notes / Math 55b / chapter 4: The Fundamental Group and Covering Spaces
14 The Fundamental Group
· Denis Auroux
Recall: composition of paths with :
On path-homotopy classes, is associative, has identity (the constant path) and inverses (, where is the reverse path).
To define the fundamental group we consider loops, i.e. paths from a fixed base point to itself.
Definition 14.1 (Fundamental group). The set of path-homotopy classes of loops based at , with the operation (composition), is called the fundamental group of , denoted .
Dependence on the base point
If are in the same path-component of , let be a path from to . Then for any loop based at , we get a loop at by taking ,
and so we get a map
Proposition 14.2. is a group isomorphism.
Proof.
If , then
So is a group homomorphism.
Let , the reverse path from to . Then . We claim and are inverses of each other. Indeed, for ,
Hence (and similarly as well), so is an isomorphism.
∎
Corollary 14.3. If is path-connected, then is independent of up to isomorphism.
Remark. When is a loop at , we get an automorphism of . This is in fact an inner automorphism, namely conjugation by :.
Induced homomorphisms
Definition 14.4 (Induced homomorphism (Definition/Proposition)). A continuous map induces a group homomorphism
Check:
If via , then via . So is well-defined.
(composition with is compatible with concatenation). So is a group homomorphism: .
as a functor
Consider the category of pointed topological spaces:
objects a topological space plus a choice of base point, ;
morphisms continuous maps preserving base points: means is continuous and such that .
A functor from this category to that of groups is an assignment of a group to each pointed topological space, and a group homomorphism to each map of pointed spaces, compatible with composition.
Proposition 14.5. Given ,
Hence is a functor (it maps the composition to the composition ).
(This is just .)
This implies:
Corollary 14.6. If is a homeomorphism, then is an isomorphism.
But we can do better!
Recall:
A retraction of onto a subset is such that , i.e. . Then, taking a base point ,
i.e. is injective.
A deformation retraction: assume moreover that is homotopic to by a homotopy that fixes . Then we claim are inverse isomorphisms .
More generally, recall a homotopy equivalence is such that and .
This follows from the fact that homotopic maps induce the same homomorphisms on :
Proposition 14.8.
Let be homotopic via a homotopy such that for all . Then .
If the homotopy doesn’t fix base points, let be the path defined by . Then
are related by the isomorphism :.
Proof.
Given a loop based at , consider
gives a path homotopy (based at ); hence .
Now consider defined by concatenating
the path ,
the loop in ,
the path .
Then is a path homotopy in from to .
∎
Proof of the theorem. If are homotopy inverses, , then by the proposition,
for some path , so this is an isomorphism. Hence is injective and is surjective.
Similarly, is an isomorphism , so is injective and is surjective (here is the map induced by with base point ).
Hence is an isomorphism, and is also an isomorphism. ∎
Notes not reached in this lecture, covered at the start of Lecture 15
Covering spaces
At some point we’d like to show . We’ll do this by introducing a key tool for the study of : the notion of covering spaces.
Definition 14.9 (Evenly covered). Let be a continuous surjective map. We say evenly covers an open subset if
where the are disjoint open subsets, and for each , is a homeomorphism. The are called slices.
(Equivalently: there is a homeomorphism , with given the discrete topology, such that :
We say the diagram of maps commutes.)
Definition 14.10 (Covering space). If every point of has a neighborhood which is evenly covered by , we say is a covering space of and is a covering map. is called the base of the covering.
Example 14.11. If is any topological space and a set with the discrete topology, then is a covering map.
Example 14.12.,, is a covering map! E.g. take (or any arc of the circle). Then
and is a homeomorphism on each slice.
Example 14.13. Consider . Then , (so ), is an -fold covering.
Theorem 14.14. If and are covering maps, then is a covering map.
Proof. Given , let and be neighborhoods such that and (slices). Then
a union of open slices homeomorphic to .∎
Example 14.15. Consider the torus : since covers , covers .
If is a covering and a subspace, then by restriction we get a covering .
Example 14.16. Let be a base point on the circle, and let . Then is the “figure eight space” .
Then we have a covering , where
The horizontal lines map to one circle and the vertical lines map to the other.
Lifting
Definition 14.17 (Lifting). Given a continuous map , a lifting of a continuous map is a map such that , i.e. the diagram
commutes.
If is a covering map, then we can locally lift: namely, if and is evenly covered, then we can lift to one of the sheets.
Key point: if is a covering, then paths and path homotopies in always lift.
Example 14.18. Consider ,, and the path ,. This has infinitely many possible lifts to paths in , depending on where gets lifted to.
Theorem 14.19 (Path lifting). Let be a covering map, a path starting at , and . Then there exists a unique lift such that .
Proof. Cover by open sets which are evenly covered by . Then the preimages are an open cover of , which is compact, so there exists a Lebesgue number such that for all , for some . Hence we can find a finite subdivision such that each lies inside one of the .
Define . Assume we have defined for . Then we define for as follows. Recall for some which is evenly covered by , slices. Let be the slice which contains . The map is a homeomorphism, so it has a continuous inverse, and we can define
which extends continuously over . Repeating the process, we obtain a continuous lift .
is unique since for each there was a unique slice containing and a unique way to lift into it. ∎
Theorem 14.20 (Homotopy lifting). Let be continuous with , a covering map, and . Then there exists a unique lift such that .
Proof. The proof is exactly the same: subdivide into squares of side length which map into open subsets of that are evenly covered; then construct the lift one square at a time. ∎