Michi’s blog » Young Topology: The fundamental groupoid

## Young Topology: The fundamental groupoid

• May 4th, 2007
• 3:26 pm

Today, with my bright topology 9th-graders, we discussed homotopy equivalence of spaces and the fundamental groupoid. In order to get the arguments sorted out, and also in order to give my esteemed readership a chance to see what I’m doing with them, I’ll write out some of the arguments here.

I will straight off assume that continuity is something everyone’s comfortable with, and build on top of that.

## Homotopies and homotopy equivalences

We say that two continuous maps, f,g:X→Y between topological spaces are homotopical, and write , if there is a continuous map such that H(x,0)=f(x) and H(x,1)=g(x). This captures the intuitive idea of step by step nudging one map into the other in formal terms.

Two spaces X,Y are homeomorphic if there are maps , such that and .

Two spaces X,Y are homotopy equivalent if there are maps , such that and .

Now, if f,g are maps X→Y and f=g, then , since we can just set H(x,t)=f(x)=g(x) for all t, and get a continuous map out of it. Thus homeomorphic spaces are homotopy equivalent, since the relevant maps are equal, and thus homotopic.

There are a couple of more properties for homotopic maps we’ll want. It respects composition – so if and h:Y→Z and e:W→X then and . This can be seen by considering h(H(x,t)) and H(e(x),t) respectively.

Denote by D2 the unit disc in , and by {*} the subset {(0,0)} in . Then . In one direction, the relevant map is just the embedding, and in the other direction, it collapses all of D2 onto {*}. One of the two relevant compositions is trivially equal the identity map, and in the other direction, the linear homotopy H(x,t)=tx will do well. Thus the disc and the one point space are homotopy equivalent.

## The fundamental groupoid

Let X be a topological space (most probably with a number of neat properties – I will not list just what properties are needed though), and consider for each pair x,y of points in X, the set [x,y] of homotopy classes of paths from the point x to the point y. A path, here, is a continuous map [0,1]→X. We can compose classes – if and , then we can consider the map
. This is a path from x to z, and so belongs to a class in [x,z]. This class is well defined from the choices of γ, γ' since homotopies and composition of maps work well together.

This gives us a composition. It is associative - on homotopy classes. What happens if we look at maps instead of homotopy classes is part of the subject of my own research. It has an identity at each point x - the constant path γ(t)=x, and for each class in [x,y] there is a class in [y,x] such that their composition is homotopic to the constant path in [x,x].

Thus, we get a groupoid. This is called the fundamental groupoid, and denoted by . If we fix a point, and consider [x,x], then this is a group, called the fundamental group with basepoint x, and denoted by .

For , a linear homotopy will make any two paths in [x,y] homotopic – and so |[x,y]|=1 in for any x,y.

For S1 – the circle – we can choose to view it as [0,1]/(0=1). Then we can consider the paths fm(t)=a(1-t)+bt+nt. This is a path from a to b, and it winds n times around the circle. Each path in [a,b] is homotopic to a fm, by a linear homotopy, which just rescales the speeds through various bits and pieces, and possibly straightens out when you double back. Thus, . Furthermore, if you compose fmfn, you’ll get fn+m.

If we pick out the fundamental group out of this groupoid, we’ll get the well known fundamental group .

Now, suppose we have two homotopy equivalent spaces X and Y, with the homotopy equivalence given by f:X→Y and g:Y→X. Then consider the map f*:[x,y]X→[f(x),f(y)]Y given by f*γ(t)=f(γ(t)). I claim
1) f* is bijective.
2) f* works well with composition of classes.

For bijectivity we start with injectivity in one direction. Consider two paths in [x,y]. We need to show . If , then . However, then

which contradicts . Thus , and so also .

The proof is symmetric in the choice of direction, and so we can just repeat the same argument to get that g* is also an injection. Thus we can conclude that f* is in fact a bijection.

Now, for the second part, we consider and . We need to show that . But is the path that first runs through in half the time, then runs through in the rest of the time, and just transports this path point by point to Y. And transports point by point to Y and transports point by point to Y, and just runs through the first of these in half the time, then the rest in the rest of the time.

Thus, homotopy equivalent spaces have the same fundamental groupoid.

### 7 People had this to say...

• Creighton Hogg
• May 4th, 2007
• 16:36

Hey, I think it’s great that you’re helping to teach this stuff to kids. When I was maybe a year older than them, I started teaching myself topology & it was a lot harder than it would have been with other people to talk to. So, on behalf of my younger self, thanks for teaching interested kids.

• Michi
• May 4th, 2007
• 16:39

Creighton: For one of my reasons to take interested kids VERY seriously, take a look at my post on why I keep organizing congresses – I have a personal history as the interested kid who got taken seriously, and this is what placed me where I am today.

That said, these kids are quite a bit more skilled than I am. I am constantly amazed with the mere fact that they keep up with it.

• Creighton Hogg
• May 4th, 2007
• 16:54

Cool.
Yeah, I never really got taken that seriously but otherwise we have pretty similar histories with suddenly discovering how great real math is. I found an old book for a couple of dollars in a used bookstore that went from basic addition to basic multi-variate calculus, and devoured the book after school during the beginning of my sophmore year of highschool.
Mathematics is a beautiful subject, and while I’ve never done as much as you’re doing, I try to encourage kids I meet by giving them a glimpse of what real math is like. One of my wife’s teenage cousins likes her geometry class, so I spent some time and told her a bit about what happens when you leave Euclidean geometry: the difference between how cylinder’s are curved and how sphere’s are curved, why you can’t giftwrap a basketball, and how it all ties to gravity and the early universe.

[...] but it seems that the majority of contributors have explanations of topics for this edition. Mikki, a professor, visits a high school once a week, and describes the super-challenging problems he [...]

• Jon
• May 23rd, 2007
• 7:13

I’m a bit confused. If h: Y -> Z, and g: X -> Y, then h(f(x)) is in Z, and g(f(x)) makes no sense, since f(x) is in Y, not X. Then saying h(f(x)) is homotopic to g(f(x)) makes no sense. Did you mean h(f(x)) is homotopic to h(g(x))?

• Michi
• May 23rd, 2007
• 8:51

Jon: Doh! Of course I mean hf homotopic to hg.

• Jon
• May 23rd, 2007
• 9:53

Ah, thanks.
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