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Background

Let $(X,x)$ be a pointed topological space. Then the fundamental group $\pi_1(X,x)$ becomes a topological space: Endow the set of maps $S^1 \to X$ with the compact-open topology, endow the subset of maps mapping $1 \to x$ with the subspace topology, and finally use the quotient topology on $\pi_1(X,x)$. This topology is relevant in some situations. A very interesting paper dealing with this topology is:

[1] Daniel K. Biss, A Generalized Approach to the Fundamental Group, The American Mathematical Monthly, Vol. 107

You can find this online. This is somehow an introduction to

[2] Daniel K. Biss, The topological fundamental group and generalized covering spaces , Topology and its Applications, Vol. 124

Question

How can we prove that $\pi_1(X,x)$ is a topological group? Clearly the inversion map $\pi_1(X,x) \to \pi_1(X,x)$ is continuous, since $S^1 \to S^1, z \mapsto \overline{z}$ is continuous and induces this map. But I don't know how to attack the continuity of the multiplication. It's not hard to see that the multiplication on $map((S^1,1),(X,x))$ is continuous, since it is induced by a fold map $S^1 \to S^1 + S^1$. In order to carry this over to $\pi_1(X,x)$, there are at least two problems which I encounter:

  • The quotient map $map((S^1,1),(X,x)) \to \pi_1(X,x)$ may be not open.
  • The product of the quotient maps $map((S^1,1),(X,x))^2 \to \pi_1(X,x)^2$ may be not a quotient map.

In [1] it is claimed that $\pi_1(X,x)$ is always a topological group, and this should be proven in [2], but I have no acecss to [2].

An example that products of quotient maps don't have to be quotient maps can be found here. Remark however that this is true in the category of compactly generated spaces.

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This is a contentless comment merely designed to alert Martin to the fact that I've updated my answer. –  Andrew Stacey Jun 1 '10 at 11:09
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Here is something that I always wonder about the counterexamples to this: Why are we using function spaces outside the compactly-generated context? Why don't we simply start with CG spaces with their internal product and function objects? Then define the topological fundamental group there, where it does not suffer from these problems? –  Tyler Lawson Jun 1 '10 at 12:52
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This is a fair question in that there may be other natural topologies on $\pi_{1}(X,x)$ better suited for "capturing" particular topological properties of $X$. In this case, many of the counterexamples and their loop spaces are compactly generated but, of course, the quotient may not be. If you insist on having a topological group, doing everything in the category of compactly generated spaces may be the way to go. But if we are interested in the homotopy of pathological spaces, this will be at the cost of forgetting some of the pathologies and ending up with a much weaker invariant. –  Jeremy Brazas Jun 1 '10 at 14:40
    
If $X$ is compact, then its loop space is a uniform space (under the uniformity of uniform convergence), and then the topology of the quotient uniformity on $\pi_1(X)$ makes into a topological group. This is shown by the argument in Andrew Stacey's answer below, using that the product of two quotient maps is a quotient map in the category of uniform spaces and uniformly continuous maps. See Isbell's "Uniform spaces" (1964), Exercise III.8(c). See also my answer at mathoverflow.net/questions/54391 –  Sergey Melikhov Feb 6 '11 at 4:09
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4 Answers

up vote 17 down vote accepted

Update: A bit of a digital paper chase led me, via David Robert's thesis (note that in the latest version, it is Chapter 5, section 2 that is most relevant), to this paper on the arXiv. The last sentence of the abstract is:

These hoop earring spaces provide a simple class of counterexamples to the claim that $\pi_{1}^{top}$ is a functor to the category of topological groups.

I recommend reading this article.

(Added later: In case it's not clear, the author of that paper is Jeremy Brazas who added an answer afterwards, so if you vote for my answer, you should definitely vote for his!)


Original Answer: These were my initial thoughts before I found the references above. These were what made me sufficiently intrigued to do the paper chase and find the above-mentioned thesis and article.

The proof given in the second paper (by Biss) that is mentioned in the question is short enough that I think it reasonable to copy it out here. I shan't copy out the obvious diagram so need to establish some notation first:

  1. $m \colon \pi_1^{Top}(X,x) \times \pi_1^{Top}(X,x) \to \pi_1^{Top}(X,x)$ is the multiplication map in question
  2. $p \colon \operatorname{Hom}((S^1,1),(X,x)) \to \pi_1^{Top}(X,x)$ is the quotient map
  3. $\overline{m} \colon \operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x)) \to \operatorname{Hom}((S^1,1),(X,x))$ is the "upstairs" multiplication map. (It's a tilde in the original, but that isn't displaying correctly for me so I daren't use it.)

The proof then proceeds:

To show that $m$ is continuous, it suffices to show that $\overline{m}$ is continuous, for then if $U \subset \pi_1^{Top}(X,x)$ is open, $(p \times p)^{-1} m^{-1}(U) = \overline{m}^{-1}p^{-1}(U)$ is open, but by the definition of a quotient map, $(p \times p)^{-1} m^{-1}(U)$ is open if and only if $m^{-1}(U)$ is.

There then follows a proof that $\overline{m}$ is continuous, a fact that I trust does not need proving.

Comments on your comments:

  1. We don't need the quotient map to be open since we are only ever dealing with preimage sets. It is certainly not always true that if $q \colon X \to Y$ is a quotient that $q(U)$ is open in $Y$ for every open $U$ in $X$. But it is true by definition that $q^{-1}(U)$ is open in $X$ if and only if $U$ is open in $Y$. This is because the topology on $Y$ is precisely that to make this true. So since we are only dealing with sets of the form $(p \times p)^{-1}(A)$ then the assertion is valid assuming that $p \times p$ is a quotient map.

  2. Here, I find myself worried. A quick back-of-envelope check seems to show that one can't simply assume that the product of quotients is again a quotient in Top (a counterexample eludes me as I don't have Counterexamples in Topology to hand and I'm too used to dealing with "nice" spaces). It may be the case that for Hom-spaces then there's some magic that can be done (though such is not mentioned in the paper); but again the best that I can do on the back of an envelope is observe that (modulo some basepoint mess) by construction $\operatorname{Hom}((S^1,1),(X,x)) \times \operatorname{Hom}((S^1,1),(X,x))$ quotients to $\pi_1^{Top}((X,x) \times (X,x))$. But to proceed, one would need to know that $\pi_1^{Top}$ was a product-preserving functor. This is morally the same as saying that it is representable - which looks good since we have an obvious representing object $S^1$! However, this can't be made into a proper argument since although we have a representing object, we don't have an enriched Hom-functor $hTop \times hTop \to Top$ which to evaluate at $S^1$.

So I would look for a counterexample to the product of quotients being a quotient, and see where that leads you. Either you'll find a proper counterexample to the proposition in question, or you'll see why in this special case, such a counterexample could not occur.

(Of course, I may well be missing something obvious!)

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Thanks for the answer. I've added a link above to a counterexample concerning products of quotient maps. So Biss just ignores this failure of compatibility? Everything is fine with locally compact spaces. –  Martin Brandenburg Jun 1 '10 at 9:09
    
I'm hesitant at drawing any conclusions. It may be that there is something I'm missing that is special about mapping spaces and that he thinks is so obvious that he doesn't bother stating it. Plus I only skimmed the rest of the article, so there may be something elsewhere. I'm sufficiently intrigued that I'm looking at some follow-up papers and will report back! –  Andrew Stacey Jun 1 '10 at 9:40
    
Here's another relevant paper: ams.org/mathscinet-getitem?mr=2457961 If you have access to it, take a look at Theorem 2.3. It claims that for these hom-spaces, the product of the quotients in this case is the quotient of the products. Note, though, that this comes after the proof that the (higher) homotopy groups are topological groups. –  Andrew Stacey Jun 1 '10 at 10:01
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The plot thickens: ams.org/mathscinet-getitem?mr=2280908 claims a counterexample to Biss's result and adds the condition of locally-path connected and metrisable to surmount it. However, there is no discussion of exactly where the supposed mistake in Biss's paper occurs. –  Andrew Stacey Jun 1 '10 at 10:07
    
You need to make it clear that "the proof in the second paper" refers to the erroneous proof in [2] from OP's formulation, and $\mathit{not}$ to the counterexample from the second (arXiv) paper that you've just mentioned. Even better, mark the second part of your answer as outdated. –  Victor Protsak Jun 2 '10 at 0:35
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Andrew's answer is right, but I'll just throw in a few comments since "topological" homotopy invariants are of great interest to me. Here Paul Fabel has shown that $\pi_{1}^{top}$ on the Hawaiian earring is not a topological group. It turns out that multiplication can fail to be continuous even for some reasonably nice spaces (like locally simply connected planar continua). This and a nice connection with free topological groups appears in a manuscript I just posted on arXiv (perhaps shameless self-promotion...but I'll post the link when it becomes available).

This quotient map mistake has appeared many places in the literature, even in an appendix by Peter May from the 70's who described the topological fundamental groupoid (giving each hom-set the quotient topology of fixed endpoint Moore path spaces). The same false assertion that products of quotient maps are again quotient maps was also used to "show" the higher topological homotopy groups are topological groups.

With this quotient topology, the topological fundamental group(oid) is discrete on spaces that have universal covers (are path connected, locally path connected, and semi-locally simple connected). When it is non-discrete, this topology is often difficult to deal with. After all...there are going to be many homotopic loops that we identify in the quotient but which "look nothing alike" in the space.

While we don't always have a topological group, we're not completely out of luck. $\pi_{1}^{top}(X)$ is always a quasitopological group in the following sense:

definition: A quasitopological group is a group $G$ with topology such that inversion is continuous and multiplication $G\times G\rightarrow G$ is continuous in each variable.

A basic theory of these objects can be found in "Topological Groups and Related Structures" some of which you can get on google books.

Edit: The topological fundamental group and free topological groups is the new paper I mentioned. It is a bit denser than the one from Andrew's answer but if anyone is interested enough to read it, I'd greatly appreciate any comments, suggestions, or corrections.

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Thank you for mentioning that the topology on $\pi_1(X)$ is discrete if $X$ has a universal cover - this is good to hear. –  Peter Samuelson Sep 12 '13 at 0:19
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Let $X$ be a topological space and let $a$ be a point of $X$. There are three interesting topologies on $\pi_1(X,a)$: the quotient topology of the topology of compact convergence, and two other one which you'll let me call admissible and adequate.

In all three topologies, the classes of open normal subgroups coincide; moreover they are simultaneously Hausdorff (or not).

1) As observed, the topology of compact convergence is not a group topology, although the multiplication is separately continuous.

2) Say a subgroup $H$ of $\pi_1(X,a)$ is admissible if, any $b\in X$ has a neighborhood $V$ such that forany path class $\gamma\in\pi(X,a,b)$ (fundamental groupoid) and any loop $c$ in $V$ based at $b$, $\gamma c \gamma^{-1}$ belongs to $H$.

There is a unique group topology on $\pi_1(X,a)$ such that the normal admissible subgroups form a basis of neighborhoods of the identity. For this topology, the open subgroups are the admissible subgroups.

If $X$ is locally arcwise connected, then a subgroup is admissible if and only if it is the stabilizer of a point above $a$ in a covering of $X$.

The admissible topology is coarser than the topology of compact convergence, and may be strictly coarser.

3) Let us say that a subgroup $H$ of $\pi_1(X,a)$ is adequate if, for any $b\in X$ and any path class $\gamma\in\pi(X,a,b)$, there exists a neighborhood $V$ of $b$ such that for any loop $c$ in $V$ based at $b$ $\gamma c \gamma^{-1}$ belongs to $H$. (Note the switch in the order of quantifiers.)

There exists a unique group topology on $\pi_1(X,a)$ for which adequate subgroups form a basis of open subgroups.

This topology is finer than the topology of compact convergence, to which it coincides if $X$ is locally arwise connected.

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It seems that for arc connected $X$, your admissible topology is discrete if and only if $X$ is semi-locally simply connected. But $\pi_{1}^{top}(X)$ is not always discrete for semi-locally simply connected $X$. This makes me think 1) and 2) are not always comparable. Am I missing something here? –  Jeremy Brazas Jun 1 '10 at 19:21
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According to Fabel's Theorem 2 in Metric spaces with discrete topological fundamental groups, Topology and its Applications 154 (2007), 635-638, $\pi_1^{top}(X)$ is discrete for connected and locally path connected spaces that are semi-locally simply connected. In that case, all three topologies are discrete. –  ACL Jun 1 '10 at 21:29
    
Yes, all three are discrete when a universal cover exists. Example 1 in the same paper is locally simply connected but not locally path connected. It has non-discrete topological fundamental group but the topology generated by admissible subgroups will be discrete. Thus the admissible topology is not always courser than the quotient topology. –  Jeremy Brazas Jun 1 '10 at 22:11
    
Additionally, the Hawaiian earring $\mathbb{HE}$ is locally arc connected but $\pi_{1}^{top}(\mathbb{HE})$ is not a topological group. So the group topology generated by adequate subgroups will not not coincide with with the quotient topology (which is not a group topology) even in some of the simplest non-discrete cases. These two other topologies are very interesting alternatives, I am just hesitant about some of the comparison statements. –  Jeremy Brazas Jun 1 '10 at 22:18
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The fundamental groupOID of any path-connected, locally path-connected and semi-locally simply connected space is again a topological groupoid (now you don't need to be pointed). The topologized fundamental group at $x \in X$ is now just the stablizer group of $x$ which inherits a topology from the arrow space of the groupoid. There is some on this in n-lab. Look under the heading "Topologizing the fundamental groupoid" in http://ncatlab.org/nlab/show/fundamental+groupoid.

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I'm embarrassed that I didn't spot the nlab link! However, note that: 1) there are conditions on the space which don't appear to be present in the paper in question; 2) the topology is not (always?) the same as that discussed in the paper in question (see the green box); 3) there is the statement "composition is no longer continuous" (and this may be using the quotient topology). –  Andrew Stacey Jun 1 '10 at 10:35
    
Good point. I was wondering if this perhaps lines up with why "the plot thickens". –  David Carchedi Jun 1 '10 at 11:02
    
I just found what looks like a relevant paper on the arXiv and added the link to my answer. –  Andrew Stacey Jun 1 '10 at 11:08
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As I wrote the material on the nlab, I should point out that 'composition is no longer continuous' does refer to the quotient topology of the path space. And the stabiliser groups of this topologised fundamental groupoid are, under the stated conditions, discrete, because then the universal (connected) covering space is the source fibre at any point, and the fundamental group has then underlying space the fibre of the covering space at the basepoint chosen. –  David Roberts Jun 2 '10 at 7:51
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