MATH730 NOTES WEEK 8

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1 MATH730 NOTES WEEK 8 1. Van Kampen s Theorem The main idea of this section is to compute fundamental groups by decomposing a space X into smaller pieces X = U V where the fundamental groups of U, V, and U V are easier to understand. Proposition 1.1. Suppose that X = U V, where U and V are path connected open subsets and both contain the basepoint x 0. If U V is also path connected, then the smallest subgroup of π 1 (X) containing the images of both π 1 (U) and π 1 (V ) is π 1 (X) itself. That is, π 1 (X) = π 1 (U)π 1 (V ). Note: The path connected condition on U V is important: Let U be S 1 (0, 1) and let V be S 1 (0, 1), which are both simply connected, but the intersection U V is not path connected and the product of two trivial groups cannot be π 1 (S 1 ) Z. Proof. Let γ I X be a loop at x 0. By the Lebesgue number lemma, we can subdivide the interval I into smaller intervals [s i, s i+1 ] such that the image of each subinterval under γ is in either U or V. Write γ i for the restriction of γ to the ith subinterval. The endpoints of each γ i must be in U V. Since U V is path connected, there is a path α i from γ i (1) to x 0 for each i. Then [γ] = [γ n ] [γ 2 ] [γ 1 ] = [γ n α 1 n 1] [α 2 γ 2 α 1 1 ] [α 1 γ 1 ] This expresses the loop γ as a product of loops in U and loops in V. If we want to calculate π 1 (X), then it is not very helpful to think about the product of π 1 (U) and π 1 (V ) inside π 1 (X). We want an external group in terms of the simpler fundamental groups. There are homomorphisms π 1 (U) π 1 (X) and π 1 (V ) π 1 (X) induced by inclusions, and we want some sort of product of these with a map to pi 1 (X). Note that the direct product of groups π 1 (U) π 1 (V ) can t work because elements of π 1 (U) commute with the elements of π 1 (V ), so this would also hold in the image of any homomorphism π 1 (U) π 1 (V ) π 1 (X). We want a group built freely out of π 1 (U) and π 1 (V ), and we can use the free product of groups π 1 (U) π 1 (V ). The elements of the free product G 1 G 2 are finite length words g 1 g 2 g n where each g i is in either G 1 or G 2, none are identity elements, and if g i G 1 then g i+1 G 2. That is, reduced words. Example. The free group on two generators is a free product. F 2 = Z Z. Example. The free group on n generators is the free product n Z. Example. Let C 2 be cyclic group of order 2. The free product C 2 C 2 is an infinite group. If a, b are the nonidentity elements of the two copies of C 2, then elements of C 2 C 2 are a, ab, aba, abab,..., b, ba, bababababa,..., etc. There is a homomorphism C 2 C 2 C 2 that sends both a and b to the nontrivial element. The kernel of this map is all words of even length, the subgroup generated by the word ab. So C 2 C 2 is an extension of C 2 by the infinite cyclic group Z. (Or C 2 C 2 is a semidirect product of C 2 with Z.) 1

2 2 MATH730 The free product has a universal property! For any groups G 1, G 2, there are inclusion homomorphisms G i G 1 G 2. Proposition 1.2. Suppose that K is any group with homomorphisms φ i G i K. Then there is a unique homomorphism Φ G 1 G 2 K which restricts to the given homomorphisms φ i. In other words, the free product is the coproduct in the category of groups. Then we can restate Proposition 1.1 as follows: Proposition 1.3. Suppose that X = U V, where U, V are path connected open subsets of X containing the basepoint. If U V is also path connected, then the natural homomorphism Φ π 1 (U) π 1 (V ) π 1 (X) is surjective. So we need to understand the kernel N, since the first isomorphism theorem implies that π 1 (X) (π 1 (U) π 1 (V ))/N. Consider a loop γ U V. It has images γ U π 1 (U) and γ V π 1 (V ) and these map to the same element in π 1 (X), so γ U γv 1 is in the kernel. Proposition 1.4. With the same assumptions as before, the kernel of π 1 (U) π 1 (V ) π 1 (X) is the normal subgroup N generated by elements of the form γ U γ 1 V. The normal subgroup generated by the elements of the form γ U γv 1 can be characterized either as (1) the intersection of all normal subgroups containing γ U γv 1 or (2) the subgroup generated by all conjugates gγ U γv 1g 1. I will skip this proof. See Thm 1.20 in the text if you wish to know the details. We can now assemble these results to get Van Kampen s theorem. Theorem 1.5 (Van Kampen). Suppose that X = U V, where U, V are path connected open subsets containing x 0. If U V is also path connected, then π 1 (X) (π 1 (U) π 1 (V ))/N, where N is the normal subgroup of pi 1 (U, x 0 ) π 1 (V, x 0 ) generated by elements of the form ι U (γ)ι V (γ) 1 for γ π 1 (U V, x 0 ). Definition. Let φ i H G i be group homomorphisms. The amalgamated free product is the quotient G 1 H G 2 = (G 1 G 2 )/N, where N is the normal subgroup generated by elements of the form φ 1 (h)φ 2 (h) 1. The amalgamated free product satisfies the universal property of the pushout in the category of groups. So we can restate Van Kampen s theorem categorically. Theorem 1.6. Suppose that X = U V, where U, V are path connected open subsets containing x 0 with path connected intersection U V. (That is, X is the pushout U U V V of open path connected subsets.) Then the inclusions of U and V into X induce an isomorphism π 1 (U) π1(u V ) π 1 (V ) π 1 (X). (That is, the fundamental group construction takes a pushout of spaces to a pushout of groups.) U V U π 1 (U V, x 0 ) π 1 (U, x 0 ) V X π 1 (V, x 0 ) π 1 (X, x 0 )

3 MATH applications of VK theorem. We start with some fundamental groups we know already... Example. Let X = S 1 S 1, let U be an open set containing one of the circles, plus an ɛ-ball around the basepoint in the other circle, and let V be the symmetric set for the other circle. Then the intersection U V is an X and is contractible. Both U and V are equivalent to S 1. Thus π 1 (S 1 S 1 ) π 1 (S 1 ) π1( ) π 1 (S 1 ) Z Z F 2. Example. Let X = S 1 S 1 S 1, let U be a neighborhood of S 1 S 1 and V a neighborhood of the remaining S 1. Then π 1 (S 1 S 1 S 1 ) F 3. Example. Let X = S 1 S 2 and take U to be a neighborhood of S 1, and V a neighborhood of S 2. Then π 1 (S 1 S 2 ) π 1 (S 1 ) π 1 (S 2 ) Z. Is π 1 (X Y ) always the free product of π 1 (X) and π 1 (Y )? Almost! Need that the neighborhoods U and V were homotopy equivalent to X and Y and that intersection is contractible. Definition. Recall a retraction of a space X onto a subspace A is a map r X A which restricts to the identity on A. We say X deformation retracts onto A is there exists a retraction r X A and a homotopy h X I X with h(x, 0) = ir(x) and h(x, 1) = x and h(a, t) = a for all a A and all t I. ************************ A deformation retraction is a special case of a homotopy equivalence. Our definition is sometimes called a strong deformation retraction. Example. S n 1 R n {0} is a deformation retract, with r(x) = x/ x and h(x, t) = tx + (1 t)x/ x. Example. We showed that there is no retraction D 2 S 1, so there is no deformation retraction. Hatcher defines a retract as a map r X X such that r(x) = A and r A = id. Then a deformation retraction of X onto A is a homotopy from the identity of X to a retraction r X X of X onto A. Note that every space retracts onto a point, but not every space deformation retracts onto a point. (The homotopy gives a path from every point to the chosen point, so the space must be path connected at least.) Definition. We say that x 0 X is a nondegenerate basepoint for X if x 0 has a neighborhood U such that x 0 is a deformation retract of U. Proposition 1.7. Let x 0 and y 0 be nondegenerate basepoints for X and Y, respectively. Then π 1 (X Y ) π 1 (X) π 1 (Y ). Let X be a space and φ S 1 X a map. Let X = X φ D 2, the space obtained by attaching a disk along the map φ. The inclusion of the boudary S 1 D 2 is null homotopic, so the composition φ S 1 X X is also null homotopic. So attaching the 2-cell kills off the class [φ] π 1 (X). Proposition 1.8. Let X be path connected and let φ S 1 X be a loop at x 0 in X. Let X = X φ D 2. Then π 1 (X, ι(x 0 )) π 1 (X, x 0 )/ [φ] where [φ] is the normal subgroup generated by [φ].

4 4 MATH730 Proof. Consider the open subsets U, V of D 2, where U = D 2 B 1/3 and V = B 2/3. The map ι D 2 D 2 X restricts to a homeomorphism (with open image) on the interior of D 2, so the image of V in X is open and path-connected. Let U = X U, which is open in X because it is the image under the quotient X D 2 X of the open set X U. The set U is also path-connected. The sets U and V cover X, and U deformation retracts onto X, since U deformation retracts onto the boundary S 1. The set V is contractible, and the path connected subset U V deformation retracts onto the circle of radius 1/2. The map Z π 1 (U V ) π 1 (U ) π 1 (X) sends the generator to [φ]. The van Kampen theorem implies that π 1 (X ) π 1 (X)/ φ. Did you catch that we cheated in that proof? We need a basepoint in U V, so a careful proof would have a change-of-basepoint discussion. Higher dimensional cells? Proposition 1.9. Let X be pathconnected and let φ S n 1 X be an attaching map for an n-cell in X, based at x 0. Let X = X φ D n. If n 3, π 1 (X, ι(x 0 )) π 1 (X). Proof. Same proof as for 2-cell, but now U V S n 1 is simply connected. Example. Attach a 2-cell to S 1 along the double cover p 2 S 1 S 1. This loop corresponds to the element 2 in π 1 (S 1 ) Z, so X = RP 2 has π 1 (X ) Z/2Z. We saw RP n has a CW structure with 1 i-cell for i n. Attach a 3-cell to RP 2 along the double cover S 2 RP 2. By the proposition, this doesn t change the fundamental group, so π 1 (RP 3 ) Z/2Z, and similarly for higher RP n. Example. Attach a 2-cell to S 1 S 1 along the map aba 1 b 1, where a, b are standard inclusions S 1 S 1 S 1. The resulting CW complex is the torus S 1 S 1 and so π 1 (S 1 S 1 ) F 2 / aba 1 b 1 Z 2... Proposition The homomorphism φ F (a, b) Z 2 defined by φ(a) = (1, 0) and φ(b) = (0, 1) induces an isomorphism F (a, b)/ aba 1 b 1 Z 2. Proof. Let N F (a, b) be the normal subgroup generated by aba 1 b 1. By the first isomorphism theorem, F (a, b)/ker(φ) Z Z, so it suffices to show that N = ker(φ). Since aba 1 b 1 ker(φ), N ker(φ). Since N K, it suffices to show that K/N is trivial. Let gn K/N, so g = a n1 b m1 a n2 b m2 a n k. Since abn = ban, gn = a n1+ +n k b m1+ +m k N, but g K, so Σn i = 0 and Σm i = 0, so gn = en and K/N is tirivial. Example. The Klein bottle K is a quotient of I 2. Then π 1 (K) F (a, b)/ aba 1 b. Letting c = a 1 b allows us to rewrite this as π 1 (K) F (a, c)/ a 2 c 2. You can see this geometrically by cutting the square along the diagonal and repaste the triangles along the previously flip-identified edges graphs. Definition. A graph is a 1-dimensional CW complex. A tree is a connected graph such that it is not possible to start at vertex v 0, travel along successive edges, and arrive back at v 0 without using the same edge twice. Example. some examples and nonexamples, K 5, K 3,3, etc Proposition Any tree is contractible. If v 0 is a vertex of the tree T, then v 0 is a deformation retract of T. Proof. Suppose the tree is finite and induct on the number of edges. If T has one edge, it is homeomorphic to I. Assume any tree with n edges deformation retracts onto any vertex and let T be a tree with n + 1 edges. Let v 0 T and let v 1 T be a vertex that is maximally far away from v 0 in terms of number of edges traversed. Then v 1 is the endpoint of a unique

5 MATH730 5 edge e, which we can deformation retract onto its other endpoint. The result is a tree with n edges, which deformation retracts onto v 0 by the IH. Corollary Any tree is simply connected. Definition. If X is a graph and T X is a tree, we say that T is a maximal tree if it is not contained in any other larger tree. By Zorn s lemma, any tree is contained in some maximal tree. (see 1A.1) Theorem Let X be a connected graph and let T X be a maximal tree. The quotient space X/T is a wedge of circles, one for each edge of X T. The quotient q X X/T is a homotopy equivalence. Proof. The maximal tree T contains every vertex, so all edges in the quotient are loops. Define b X/T S 1 X by defining maps out of each wedge summand. Fix a vertex v 0 T and pick a deformation retraction of T to v 0. Then for each vertex v, the homotopy provides a path α v from v 0 to v. Given an edge in X T from v 1 to v 2, send the summand S 1 to the loop αv 1 2 eα v1. The composition q b on a wedge summand S 1 looks like c v0 id c v0 which is homotopic to the identity. The other composition b q is more complicated. (skip) Corollary The fundamental group of a graph is a free group. Theorem Let p E B be a covering map, where B is a connected graph. Then E is also a graph. Proof. A graph is the space obtained by gluing a set of edges to a set of vertices. Let B 0 be the set of vertices of B and let B e be the set of edges. Let E 0 E be p 1 (B 0 ) and define E e B e E 0 to be the set of pairs {(α {0, 1} B 0, v)} such that α(0) = p(v). Then we have compatible maps E 0 E and Ee I E (given by unique path lifting). These then give a map from the pushout E 0 I Ee I E. The pushout is a 1-dimensional CW complex and we will show this map is a homeomorphism. (surjective) Let x E. Then p(x) is in some 1-cell β of B. Pick a path γ in B from p(x) to a vertex b 0 which remains in β. Then γ lifts uniquely to a path γ starting at x in E. If v = γ(1), then x lives in the 1-cell (β, v), so f is surjective. (injective) (homeo) The map from the pushout to E is a map of covers which induces a bijection on fibers, so it is an isomorphism of covers. Theorem Any subgroup H of a free group G is free. Proof. Define B to be a wedge of circles, one for each generator of G. Then π 1 (B) G. Let H G and let p E B be a covering such that p (π 1 (E)) = H. By the previous result, E is a graph, so π 1 (E) is a free group.