notes / Math 55b / chapter 4: The Fundamental Group and Covering Spaces
18 The Seifert–Van Kampen Theorem
· Denis Auroux
Universal covering space
(Left over from last lecture on the classification of covering spaces.)
Definition 18.1. If is a covering and is simply connected, say is a universal covering of .
Note. This corresponds to the trivial subgroup ; it is unique up to equivalence by the classification theorem.
Example 18.2.
; torus.
An infinite tree covering the figure eight (horizontal edges , vertical edges ):
(The following theorem and its discussion were skipped in class.)
Theorem 18.3. Let be a universal covering and any path-connected covering. Then there exists a covering map such that ; and is a universal covering of .
is constructed by lifting:
(it exists since ), and one can show it’s a covering map as well.
So, in fact, if has a universal covering, all other coverings can then be obtained as quotients!
Some spaces have no universal covering!
Example 18.4 (Hawaiian earrings). The “Hawaiian earrings” , where is the circle of radius centered at , inside .
Any covering space must evenly cover a neighborhood of the origin, which prevents it from being simply connected (for sufficiently large, the loop around lifts to a loop).
If one avoids such pathological examples (assuming is (semi-)locally simply connected), one can build the universal cover as a space of pairs where and is a homotopy class of paths . This has a preferred topology for which any simply connected neighborhood is evenly covered: if , adding a path inside (or its inverse) gives a preferred bijection
independent of the choice of path inside since is simply connected.
Seifert–Van Kampen
Question 18.5. Assume , with and open subsets, and we know and . Can we find ?
Example 18.6.
, with and trivial.
Figure 8 , where each of and has the homotopy type of .
The Seifert–Van Kampen theorem gives a general way to calculate in this situation. First we state and prove a weaker (and easier) result.
Theorem 18.7. Suppose , and open, path-connected, . Let and be the inclusion maps. Then the images of and generate .
I.e. every element of can be expressed as a product of elements in and ; i.e. every loop in is path-homotopic to a composition of loops entirely contained in either or .
Proof. Let be a loop based at . is an open cover and is compact, so using the Lebesgue number lemma, we can subdivide into such that is contained in either or . Eliminating unnecessary from the list, we can assume and alternate along the way, and in particular for all .
Let , so that . For each , choose a path in from to (take constant path at ). Then
where each factor is a loop at entirely contained in or in .∎
Corollary 18.8. If with and open and simply-connected, and path-connected, then is simply-connected.
Beware: path-connected is crucial. Otherwise, (arc) (arc) is a counterexample.
Example 18.9. Let ,, and (: north pole), (: south pole).
Then and are homeomorphic to via stereographic projection, mapping each point to the point where the line in through and intersects the equatorial plane .
I.e.
(exercise: check this is a homeomorphism). Change the sign to in the denominator for .
Hence: and , homeomorphic to , are simply connected; , homeomorphic to , is path-connected (!).
Corollary 18.10. is simply connected for .
Corollary 18.11. An open subset of ,, cannot be homeomorphic to an open subset of .
Proof. Let be open and . Then there exists an open ball , and deformation retracts onto a sphere, so is simply connected. Whereas if with open, then for every open with , can’t be simply connected (it retracts to a circle).
(The argument for vs. is easier: it only uses connectedness.) ∎
Example 18.12. Let the figure 8 space, with loops and around the two circles.
We can cover by open sets which have deformation retractions to , with a small cross-shaped neighborhood of the crossing point (connected). By the theorem, is generated by the images of two maps from , i.e. we can express every loop in terms of powers of and ( loops around each , generators of ,), i.e. every element is a product of ’s and ’s. But we don’t know the relations between and .
One can show that and don’t commute: . In fact, free group on 2 generators.
Free products of groups
To formulate SvK, we need to discuss the notion of free product of groups.
Given groups , consider words(; finite). Say is a reduced word if
no (else delete it from the list);
consecutive elements aren’t in the same (else rewrite to ).
Define the product rewrite to a reduced word; the inverse ; identity the empty word.
Definition 18.13. The free product is the set of reduced words, with this operation.
Example 18.14.
which start and end with either or .
More generally, the free group on generators is defined to be the free product of the cyclic groups (so its elements are all the reduced words ).
Observe: contains subgroups (words of length 1; also denoted ) such that , generate , and there are “no relations among elements of these subgroups”.
In particular: if have presentations
(generators | relations, which are words in the generators; i.e. the quotient of the free group on by the smallest normal subgroup containing the ’s), then
where only involve the and only involve the .
Alternative characterization (★): for any group and any homomorphisms , there exists a unique homomorphism such that
commutes for all . Indeed: expressing elements of as reduced words, we must have (each ).
The Seifert–Van Kampen theorem
Let , with and open in , and path-connected, . The inclusions
induce homomorphisms on . By the universal property of the free product, there exists a unique homomorphism such that
commutes. (Define on words in elements of and using and on each component of the word!)
The above theorem (every element of can be written as a product of elements in ,) implies that is surjective.
Theorem 18.15 (Seifert–Van Kampen). The homomorphism defined above is surjective, and its kernel is the smallest normal subgroup of which contains all elements of the form for . I.e.
Corollary 18.16. If is simply connected then .
Corollary 18.17. If is simply connected then , where is the smallest normal subgroup containing the image of .
Example 18.18 (Figure 8 (using Corollary 1)).
deformation retract onto circles, and is contractible. Hence , the free group generated by loops around the two circles.
Example 18.19 (Wedge of circles). By induction, for a wedge of circles, , with homeomorphic to for all and :
the free group on generators loops generating . (Similarly for a finite graph with loops.)
Fundamental groups of surfaces
Fundamental groups of surfaces can also be calculated using Van Kampen!
Example 18.20 (Torus). We can now calculate of the torus in a different way (the easiest is still the universal cover ).
, where for all and for all . Let and the open ball of radius around .
deformation retracts onto a wedge of two circles:
is simply connected.
has the homotopy type of .
Using Corollary 2 above: , where is normally generated by the image of the loop which generates (i.e. generated by it and its conjugates).
is a free group on generators , and the image of under the inclusion is .
[The “obvious” picture (push out to the boundary of the square, where it becomes the loop around the edges, based at a corner) needs to be corrected slightly: the base point should be fixed !]
So we set , i.e. , and get an abelian group :
(The following computation for was skipped in class.)
Example 18.21 (Projective plane). Similarly for , using .
Now write , with and a disc centered at .
deformation retracts onto the boundary , so with generator .
is simply connected.
has the homotopy type of .
, where is the normal subgroup generated by the image of the generator under inclusion, which is . So
Example 18.22 (Klein bottle). Recall , where and .
Again write , with and a disc centered at ; then .
retracts onto the boundary figure 8 space (with loops and ), so the free group . has the homotopy type of ; the generator maps to under inclusion. So
which is not abelian: , not ! I.e. : is conjugate to its inverse!
But this contains an index 2 subgroup generated by and , which commute! (; taking inverses, , so
So ✓.) (So the subgroup .)
(One can show, by rearranging letters via , that this subgroup contains all words with an even number of ’s, so it is an index 2 subgroup.)
This subgroup corresponds to a degree 2 covering map by the torus, !
It is the quotient by . I.e. map to
Cool fact that this relates to: if you coat a Klein bottle in paint all over, the paint forms a torus.