# Young Topology: The fundamental groupoid

Published: Fri 04 May 2007

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 $$f\simeq g$$, if there is a continuous map $$H\colon X\times[0,1]\to Y$$ 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 $$f\colon X\to Y$$,$$f^{-1}\colon Y\to X$$ such that $$ff^{-1}=\operator{Id}_Y$$ and $$f^{-1}f=\operator{Id}_X$$.

Two spaces X,Y are homotopy equivalent if there are maps $$f\colon X\to Y$$,$$f^{-1}\colon Y\to X$$ such that $$ff^{-1}\simeq\operator{Id}_Y$$ and $$f^{-1}f\simeq\operator{Id}_X$$.

Now, if f,g are maps X→Y and f=g, then $$f\simeq g$$, 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 $$f\simeq g\colon X\to Y$$ and h:Y→Z and e:W→X then $$hf\simeq hg$$ and $$fe\simeq ge$$. This can be seen by considering h(H(x,t)) and H(e(x),t) respectively.

Denote by D2 the unit disc in $$\mathbb R^2$$, and by {*} the subset {(0,0)} in $$\mathbb R^2$$. Then $$D^2\simeq\{*\}$$. 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 $$\gamma\in[x,y]$$ and $$\gamma'\in[y,z]$$, then we can consider the map
$$\gamma\gamma'(t)=\begin{cases}\gamma(2t)&0\leq t<1/2\\\gamma'(2t-1)&1/2\leq t\leq1\end{cases}$$. 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 $$\pi_1(X)$$. If we fix a point, and consider [x,x], then this is a group, called the fundamental group with basepoint x, and denoted by $$\pi_1(X,x)$$.

For $$\mathbb R^n$$, a linear homotopy will make any two paths in [x,y] homotopic - and so |[x,y]|=1 in $$\pi_1(\mathbb R^n)$$ 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, $$[a,b]=\mathbb Z$$. 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 $$\pi_1(S^1,p)=\mathbb Z$$.

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]:sub: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 $$\gamma\not\simeq\gamma'$$ in [x,y]. We need to show $$f_*\gamma\not\simeq f_*\gamma'$$. If $$f_*\gamma\simeq f_*\gamma'$$, then $$g_*f_*\gamma\simeq g_*f_*\gamma'$$. However, then
$$\gamma\simeq g_*f_*\gamma\simeq g_*f_*\gamma'\simeq\gamma'$$
which contradicts $$\gamma\not\simeq\gamma'$$. Thus $$g_*f_*\gamma\not\simeq g_*f_*\gamma'$$, and so also $$f_*\gamma\not\simeq f_*\gamma'$$.

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 $$\gamma\in[x,y]$$ and $$\delta\in[y,z]$$. We need to show that $$f_*(\gamma\delta)=f_*\gamma f_*\delta$$. But $$\gamma\delta$$ is the path that first runs through $$\gamma$$ in half the time, then runs through $$\delta$$ in the rest of the time, and $$f_*(\gamma\delta)$$ just transports this path point by point to Y. And $$f_*\gamma$$ transports $$\gamma$$ point by point to Y and $$f_*\delta$$ transports $$\delta$$ point by point to Y, and $$f_*\gamma f_*\delta$$ 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.