Fundamental Group

The first nontrivial homotopy class of topological spaces is represented by $\mathbb{S} ^0,$ since it has two (disjoint connected) components; indeed, it is elementary to show that connectedness may be characterised by the nonexistence of continuous maps from a space onto $\mathbb{S} ^0,$ and actually, `up to homotopy' is good enough for such characterization. Next comes $\mathbb{S} ^1$ which has only one component but it admits also loops which cannot be `homotopied' to a point. We denote by $[\mathbb{S} ^1,\mathbb{S} ^1]$ the collection of homotopy equivalence classes of continuous maps from $\mathbb{S} ^1$ to itself (ie loops in $\mathbb{S} ^1$) which preserve a given basepoint $\ast.$ Intuitively, the set of classes is parametrized by $\mathbb{Z} ,$ since the loops either go forwards a net integer number of times round the circle, or backwards, or are the trivial loop, up to homotopy. The fact that $[\mathbb{S} ^1,\mathbb{S} ^1]\equiv \mathbb{Z}$ as a set (and, later, as a group) and is not a singleton tells us that $\mathbb{S} ^1$ bounds a 2-dimensional `hole', whereas $\mathbb{R} ^1$ and $\mathbb{S} ^0$ do not. For any pointed X, that is a topological space with a chosen basepoint $\ast,$ its fundamental group or first homotopy group is the set of homotopy classes of loops based at $\ast:$

$$\pi_1 (X) \ =\ [\mathbb{S} ^{1}, X]$$ (1)

which has a natural group structure by composition of curves. Also, we denote by $\pi _{0} (X)=[\mathbb{S} ^0,X]$ the (pointed) set of path components of X; so X is connected if $\pi_0(X)$ is a singleton.

The higher homotopy groups, $[\mathbb{S} ^n,X],$ are extremely powerful tools in analysis, geometry and topology [4]; we shall not need that theory here but to whet your appetite for it we note the beautiful and surprising result of Hopf which led to homotopy theory (cf [4] p 100):

$$[\mathbb{S} ^3,\mathbb{S} ^2]\cong \mathbb{Z} , \ \ {\rm so} \ \pi_3(\mathbb{S} ^2)=\mathbb{Z}$$

in contrast to

$$\pi_m(\mathbb{S} ^1)=0 \ \ {\rm for} \ m>1$$

Find out more about Hopf by clicking about Hopf

If $\pi _{1} (X)$ is the trivial group we write $\pi _{1}(X)=0,$ and if $\pi _{1}(X) = 0$ and $\pi_0(X)$ is a singleton, then we say that X is simply-connected--which is independent of the choice of basepoint. When the basepoint must be denoted, because X has more than one connected component, we write $\pi_1(X,x_0)$ for example. A common way to make use of this construction is to show that two spaces have different fundamental groups; then it follows that they must have different homotopy types and hence cannot be homeomorphic. Continuous maps between spaces induce group homomorphisms between their fundamental groups, a powerful way to study families of spaces. Exercises

1.

The result $\pi _{1} (\mathbb{S} ^{1}) \cong \mathbb{Z}$ can be approached as follows.

(a)

$p : \mathbb{R}\rightarrow \mathbb{S} ^1 : x \mapsto e^{2\pi ix}$ is a continuous surjection.

(b)

$\phi : \mathbb{Z}\rightarrow \pi _{1} (\mathbb{S} ^{1} ) : n \mapsto [\,p \sigma _{n}]$ where $\sigma _{n} : I \rightarrow \mathbb{R} : t \mapsto nt$ is a group homomorphism.

(c)

Paths in $\mathbb{S} ^1$ admit unique lifts to $\mathbb{R}$.

(d)

$\mathbb{R}\stackrel{p}\rightarrow \mathbb{S} ^1$ has the homotopy lifting property.

(e)

$\phi$ is an isomorphism.

2.

If $X = U\cup V$ for some open 1-connected subsets U,V, and $U\cap V$ is 0-connected, then X is 1-connected since loops in X are homotopic to a product of loops in U or in V. Hence $\pi _{1} (\mathbb{S} ^{n} ) = 0$ for $n \geq 2$.

3.

Consider the two paths c and a going half counterclockwise and half clockwise respectively round $\mathbb{S} ^1$ as the unit circle in ${\mathbb{C} }$ :

$$c : [0,1]\longrightarrow \mathbb{S} ^{1} : s\longmapsto e^{1\pi s}\,,$$

$$a : [0,1]\longrightarrow \mathbb{S} ^{1} : s\longmapsto e^{-1\pi s}\,.$$

Show they induce the same isomorphisms between $\pi_{1}(\mathbb{S} ^{1},1)$ and $\pi_{1}(\mathbb{S} ^{1},-1)$.

4.

The fundamental group of $\mathbb{S} ^1\vee \mathbb{S} ^1$ is $\mathbb{Z}\ast \mathbb{Z}$ and hence is nonabelian. The paths in $\mathbb{S} ^1\vee \mathbb{S} ^1$ corresponding to a,c in the previous example do not induce the same isomorphisms.

5.

No continuous map from the unit disk to its boundary can restrict to the identity on the boundary--simply consider the fundamental groups and the homomorphism induced by the inclusion map of the boundary.

6.

Show that

$$\begin{eqnarray*}\pi_1(\mathbb{E} ^3) &=& 0, \\ \pi_1(\mathbb{S} ^1\times \mathbb{S} ^1) &=& \mathbb{Z}\times \mathbb{Z} . \end{eqnarray*}$$