notes / Math 55b / chapter 4: The Fundamental Group and Covering Spaces
15 Covering Spaces and Path Lifting
· Denis Auroux
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.
Covering spaces
Definition 15.1 (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 15.2 (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 15.3. If is any topological space and a set with the discrete topology, then is a covering map.
Example 15.4.,, is a covering map! E.g. take (or any arc of the circle). Then
and is a homeomorphism on each slice.
Example 15.5. Consider . Then , (so ), is an -fold covering.
Theorem 15.6. 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 15.7. Consider the torus : since covers , covers .
If is a covering and a subspace, then by restriction we get a covering .
Example 15.8. 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 15.9 (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 15.10. Consider ,, and the path ,. This has infinitely many possible lifts to paths in , depending on where gets lifted to.
Theorem 15.11 (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 15.12 (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. ∎
Proposition 15.13 (Observation). If is a path-homotopy from to (in ), then is a path homotopy (in ) from to .
Indeed, if for all , then , which is a discrete subset of (one point in each slice), so we must have for all (always the same point). Similarly for the other endpoint .
On the other hand, loops don’t always lift to loops!
Example 15.14. A loop going once around the circle from lifts to a path in starting at and ending at , which is not a loop.
But since path lifting is unique, given a starting point , the endpoint is uniquely determined. This leads to a key notion:
Notes not reached in this lecture, covered at the start of Lecture 16
The lifting correspondence
Definition 15.15 (Lifting correspondence). For a covering , the lifting correspondence
is defined by , where is the lift of such that .
Question 15.16. Why is well-defined? (I.e. independent of the choice of in its homotopy class?)
Show solution
If is a path homotopy , then its lift starting at is a path homotopy between and , so .
Example 15.17. For the covering , taking and : if loops around the circle times (counting counterclockwise), then its lift ends at . This gives a map (surjective).
Now we know, at last, that isn’t simply connected!
Proposition 15.18. If is path connected, then is surjective.
Proof. Let , and a path from to . Then is a loop at whose lift starting at is . So .∎
Recall:
Proposition 15.19. If is simply connected, then any two paths from to are path-homotopic.
Proof. is a loop at , so ( simply connected). Then
∎
So:
Theorem 15.20. If is a covering and is simply connected, then is a bijection.
Proof. By the above, is surjective. If , then are paths in starting at and ending at the same point . Since is simply connected, . Hence , i.e. , so . So is injective. ∎
Theorem 15.21..
Proof. Consider the covering map ,. Since is simply connected, by the above theorem the lifting correspondence
is a bijection. We just need to show it is a group homomorphism.
Let and let ,; i.e. the lifts and starting at end at and . Define a new path by : this is the lift of starting at . Then is a well-defined path in from to , and it is the lift of starting at . So
∎
(One can show similarly that for the torus, , using the covering .)
The Brouwer fixed point theorem
Next we study applications of .
Let denote the closed ball of radius in , with boundary the unit sphere . Recall that if , a retraction is a continuous map such that for all .
Theorem 15.22. There is no retraction of onto .
Proof. If is a retraction, then , so
where because is convex in (straight-line homotopy). Then is the trivial homomorphism, which is . Contradiction. ∎
(A more elementary way to say this: given a nontrivial loop in , is nullhomotopic in , via some homotopy from to . Then is a path-homotopy in , contradiction.)
[With more algebraic topology, similarly there is no retraction for all .]
This implies:
Theorem 15.23 (Brouwer fixed point theorem). If is continuous, then there exists such that .
[With more algebraic topology, the same holds for continuous maps for all . The case follows from the intermediate value theorem, cf. HW2.]
Proof. Assume is continuous and for all . Then define by mapping each to the point where the ray from to hits .
(Formula: , where is such that . We can solve for by the quadratic formula, so does depend continuously on .)
This gives a continuous map ; moreover, if then , so we get a retraction . Contradiction. ∎