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.