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This question is inspired by an answer of Tim Porter.

Ronnie Brown pioneered a framework for homotopy theory in which one may consider multiple basepoints. These ideas are accessibly presented in his book Topology and Groupoids. The idea of the fundamental groupoid, put forward as a multi-basepoint alternative to the fundamental group, is the highlight of the theory. The headline result seems to be that the van-Kampen Theorem looks more natural in the groupoid context.
I don't know whether I find this headline result compelling- the extra baggage of groupoids and pushouts makes me question whether the payoff is worth the effort, all the more so because I am a geometric topology person, rather than a homotopy theorist.

Do you have examples in geometric topology (3-manifolds, 4-manifolds, tangles, braids, knots and links...) where the concept of the fundamental groupoid has been useful, in the sense that it has led to new theorems or to substantially simplified treatment of known topics?

One place that I can imagine (but, for lack of evidence, only imagine) that fundamental groupoids might be useful (at least to simplify exposition) is in knot theory, where we're constantly switching between (at least) three different "natural" choices of basepoint- on the knot itself, on the boundary of a tubular neighbourhood, and in the knot complement. This change-of-basepoint adds a nasty bit of technical complexity which I have struggled with when writing papers. A recent proof (Proposition 8 of my paper with Kricker) which would have been a few lines if we hadn't had to worry about basepoints, became 3 pages. In another direction, what about fundamental groupoids of braids?
Have the ideas of fundamental groupoids been explored in geometric topological contexts? Conversely, if not, then why not?

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Natural generalizations of braids seem to be things like tangles and string links, which aren't groupoids but they are categories. In that sense they fit into a natural framework of things like cobordism categories. I suppose I don't see the fundamental groupoid as much of a simplification -- the Seifert-Van Kampen theorem is essentially just as complicated in the groupoid setting, the only thing you get to avoid is putting in all the basepoint arcs after subdivision. This "gain" comes at the cost of using groupoids. How much you value the approach depends on what you need the tool for. –  Ryan Budney Oct 3 '10 at 18:47
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I'm sorry that I don't know much about geometric topology and can't answer the question, but let me point out something that you may already know: multiple basepoints are useful in the presence of symmetries. For example, let's say you have a connected space X with a Z/2 symmetry. Then H_1(X;Z) gets a Z/2-action; shouldn't this come from an action of Z/2 on \pi_1(X) via the isomorphism \pi_1^{ab}=H_1? If X has a fixed point we can base ourselves there and be fine, but otherwise it's nice to simultaneously use two conjugate basepoints to see the Z/2-action on \pi_1. –  Dustin Clausen Oct 3 '10 at 21:44
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@Dustin, a standard strategy here would be (1) if the action has a fixed point, use the fixed point and (2) if there is no fixed point, you have a covering space, so study that. –  Ryan Budney Oct 3 '10 at 21:53
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Ryan, that sounds good -- thanks for the perspective. Here's a link: youtube.com/watch?v=kjX2MKaZ-rg –  Dustin Clausen Oct 3 '10 at 22:12
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For some Grothendieckian support for Ronnie see the comment at bangor.ac.uk/~mas010/pstacks.htm which begins "...people are accustomed to work with fundamental groups and generators and relations for these and stick to it, even in contexts when this is wholly inadequate, namely when you get a clear description by generators and relations only when working simultaneously with a whole bunch of base-points chosen with care-or equivalently working in the algebraic context of groupoids, rather than groups." –  David Corfield Oct 4 '10 at 8:51
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11 Answers 11

up vote 18 down vote accepted

Here is an interesting example where groupoids are useful. The mapping class group $\Gamma_{g,n}$ is the group of isotopy classes of orientation preserving diffeomorphisms of a surface of genus $g$ with $n$ distinct marked points (labelled 1 through n). The classifying space $B\Gamma_{g,n}$ is rational homology equivalent to the (coarse) moduli space $\mathcal{M}_{g,n}$ of complex curves of genus $g$ with $n$ marked points (and if you are willing to talk about the moduli orbifold or stack, then it is actually a homotopy equivalence)

The symmetric group $\Sigma_n$ acts on $\mathcal{M}_{g,n}$ by permuting the labels of the marked points.

Question: How do we describe the corresponding action of the symmetric group on the classifying space $B\Gamma_{g,n}$?

It is possible to see $\Sigma_n$ as acting by outer automorphisms on the mapping class group. I suppose that one could probably build an action on the classifying space directly from this, but here is a much nicer way to handle the problem.

The group $\Gamma_{g,n}$ can be identified with the orbifold fundamental group of the moduli space. Let's replace it with a fundamental groupoid. Fix a surface $S$ with $n$ distinguished points, and take the groupoid where objects are labellings of the distinguished points by 1 through n, and morphisms are isotopy classes of diffeomorphisms that respect the labellings (i.e., sending the point labelled $i$ in the first labelling to the point labelled $i$ in the second labelling).

Clearly this groupoid is equivalent to the original mapping class group, so its classifying space is homotopy equivalent. But now we have an honest action of the symmetric group by permuting the labels on the distinguished points of $S$.

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One way to see the action of $\Sigma_n$ on $B\Gamma_{g,n}$ would be to use flashy explicit bar construction models, like Paolo Salvatore's "Configuration spaces with summable labels". –  Ryan Budney Oct 3 '10 at 22:08
    
== This is nice. –  Daniel Moskovich Oct 4 '10 at 13:08
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@Ryan - I don't think a configuration space model for the bar construction helps here. The symmetric group should act via its action on the labels, but as I said, in this case it only acts by outer automorphisms. Acting on a group $G$ by outer automorphisms is the same as acting on $BG$ up to homotopy by unbased maps, and the problem is to rectify this action to a strict action on some model for $BG$. –  Jeffrey Giansiracusa Oct 4 '10 at 21:02
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Thanks Jeff. Okay, this is a place where it's not immediately clear how one could do without groupoids. It appears my role as devil's advocate has been punctured. –  Ryan Budney Oct 8 '10 at 16:39
    
@Jeff: This is a situation where passing to homotopy classes, as you did in your very first sentence, makes the action less clear. The classifying space $B\text{Diff}^+(\Sigma_{g,n})$ (which is the same as $B\Gamma_{g,n}$) is the space of submanifolds of $\mathbb{R}^\infty$ diffeomorphic to $\Sigma_g$, together with a sequence of $n$ points lying on the submanifold. The action of $S_n$ on the classifying space is just to permute these points; this is a perfectly nice action. –  Tom Church Oct 14 '10 at 17:24
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The following picture illustrates a not unusual situation. nonconn

It helps here to have a theorem which immediately determines the fundamental groupoid of the union on these base points. Then one uses algebra and combinatorics to work out particular fundamental groups, if one wants them.

I came into groupoids by trying to find a new proof of the fundamental group of the circle. It turned out that one could do this using the fundamental groupoid on two base points. Writing the 1968 edition of my book now called `Topology and Groupoids' (T&G) (available on amazon.com and e-version from www.kagi.com) convinced me that all of 1-dimensional homotopy theory was better expressed in terms of groupoids rather than groups, in that one obtained more powerful theorems with simpler proofs. Later results on the fundamental groupoid of orbit spaces (Chapter 11 of T&G) are more awkward to express in terms of groups; this elaborates on the point by Dustin Clausen. See further details below.

Henry Whitehead answered the question of "Why not restrict to CW-complexes with just one vertex?" by considering covering spaces. Philip Higgins gave a considerable generalisation of Grusko's theorem by considering covering morphisms of groupoids, see his 1971 book `Categories and groupoids' available as a TAC Reprint, 2005.

In 1966 I thought about prospective uses of groupoids in higher homotopy theory, and this led over many years to higher dimensional Seifert-van Kampen Theorems, with a range of new nonabelian calculations of second relative homotopy groups and triad homotopy groups (for the latter, see the "nonabelian tensor product of groups"). That sounds relevant to geometric topology!

So one answer to the original question is that the use of groupoids opens new worlds of possibilities.

Actually the idea of `change of base point for the fundamental group' is a bit bizarre: one does not describe a railway timetable in terms of return journeys and change of starting point for these! Why is this still taught to students?

In the end, an aesthetic viewpoint implies more power!

Thanks to those above who give me additional examples.

More information on my page From groups to groupoids.

September 2012: I forgot to add to this answer more information on orbit spaces, with particular reference to "two base points".

Ross Geoghegan in his 1986 review (MR0760769) of two papers by M.A. Armstrong on the fundamental groups of orbit spaces wrote: "These two papers show which parts of elementary covering space theory carry over from the free to the nonfree case. This is the kind of basic material that ought to have been in standard textbooks on fundamental groups for the last fifty years." At present, to my knowledge, "Topology and Groupoids" is the only topology text to cover such results.

Consider the action of the cyclic group of order 2, $Z_2$ on the unit circle $S$ by complex conjugation. Take $1$ as base point. The induced action of $Z_2$ on the fundamental group $\pi_1(S,1)$ is $n\mapsto -n$, and the quotient by this action is $Z_2$. But the quotient of $S$ by the action is a semicircle, which is contractible. What has gone wrong?

The problem is there are two fixed points of the action. The quotient of the action of $Z_2$ on the groupoid $\pi_1(S, A)$, where $A$ consists of the points $\pm 1$, is indeed correct.

The point is that a group acting on a space $X$ acts also on the fundamental groupoid $\pi_1 X$. If $X$ is Hausdorff, the action is properly discontinuous, and $X$ has a universal cover, then the fundamental groupoiud of the orbit space $X/G$ is the orbit groupoid of $\pi_1 X$. This is the groupoid expression of Armstrong's results. See Chapter 11 of Topology and Groupoids.

April 21,2013: The book Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids gives an account of this new approach to basic algebraic topology at the border between homology and homotopy, without using singular homology theory, or simplicial approximation, but relying on the idea of multiple compositions of cubes. This also allows for results on second relative homotopy groups, results which, being essentially nonabelian, are not obtainable by traditional algebraic topology. It also avoids the "trick" of taking the free abelian group on ordered or oriented simplices in order to define chain groups, and the boundary map.

All this comes from considering the question:

if groupoids are useful in $1$-dimensional homotopy theory, how useful can they be in higher homotopy theory?

One quickly notices that whereas group objects internal to groups are abelian groups, group objects internal to groupoids are in some sense "more nonabelian" than groups, as are groupoid objects internal to groupoids. So one looks to such objects to model higher homotopy properties: and this has been achieved.

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One situation in which it is essential to use groupoids is the study of orbifolds.

Slogan: The set of points of an orbifold is a groupoid.


Here's a concrete problem that is illuminated by the language of groupoids. Suppose I have an orbifold $M$ with a singuar stratum $X$. The stratum $X$ is isomorphic to $S^1$, and its isotropy group is some finite group $G$. Let's also assume that $X$ is oriented.

Question: What is the "monodromy" of going around that stratum?

At first glance, one might guess that it's an element of $Aut(G)$.
That's wrong! The monodromy is an element of $Out(G)$.
So we have a somewhat paradoxical situation in front of us: there is a group associated to every point of $X$. Yet, the monodromy is not acting by automorphisms of that group.

Here's an example of orbifold that nicely illustrates the kind if situation that can occur: $$ M = (S^1\times V )/S_n, $$ where $S_n$ is the symmetric group and $V$ is a faithful representation. The group $S_n$ acts on the circle $S^1$ via the projection $S_n\twoheadrightarrow\mathbb Z_2$, and then the antipodal map. The representation $V$ of $S_n$ is just put there so that the orbifold isn't too degenerate (it can be omitted if you don't mind working with non-effective orbifolds).

In that example, the manifold $X$ is $S^1/\mathbb Z_2$. The isotropy group is the alternating group $A_n$. The monodromy is computed in the following way. Go half way around $S^1$, and then identifying "$A_n$ at point -1" with "$A_n$ at point +1" via any element of $S_n$ that sends -1 to +1. A choice of such an element yields an automorphism of $A_n$. But since there is no best way making such a choice, the only canonical thing is its class in $Out(A_n)$.

Ok. Maybe now is good moment to try to remove some of the confusion.
It all becomes more clear once you realize that the thing that is associated to a point of $X$ is not a group. It's a groupoid:

If $[M/G]$ is an orbifold and $x$ is a point in $M/G$, then the groupoid that lives above $x$ has objects given by points $m\in M$ mapping to $x$. An arrow from $m$ to $m'$ is given by a element of $G$ that sends $m$ to $m'$.

The monodromy is then simply an automorphism of that groupoid (so now there's nothing weird any more). But this automorphism might fail to fix any of the objects of the groupoid. And so it can't be viewed as an automorphism of the corresponding group, unless you make some unnatural choices.

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In low-dimensional topology, rather than using the terminology orbifold people would just talk about marked surfaces or manifolds with trivalent graphs embedded, etc. Orbifolds (and your groupoid formalism) would have been talked about as a type of labelled stratified space. This is symbiotic with the graph-of-groups terminology, described below. So the natural question to ask is, what do you "gain" by using groupoid terminology instead of more old-fashioned terminology -- other than being fashionable? I think that's why Daniel is asking about new theorems, etc. –  Ryan Budney Oct 3 '10 at 20:47
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The notion of groupoid is closed enough to the notion of group that I don't expect that there are any results that really require groupoids in their proofs. In other words, if you're willing to pick a base points in every connected component of every groupoid that you encounter, then you'll be able to translate any argument into the world of groups. –  André Henriques Oct 3 '10 at 21:06
    
At least a convincing argument I've heard is that by fixing a basepoint and throwing out the rest of the data, we're losing actual information about the space. It's only okay to pick a representative if you can show that any other representative will not only be isomorphic but uniquely so. To put it in topological terms (and in a way made precise by looking at simplicial localizations), we can only pick a basepoint without throwing away information if the groupoid is contractible. –  Harry Gindi Oct 3 '10 at 21:27
    
What's the relationship between the classifying space of a groupoid and its subgroups? It seems like there should be a fibration linking the two. So the reduction to groups in general seems to involve some effort. –  Ryan Budney Oct 3 '10 at 21:36
    
I've thought about this some more, and this well-written answer makes a lot of sense. But is this approach really all that much simpler and more natural than what we usually do, which is to work with covering spaces, and to mod out later by the indeterminancy induced by choosing a basepoint? –  Daniel Moskovich Oct 4 '10 at 12:49
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A short example.

The family of pure braid groups does not possess a symmetric operad structure.

But the fundamental groupoid of the little 2-discs operad is naturally a symmetric operad.

Although the fundamental groups of the little 2-discs operad are the pure braid groups, there is no way to choose basepoints consistent with the operad structure.

The moral is that groupoids are not naturally pointed, whilst groups are. If you're working with fundamental groups you should really be working with pointed spaces. Ofcourse you can ignore this and you'll only run into trouble if your mathematics doesn't work with pointed spaces, see the example above.

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I'd like to understand this example in more detail- could you explain a bit more (or give a reference) to how this construction works? –  Daniel Moskovich Oct 4 '10 at 12:38
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Steenrod defined a local coefficient system as a functor from the path-groupoid of your space to a category. It is tricky to define homology/cohomology with local coefficients by just picking base points since the identifications given by the paths are needed as well. Calculations of induced maps are especially prone to error. Perhaps my favorite example is the following.

Let $\tau\colon S^{2n} \to S^{2n}$ be the antipodal map with quotient $RP^{2n}$ and quotient map $\pi\colon S^{2n} \to RP^{2n}$. There is a twisted coefficient system $Z^w$ on $RP^{2n}$ so that $H_{2n}(RP^{2n};Z^w) \cong Z$ which is used in the non-orientable Poincare duality theorem.There is a natural notion of the pull-back coefficient system and the system $\pi^\ast Z^w$ is equivalent to the usual trivial system. Hence $H_{2n}(S^{2n};\pi^\ast Z^w) \cong Z$ but the equivalence involves a choice.

Since $\pi\circ \tau = \pi$, there is a ``natural'' identification of $(\pi\circ\tau)^\ast Z^w$ with $\pi^\ast Z^w$. With these choices, the antipodal map has degree 1. If you make your choices so that the antipodal map has degree $-1$, then the two maps induced by $\pi$ have different degrees.

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I sort-of follow, but not quite. What, explicitly, are the choices when you say "with these choices"? –  Daniel Moskovich Oct 4 '10 at 13:02
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I'm not following, either. Steenrod's local coefficient system formulation isn't the only formalism available. A lot of people think of a system of coefficients as a "bundle of groups". This avoids mentioning the fundamental group(oid) completely. –  Ryan Budney Oct 4 '10 at 15:27
    
@ Ryan: Isn't a bundle of coefficient groups equivalent to a map $\pi_1(M,*) \rightarrow G$? If that's right, then you could refer to a bundle of groups without reference to the fundamental group, but it'd still be lurking in there somewhere. I.e., fundamental-groupy stuff is exactly the thing that makes local coefficients interesting. And, to bring it back to the original question, I'd imagine that this should induce a map (functor?) $\Pi_1(M) \rightarrow G$? –  Aaron Mazel-Gee Oct 8 '10 at 16:20
    
That's very much my point and the point of the thread -- is usage of groupoids more than a terminology preference? –  Ryan Budney Oct 8 '10 at 16:27
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@Ryan: "terminology preferences" can be very useful. An extreme example is Roman numerals versus Arabic numerals. I also tend to take the line expressed by the question: "Is there compelling evidence that that one can do better without one hand strapped behind ones back?" Or, as Gian-Carl Rota wrote on another matter: "Exterior algebra is not meant to prove old facts, it is meant to disclose a new world. Disclosing new worlds is as worthwhile a mathematical enterprise as proving old conjectures.” The idea of groupoids has led me to new worlds, and modes of expression. What fun! –  Ronnie Brown Jan 29 at 21:24
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In my proof that mapping class groups are automatic, Ann. of Math. (2) 142 (1995), no. 2, 303–384, I used a theorem from ECHLPT "Word Processing in Groups" which says that if a groupoid is automatic then the corresponding group is automatic.

That theorem was applied in the situation of a finite type surface $S$ with one or more punctures, using the groupoid mentioned in Bruno Martelli's answer which has come to be called the "Ptolemy groupoid" of $S$, due to connections with work of Robert Penner. That groupoid needs to be altered slightly for purposes of my proof, by adding data which breaks the finite symmetry group of an ideal triangulation. The data I added was an enumeration of the prongs of the triangulation, so the objects of the resulting groupoid are "ideal triangulations with enumerated prongs". The generating morphisms of this groupoid are of two types: permutations of the enumeration; and the flip relators mentioned by Bruno Martelli, called "elementary moves" in my paper, together with some rule for enumerating the prongs of the new ideal triangulation resulting from the elementary move.

The group corresponding to this groupoid turns out to be the mapping class group of $S$, and hence the theorem from ECHLPT is applicable.

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I'd like to expand on Dustin's point. There is simply no way to think sensibly about equivariant topology, whether algebraic or geometric, without taking account of multiple basepoints. Even taking account of them one runs into subtle difficulties invisible without them (see eg [65] on my web page). I'll give examples from algebraic topology, since that is what I know best, but examples from geometric topology must abound, as illustrated in other answers.

Take a compact Lie group, or even just a finite group, and consider a smooth closed $G$-manifold $M$. What does it mean for $G$ to be orientable, and what is an orientation? These are seriously interesting questions, necessary to make sense of equivariant Poincar\'e duality, and they are difficult except in the boringly simple-minded case (treated in [53] on my website) when the tangent $G_x$-representation $T_x$ is isomorphic to the restriction to $G_x$ of an ambient $G$-representation $V$ for all $x\in M$. Usually there is no such $V$, and then I can't imagine answers that do not use functors defined on equivariant fundamental groupoids (which themselves are not altogether obvious to define.) Three references which give rather different answers to these questions are [93] and [100] on my web site, and http://front.math.ucdavis.edu/0310.5237, by Costenoble and Waner. I actually do not know how to compare these answers or to calculate with them.

Again, while one can (twistedly) escape explicit use of fundamental groupoids when setting up the Serre spectral sequence with local coefficients nonequivariantly, one cannot do so equivariantly.

Perhaps invoking equivariant theory is overkill, but the fundamental groupoid is such a natural thing, and so elementary, that it seems a little perverse to try to avoid it!

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Following on from Peter's comment, it should be pointed out that the book on Nonabelian algebraic topology deals with the case of chain complexes with a groupoid of operators, and this is useful, even necessary, in relating the fundamental crossed complex $\Pi X_*$ of a CW-complex with its skeletal filtration to the cellular chains of the universal covers at various base points, see Section 8.4. Also equivariant crossed complexes are discussed in papers [94 (1997), 114 (2001)] on my web site, joint with Golasinski, Porter, and Tonks. –  Ronnie Brown Apr 21 '13 at 21:18
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Thank you for this answer! This past week, I just finished teaching a course out of "A concise course", following which I am convinced that indeed "the fundamental groupoid is such a natural thing, and so elementary, that it seems a little perverse to try to avoid it!", to the point that I used groupoids arguments to a greater degree that the "concise course" does in some parts of chapter 3, for example. –  Daniel Moskovich Apr 22 '13 at 1:05
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Thank you, Daniel! Toronto must have brave students :) –  Peter May Apr 22 '13 at 1:48
    
See also the course at Harvard on "Groupoids in topology" math.harvard.edu/~oantolin/groupoids/index.html I believe that students have less "bagage" than older workers and so can often easily catch on to the "right" concept. I first heard about groupoids in a course on homotopy theory by Michael Barratt in 1955-56. (But did not take it up till 1965.) –  Ronnie Brown Apr 22 '13 at 10:38
    
I would also like to add that the realisation that all of $1$-dimensional homotopy theory seemed to me better expressed in terms of groupoids rather than groups led me to look for the potential use of higher groupoids, from a cubical viewpoint, in higher homotopy theory; Chris Spencer, Philip Higgins and Jean-Louis Loday helped hugely to develop this idea, and its applications to new calculations in homotopy theory. –  Ronnie Brown Oct 1 '13 at 17:41
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I just wanted to add something to the discussion about the utility of adding additional basepoints. It turns out this is crucial for understanding certain aspects of embedding theory. See the bottom of this answer for an explanation.

For a map of spaces $A \to B$, let $T(A\to B)$ be the category of spaces which factorize this map. This has objects given by factorizations $A \to X \to B$ and morphisms maps $X \to X'$ which are factorization compatible in the obvious sense. Let's consider the case of the constant map $S^0 \to \ast$. Clearly, an object of $T(S^0\to \ast)$ is just a space with a preferred pair of basepoints.

Then unreduced fiberwise suspension can be regarded as a functor $$ S: \text{Top}(\emptyset \to \ast) \to \text{Top}(S^0 \to *) . $$ That is, the functor which assigns to an unbased space its unreduced suspension, considered as a space with two basepoints.

Now a desuspension question in this context asks given an object $X \in \text{Top}(S^0 \to *)$, is there an object $Y \in \text{Top}(\emptyset \to \ast) $ and a weak equivalence $$ SY \simeq X ? $$ More generally, I've gotten a lot of mileage out of the fiberwise version of this question.

Given a space $B$ we can consider the unreduced fiberwise suspension of $\emptyset \to B$ as the projection map $B \times S^0 \to B$ (here unreduced fiberwise suspension of $Y\to B$ means the double mapping cylinder of the diagram $B \leftarrow Y \to B$, or concretely, it's $B \times 0 \cup Y \times [0,1]\cup B \times 1$.

Unreduced fiberwise suspension is then a functor $$ S_B: \text{Top}(\emptyset \to B) \to \text{Top}(B\times S^0 \to B) , $$ and one can consider the problem of whether an object $X \in \text{Top}(B\times S^0 \to B)$ can be written as $S_B Y$ up to weak equivalence.

Why I care about this problem

This problem naturally arises in embedding theory: if $P \to N \times [0,1]$ is an embedding, where $P$ and $N$ are closed manifolds and if $W$ is the complement of $P$ in $N \times [0,1]$ then $W$ is an object of the category $\text{Top}(N\times S^0 \to N$) and a necessary obstruction to compressing $P$ as an embedding into $N$ is that $W$ should fiberwise desuspend over $N$. Furthermore, in certain instances the existence of fiberwise unreduced desuspension suffices to finding the compression of the embedding. (This story is explained in detail in the paper: Poincaré duality embeddings and fiberwise homotopy theory, Topology 38, 597–620 (1999).)

Postscript

In the fiberwise context there is a real difference between the reduced and unreduced cases of the desuspension problem. For example, in the case of the compression problem $P \to N \times I$ described above, the two inclusions $N \times i \to P$ for $i = 0,1$ might have distinct (fiberwise) homotopy classes. If this is the case, then there's no chance that the complement data $W$ can underly a reduced fiberwise suspension, for if it did, then the map $N \times S^0 \to P$ would factor through $S_N N \cong N \times D^1$, giving a homotopy of the two inclusions $N \times i \to P$.

(For $Y \in \text{Top}(\text{id}:B \to B)$, the reduced fiberwise suspension $\Sigma_B Y$ is given by $$ \Sigma_B Y = \text{colim}(B \leftarrow S_B B \to S_B Y) . $$ This is an endo-functor of $\text{Top}(\text{id}:B \to B)$.

An even more mundane example is this: when $B = \ast$, we can consider $S^0$ with its two distinct basepoints. Clearly $S^0 = S\emptyset$, but $S^0$ is not, even up to weak equivalence, the reduced suspension of any based space.

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Crisp and Paris's proof of Tits's conjecture that the subgroup of an Artin group generated by the squares of the generators is itself a right-angled Artin group uses groupoids in an essential way. For each small type Artin group, they construct a surface with boundary by gluing annuli along squares and mark each square with a base point, and then then study the action of the Artin group on the fundamental groupoid (with respect to a basepoints which correspond to the gluing squares) of the obvious graph which is a deformation retract of this surface. In this way, they construct a representation of such an Artin group into automorphisms of the fundamental groupoid of a graph. Here is the reference:

Crisp and Paris, ``The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group'', Invent. Math. 145 (2001).

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Isn't this just a re-labelling of an older technique that combinatorial and geometric group theorists call a "graph of groups" argument/construction or Bass-Serre theory? –  Ryan Budney Oct 3 '10 at 19:31
    
A graph of groups is to a graph what an orbifold is to a manifold. So a graph of groups is very much related to groupoids, just like orbifolds are. –  André Henriques Oct 3 '10 at 20:17
    
Following up the last comment, see the paper Higgins, P.J. The fundamental groupoid of a graph of groups. J. London Math. Soc. (2) 13~(1) (1976) 145--149. which gives a normal form without any choice of base point or tree, and has been ignored by the specialists. One can regard the method as "distributed computing", with one computer at each vertex, analysing a word as it passes through. Emma Moore at Bangor did a nice thesis on this (2001). –  Ronnie Brown Jan 25 '12 at 7:44
    
Ronnie, this last comment begs the question. The reason Higgins' paper 'has been ignored by the specialists' is that it isn't particularly useful: the groupoid normal form isn't importantly different from the usual group normal form. (Also, the groupoid normal form is already implicit in Serre's work.) –  HJRW Apr 22 '13 at 14:33
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My knowledge of algebraic topology is limited, but I have found one particular example where groupoids are almost imposing themselves: a purely algebraic proof that every subgroup of a free group is free.

The usual geometric proof that a subgroup $H$ of the free group $F_2$ on two generators is free goes like this: one represents $F_2$ as the fundamental group $F_2 = \pi_1(S^1\vee S^1)$. Then, $H$ is the fundamental group of a covering of $p : X \to S^1\vee S^1$. But $X$ is a graph and the fundamental group of a graph is always free.

Of course, it should be possible to translate the above into a purely algebraic proof, but this is very difficult if you don't introduce groupoids or a related notion. I haven't done it in detail, but it appears to me that one has to treat the group $F_2$ as a groupoid "inside" the subgroup $H$. The objects of this groupoid are equivalence classes

$$ g \sim g' \iff \exists h \in H. gh = g' $$

and the morphisms are given by $\lbrace a : [g] \to [ga] \rbrace \cup \lbrace b : [g] \to [gb] \rbrace$ where $a$ and $b$ are the generators of $F_2$. This groupoid is a purely algebraic model of the covering space $X$. The group $H$ corresponds to the subgroupoid of morphisms of the form $\lbrace h : [e] \to [e] \rbrace$. It remains to be shown that $X$ can be contracted along a spanning tree while $H$ is left unchanged.

I think the point is that this last contraction rips apart the group structure of $F_2$.

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The theorem that subgroups of free groups are free can be proved purely algebraically (well, I consider combinatorial group theory as done by manupulating words to be algebra...) without even mentioning groupoids, and I think that the notion does not quite impose itself in the argument. Potatos potatos? –  Mariano Suárez-Alvarez Oct 4 '10 at 12:43
    
Well, you can shuffle cards or perform other permutations without ever mentioning groups, too... Do you have a proof in mind that uses a different construction? I would give the label "groupoid" to anything that looks like the above; more precisely, my criterion for the label "groupoid" is that $h \in H$ is represented as a "composition of edges in a graph". –  Greg Graviton Oct 4 '10 at 15:45
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The point is that group theory has a sufficient body of non-trivial theorems where the terminology becomes a useful conceptual macro. In almost every mathematical paper there are "local definitions" of concepts useful for some stage in a proof. These ideas only earn broad acceptance once people see how convienient the theory is. What I think Daniel is getting at is, "is there such body of theorems for groupoids"? Or are groupoids just a terminology preference of authors -- some choose to use them, some choose to stick to concepts that do not explicitly invoke groupoids, but are equivalent. –  Ryan Budney Oct 4 '10 at 19:16
    
Fair enough. (I don't know enough about groupoids to judge them). Interestingly, there are counterexamples to the "body of theorems" requirement, though. For instance, there are no non-trivial theorems about functors in general, but they are a really slick way of talking about one ever reappearing pattern. –  Greg Graviton Oct 5 '10 at 7:58
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Mariano, I don't think that's entirely fair. The 'usual' purely algebraic proof involves Nielsen reduction (see, for instance, Lyndon and Schupp). Although we think of it as morally similar to the topological proof, there are technicalities that make it a little different. (For instance, the algebraic proof naturally deals with ordered bases, whereas the topological proof spits out a basis as a set.) Greg's proposal does seem like a closer translation into algebra of the topological proof. +1 to Greg. –  HJRW Oct 9 '10 at 14:16
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If you study the set of ideal triangulations of a fixed punctured surface you find out that:

  • ("generators") by using sequences of flips you can relate any pair of triangulations,
  • ("relators") two such sequences are related by a well-understood set of moves (the most important one is the pentagon relation)

The object you get with these "generators" and "relators" is not really a group, it is just a groupoid, called the Ptolemy groupoid. See for instance the paper of Chekhov and Fock introducing quantum Teichmuller space.

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of what is it the fundamental groupoid? –  Daniel Moskovich Oct 8 '10 at 16:15
    
Oops, I skipped the word "fundamental". Luckily, this is indeed the fundamental groupoid of some topological space. The topological space is the arc complex. Every higher dimensional simplex in the arc complex corresponds to an ideal triangulation. Take one point in the interior of each such simplex: the resulting groupoid is what we need here. –  Bruno Martelli Oct 8 '10 at 16:54
    
(To be more precise, the topological space should be the quotient of the arc complex by the action of the mapping class group. The result is an orbifold. The orbifold fundamental group is the mapping class group. I don't know the details, however the groupoid notion here fits very well) –  Bruno Martelli Oct 8 '10 at 20:04
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See my very recent (April 2013) answer for connections with this answer. –  Lee Mosher Apr 21 '13 at 14:38
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