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

1. Note that $w_{n,n}=e$ and the word length otherwise is $|w_{m,n}|=4m$.
2. Fix $n$ and suppose $n. 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. 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:

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

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 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 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 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.

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 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.

## Topologized Fundamental Groups: The Quotient Topology Part 2 (Discreteness)

In Part 1, I described the construction of $\pi_{1}^{qtop}(X,x_0)$, the fundamental group equipped with the quotient topology and some of the drama around $\pi_{1}^{qtop}(X,x_0)$ failing to always be a topological group. In this second post, I plan to connect $\pi_{1}^{qtop}(X,x_0)$ back to spaces with “nicer” local structure by discussing when $\pi_{1}^{qtop}(X,x_0)$ is discrete. The theorem we’ll prove is the main result from [1].

If you’ve got a CW-complex, manifold, simplicial complex, or some other locally contractible space, you should very much hope that $\pi_{1}^{qtop}(X,x_0)$ is a discrete group. If $\pi_{1}^{qtop}(X,x_0)$ is NOT discrete, then it’s because the topological part of $\pi_{1}^{qtop}$ is detecting some non-trivial local structures in $X$. In this post, we’ll explore what non-discreteness is really telling you.

Reminder: $\Omega(X,x_0)$ is the loop space with the compact-open topology and $\pi_{1}^{qtop}(X,x_0)$ is the fundamental group with the quotient topology inherited from the map $q:\Omega(X,x_0)\to\pi_{1}(X,x_0)$, $q(\alpha)=[\alpha]$ identifying homotopy classes of loops. We’ll need to use the description of a basis for the compact-open topology from Part 1.

First, let’s identify a clear case where $\pi_{1}^{qtop}(X,x_0)$ is not discrete. Recall that a space $X$ is semilocally simply connected at $x\in X$ if there exists an open neighborhood $U$ of $x$ such that the homomorphism $pi_1(U,x)\to \pi_1(X,x)$ induced by the inclusion map $U\to X$ is trivial, that is, if every loop in $U$ based at a $x$ contracts by a null-homotopy in $X$. We say $X$ is semilocally simply connected if it has this property at all of its points.

Lemma 1: If $X$ is path connected and $\pi_{1}^{qtop}(X,x_0)$ is discrete, then $X$ is semilocally simply connected.

Proof. Suppose $X$ is not semilocally simply connected at some point $x\in X$. Fix a path $\alpha:[0,1]\to X$ from $x_0$ to $x$ and let $\mathscr{N}$ be the set of open sets in $X$ containing $x$. Since $X$ is not semilocally simply connected at, for every $U\in \mathscr{N}$, there exists a loop $\beta_{U}:[0,1]\to U$ based at $x$ such that $\beta_U$ is not null-homotopic in $X$, that is, $[\beta_{U}]\neq 1$ in $\pi_1(X,x)$. Thus $[\alpha\cdot\beta_{U}\cdot\alpha^{-}]\neq 1$ in $\pi_1(X,x_0)$. However, $\mathscr{N}$ is a directed set (by subset inclusion) and so $\{\alpha\cdot\beta_{U}\cdot\alpha^{-}\}_{U\in\mathscr{N}}$ is a net in $\Omega(X,x_0)$. Moreover, $\{\alpha\cdot\beta_{U}\cdot\alpha^{-}\}_{U\in\mathscr{N}}$ converges to $\alpha\cdot c_x\cdot \alpha^{-}$ in the compact-open topology where $c_x$ denotes the constant loop at $x$. Since $q:\Omega(X,x_0)\to \pi_{1}^{qtop}(X,x_0)$ is continuous, the net of non-trivial homotopy classes $\{[\alpha\cdot\beta_{U}\cdot\alpha^{-}]\}_{U\in\mathscr{N}}$ converges to the identity element $1=[\alpha\cdot c_x\cdot \alpha^{-}]$ in $\pi_{1}^{qtop}(X,x_0)$. Since a net of non-identity elements converges to the identity element, there is no way the trivial subgroup $\{1\}$ can be open. Thus $\pi_{1}^{qtop}(X,x_0)$ is not discrete. $\square$

So immediately, all of the regular suspects on this blog, e.g. the earring space, harmonic archipelago, Menger cube, etc. have non-discrete fundamental group.

Now, the above result doesn’t say anything about local path connectivity. There a plenty of simply connected spaces that are not locally path connected, e.g. the Warsaw circle. But to prove a converse to Lemma 1, we need to add a local path connectivity condition and, in the, end we’ll see this actually is necessary if we’re looking for fully classify when $\pi_{1}^{qtop}(X,x_0)$ is discrete.

Lemma 2: If $X$ is locally path connected and semilocally simply connected, then $\pi_{1}^{qtop}(X,x_0)$ is a discrete group.

Proof. To show that $\pi_{1}^{qtop}(X,x_0)$ is discrete, we must show that for any loop $\alpha:[0,1]\to X$, the 1-point set $\{[\alpha]\}$ is open. Since $q:\Omega(X,x_0)\to\pi_{1}^{qtop}(X,x_0)$ is a quotient map, $\{[\alpha]\}$ is open in $\pi_{1}^{qtop}(X,x_0)$ if and only if $q^{-1}(\{[\alpha]\})=[\alpha]$ is open in $\Omega(X,x_0)$. Remember that $[\alpha]$ is the homotopy class of $\alpha$, that is, the set of all loops path-homotopic to $\alpha$. In order to do this we must show that all loops “nearby” $\alpha$ are homotopic to $\alpha$ where “nearby” really means “in some compact-open neighborhood.”

For each $t\in [0,1]$, let $U_t$ be an open neighborhood of $\alpha(t)$ such that every loop in $U_t$ is null-homotopic in $X$ (here, we are using the semilocally simply connected property). Using the Lebesgue Number Lemma, we may find an integer $n\geq 1$ such that if $K_{n}^{j}=\left[\frac{j-1}{n},\frac{j}{n}\right]$ (as in Part 1), then for each $j\in \{1,2,\dots,n\}$, we have $\alpha(K_{n}^{j})\subseteq U_{s_j}$ for some $s_j$. To simplify notation, we’ll write $U_j$ for $U_{s_j}$.

Now we have $\alpha(t_j)\in U_{j}\cap U_{j+1}$. However, it may not be the case that the intersection $U_{j}\cap U_{j+1}$ is path connected. We can address this issue in the following way. Since $X$ is locally path connected, for each $j\in \{1,2,\dots ,n-1\}$, find a path-connected neighborhood latex $V_j$ of $\alpha(\frac{j}{n})$ such that $V_j\subseteq U_{j}\cap U_{j+1}$.

Now we are ready to define the neighborhood

$\mathscr{U}=\bigcap_{j=1}^{n}\langle K_{n}^{j},U_j\rangle\cap \bigcap_{j=1}^{n-1}\langle \{\frac{j}{n}\},V_j\rangle$

Recall that the notation $\langle C,U\rangle$ denotes the set consisting of all loops that map the compact set $C$ into the open set $U$. So we can think of $\mathscr{U}$ above as the set of all loops that follow an ordered list of instructions. If $\beta\in \mathscr{U}$, then $\beta$ must first proceed through $U_1$ and end somehwere in $V_1$. It must then proceed through $U_2$ and end in $V_2$, etc.

The neighborhood $\mathscr{U}$ of $\alpha$.

Our remaining job is to show that $[\alpha]$ is open and this will be done if we can show that $\mathscr{U}\subseteq [\alpha]$, that is every loop in $\mathscr{U}$ is homotopic to $\alpha$.

Let $\beta\in \mathscr{U}$. We’ll construct a homotopy $\alpha\simeq\beta$. For $j\in\{1,2\dots,n-1\}$, both $\alpha(t_j)$ and $\beta(t_j)$ lies in $V_j$ and so we may find a path $\gamma_j:[0,1]\to V_j$ from $\alpha(t_j)$ to $\beta(t_j)$.

The path $\beta$ and the connecting paths $\gamma_j$.

Notice that

• $\alpha|_{K_{n}^{1}}\cdot \gamma_1\cdot \beta|_{K_{n}^{1}}^{-}$ is a loop in $U_1$,
• $\gamma_{j-1}^{-}\cdot \alpha|_{K_{n}^{j}}\cdot \gamma_{j}\cdot \beta|_{K_{n}^{j}}^{-}$ is a loop in $U_j$ when $2\leq j\leq n-1$,
• $\gamma_{n-1}^{-}\cdot \alpha|_{K_{n}^{n}}\cdot \beta|_{K_{n}^{n}}^{-}$ is a loop in $U_n$.

A close-up view of the general case where we have create a loop in $U_j$ with corresponding portions of $\alpha$ and $\beta$. Since this loop lies in $U_j$, it is null-homotopic in $X$.

By our choice of the sets $U_j$, all of these loops are null-homotopic in $X$. In particular, this means that

• $\alpha|_{K_{n}^{1}}\simeq \beta|_{K_{n}^{1}}\cdot \gamma_{1}^{-}$,
• $\alpha|_{K_{n}^{j}}\simeq\gamma_{j-1}^{-}\cdot \beta|_{K_{n}^{j}}\cdot \gamma_{j}^{-}$ when $2\leq j\leq n-1$,
• $\alpha|_{K_{n}^{n}}\simeq \gamma_{n-1}\cdot\beta|_{K_{n}^{n}}$.

Composing these homotopies “horizontally” gives that $\alpha=\prod_{j=1}^{n}\alpha|_{K_{n}^{j}}$ is homotopic to

$(\beta|_{K_{n}^{1}}\cdot \gamma_{1}^{-})\cdot \left(\prod_{j=2}^{n-1}\gamma_{j-1}^{-}\cdot \beta|_{K_{n}^{j}}\cdot \gamma_{j}^{-}\right)\cdot (\gamma_{n-1}\cdot\beta|_{K_{n}^{n}})$

cancelling the inverse pairs $\gamma_{j}^{-}\cdot\gamma_{j}$ gives a homotopy with $\prod_{j=1}^{n}\beta|_{K_{n}^{j}}=\beta$. Thus $\alpha\simeq\beta$. We conclude that $\mathscr{U}\subseteq [\alpha]$. $\square$

Lemmas 1 and 2 tell us that for locally path-connected spaces, being semilocally simply connected is equivalent to $\pi_{1}^{qtop}(X,x_0)$ being a discrete group.

Theorem: Suppose $X$ is locally path connected. Then $\pi_{1}^{qtop}(X,x_0)$ is discrete if and only if $X$ is semilocally simply connected.

This tells us that, for locally path-connected spaces, non-discreteness of $\pi_{1}^{qtop}(X,x_0)$ as a topological invariant, really is detecting the existence of local 1-dimensional wildness in a space. When our space in question is not locally path connected, things get a bit trickier.

Example: The following “hoop earring” space $Y$ is semilocally simply connected but is not locally path-connected. This space includes the sequence of circles that all meet at a point $y_0$ and includes the limit circle too. If $\ell_n:[0,1]\to Y$ is the loop going once around the n-th circle (parameterized in a standard way) and $\ell_{\infty}:[0,1]\to Y$ goes once around the limit circle, then $\{\ell_n\}\to\ell_{\infty}$ in the compact-open topology and so $\{[\ell_n]\}\to [\ell_{\infty}]$ in $\pi_{1}^{qtop}(Y,y_0)$ even though none of these homotopy classes are the same. Since $\pi_{1}^{qtop}(X,x_0)$ contains a non-trivial convergent sequence, it can’t be discrete. In fact, this group is isomorphic to something called a free topological group (but that’s much harder to show)!

The space $Y$: a wedge of converging circles has non-discrete fundamental group.

## References.

[1] J. Calcut, J. McCarthy, Discreteness and homogeneity of the topological fundamental group, Topology Proc.  34 (2009) 339-349.

## Topologized Fundamental Groups: The Quotient Topology Part 1

Next up for topologies on the fundamental group is what I’d consider the most “natural” one. It’s almost certainly the topology you’d most often get if you asked random topologists on the street to construct one for you. This is the natural quotient topology. I’m excited about this one. I studied this one a lot as a grad student so it’s near and dear to my heart.

Now, even though it’s easy to construct and the added topology often contains way more information than the ordinary non-topologized fundamental group, it is actually quite tricky to work with. I also mentioned in my primer post that this topology comes with historical baggage and I’ll get to that later in this post.

Let’s go ahead and define it. Throughout, $X$ will be a path-connected space with basepoint $x_0\in X$. Let $\Omega(X,x_0)$ denote the space of loops in $X$ based at $x_0$, that is, maps $\alpha:[0,1]\to X$ with $\alpha(0)=\alpha(1)=x_0$. We give $\Omega(X,x_0)$ the usual compact-open topology, which is generated subbasis sets $\langle K,U\rangle=\{\alpha\mid \alpha(K)\subseteq U\}$ for compact $K\subseteq [0,1]$ and open $U\subseteq X$.

Every based loop $\alpha\in\Omega(X,x_0)$ has a corresponding path-homotopy class $[\alpha]\in \pi_1(X,x_0)$. This defines a surjection $q:\Omega(X,x_0)\to\pi_1(X,x_0)$, $q(\alpha)=[\alpha]$ which identifies homotopy classes.

Definition: Let $\pi_{1}^{qtop}(X,x_0)$ denote $\pi_1(X,x_0)$ equipped with the quotient topology with respect to the map $q:\Omega(X,x_0)\to\pi_1(X,x_0)$. We refer to this topology as the natural quotient topology on $\pi_{1}(X,x_0)$.

What this means: Remember that a surjective function $q:X\to Y$ is a quotient map if $U$ is open in $Y$ if and only if it’s a preimage $q^{-1}(U)$ is open in $X$. Hence, a set $A\subseteq \pi_{1}^{qtop}(X,x_0)$ is open (closed) if and only if the set $q^{-1}(A)$ of all loops representing the homotopy classes in $A$ is open (closed) in $\Omega(X,x_0)$.

The “qtop” superscript is actually going to do double duty here. Yes, it stands for “quoteint topology” but it also invokes the term “quasitopological” as in quasitopological group. The early literature on the quotient topology is messy but has corrected itself and has seen a lot of growth in the past 10 years. A summary appears in the following paper.

J. Brazas, P. Fabel, On fundamental groups with the quotient topology, J. Homotopy and Related Structures 10 (2015) 71-91

Terminology: Occasionally, authors will call the quotient topology on $\pi_1$ the “compact-open topology.” Tisk tisk. Look…I get it. The quotient topology does descend from the compact-open topology. Fine. But it’s not the compact-open topology itself so why call it the compact-open topology. “c.o.-quotient” topology would be better. I get a little hot about this because other topologies on $\pi_1$ like the tau-topology also depend very closely on the compact-open topology on loop spaces so when authors do this, I see it as an example of choosing to confuse terminology that avoids readily available descriptive terms. It’s a quotient topology… just call it what it is. Sometimes I’ve called it the quasitopological fundamental group, which is not ideal for more subtle reasons. At this point just “the natural quotient topology” is probably best. Sorry* for the rant.

In this sequence of posts, I’m going to detail some of the history, properties, and uses of the natural quotient topology on the fundamental group.

## Understanding the compact-open topology

Since we’re form a quotient space from the compact-open topology, let’s briefly unpack what the compact-open topology is really about. A subbasic set of the form $\langle K,U\rangle$ in $\Omega(X,x_0)$ can be thought of as a single instruction. It contains all of the paths that do something similar, namely map $K$ into $U$. A basis set is an intersection of subbasic sets, say $\bigcap_{i=1}^{n}\langle K_i,U_i\rangle$, which can be thought of as a finite set of instructions. If we have $\alpha\in \bigcap_{j=1}^{n}\langle K_j,U_j\rangle$, then $\alpha$ must map $K_1$ into $U_1$, and $K_2$ into $U_2$, and so on. Provided the compact sets $K_i$ cover $[0,1]$, we have some kind of restriction on every point in the domain. For example, let $K_{n}^{j}=\left[\frac{j-1}{n},\frac{j}{n}\right]$. Then $\mathscr{U}=\bigcap_{j=1}^{n}\langle K_{n}^{j},U_j\rangle$ is a basic open set and every element of $\mathscr{U}$ must be a path which proceeds sequentially through the sets $U_1,U_2,\dots,U_n$ in $X$ at a certain rate.

A basic open set in the compact open topology

It’s a folklore lemma that’s a bit tedious to prove that the sets of the form $\mathscr{U}$ actually form a basis for the compact-open topology on $\Omega(X,x_0)$.

The compact-open topology is the one most often used on loop spaces because

1. it usually has nice categorical properties,
2. it generalizes the topology of uniform convergence. In particular, if $X$ is a metric space, then $\{\alpha_n\}\to \alpha$ in $\Omega(X,x_0)$ with the compact-open topology if and only if $\{\alpha_n\}\to \alpha$ uniformly in $X$,
3. standard operations on loops are continuous.

Let’s expand upon 3. The usual concatenation of loops gives us an operation $c:\Omega(X,x_0)\times\Omega(X,x_0)\to \Omega(X,x_0)$, $c(\alpha,\beta)=\alpha\cdot\beta$ where $\alpha\cdot\beta$ is the loop that does $\alpha$ on $[0,1/2]$ and $\beta$ on $[1/2,1]$. We also have a reverse path operation $rev:\Omega(X,x_0)\to \Omega(X,x_0)$, $rev(\alpha)=\alpha^{-}$. Here, $\alpha^{-}(t)=\alpha(1-t)$ simply does $\alpha$ in the opposite orientation. It’s a nice exercise to check that $c$ and $rev$ are continuous. One should beware that $c$ is not stricly unital or associative.

We can also restrict the concatenation map $c$ to continuous right- and left-concatenation maps:

• $\rho_{\beta}:\Omega(X,x_0)\to\Omega(X,x_0)$, $\rho_{\beta}(\alpha)=\alpha\cdot\beta$;
• $\lambda_{\beta}:\Omega(X,x_0)\to\Omega(X,x_0)$, $\lambda_{\beta}(\alpha)=\beta\cdot\alpha$.

Both of these are continuous as they can be identified with restrictions of $c$.

## What kind of thing is $\pi_{1}^{qtop}(X,x_0)$?

The definition of $\pi_{1}^{qtop}(X,x_0)$ is so natural and simple that this topologized group should to be a pretty nice thing, right? Let’s see what we can do. We really only need one tool from general topology.

University property of quotient maps: Consider spaces $A,B,C$ and functions $q,f,g$ that make the following diagram commute (so $f\circ q=g$).

If $q$ is a quotient map and $g$ is continunous, then $f$ is continuous too.

The universal property is a powerful tool for showing that functions are continuous. Note that the universal property might appear in slightly different ways. For instance, we might instead have the following commutative square.

In a situation like this, the universal property becomes useful if we’re not sure if $h$ is continuous or not. If the left map $q$ is a quotient map and the upper composition $g\circ f$ is continuous, then $h$ will also be continuous. Of course, this is just a special case of the triangle above, but for the sake of applications it’s good to be ready to use it in diagrams of various shapes.

Let’s try our best to prove that $\pi_{1}^{qtop}(X,x_0)$ is a topological group. First, we’ll prove that inversion is continuous.

Proposition: Group inversion $inv:\pi_{1}^{qtop}(X,x_0)\to \pi_{1}^{qtop}(X,x_0)$, $inv([\alpha])=[\alpha]^{-1}$ is continuous.

Proof. Consider the following commutative square.

Here, $rev(\alpha)=\alpha^{-}$ is the reverse map. Since $inv(q(\alpha))=[\alpha]^{-1}=[\alpha^{-}]=q\circ rev(\alpha)$, the diagram does indeed commute. Now both vertical maps are the quotient map $q$ and as noted above, $rev$ is continuous. Since the left map is quotient and the upper composition $q\circ rev$ is continuous, the bottom maps $inv$ is continuous by the universal property of quotient maps. $\square$.

Ok, with the universal property in hand, that wasn’t so bad! Let’s see if we can do the same thing for group multiplication $\mu:\pi_{1}^{qtop}(X,x_0)\times \pi_{1}^{qtop}(X,x_0)\to \pi_{1}^{qtop}(X,x_0)$. WARNING: I’m about to propose an incorrect proof. It’s going to have an error in it. See if you can find it.

Proposed Proof. Consider the following commutative diagram where $c$ is the concatenation map and $\mu$ is the group operation in the fundamental group.

Note that $\mu(q\times q(\alpha,\beta))=[\alpha][\beta]=[\alpha\cdot\beta]=q(c(\alpha,\beta)$ so the diagram does indeed commute. By definition $q$ is a quotient map and as noted above $c$ is continuous. Since the left vertical map $q\times q$ is quotient and the upper composition $q\circ c$ is continuous, the bottom map $\mu$ is continuous by the universal property of quotient maps. (QED?)

Seem ok? See any problems? The diagram commutes just fine. The upper composition is continuous. The error is in the phrase: “Since the left vertical map $q\times q$ is quotient.” Here’s a valuable lesson friends. If we’re using the product topology (which is what you use for a topological group $G$ to have continuous multiplication $G\times G\to G$), the direct product of two quotient maps is not always a quotient map. There are counterexamples in most introductory general topology books.

Just because a proof is wrong doesn’t always mean the claim is false. However…our wish here is false. Spoiler Alert: $\pi_{1}^{qtop}(X,x_0)$ is NOT always a topological group. In fact, I’d say it’s rarely a topology group when it’s Hausdorff and non-discrete. Paul Fabel gave the first counterexample [4] and shortly after I published one [2] that connects to other structures from topological group theory.

## Oops

The above logical error is an easy one to make and miss. The most notable place this mistake was made is in the 2002 paper:

[1] D. Biss, The topological fundamental group and generalized covering spaces,
Topology Appl. 124 (2002), 355–371. RETRACTED.

This paper had some really cool global ideas. However, there are a ton of independent errors in it…. I mean a ton. Very few results in it are correct. As you can see it is now retracted – and it even made retraction watch. According to Google Scolar, as of 8/25/2022, this paper has 90 citations! The errors within created something of a mess for a while because some papers called upon Biss’ false claims. For one, we no longer called it the “topological fundamental group” because we now know it’s not always “topological” in the sense of being a “topological” group. Unfortunately, several papers have cited [1] without mentioning the many errors or retraction.

The same error was used in [6] to claim the higher homotopy groups with the natural quotient topology are always topological groups. Paul Fabel constructed counterexamples to this higher-dimensional claim in [5].

Actually, the same mistake was made 12 years earlier in an appendix [7] written by my mathematical grandfather J.P. May. This short note involves a groupoid version of $\pi_{1}^{qtop}(X,x_0)$ but it includes essentially the same error. However, unlike in Biss’ paper, the overall results are not damaged by this error because of the kinds of spaces being used.

I think most professional mathematicians, incluidng myself, have published some false arguments or statements. Sometimes our intuition guides our writing more than formal logic does. This can lead to a correct result but a proof with an easily fixable logical gap. Occasionally, like in this case, a published error might be unfixable.

Nobody wants to have their published mathematics end up being wrong. The point of publicly mentioning these specific papers is not to be hard on the people making the mistakes. I mean…. J.P. May, who asked me about this situation back in 2010, is one of the most influential and prolific algebraic topologists in the history of the subject. Also, Biss’ paper did end up being very influential even if it’s not in the intended way. That’s something to consider. Rather, the point of this part of the post is to:

1. Encourage awareness about published errors so others don’t fall into the trap of doing mathematics that depends on false claims.
2. Emphasize to young mathematicians that incorrect math can lead to new (and correct) ideas and areas of research and that even great mathematicians make mistakes here and there. It’s possible to own up to a mistake without getting defensive or letting it defeat you personally.
3. Show that it is possible to point out mistakes of others at the research level in a respectful and kind manner.

## How I learned to love horrifyingly complicated things

Now, look…sometimes the direct product of two quotient maps is a quotient map. And it’s true that for some spaces $X$, this works and $\pi_{1}^{qtop}(X,x_0)$ does end up being a topological group. But this is really only guaranteed when the domain and codomain satisfy some (local) compactness criteria. This apparent failure of the topological category $\mathbf{Top}$ is closely related to the fact that it is not Cartesian closed. Are there ways to cheat the system? Sure. This apparent deficiency goes away if you replace the usual category of spaces with a coreflective Cartesian closed category like compactly generated spaces, sequential spaces, delta-generated spaces, etc. However, if you work internal to one of these categories, the group-objects in those categories may not be true topological groups.

Before, you demand that we switch to one of these categories consider where this is going. Maybe consider the question “what is the fundamental group good for?” I’d say its utility is that it is an invariant, which creates a symbiotic relationship between topology and algebra. You can study and classify spaces using the algebra and functorality of fundamental groups and, on the flip side, if you want to prove things about a collection of groups it’s often a good idea to realize them as fundamental groups of spaces with some common features and then use the topology to prove things about the groups. For example, fundamental groups can be used to classify surfaces and, in the other direction, covering space theory provides nice proofs of results like the Nielsen-Schreier and Kurosh Theorems.

What can a topologized fundamental group like $\pi_{1}^{qtop}(X,x_0)$ be good for? Potentially, it could create an extended symbiotic relationship between spaces with complicated local structure and topologized groups. I’d say this has been carried out in a successful way and the progress is ongoing. For example, long-standing gaps in free topological group theory have been filled using topologized fundamental groups. The only known proofs rely on working with $\pi_{1}^{qtop}(X,x_0)$. Such things could not have been found if everyone had just decided to shove everything into another category.

Also, from a pragmatic viewoint…there are international communities of general topologists studying the topology of actual topological groups. There are not large communities who just study groups internal to, say, compactly generated spaces. Not that there shouldn’t be…but there just aren’t.

Personally, the fact that $\pi_{1}^{qtop}$ ends up being really complicated taught me a valuable lesson early on in my research career. If I became unwilling to struggle to work with complicated structures that some mathematicians might find ugly or terrifying, my mathematical world would remain small and leave me with less potential to reveal fascinating and beautiful possibilities.

## $\pi_{1}^{qtop}(X,x_0)$ is a quasitopological group

If $\pi_{1}^{qtop}(X,x_0)$ not being a topological group is a castle in ruins, let’s pick up some of rubble and build a tiny house out of it.

Definition: a group $G$ with topology is a quasitopological group if inversion $g\mapsto g^{-1}$ is continuous and if multiplication $G\times G\to G$ is continuous in each variable, that is, if all left and right translation maps $G\to G$ given by $h\mapsto gh$ and $h\mapsto hg$ are continuous for all $g\in G$.

For $[\beta]\in\pi_{1}^{qtop}(X,x_0)$, let $\rho_{[\beta]}:\pi_{1}^{qtop}(X,x_0)\to \pi_{1}^{qtop}(X,x_0)$, $\rho_{[\beta]}([\alpha])=[\alpha][\beta]$ be right multiplication by $[\beta]$, and $\lambda_{[\beta]}:\pi_{1}^{qtop}(X,x_0)\to \pi_{1}^{qtop}(X,x_0)$, $\lambda_{[\beta]}([\alpha])=[\beta][\alpha]$ be left multiplication by $[\beta]$.

Lemma: For any based space $(X,x_0)$, $\pi_{1}^{qtop}(X,x_0)$ is a quasitopological group.

Proof. We have already seen above that inversion in $\pi_{1}^{qtop}(X,x_0)$ is continuous. Fix $[\beta]\in\pi_{1}^{qtop}(X,x_0)$  and consider the following commutative diagrams involving left and right concatenation/multiplication.

In both squares, the left map is quotient and the upper composition is continuous. Therefore the bottom maps are continuous by the universal property of quotient maps. $\square$.

Theorem: $\pi_{1}^{qtop}:\mathbf{Top_{\ast}}\to \mathbf{qTopGrp}$ is a functor from the category of based topological spaces to the category of quasitopological groups and continuous homomorphisms.

Proof. The previous lemma, tells us that $\pi_{1}^{qtop}$ is well-defined on objects. The underlying algebraic structure of $\pi_{1}^{qtop}$ is the usual functor $\pi_{1}$ so we only need to check that if $f:(X,x_0)\to (Y,y_0)$ is a based map, then the induced homomorphism $f_{\#}: \pi_{1}^{qtop}(X,x_0)\to \pi_{1}^{qtop}(Y,y_0)$ is continuous. First, note that there is an induced loop-space function $\Omega(f):\Omega(X,x_0)\to \Omega(Y,y_0)$, $\Omega(f)(\alpha)=f\circ \alpha$. If $\langle K,U\rangle$ is a subbasic open set in $\Omega(Y,y_0)$, then $\Omega(f)^{-1}(\langle K,U\rangle)=\langle K,f^{-1}(U)\rangle$ is a subbasic open subset of $\Omega(X,x_0)$. Thus $\Omega(f)$ is continuous. The map $f$ also induces the homomorphism $f_{\#}:\pi_1(X,x_0)\to\pi_1(Y,y_0)$, $f_{\#}([\alpha])=[f\circ \alpha]$. Now consisder the following diagram.

Since $f_{\#}(q(\alpha))=f_{\#}([\alpha])=[f\circ \alpha]=q\circ \Omega(f)(\alpha)$, the diagram commutes. Moreover, the left vertical map is quotient and the upper composition is continuous. Therefore, the bottom function is continuous by the universal property of quotient maps. $\square$.

If $f:(X,x_0)\to (Y,y_0)$ and $g:(Y,y_0)\to (X,x_0)$ are based homotopy inverses, then the homomorphisms $f_{\#}$ and $g_{\#}$ they induce are continuous inverses and therefore are homeomorphisms. This means that $\pi_{1}^{qtop}$ is an invariant of based homotopy type.

Exercise: Use the universal property of quotient maps to prove that if $\beta:[0,1]\to X$ is a path from $x_0$ to $x_1$, then the basepoint-change isomorphism $\varphi_{\beta}:\pi_{1}^{qtop}(X,x_0)\to\pi_{1}^{qtop}(X,x_1)$, $\varphi([\alpha])=[\beta^{-}\cdot\alpha\cdot\beta]$ is a homeomorphism.

From this exercise, the same argument that $\pi_1$ is an invariant of unbased homotopy type can be used to show that $\pi_{1}^{qtop}$ is an invariant of unbased homotopy type.

Corollary: If $X$ and $Y$ are path connected and $X\simeq Y$, then $\pi_{1}^{qtop}(X,x_0)\cong \pi_{1}^{qtop}(Y,y_0)$ as quasitopological groups for any choice of $x_0\in X$ and $y_0\in Y$.

There’s a lot more to say about $\pi_{1}^{qtop}$ and I’ll get to some of it in future posts. For now we can take away the fact that even though $\pi_{1}^{qtop}(X,x_0)$ is not always a topological group, it is still pretty close to being a topological group. Moreover, the natural quotient topology gives us a homotopy invariant, which is much stronger invariant than the usual fundamental group, particularly when it comes to spaces with complicated local structures.

## References

[1] D. Biss, The topological fundamental group and generalized covering spaces,
Topology Appl. 124 (2002), 355–371.

[2] J. Brazas, The topological fundamental group and free topological groups, Topol. Appl. 158 (2011) 779–802.

[3] J. Brazas, P. Fabel, On fundamental groups with the quotient topology, J. Homotopy Relat. Struct. 10 (2015) 71–91.

[4] P. Fabel, Multiplication is discontinuous in the Hawaiian earring group, Bull. Pol. Acad. Sci., Math. 59 (2011) 77–83.

[5] P. Fabel, Compactly generated quasitopological homotopy groups with discontinuous multiplication, Topol. Proc. 40 (2012) 303–309.

[6] H. Ghane, Z. Hamed, B. Mashayekhy, and H. Mirebrahimi, Topological
homotopy groups, Bull. Belgian Math. Soc. 15 (2008), 455–464.

[7] J.P. May, G-spaces and fundamental groupoids, appendix, K-Theory 4
(1990), 50–53.

*I’m not really sorry. It does bug me.