Topologized Fundamental Groups: The Whisker Topology, Part 1

This post follows the primer: How to “topologize” the fundamental group. I’m starting with the “whisker topology” because it comes from a very familiar construction. Also, with such a simple definition, a lot of the questions one could ask about it are pretty straightforward to answer. I found writing this to be a string of fun little exercises.

Despite the apparent simplicity, compact sets in the whisker topology are not always so easy to understand. To my own bewilderment, I’ve been needing to understand exactly such things is some recent projects on wild higher homotopy groups. So I’m also hoping these posts will help me clarify some of my own thoughts about it.

Let’s jump right into it. If you have seen covering space theory, you are likely to recognize the following construction.

A construction from covering space theory

Let X be a path-connected space with basepoint x_0\in X. If we were trying to construct a universal covering space for X (whether it actually exists or not), we might try the following “standard” construction: Let \widetilde{X} be the set of path-homotopy classes [\alpha] of paths \alpha:[0,1]\to X starting at x_0. We define a topology on \widetilde{X} by determining a basis. A basic open neighborhood of [\alpha] is of the form N([\alpha],U)=\{[\alpha\cdot\delta]\mid \delta([0,1])\subseteq U\} where U is an open neighborhood of \alpha(1) in X. Here, \alpha\cdot\delta denotes path concatenation. For later on, \alpha^{-} will denote the reverse path of \alpha and c_{x_0} will be denote the constant path at x_0 so that 1=[c_{x_0}] is both the identity element of \pi_1(X,x_0) and the basepoint of \widetilde{X}.

Visualizing a basic neighborhood N([\alpha],U) of [\alpha] in the whisker topology

Given a path \alpha:([0,1],0)\to (X,x_0) and a neighborhood U of \alpha(1) in X, the neighborhood N([\alpha],U) of [\alpha] in \widetilde{X} consists of only homotopy classes that differ from [\alpha] by a “small” change or “whisker” at it’s end. This is why this topology is often referred to as the whisker topology. The directionality in the definition of the whisker topology is important! Nearby homotopy classes only differ by small differences on the right side.

This definition can be found in most textbooks that include covering space theory but it’s studied in detail in [5] and [6]. I’m not sure who first named it the “whisker” topology. In many settings, it does not need a name but since we are studying and comparing several different topologies, a name is appropriate for this discussion.

We define a function p:\widetilde{X}\to X as the endpoint projection p([\alpha])=\alpha(1). This function is certainly continuous since p(N([\alpha],U))\subseteq U for any neighborhood U and path \alpha with \alpha(1)\in U.

If you recognize this construction, you might also recall the following theorem, which is often used to cover the existence part of the classifciation of covering maps: If X is locally path connected and semilocally simply connected, then p:\widetilde{X}\to X is a universal covering map.

The space \widetilde{X} can be constructed for an arbitrary based space (X,x_0). Even when p:\widetilde{X}\to X does not end up being a universal covering map, this construction is often useful.

Since \widetilde{X} is often a covering space, our minds quickly want to identify \widetilde{X} with some familiar space. For example, if X is a torus, we’d identify \widetilde{X} with \mathbb{R}^2. But formally, \widetilde{X} is still a set of homotopy classes that contains the fundamental group \pi_1(X,x_0) as a genuine subset. Specifcially, the fundamental group is the basepoint fiber p^{-1}(x_0)=\pi_1(X,x_0).

Therefore, we may give \pi_1(X,x_0) the subspace topology inherited from \widetilde{X}. When we intersect the basic neighborhoods of \widetilde{X} with \pi_1(X,x_0), all of the small “whiskers” end up being loops.

Definition: The whisker topology on \pi_1(X,x_0) is generated by basic open sets

B([\alpha],U)=N([\alpha],U)\cap \pi_1(X,x_0)

where [\alpha]\in \pi_1(X,x_0) and U is an open neighborhood of x_0 in X. We will write \pi_{1}^{wh}(X,x_0) to denote the fundamental group equipped with the whisker topology.

A basic neighborhood B([\alpha],U) of an element [\alpha] in the fundamental group with the whisker topology consists of homotopy classes of the form [\alpha\cdot\epsilon].

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

It’s tempting to think that because the whisker topology comes from such a standard and simple construction that \pi_{1}^{wh}(X,x_0) will be a topological group. However, it turns out that this very often not true. This is the functorial topology on \pi_1 that is, in a sense, the furthest from being a group topology.

Definition: Let G be a group with a topology. Let \lambda_{g}:G\to G, \lambda_{g}(a)=ga denote left translation by g. If \lambda_{g} is continuous for all g\in G, then G is a left topological group. Similarly, if the right translations \rho_{g}(a)=ag are continuous for all g, then G is a right topological group. If G is both a left and right topological group and inversion is continuous, then G is a quasitopological group.

If G is a left topological group, then all of the left-translation maps \lambda_{g} will be homeomorphisms since \lambda_{g^{-1}} is the inverse of \lambda_{g}.

Proposition: \pi_{1}^{wh}(X,x_0) is a left topological group.

Proof. Fix [\beta]\in\pi_{1}^{wh}(X,x_0) and consider the left translation map \lambda_{[\beta]}:\pi_{1}^{wh}(X,x_0)\to \pi_{1}^{wh}(X,x_0), \lambda_{[\beta]}([\gamma])=[\beta][\gamma] is continuous. A straightforward check gives that \lambda_{[\beta]}(B([\alpha],U))=B([\beta\cdot\alpha],U). This makes it clear that \lambda_{[\beta]} is a homeomorphism. \square

Consequently, \pi_{1}^{wh}(X,x_0) is a homogeneous space. But what about the other operations? For this, let’s explore an example.

Example: Let \mathbb{E}=\bigcup_{n\geq 1}C_n be the usual earring space where C_n\subseteq \mathbb{R}^2 is the circle of radius 1/n centered at (1/n,0). We take x_0=(0,0) to be the wild point and and let \ell_n:[0,1]\to \mathbb{E} denote a loop that traverses the n-th circle once counterclockwise. Let U_n be a neighborhood of x_0 that contains \bigcup_{k\geq n}C_k and meets C_k, k<n in an open arc.

The earring space

Then notice that \{[\ell_n]\}_{n\geq 1}\to 1 where 1 denotes the identity element. Also, \{[\ell_1\cdot\ell_n]\}_{n\geq 1} converges to [\ell_1]. However, \{[\ell_n\cdot\ell_1]\}_{n\geq 1} does NOT converge to [\ell_1] because N([\ell_1],U_2) doesn’t contain any of the elements [\ell_n\cdot\ell_1]; it can only contain elements of the form [\ell_1][\beta] where \beta has image in \bigcup_{k\geq 2}C_k.

As a consequence, we see that \pi_{1}^{wh}(\mathbb{E},x_0) is NOT a right topological group and certainly is not a topological group. Moreover, inversion is NOT always continuous. If inversion was continuous, \{[\ell_1\cdot\ell_n]\}_{n\geq 1}\to [\ell_1] would give \{[\ell_{n}]^{-1}[\ell_{1}]^{-1}\}_{n\geq 1}\to [\ell_{1}]^{-1} but this can’t occur because the sequence \{[\ell_{n}]^{-1}[\ell_{1}]^{-1}\}_{n\geq 1} never enters the open set B([\ell_{1}]^{-1},U_2).


I won’t say to much about the details here but \pi_{1}^{wh} is a functor. Let \mathbf{LTopGrp} be the category of left topological groups and continuous group homomorphisms.

Exercise: If f:(X,x_0)\to (Y,y_0) is a based map, prove that the induced homomorphism f_{\#}:\pi_{1}^{wh}(X,x_0)\to \pi_{1}^{wh}(Y,y_0), f_{\#}([\alpha]=[f\circ\alpha] is continuous.

Once the exercise is done, the fact that \pi_{1} is a a functor, direclty implies that \pi_{1}^{wh}:\mathbf{Top_{\ast}}\to \mathbf{LTopGrp} is a functor. 

In fact, since f_{\#}=g_{\#} whenever f,g:(X,x_0)\to (Y,y_0) are homotopic rel. basepoint, \pi_{1}^{wh} induces a functor \pi_{1}^{wh}:\mathbf{HTop_{\ast}}\to \mathbf{LTopGrp} on the pointed-homotopy category (objects are based spaces and morphisms are basepoint-preserving homotopy classes of based maps).

When CAN we expect continuous operations?

Some of us might be a little unhappy that so many operations in \pi_{1}^{wh}(X,x_0) are going to be discontinuous so often. You might think, well…what if we allowed for whiskers to appear on both ends instead of just on the right. Here, I mean define B([\alpha],U)=\{[\epsilon\cdot\alpha\cdot\delta]\mid \epsilon([0,1])\cup \delta([0,1])\subseteq U\} instead. It’s not a bad idea to consider this for a moment but remember that \pi_{1}^{wh}(X,x_0) is a left topological group. Right multiplication is often discontinuous precisely because we have the whiskers on the right. So if we allow for whiskers on the left and right, then left multiplication won’t be continuous either.

It may seem like this could not possibly be useful until we recall that this topology came from \widetilde{X}. So the whisker topology of \pi_{1}^{wh}(X,x_0) is really meant to be more like the vertex set some kind of graph or tree that branches off in infinite and topologically non-trivial ways (maybe think of something that you could consider a non-discrete vertex set in an \mathbb{R}-tree if that idea is familiar to you).

However, if you’re really hung up on the whole “fails to be a topological group” thing, we can work to understand this failure better through the following theorem. I’ve never seen this theorem in the literature before but it wasn’t that hard to figure out or prove. If anyone know of a reference, please let me know so I can provide some attribution.

Theorem: Let G=\pi_{1}^{wh}(X,x_0). Then the following are equivalent:

  1. Group inversion in:G\to G, in(g)=g^{-1} is continuous,
  2. G is a topological group,
  3. For every g\in G, conjugation c_{g}:G\to G, c_g(h)=ghg^{-1} is continuous.

Proof. The hardest direction is 1. \Rightarrow 2 so we’ll prove that first. Suppose group inversion in:G\to G, in(g)=g^{-1} is continuous. To show that G is a topological group, it’s enough to show that multiplication ([\alpha],[\beta])\mapsto [\alpha\cdot\beta] is continuous. Let U be a neighborhood of x_0 in X so that B([\alpha\cdot\beta],U) is a basic open neighborhood of [\alpha\cdot\beta]. Since inversion is continuous, specifically at [\beta^{-}]\in G, we may find an open neighborhood V of x_0 such that V\subseteq U and in(B([\beta^{-}],V))\subseteq B([\beta],U). The second inclusion means that whenever \delta is a loop in V based at x_0, we have [\delta^{-}][\beta]=[\beta][\gamma] for some loop \gamma in U.

We will show that group multiplication maps B([\alpha],V)\times B([\beta],V) into B([\alpha\cdot\beta],U). Let [\alpha\cdot\delta]\in B([\alpha],V) and [\beta\cdot\epsilon]\in B([\beta],V) for loops \delta,\epsilon in V. Since \delta^{-} has image in V, we have [\delta][\beta]=[\beta][\gamma] for some loop \gamma in U. Thus


where \gamma\cdot\epsilon has image in U. Thus the product [\alpha\cdot\delta][\beta\cdot\epsilon] lies in B([\alpha\cdot\beta],U), completing the proof of the first direction.

The implication 2. \Rightarrow 3. is clear so it is suffices to prove 3. \Rightarrow 1. Suppose for every g\in G, that conjugation by g is continuous. Let [\beta]\in G. We will check that inversion is continuous at [\beta]. Let U be an open neighborhood of x_0 in X so that B([\beta^{-}],U) is an open neighborhood of in([\beta])=[\beta^{-}]. Since conjugation c_{[\beta]} is continuous at the identity element, we may find a neighborhood V of x_0 such that [\beta]B(1,V)[\beta^{-}]\subseteq B(1,U). In other words, if \delta is any loop in V based at x_0, then there is a loop \epsilon in U such that [\beta][\delta][\beta^{-}]=[\epsilon]. Now it suffices to show that in(B([\beta],V))\subseteq B([\beta^{-}],U). Let \delta be a loop in V based at x_0. Since \delta^{-} is also in V, we may find a loop \epsilon in U such that [\beta][\delta^{-}][\beta^{-}]=[\epsilon]. Then in([\beta\cdot\delta])=[\delta^{-}\cdot\beta^{-}]=[\beta^{-}][\epsilon]\in B([\beta^{-}],U). \square

The whisker topology becomes a group topology if and only if for every \beta and neighborhood U, there exists a neighborhood V such that for any loop \epsilon in V, the conjugate \beta\cdot\epsilon\cdot\beta^{-} is homotopic to some loop in U.

What I like about this theorem is that the proof really uses the whisker topology (showing how it depends on the topology of X at x_0) and can’t just be proven by analyzing compositions of operations. I made sure to include Condition 3. in the theorem because even if we know \pi_{1}^{wh}(X,x_0) is abelian, it’s not immediately clear that inversion is continuous. However, we do know that these conjugation maps in abelian groups are always the identity! 

Corollary: If \pi_1(X,x_0) is abelian, then \pi_{1}^{wh}(X,x_0) is a topological group.

But just being non-commutative doesn’t guarantee that \pi_{1}^{wh}(X,x_0) will fail to be a topological group.

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

Example: If X=\bigvee_{a\in A}S^1 is an ordinary wedge of circles with the weak topology, then \pi_{1}(X,x_0) is a free group but \pi_{1}^{wh}(X,x_0) will be discrete and all discrete groups are topological groups.

So, being a topological group here is not just about purely algebraic things like commutativity or conjugation. What’s really going on? If you read the proof closely you can see that for Condition 3. you really only need to use that the conjugation maps are continuous at the identity element. This idea is exactly what is described in the above illustration. So for \pi_{1}^{wh}(X,x_0) to be a topological group, all elements g, no matter how “large,” have to conjugate small loops to small loops. Or maybe think about the negation this way…If there is some g\neq 1 and also arbitrarily small a\neq 1 with gag^{-1} remaining “large,” then you’re in trouble.

Example: The harmonic archipelago gives an example where \pi_{1}^{wh}(X,x_0) is non-commutative but is a topological group because it is indiscrete

So now here’s a question. I haven’t thought too deeply about it so I don’t exactly know how hard it is. If you know of an answer, feel free to share.

Question: Is there a space X where \pi_{1}^{wh}(X,x_0) is a non-discrete, Hausdorff, and non-commutative topological group?

In the next post, I’ll discuss more about separation axioms where we’ll run into the familiar term “homotopically Hausdorff” and see that when this happens the group \pi_{1}^{wh}(X,x_0) is always zero dimensional!


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.

[6] Z. Virk, A. Zastrow, The comparison of topologies related to various concepts of
generalized covering spaces, Topology Appl. 170 (2014) 52–62.

This entry was posted in Covering Space Theory, Fundamental group, topological fundamental group, Topological groups. Bookmark the permalink.

3 Responses to Topologized Fundamental Groups: The Whisker Topology, Part 1

  1. Pingback: Topologized Fundamental Groups: The Whisker Topology, Part 2 | Wild Topology

  2. Pingback: Topologized Fundamental Groups: The Whisker Topology, Part 3 | Wild Topology

  3. Pingback: How to “topologize” the fundamental group: a primer | Wild Topology

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s