## 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.

## Topologized Fundamental Groups: The Whisker Topology, Part 3

In Part 1 and Part 2, I gave detailed introductory exposition about the whisker topology on the fundamental group. In general, this topologized fundamental group $\pi_{1}^{wh}(X,x_0)$ is a left topological group and therefore a homogeneous space. Moreover, whenever this group is $T_1$, it’s also zero-dimensional. This bit about zero-dimensionality is a huge restriction which tells us the whisker topology is a very fine topology. Most often, we’re interested in metrizable spaces so it’s natural to ask what we can acheive in this situation.

The proof I’m going to give isn’t too hard but it does require working through several individual steps. If nothing else, this proof is basically the same as a certain proof that a universal covering space of a locally path-connected, metrizable space is metrizable.

## (Pseudo)metrizability of the whisker topology

Supposing that $X$ is metrizable, pick a metric $d$ that induces the topology of $X$. Let’s define a distance function $\rho$ on $\pi_{1}^{wh}(X,x_0)$. For loops $\alpha,\beta$ based at $x_0$, set

$\rho([\alpha],[\beta])=\inf\{diam(\epsilon)\mid [\epsilon]=[\alpha^{-}\cdot\beta]\}$

Now, it’s possible that $\rho([\alpha],[\beta])=0$ even when $[\alpha]\neq [\beta]$ so we should only expect that $\rho$ will be a pseudometric in general. Symmetry is pretty clear from the definition. The triangle inequality is maybe a little less clear but working it out is a nice exercise. Let’s do it. There are some different ways to write proofs like this involving infimums and supremums that use a variety of established/known facts but these are pretty efficient (though they are by contradiction).

Lemma: If $\alpha,\beta$ are loops in $X$ based at $x_0$ and $\gamma=\alpha\cdot\beta$, then $diam(\gamma)\leq diam(\alpha)+diam(\beta)$.

Proof. Recall the supremum definition of the diameter of a loop. We’ll use the general fact that $\sup(S)>a$ if and only if there exists $s\in S$ with $s>a$. If $diam(\gamma)> diam(\alpha)+diam(\beta)$, then there exists $a,b\in [0,1]$ with $d(\gamma(a),\gamma(b))>diam(\alpha)+diam(\beta)$. If $a,b\in [0,1/2]$, then $d(\gamma(a),\gamma(b))=d(\alpha(a/2),\alpha(b/2))>diam(\alpha)$, which is a contradiction. If $a,b\in [1/2,1]$, then $d(\gamma(a),\gamma(b))=d(\beta(2a-1),\alpha(2b-1))>diam(\beta)$. Without loss of generality, suppose $a\in[0,1/2]$ and $b\in [1/2,1]$. Then

$d(\alpha(2a),\beta(2b-1))= d(\gamma(a),\gamma(b))>diam(\alpha)+diam(\beta)\geq d(\alpha(2a),x_0)+d(x_0,\beta(2b-1))$

but this is a violation of the triangle inequality in $X$. $\square$

Lemma: Given $[\alpha],[\beta],[\gamma]\in\pi_{1}^{wh}(X,x_0)$, $\rho([\alpha],[\gamma])\leq \rho([\alpha],[\beta])+\rho([\beta],[\gamma])$.

Proof. We’ll play a similar game in this lemma using the analagous elementary fact for infimums:  $a>\inf(S)$ if and only if there exists $s\in S$ with $a>s$,

Suppose $\rho([\alpha],[\gamma])>\rho([\alpha],[\beta])+\rho([\beta],[\gamma])$. Since $\rho([\alpha],[\gamma])-\rho([\beta],[\gamma])>\rho([\alpha],[\beta])$, there exists a loop $\epsilon$ with $[\epsilon]=[\alpha^{-}\cdot\beta]$ and $\rho([\alpha],[\gamma])-\rho([\beta],[\gamma])>diam(\epsilon)$. Since $\rho([\alpha],[\gamma])-diam(\epsilon)>\rho([\beta],[\gamma])$, there exists a loop $\delta$ with $[\delta]=[\beta^{-}\cdot\gamma]$ and $\rho([\alpha],[\gamma])-diam(\epsilon)>diam(\delta)$. Now $[\epsilon\cdot\delta]=[\alpha^{-}\cdot \beta][\beta^{-}\cdot\gamma]=[\alpha^{-}\cdot\gamma]$ and $\rho([\alpha],[\gamma])>diam(\epsilon)+diam(\delta)$. We have $diam(\epsilon)+diam(\delta)\geq diam(\epsilon\cdot\delta)$ from the previous lemma so $\rho([\alpha],[\gamma])>diam(\epsilon\cdot\delta)$. Since $[\epsilon\cdot\delta]=[\alpha^{-}\cdot\gamma]$, this is a contradiction. $\square$.

With the triangle inequality in hand, we have a pseudometric! Now we should check that it induces the topology $\pi_{1}^{wh}(X,x_0)$. In general, if $d$ is a (pseudo)metric on a set $S$, $O_d(s,r)=\{t\in S\mid d(t,s) will denote the open ball of radius $r$ about $s$.

Lemma: The pseudometric induces the topology of $\pi_{1}^{wh}(X,x_0)$.

Proof. Let $B([\alpha],U)$ be a basic neighborhood of $[\alpha]$ in $\pi_{1}^{wh}(X,x_0)$. Find $r>0$ such that $O_{d}(x_0,r)\subseteq U$ where $O_{d}(x_0,r)$ is the open $r$-ball about $x_0$. Suppose $[\beta]\in O_{\rho}([\alpha],r)$. Since $\rho([\alpha],[\beta]), there exists a loop $\gamma$ based at $x_0$ with $[\gamma]=[\alpha^{-}\cdot\beta]$ and $diam(\gamma). Since $\gamma$ has image in $O_{d}(x_0,r)$, it has image in $U$. This gives $[\beta]=[\alpha\cdot\epsilon]\in B([\alpha],U)$.

For the other direction, suppose $r>0$ and consider the neighborhood $O_{\rho}([\alpha],r)$. We’ll show that $B([\alpha],O_d(x_0,r/3))\subseteq O_{\rho}([\alpha],r)$. Suppose $[\beta]\in B([\alpha],O_d(x_0,r/3))$. Write $[\beta]=[\alpha\cdot\epsilon]$ for loop $\epsilon$ in $O_d(x_0,r/3))$. Then $[\epsilon]=[\alpha^{-}\cdot\beta]$ and $diam(\epsilon)\leq \frac{2r}{3}$. Thus $\rho([\alpha],[\beta])\leq \frac{2r}{3}, showing $[\beta]\in O_{\rho}([\alpha],r)$. $\square$

Theorem: If $X$ is metrizable, then $\pi_{1}^{wh}(X,x_0)$ is pseudometrizable.

Remark: This proof also goes through for the space $\widetilde{X}$ of all path-homotopy classes of paths starting at $x_0$ with the whisker topology of which $\pi_{1}^{wh}(X,x_0)$ is a subspace. Indeed, whenever $X$ is metrizable, $\widetilde{X}$ is pseudometrizable (See [5,Lemma 2.12]). Why I’m taking a second to do the proof is to emphasize that it works for ALL metrizable spaces, not just those for those having the usual assumptions of covering space theory.

But…can we get rid of the “pseudo?” In a pseudometric space it’s possible to have distinct points that have zero-distance from each other. Such a space is not even $T_0$.  This can certainly happen for $\pi_{1}^{wh}(X,x_0)$ and we saw an instance of it in an earlier post. But it’s also true that a pseudometrizable space is metrizable if and only if it’s Hausdorff and we characterized the Hausdorff property for $\pi_{1}^{wh}(X,x_0)$ in Part 2. Moreover, we showed that $\pi_{1}^{wh}(X,x_0)$ is zero-dimensional whenever it’s Hausdorff.

Let’s summarize. Taking several of our established results, we have the following.

Theorem: Let $X$ be a path-connected, metrizable, homotopically Hausdorff space and $x_0\in X$. Then $\pi_{1}^{wh}(X,x_0)$ is a homogeneous, zero-dimensional, metrizable space.

This applies in lots of cases, include all one-dimensional and planar spaces.

## Separability

Now I’m wondering about separability. Here, I’m still assuming $(X,d)$ is a homotopically Hausdorff metric space so that we’re in the situation of the previous theorem. Clearly $\pi_{1}^{wh}(X,x_0)$ is separable if it is countable. That’s not so interesting. What about in other cases? Honestly, I’m not too sure how to characterize this in general. Here’s how I think we can understand it. Let $U_n$ be the $1/n$-ball about $x_0$ in $X$. If $S=\{s_1,s_2,s_3,\dots\}$ was a dense countable subset of $\pi_{1}^{wh}(X,x_0)$, then for every homotopy class $g\in \pi_{1}^{wh}(X,x_0)$ and neighborhood $n\in\mathbb{N}$, we’d have $s_k\in B(g,U_n)$ for some $k=k(g,n)$. Another way to say this is that there is a sequence $s_{k(g,1)},s_{k(g,2)},s_{k(g,3)},\dots$ in $S$ where the products $gs_{k(g,1)}^{-1},gs_{k(g,3)}^{-1},gs_{k(g,3)}^{-1},\dots$ have representatives that get arbitrarily small. So we can take any element of the fundamental group and make it small by multiplying on the right by inverses of the elements of $S$.

In the earring group, a Cantor diagonalization argument should be possible to see it’s not separable. I don’t want to get hung up on this here and get caught up in examples, but it’s a curious question and maybe someone else wants to go down this rabbit hole. This could make a nice student project.

Question: If $X$ is a homotopically Hausdorff metric space, when is $\pi_{1}^{wh}(X,x_0)$ separable? What are some nice examples illustrating when it is and isn’t separable?

To finish the post, I’m going to go back to why I actually spent 3 posts writing about a $\pi_1$  topology I’ve never really cared much about.

## Why am I starting to care more about the whisker topology?

I’ll be honest… I used to think the whisker topology wasn’t too interesting or useful. Here, I’ll spend a little time trying to explain why I’ve changed my tune on this. I want to be clear that I’m not so much trying to publicize my own results as must as I want to publicize the fact that I don’t really understand a certain naturally occuring group as well as I’d like to.

Consider the one-point union $Y=\mathbb{E}_1\vee\mathbb{E}_n$ of the earring space and the n-dimensional earring as seen below for $n=2$.

The one-point union $\mathbb{E}_1\vee\mathbb{E}_2$.

The group $\pi_n(Y)$ is fairly complicated because $\pi_1(Y)=\pi_1(\mathbb{E}_1)$ is uncountable and infinitary. Unlike with $S^1\vee S^2$, the usual $\pi_1$-action is not enough to really understand $\pi_2$. Recently, I proved that $\pi_n(Y)$ canonically embeds into the group $\prod_{j\in\mathbb{N}}\oplus_{\pi_1(\mathbb{E}_1)}\mathbb{Z}$. I’d like to understand the image in this product of $\mathbb{Z}$‘s better.

We can think of elements of $G= \prod_{j\in\mathbb{N}}\oplus_{\pi_1(\mathbb{E}_1)}\mathbb{Z}$ as functions $g:\mathbb{N}\times \pi_1(\mathbb{E}_1)\to\mathbb{Z}$, which has finite support in the second variable: for each $j\in\mathbb{N}$, the set $\{[\alpha]\in\pi_1(\mathbb{E}_1)\mid g(j,[\alpha])\neq 0\}$ is finite.

The canonical homomorphism $\Psi:\pi_n(Y)\to G$ is not onto! Here’s the characterization of the image: an element $g\in G$ is in the image of $\Psi$ and therefore uniquely represents a homotopy class in $\pi_n(Y)$ if and only if $\{[\alpha]\in \pi_1(\mathbb{E}_1)\mid g(j,[\alpha])\neq 0\}$ is countable and has compact closure in $\pi_{1}^{wh}(\mathbb{E}_1)$. So elements of $\pi_n(Y)$ are classified by functions $g:\mathbb{N}\times\pi_1(\mathbb{E}_1)\to\mathbb{Z}$ with countable support and such that that $g$ is non-trivial on only a “bounded” set in the second variable (but only with the whisker topology!). Even though I have a geometric understanding of this group and feel this description is good enough for most any use I can imagine, there is topology embedded in the indexing sets. My instinct tells me that it’s built-in topological nature can’t be done away with but this idea is not really something I know how to formalize. Anyway, I never would have guessed the whisker topology would become relevant to wild higher homotopy theory.

## So what are the compact subsets?

That stuff about homotopy groups is why I was interested in understanding compact subsets of $\pi_{1}^{wh}(X,x_0)$. The theorem we worked out in these posts says that $\pi_{1}^{wh}(X,x_0)$ is often zero-dimensional and metrizable. In this case, a compact subset $K$ of $\pi_{1}^{wh}(X,x_0)$ will be a zero-dimensional compact metric space.

This leaves open the possibility $K$ could be homeomorphic to a Cantor set, which does seem reasonable. Here’s an example.

Example: $X=\prod_{n=1}^{\infty}\mathbb{RP}_2$ is the infinite direct product of copies of the projective plane, then $\pi_1(X,x_0)$ is abelian and therefore $\pi_{1}^{wh}(X,x_0)\cong \prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$ is a topological group by results in Post 1. But we can say more. We can identify $\pi_{1}^{wh}(\mathbb{RP}^2,x)$ with the discrete group $\mathbb{Z}/2\mathbb{Z}$ for any choice of basepoint. Now, a basic neighborhood of the basepoint in $\prod_{n=1}^{\infty}\mathbb{RP}_2$ can be taken to be of the form $V= \prod_{i=1}^{k}U\times \prod_{i > k}\mathbb{RP}_2$ where $U$ is a contractible neighborhood of the basepoint in $\mathbb{RP}_2$ .Therefore, if $\alpha=(\alpha_i)$ is a loop in $X$, then $B([\alpha],V)=\prod_{i=1}^{k}\{[\alpha_i]\}\times \prod_{i > k}\mathbb{Z}/2\mathbb{Z}$. This is the key idea needed to check that $\pi_{1}^{wh}(X,x_0)$ is canonically isomorphic to the topological group $\prod_{n=1}^{\infty}\mathbb{Z}/2\mathbb{Z}$ with the product topology. In particular, $X$ is a metric space for which $\pi_{1}^{wh}(X,x_0)$ is homeomorphic to a Cantor set. $\square$

So, yes, a compact subset $K$ of $\pi_{1}^{wh}(X,x_0)$ can be as complicated as a Cantor set. But descriptive set theory tells us it can’t be much worse. Every Polish space is the disjoint union of a countable scattered subspace and (possibly) a perfect set. In our situation, zero-dimensionality ensures that if there is a perfect subset, it must be a Cantor set.

Theorem: Let $X$ be a path-connected, metrizable, homotopically Hausdorff space and $x_0\in X$. If $K\subseteq \pi_{1}^{wh}(X,x_0)$ is a compact, then $K$ is either homeomorphic to a countable compact ordinal or the union of a countable scattered space and a Cantor set.

I don’t know that this solves the entirety of my lack of understanding of my representation of $\pi_n(Y)$ using the whisker topology but it does give me a way to handle compact sets in $\pi_{1}^{wh}(X,x_0)$ if I need to.

## References

The references here include several papers that involve the whisker topology from a group of Iranian researchers who have done a lot of research in this area.

[1] N. Jamali, B. Mashayekhy, H. Torabi, S.Z. Pashaei, M. Abdullahi Rashid, On
topologized fundamental groups with small loop transfer viewpoints, Acta Math. Vietnamica, 44 (2019) 711–722.

[2] M. Abdullahi Rashid, N. Jamali, B. Mashayekhy, S.Z. Pashaei, H. Torabi, On subgroup topologies on the fundamental group. Hacettepe Journal of Mathematics & Statistics 49 (2020), no. 3, 935 – 949.

[3] M. Abdullahi Rashid, B. Mashayekhy, H. Torabi, S.Z. Pashaei, On subgroups of
topologized fundamental groups and generalized coverings, Bull. Iranian Math. Soc.
43 (2017), no. 7, 2349–2370.

[4] M. Abdullahi Rashid, S.Z. Pashaei, B. Mashayekhy, H.Torabi, On the Whisker Topology on Fundamental Group. Conference Paper from 46th Annual Iranian Mathematics Conference 46 (2015). Note: easily found through a google search but I can’t get a link to work.

[5] H. Fischer, A. Zastrow, Generalized universal covering spaces and the shape
group, Fund. Math. 197 (2007) 167-196.

## Topologized Fundamental Groups: The Whisker Topology, Part 2

This post is the second on the “whisker topology” on fundamental groups, in a larger sequence of posts on topologized fundamental groups. If you haven’t seen it, you’ll probably want to start with the first post to get the basic definitions and terminology. The topology on $\pi_{1}^{wh}(X,x_0)$ is generated by the sets $B([\alpha],U)=\{[\alpha\cdot\epsilon]\in \pi_1(X,x_0)\mid \epsilon([0,1])\subseteq U\}$ where $U$ is an open neighborhood of $x_0$.

One thing I kind of glossed over earlier is the fact that changing the basepoint can change the whisker topology. To convince you of this, let’s take a look at the following question, which I left as an exercise in the previous post.

## When is the Whisker Topology discrete?

In the first post, we saw that $\pi_{1}^{wh}(X,x_0)$ need not be a topological group but that it is still a homogeneous space. Consequently, if any one-point set is open, including the trivial subgroup, then the space will be discrete. Based on this observation, we know that $\pi_{1}^{wh}(X,x_0)$ will be discrete if and only if some basic neighborhood $B(1,U)$ of the identity contains only the identity element, i.e. $B(1,U)=\{1\}$. This will happen if and only if $[\alpha]=1$ for any loop $\alpha$ in $U$, that is, if any loop in $U$ based at $x_0$ is null-homotopic in $X$. This is precisely the definition of being semilocally simply connected at $x_0$!

Here, we’re using the “based” version of this property: $X$ is semilocally simply connected at $x\in X$ if there exists an open neighborhood of $U$ such that the homomorphism $\pi_1(U,x)\to \pi_1(X,x)$ induced by inclusion is trivial.

Proposition: $\pi_{1}^{wh}(X,x_0)$ is discrete if and only if $X$ is semilocally simply connected at $x_0$.

So the whisker topology only detects wildness only the basepoint. Algebraically, if $x_1\in X$, then there is a basepoint-change (group) isomorphism $\pi_1(X,x_0)\cong \pi_1(X,x_1)$. However, the topologized groups $\pi_{1}^{wh}(X,x_0)$ and $\pi_{1}^{wh}(X,x_0)$ will not always be homeomorphic!

Example: If $y_0$ is a point in the earring space $\mathbb{E}$ other than the wild point, then $\pi_{1}^{wh}(\mathbb{E},y_0)$ will be an uncountable discrete group. But if $x_0$ is the wild point, we saw in the last post that $\pi_{1}^{wh}(\mathbb{E},y_0)$ is not discrete. So these two fundamental groups (with the same space but different basepoints) are isomorphic as groups but not homeomorphic as spaces.

That changing the basepoint can change the whisker topology emphasizes, once again, that this topology really only tells us about the topology at the basepoint.

## How Connected is the Whisker Topology?

One of the interesting things about the whisker topology is the following lemma, which tells us that for a fixed neighborhood $U$ of $x_0$, the basic neighborhoods $B([\alpha],U)$ partition $\pi_{1}^{wh}(X,x_0)$. Recall that $\widetilde{X}$ is the set of homotopy classes of paths starting at $x_0$ and $N([\alpha],U)$ denotes the corresponding basic neighborhoods in $\widetilde{X}$.

Neighborhood Lemma: In $\widetilde{X}$, if $[\beta]\in N([\alpha],U)$, then $N([\alpha],U)=N([\beta],U)$.

Proof. If $[\beta]\in N([\alpha],U)$, write $[\beta]=[\alpha\cdot\epsilon]$ for $\epsilon$ with image in $U$. Then $[\alpha]=[\beta\cdot\epsilon^{-}]$. With $\epsilon$ fixed, we prove both subset inclusions. Given $[\gamma]\in N([\alpha],U)$, write $[\gamma]=[\alpha\cdot\delta]$ for $\delta$ with image in $U$. Thus $[\gamma]=[\beta\cdot(\epsilon^{-}\cdot\delta)]$ where $\epsilon^{-}\cdot\delta$ has image in $U$ (see the illustration below). Thus $[\gamma]\in N([\beta],U)$. This proves $N([\alpha],U)\subseteq N([\beta],U)$. The other inclusion is similar. $\square$

Proof of the Neighborhood Lemma in $\widetilde{X}$.

Corollary: Given an open neighborhood $U$ of $x_0$ in $X$, two sets $B([\alpha],U)$ and $B([\beta],U)$ in $\pi_{1}^{wh}(X,x_0)$ are either equal or disjoint.

Proof. If $B([\alpha],U)$ and $B([\beta],U)$ are not disjoint for loops $\alpha,\beta$ based at $x_0$, then we have $[\alpha\cdot\delta]=[\beta\cdot\epsilon]$ for loops $\delta,\epsilon$ with image in $U$. Then $[\alpha]=[\beta\cdot(\epsilon\cdot\delta^{-})]$ where $epsilon\cdot\delta^{-}$ has image in $U$ and so $[\alpha]\in B([\beta],U)$. Recalling that $B([\alpha],U)=\pi_1(X,x_0)\cap N([\alpha],U)$, it follows from the Neighborhood Lemma that $B([\alpha],U)=B([\beta],U)$. $\square$

This corollary implies that all basic open sets in $\pi_{1}^{wh}(X,x_0)$ are always clopen, that is, the small inductive dimension of $\pi_{1}^{wh}(X,x_0)$ is zero.

Theorem: Whenever $\pi_{1}^{wh}(X,x_0)$ is Hausdorff, it is zero-dimensional (in the sense of small inductive dimension).

Another way to see what’s going on here is to notice that the neighborhoods $B(1,U)$ of the identity element $1$ are all clopen. Therefore, if $\mathscr{U}(x_0)$ is the set of open neighborhoods of $x_0$, then $\bigcap_{U\in\mathscr{U}(x_0)}B(1,U)$ is equal to the closure of the trivial subgroup $\overline{\{1\}}$.

## When is the Whisker Topology Hausdorff?

At this point, we’ve seen that whenever $\pi_{1}^{wh}(X,x_0)$ is Hausdorff, is pretty highly structured – zero dimensional homogeneous spaces are only so complicated. So what would it take for $\pi_{1}^{wh}(X,x_0)$ to be Hausdorff?

Topological group theory tells us that a topological group is completely regular (Tychonoff) if and only if the trivial subgroup is closed. In fact, we have close to the same thing for the whisker topology even though it’s often far from being a topological group.

Theorem: The following are equivalent.

1. The trivial subgroup is closed in $\pi_{1}^{wh}(X,x_0)$,
2. $\pi_{1}^{wh}(X,x_0)$ is Hausdorff,
3. $\pi_{1}^{wh}(X,x_0)$ is $T_3$,

Proof. 3. $\Rightarrow$ 2. $\Rightarrow$ 1. is clear. Also, every zero-dimensional Hausdorff space is $T_3$. So it suffices to show 1. $\Rightarrow$ 2. Suppose the trivial subgroup $\{1\}$ is closed and $[\alpha]\neq [\beta]$ in $\pi_{1}^{wh}(X,x_0)$. Then $[\beta^{-}\cdot\alpha]\neq 1$ and so there exists an open neighborhood $U$ of $x_0$ such that $1\notin B([\beta^{-}\cdot\alpha],U)$. Suppose, to obtain a contradiction, that $B([\alpha],U)$ and $B([\beta],U)$ are disjoint. The Neighborhood Lemma gives $[\alpha]\in B([\beta],U)$. Thus $[\alpha]=[\beta\cdot\delta]$ for loop $\delta$ in $U$. Then $[\beta^{-}\cdot\alpha]=[\delta]$, which means $[\beta^{-}\cdot\alpha]\in B(1,U)$. The Neighborhood Lemma then implies $1\in B([\beta^{-}\cdot\alpha],U)$; a contradiction. $\square$

While this theorem tells us that we gain a lot of ground by just knowing the trivial subgroup is closed, understanding when this happens is a different story. It really depends on the space $X$, particularly how homotopy classes interact with the local topology at $x_0$.

There’s an idea that I wrote about previously called the “Homotopically Hausdorff” property. I’ll remind you of that here.

Definition: A space $X$ is Homotopically Hausdorff at $x\in X$ if for every non-trivial element $1\neq [\alpha]\in \pi_1(X,x)$, there exists an open neighborhood such that no loop in $U$ based at $x$ is path-homotopic to $\alpha$.

This definition is stated a little differently than in my earlier post but, as noted in the remark afterward, the two are equivalent. I’d say this definition is conceptually simpler but the other one is easier to apply. I would imagine that the following theorem is apparently the original justification for the name “homotopically Hausdorff” but the origin of the definition is surprisingly difficult to track down. I first encountered this proof in [2]. It’s a pretty straightforward proof that would make a nice exercise. It’s utility is just that it gives a different characterization of the homotopically Hausdorff property, namely, one in terms of a functorial topology on $\pi_1$.

Theorem: The group $\pi_{1}^{wh}(X,x_0)$ is Hausdorff if and only if $X$ is homotopically Hausdorff at $x_0$.

Proof. The previous theorem tells us that it suffices to prove that the trivial subgroup is not closed if and only if $X$ is not homotopically Hausdorff at $x_0$. Suppose $\{1\}$ is not closed. Then there exists $1\neq [\alpha]$ in the closure of $\{1\}$. Now let $U$ be an open neighborhood of $x_0$. Since $B([\alpha],U)$ is an open neighborhood of $[\beta]$, we have $1\in B([\alpha],U)$. Thus $1=[\alpha][\epsilon]$ for some loop $\epsilon$ in $U$. It follows that $[\alpha]=[\epsilon^{-}]$, i.e. $\alpha$ is path-homotopic to a loop in $U$. This proves $X$ is not homotopically Hausdorff at $x_0$.

For the converse, suppose $X$ is not homotopically Hausdorff at $x_0$. Then there is a non-null-homotopic $\alpha$ based at $x_0$, which is path-homotopic to a loop in every neighborhood of $x_0$. We claim that $[\alpha]$ lies in the closure of $1$. Let $U$ be a neighborhood of $x_0$. Then we have $[\alpha]=[\epsilon]$ for some loop $\epsilon$ in $U$. Equivalently, $1=[\alpha\cdot \epsilon^{-}]$. Since $\epsilon^{-}$ has image in $U$, this means $1\in B([\alpha],U)$. Since $1$ lies in every neighborhood of $[\alpha]$, we conclude that $[\alpha]$ lies in the closure of the trivial subgroup. $\square$

Example: There are lots of homotopically Hausdorff spaces. In my bestiary, I almost always list whether or not this property holds – given an example space, it’s usually easy to decicde whether this property is present or not. Generally, all one-dimensional Hausdorff spaces and subsets of surfaces are homotopically Hausdorff at all of their points and so whenever $X$ is one of these spaces $\pi_{1}^{wh}(X,x_0)$ will be Hausdorff. A fairly non-trivial example is the harmonic pants space. Of course, if you’ve got a locally “nice” space like a manifold or CW-complex, then $\pi_{1}^{wh}(X,x_0)$ is trivially Hausdorff because it is discrete. Examples of spaces, which are not homotopically Hausdorff include the harmonic archipelago and Griffiths twin cone. Actually, for both of these spaces, $\pi_{1}^{wh}(X,x_0)$ is an uncountable indiscrete group.

Example: For an example that is neither Hausdorff nor indiscrete, you can take $\mathbb{G}$ to be the Griffiths twin cone and set $X=\mathbb{G}\vee S^1$ where $x_0$ is the wedgepoint (see the figure below). Let $r: X\to S^1$ be the based retraction that collapses $\mathbb{G}$ to $x_0$. We know $\pi_{1}^{wh}(S^1,x_0)$ is isomorphic to the discrete group $\mathbb{Z}$ and functorality tells us that $r_{\#}:\pi_{1}^{wh}(X,x_0)\to \pi_{1}^{wh}(S^1,x_0)$ is continuous. Therefore $N=\ker (r_{\#})$ is a proper, non-trivial open subgroup of $\pi_{1}^{wh}(X,x_0)$. So $\pi_{1}^{wh}(X,x_0)$ is not discrete. However, we may view $\pi_1(\mathbb{G},x_0)$ naturally as a subgroup of $\pi_{1}^{wh}(X,x_0)$. It’s not too hard to see that $\pi_1(\mathbb{G},x_0)$ is actually equal to the closure of the trivial subgroup in $\pi_{1}^{wh}(X,x_0)$ (but it’s not equal to $N$!). In particular, $\pi_{1}^{wh}(X,x_0)$ is not Hausdorff.

The Griffiths twin cone with an extra circle attached.

Remark: At the start of Part 1, I mentioned that I was particularly interested in compact subsets of $\pi_{1}^{wh}(X,x_0)$. At this point, we can say that if $X$ is homotopically Hausdorf,f then a compact subset $K\subseteq \pi_{1}^{wh}(X,x_0)$ will be zero-dimensional. This is helpful but it would be nice to know when $K$ is metrizable too. So in the third and final post on the whisker topology, I’ll dive into the metrizability of the group $\pi_{1}^{wh}(X,x_0)$.

## References

[1] M. Abdullahi Rashid, S.Z. Pashaei, B. Mashayekhy, H.Torabi, On the Whisker Topology on Fundamental Group. Conference Paper from 46th Annual Iranian Mathematics Conference 46 (2015). pdf available here.

[2] H. Fischer, A. Zastrow, Generalized universal covering spaces and the shape
group, Fund. Math. 197 (2007) 167-196.