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Consider the following two structure-adding refinements of the fundamental group of a topological space:

  • the set $\pi_1(X)$ inherits a quotient topology from the compact-open topology of $X^{S^1}$, under which it is sometimes a topological group. This was discussed here.

  • the topos of sheaves on X has a fundamental group, which is in general a pro-group, reducing to an ordinary group if X is locally simply connected.

Pro-groups and topological groups are not unrelated concepts; in particular, both have a common "generalization" to localic groups. Are there any known relationships between the "topological" and "toposophic" fundamental groups of a space? Do they capture similar or different information?

Edit: As Theo pointed out in a comment, one difference is that the toposophic fundamental group is defined in terms of coverings rather than paths. This makes it better-behaved for non-locally-path-connected spaces, where there may not be very many maps out of $S^1$. But if the space is "nice" enough so that there are "enough paths to retain all the information about coverings," we can still ask about comparing the quotient topology on the group of paths with the pro-structure on the group of coverings. It might be that imposing that "niceness" condition already trivializes the extra information in one or the other or both, but if so then that would be a good answer to the question!

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$\pi_1(X)$ also inherits a topology from the inverse limit topology on the Cech fundamental group; a base of this topology is given by the point inverses of all induced maps $f_*:\pi_1(X)\to\pi_1(P)$, where $P$ is a polyhedron and $f:X\to P$ is a map. With this topology, $\pi_1(X)$ is always a topological group. Its relationships with the fundamental pro-group of $X$ are rather well-understood (see e.g. arxiv.org/abs/0812.1407). I don't know though if the fundamental pro-group of $X$ (as defined e.g. in Dydak-Segal "Shape Theory") is the same as the pro-group you're interested in. –  Sergey Melikhov Feb 5 '11 at 14:15
    
The toposophic fundamental group is (or, is roughly) the etale fundamental group. For badly behaved spaces this is a different and "better" group than the one describing maps from a circle. So the question has a lot of depth: the point is that various well-behavedness conditions on your space will lead to various relationships between the groups, but that (some of) these relationships should break in badly-behaved settings. I don't have any particular examples, just a philosophy, so this is a comment, not an answer. –  Theo Johnson-Freyd Feb 5 '11 at 22:18
    
@Theo JF: Thanks! I've edited the question along those lines to clarify what I'm specifically interested in at the moment. (-: –  Mike Shulman Feb 5 '11 at 22:45
    
Mike, for semi-locally simply-connected compact metrizable spaces $\pi_1(X)$ with either topology is discrete and is isomorphic to the inverse limit of the fundamental pro-group, and contains the same information as the fundamental pro-group. So indeed a "niceness condition trivializes the extra information" on both sides. Probably "compact" and "metrizable" are not essential restrictions here. I have no idea what is toposopic fundamental group, but for me the "well-behaved $\pi_1$" for non-locally connected spaces is the Steenrod $\pi_1$; see mathoverflow.net/questions/49526 –  Sergey Melikhov Feb 5 '11 at 23:56
    
@Sergey Take the reduced suspension of the one point compactification of the natural numbers with isolated basepoint. This is a locally simply connected but non-locally path connected compact metric space. The quotient topology and the inverse limit topology on $\pi_1$ give non-discrete (but non-isomorphic since the first is not first countable) topological groups. –  Jeremy Brazas Feb 6 '11 at 6:00

2 Answers 2

I figured my comments are getting too long so I should put some of them here, even though I'm using a different topology on $\pi_1$ (often it is the same, but certainly not always) and a possibly different pro-group (which I can't tell as I'm not familiar with toposes). EDIT: Unless otherwise noted, I'm assuming that $X$ is compact metrizable to be on the safe side.

The relationship goes in 2 steps. First the fundamental pro-group can be compared with the Cech fundamental group endowed with the inverse limit topology. In particular, as noticed by Atiyah and G. Segal, they contain precisely the same information as long as the fundamental pro-group is Mittag-Leffler. For other results in this direction see lemma 3.4 in "Steenrod homotopy". The topological homomorphism between $\pi_1(X)$ (with topology as in my comment) and the topological Cech fundamental pro-group is discussed in theorem 6.1, section 5 and elsewhere in "Steenrod homotopy".

To summarize the relationship very roughly, $\pi_1(X)$ topologized as in my comment retains much of the inverse limit of the fundamental pro-group, discards all of its derived limit, but instead gets something of the derived limit of the second homotopy pro-group (exactly how much is still a subject of ongoing research, see Theorem 6.5 and remark to corollary 8.8 in "Steenrod homotopy"). Still this is not all that it contains (cf. example 5.7 in "Steenrod homotopy").

Added later: Addressing Mike's comment, let me elaborate on the choice of topology on $\pi_1$.

The Mathoverflow thread on the quotient topology on $\pi_1$, and particularly Andrew Stacey's answer there, make it clear that the only reason that this topology is not compatible with multiplication is that the product of two quotient maps need not be a quotient map. But wait, 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). In fact, the product of any (possibly infinite) collection of quotient maps is a quotient map in the uniform category. (Quotient maps are defined in any concrete category over the category of sets, as explained e.g. in The Joy of Cats.)

This means that the topology of the quotient uniformity on $\pi_1(X)$ makes it into a topological group. By this topology I mean the following. If $X$ is compact, it carries a unique uniformity, and then the space of continuous (=uniformly continuous) maps $(S^1,pt)\to (X,pt)$ is a uniform space (endowed with the uniformity of uniform conergence; if $X$ is metrizable by a metric $d$, this uniformity is metrizable by the metric $D(f,g)=\sup\limits_{x\in X}\ d(f(x),g(x))$.) If $X$ is not compact, then we need to fix some uniformity on it and the above works. To what extent the resulting topology on $\pi_1(X)$ depends on the choice of uniformity is a good question.

Added still later: With this in mind, I no longer see a good reason for myself to think at all about the quotient topology on $\pi_1(X)$. As you can see for instance from my comment here, I have good reasons to believe that quotient topology is a sensible notion only when it coincides with the topology of quotient uniformity (i.e. for quotients of compact spaces), and otherwise the topology of quotient uniformity is the way to go. The present situation does not seem to be an exception.

Now does the topology of the quotient uniformity coincide with the "inverse limit" topology, at least when $X$ is compact? I find it easier to go down one dimension, and compare the two similar topologies on $\pi_0(X)$, the set of path-components. The "inverse limit" topology is again induced from the inverse limit topology on the Cech $\pi_0$ (which is the inverse limit of $\pi_0(P_i)$, where $X$ is the inverse limit of the compact polyhedra $P_i$), and alternatively its basic open sets are those collections of path components whose unions are the point-inverses of maps of $X$ to finite sets. And as long as $X$ is compact (or is endowed with a uniformity), we have the topology of the quotient uniformity, viewing $\pi_0(X)$ as the quotient of $X$ by the equivalence relation of being in the same path-component.

Firstly let me say that (often or not) the two topologies do share something. For instance if $X$ is the $p$-adic solenoid then both are anti-discrete, for apparently very different reasons: the inverse limit topology simply because $X$ is connected; and the topology of the quotient uniformity is antidiscrete because every point lies in the closure of the path-component of every other point. If you take a wedge of two solenoids or even a "garland" of solenoids obtained from their disjoint union by identifying pairs of points taken from distincts solenoids, both topologies are still antidiscrete. In the case of the topology of the quotient uniformity, this is so because every two points $x,y$ are the endpoints of a chain $x=p_0,\dots,p_n=y$ such that each $p_{i+1}$ lies in the closure of the path-component of $p_i$.

On the other hand, let us do the pseudo-arc. Its path-components are single points, so $\pi_0(X)=X$. The topology of the quotient uniformity is the original topology of $X$; and the "inverse limit" topology is anti-discrete because $X$ is connected.

I haven't really thought about the two topologies on $\pi_1$ but I see no reason why this should be terribly different from the case of $\pi_0$.

Still more: While the relation of the two topologies on $\pi_1$ might be not so easy to comprehend, the topology of the quotient uniformity on $\pi_1$ relates in a seemingly more comprehensible way directly to the topologized Steenrod $\pi_1$ (whose relation to the fundamental pro-group is clear). The Steenrod $\pi_1(X)$ can be defined as $\pi_1(holim P_i)$, where the compact metrizable space $X$ is the inverse limit of the polyhedra $P_i$ and $P_0=pt$. (If $P_0\ne pt$ we can shift the indices by one and add a new $P_0$, namely $pt$.) If you think about the construction of $holim$, you will see that $\pi_1(holim P_i)$ is the set of equivalence classes of level-preserving base-ray-preserving maps $S^1\times [0,\infty)\to P_{[0,\infty)}$ under the relation of level-preserving base-ray-preserving homotopy, where $P_{[0,\infty)}$ denotes the mapping telescope of the inverse sequence $\dots\to P_1\to P_0$ (glued out of the mapping cylinders $P_{[i,i+1]}$ of the individual bonding maps). Since $P_0=pt$, "level-preserving" can in fact be replaced by "proper" (see Lemma 2.5 in "Steenrod homotopy"). In other words, the Steenrod $\pi_1(X)$ is the set of equivalence classes of base-ray preserving maps from $D^2$ to the one-point compactification $P_{[0,\infty)}^+$ of the mapping telescope such that the preimage of the point at infinity is precisely the center of the disk (and nothing else).

Now it is not hard to see that the topology on the Steenrod $\pi_1(X)$ (induced from the inverse limit topology on the Cech $\pi_1(X)$ ) is precisely the topology of the quotient uniformity. The uniformity is on a subspace of the space of all continuous (=uniformly continuous) maps from $D^2$ to $P_{[0,\infty)}^+$, and so it is the subspace uniformity (of the uniformity of uniform convergence).

The continuous map from $\pi_1(X)$ with the topology of the quotient uniformity to the Steenrod $\pi_1(X)$ is given by Milnor's lemma (see Lemma 2.1 in Steenrod homotopy): every map $S^1\to X$ extends to a level-preserving map $S^1\times [0,\infty]\to P_{[0,\infty]}$, where $P_{[0,\infty]}$ is Milnor's compactification of the mapping telescope by a copy of $X$, and this extension is well-defined up to homotopy through such extensions. Also close maps have close extensions (from the proof of Milnor's lemma). The quotient $P_{[0,\infty]}/X$ is homeomorphic to $P_{[0,\infty]}^+$, so the close extensions descend to close representatives of elements of the Steenrod $\pi_1$.

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Thanks, this is interesting, although not really the question I was asking since the topology is different. Is it really true that the inverse-limit topological group contains all the information of the fundamental pro-group if the latter is Mittag-Leffler? I'm pretty sure it's not true for the inverse-limit topological group of an arbitrary pro-group; there's a counterexample due to Higman & Stone which I summarized here. –  Mike Shulman Feb 5 '11 at 23:03
    
If you can say anything about why and when the inverse-limit topology is "often the same" as the quotient topology, though, then I think that together with your remarks above (which I have yet to digest) that would go a ways towards answering the original question. –  Mike Shulman Feb 5 '11 at 23:30
    
Sorry, I meant this only for pro-groups indexed by the positive integers. I should have said that I'm only really addressing the case where $X$ is a compact metrizable space. For such $X$ the fundamental pro-group is isomorphic to one indexed by the positive integers. –  Sergey Melikhov Feb 5 '11 at 23:35
    
Regarding the two topologies on $\pi_1$: I've looked at some examples a couple years ago, and unfortunately I can't remember anything now and I'm kind of short of time packing up for a flight tomorrow morning. Frankly speaking, I don't know any good reason to consider the quotient topology (other than it being easy to define). The "inverse limit" topology agrees with the group structure and its Hausdorff quotient is metrizable as long as $X$ is compact metrizable. Nothing like this holds for the quotient topology. –  Sergey Melikhov Feb 6 '11 at 0:14
    
The quotient topology on $\pi_1$ is finer (in many cases strictly finer) than the one Sergey is talking about by the universal property of quotient spaces. –  Jeremy Brazas Feb 6 '11 at 1:36

I don't really have a great feeling for the information captured by the "toposophic" fundamental group so this answer is a bit one-sided towards what I can say about the quotient topology on the fundamental group and what it has to do with covering maps. Maybe you can form something useful from this.

First, I would say quotient $\pi_{1}(X)$ is "rarely" a topological group but is always "quasi"topological group in that you know multiplication is separately continuous. It comes with some serious topological baggage. The requirement that $X$ be locally simply connected (in the sense I define in the comments) does rule out a good number of "wild" spaces of interest but there are some very interesting locally simply connected, non-locally path connected examples.

The connection to coverings is: If $p:Y\rightarrow X$ is a covering map, the induced homomorphism $p_{\ast}:\pi_{1}(Y)\rightarrow \pi_{1}(X)$ is an open embedding of quasitopological groups.

This also implies that if $X$ is nice enough (for instance, locally simply connectivity in the sense of Wikipedia) to have a universal cover, then $\pi_{1}(X)$ is a discrete group and the topology gives no new information.

Constructing covers of non-locally path connected spaces is tricky so in general there will not be a complete Galois correspondence between open subgroups and covers. There should be a correspondence in the locally path connected case.

Added:

An enlightening difference between quotient and inverse limit topology is in the case of the Hawaiian earring.

It has been known for less than two years that quotient $\pi_1$ is not always a topological group. Besides that, its theory is not very developed. I suppose it is "wrong" if you demand a functor that takes values in the category topological groups. But as long as we are discussing other potential topologies...

There is a natural "fix" to the quotient topology using a reflection from topological algebra. It is defined in The fundamental group as topological group on my personal page. The resulting topologized fundamental group is defined for all spaces, takes values in topological groups, and is universal with respect to continuous homomorphisms from quotient $\pi_1$ to topological groups. It is well-behaved in that it admits a number of topological analogues: On "non-discrete wedges of circles" $X_+\wedge S^1$ it is free topological, every topological group is realized as a fundamental group with this topology by attaching 2-cells to one of these "wedges," van Kampen theorems involving pushouts of topological groups are possible, etc.

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Ah, that implication in your penultimate paragraph is nice! The open embedding $\langle e \rangle \to \pi_1(X)$ forcing the topology on $\pi_1(X)$ to be discrete - sweet. –  David Roberts Feb 6 '11 at 0:43
    
Well, I'm happy to say that defining $\pi_1$ in terms of maps out of $S^1$ at all is "obviously wrong" in the non--locally-path-connected case. The connection to open subgroups is intriguing because the open subgroups of a profinite group also correspond to its discrete quotients, hence to covering spaces for the profinite fundamental group. Although of course the profunite fundamental group is an honest topological group. –  Mike Shulman Feb 7 '11 at 5:53
    
Also, how can a space be locally simply connected without being locally path-connected? To me "simply connected" implies "path connected," and how can that not be true locally as well? –  Mike Shulman Feb 7 '11 at 5:57
    
This is what I am familiar with: A space is locally simply connected at a point $x$ if there is a neighborhood base at $x$ consisting of open sets $U$ with $\pi_{1}(U,x)=1$ (forgetting path components of $U$ not containing $x$. The space is locally simply connected if it is so at all of its points. –  Jeremy Brazas Feb 7 '11 at 20:17
    
Wikipedia says that a locally simply connected space is one that admits a basis of simply connected sets, which implies local path-connectedness: en.wikipedia.org/wiki/Locally_simply_connected_space –  Mike Shulman Feb 9 '11 at 6:24

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