bonding retractions , which collapse to . These induce maps on the discrete free groups. The inverse limit is topologized as a subspace of the product . Since this product is Hausdorff, so is . Finally, note that since , we have an induced continuous homomorphism given by . We proved in a previous post that is injective (warning! is *not *a topological embedding). Therefore, since continuously injects into a Hausdorff group, must be Hausdorff.

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 is closed in .

Proof. Since is the quotient of the sequential space , is sequential. Therefore, it suffices to prove that is sequentially closed. Let be a convergent sequence in such that for all . In particular, we have

for integers with . We must check that . Note that by Lemma 4, there exists integers and a convergent sequence of loops in such that and .

**Case I:** Suppose both sequence of integers and are bounded. Then is a finite subset of . Since is Hausdorff by Theorem 5, is closed and it follows that .

**Case II:** Suppose is unbounded. Then is unbounded and the (possibly unreduced) loop must go around the circle in one-direction or the other at least -many times (this is contributed by the subword of ). But this makes it impossible for the sequence to converge uniformly to any loop; a contradiction. Therefore, this case is impossible.

**Case III:** Suppose is unbounded and is bounded. Then is unbounded and is bounded. By the Pigeon-Hole Principle, there must exist some such that for infinitely many . By replacing with the subsequence of for, which , we may assume that for all . Recall that (the possibly unreduced loop) represents and so must go around the circle in one-direction or the other at least -many times (this is contributed by the subword of ). But this makes it impossible for the sequence to converge uniformly to any loop in ; a contradiction. Therefore, this case is impossible.

Thus only Case I is possible and in this case, we verified that .

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

*Proof*. By Lemma 6, is closed in . However, by Proposition 1, is not closed in where is the group operation. Thus is not continuous and is not a topological group.

Despite not being a topological group, we discussed in Part 1 that is always a *quasitopological group*. Thus 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 fails to be a topological group is because and fail to be locally compact in a certain way.

Our initial observations about the words tells us that we can topologically identify with the planar subset . Here I’m fudging the fact that we needed to get rid of the diagonal and the first row and column in but that’s superficial.

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

The elements of work a little differently. If we lay out these points on the integer grid in the same way (so is at ), 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 , converges to the identity . This essentially means that is a copy of the Frechet Uryshon Fan, which I’ll denote by . The Frechet Urysohn Fan can be constructed by infinitely many disjoint convergent sequences with their limits points identified. We then give the weak topology with respect to the subspaces .

The set is a copy of inside of . We’ve managed to play these embeddings of and off of each other to win the game. In particular, we’ve embedded these non-locally compact subspaces in in a way that (1) doesn’t allow for them to algebraically intermingle and (2) lifts to similar behavior in . 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 -action [3] all kind of look like this one. So, I suspect there is a deeper thread here to unravel.

## References

[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.