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

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 $m,n\in\mathbb{N}$ .

Note that $w_{n,n}=e$ and the word length otherwise is $|w_{m,n}|=4m$ .
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 $L$ .
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.
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 $m,n\in\mathbb{N}$ .

Note that $v_{1,n}=e$ is trivial and the word length otherwise is $|v_{m,n}|=4n$ .
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 $L$ .
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 $G$ .
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:

$w_{m,n}v_{m,n}=[x_m,x_n]^{m}[x_1,x_m]^{n}$ .

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

$g_k=w_{m_k,n_k}v_{m_k,n_k}=[x_{m_k},x_{n_k}]^{m_k}[x_1,x_{m_k}]^{n_k}$

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.

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 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.

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.

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