The Fundamental Group

Group Structure on Loops

Fundamental Group

The fundamental group of a topological space at a base point is the group of (path) homotopy classes of loops based at , with the multiplication of concatenation of loops, inverse being the reversed loop, and identity being the constant loop at .

Lemma

In , any two paths are homotopic if and only if they have the same endpoints.

Proof For any two paths and with same endpoints, gives a valid homotopy.

Proposition

.

Proof The ideal is that looks locally like . Define , , paths , , and , . Note that . We claim that sending integer to the homotopy class of forms the desired isomorphism. First show that it is a group homomorphism, i.e. . Define , . Observe that so . Note that both and are paths in from to , hence there exists a homotopy from to . Now let , then is a homotopy from to . This shows that is a homomorphism. Now we show the surjectivity. Fixing some loop . Define and . We can partition the preimage of as following: Then the restrictions of , say and are homeomorphism, with inverses and respectively. There is a partition of , with each either a subset of or a subset of . Now define satisfying , and then inductively on by where and are the unique integer for which or . This gives a path starting at and ending at some . Thus , hence, by applying to the both sides, and so . Finally, we show that is injective. Suppose , then , there exists a homotopy from to . Similarly, there exists a partition of , and , such that is either a subset of or a subset of . Now define similarly as above, which gives a homotopy from to , hence implies . Thus is injective.

Remark

In fact, the distributive law holds in general: Suppose is continuous, and are paths, then

Change of Basepoints

Proposition

Suppose is a path from to , then the map defined by is an isomorphism.

Proof We first check that it is well-defined. Suppose are homotopic paths of with homotopy . Then via . Secondly, it is a homomorphism because Moreover, is a inverse of (here means the reversed path), thus is an isomorphism.

Remark

Note that there is no ambiguity when we defined . Because actually .

Fundamental groups are powerful in many (other) realms of mathematics.

• It can even be used to prove the fundamental theorem of algebra: Proof For the sake of contradiction, suppose some non-constant polynomial has no roots. Then for each , is a loop in based at . Now fix some large . Then is a homotopy from to . Let . Define Note that since is sufficiently large is never zero, is a valid homotopy from to . Therefore, which implies since . This contradicts the assumption that is non-constant.
• We can also show Brouwer’s fixed-point theorem by the fundamental group. It states that any continuous has a fixed point, where is the closed disk in . Proof Suppose has no fixed points. Then we can define by sending to the intersect point of and the ray from to . Note that for . Consider the inclusion , we have via some homotopy . Then via , yielding a contradiction.

Proposition

Suppose and are topological spaces, then the fundamental group is isomorphic to .

Proof This is simply because the universal property of product spaces implies that there is a one-to-one correspondence between the loops at in and the pairs of loops in and at . Then taking homotopy classes for the both sets gives the two groups.

e.g. We can directly deduce that the fundamental group of a torus .

Proposition

Suppose . If retracts onto , then the inclusion induced is injective. Moreover, if deformation retracts to , then is an isomorphism.

Proof Suppose is the retraction and is the inclusion, then , so , which means , hence has to be injective. Moreover, if through the homotopy , then for any , pick a representative , let , then is a homotopy from to , so . Thus is surjective, hence an isomorphism.

Theorem

Suppose is a homotopy equivalence. Then is an isomorphism.

Proof

Simple Connectedness

A topological space is simply-connected if it is path-connected and its fundamental group is trivial.