Math 440 Problem Set 2

Math 440 Problem Set 2 Problem 4, p. 52. Let X R 3 be the union of n lines through the origin. Compute π 1 (R 3 X). Solution: R 3 X deformation retracts to S 2 with 2n points removed. Choose one of them. Stereographic projection away from the chosen point gives a homeomorphism between S 2 with 2n points removed and R 2 with 2n 1 points removed. We will show that this space is homotopy equivalent to a wedge of 2n 1 circles. First, note that R 2 with 2n 1 points removed is homeomorphic to R 2 with any other set of 2n 1 points removed. This is because whenever we have a smooth non-intersecting path ϕ: [0, 1] R 2 between two points x and y, we can find a homeomorphism between R 2 {x} and R 2 {y} that fixes all points outside an arbitrarily small neighborhood around ϕ([0, 1]). In particular, if n = 1, then we can choose our missing point to be the origin. As mentioned in class, the map sending each point x R 2 {0} to the point x/ x S 1 is part of a homotopy equivalence between R 2 {0} and S 1. We use a similar idea when we re dealing with multiple missing points: we can choose our set of 2n 1 points to be the points C = {(1, 0), (2, 0),..., (2n 1, 0)}, and the space R 2 C deformation retracts to the collection of circles with radii 1/2 centered at the points of C. So, R 3 X is homotopy equivalent to the (2n 1)-fold wedge sum S 1 S 1 S 1, whose fundamental group is the free group on 2n 1 generators. Problem 6, p. 53. Suppose a space Y is obtained from a path-connected subspace X by attaching n-cells for a fixed n 3. Show that the inclusion X Y induces an isomorphism on π 1. Apply this to show that the complement of a discrete subspace of R n is simply-connected if n 3. Solution: We will argue that at each step of adding a n-cell to X we obtain a space Y which has an isomorphic homotopy group to that of X. We start with the case n = 3. Let f : D 3 X be the glueing map between the 3-cell and the space X. Our glued space will be denoted as Y = X f D 3 We define open sets: U = X f D 3 \ x 0 and V = S 2 [0, 1], where x 0 is a point of D 3 not contained in X; and V is a slice of D 3 not intersecting X and containing x 0. We have the following isotopies: (i) U X, by deformation retract of D 3 minus a point to the disk f(d 3 ) X; (ii) V S 2

by deformation retract of its [0, 1] factor to a point and (iii) U V S 2 by the same argument as (ii). Now since we have U V = Y, we can apply Van Kampen s theorem to calculate its fundamental group. π 1 (Y ) π 1 (U) π 1 (V ), and we remark the subgroup by which one need to quotient is the trivial group, because π 1 (U V ) π 1 (S 2 ) e. But π 1 (V ) π 1 (S 2 ) e, so we conclude: π 1 (Y ) π 1 (U) π 1 (X), and that is what we wanted to prove. For the general case, we just repeat this argument changing S 2 by S (n 1) and D 3 by D n. All works fine, because the homotopy groups π 1 (D n \ x 0 ) and π 1 (S (n 1) ) are trivial for n 3. Thus, applying this reasoning for each cell attached to X and composing the isomorphims we obtain the result that was to be proved. The last statement is proved this way. Let {x i } i I be the discrete subset of R n. We start with X the deformation retract of the complement of x 0 to S (n 1). Since n 3 its homotopy group is trivial. Now we attach for each x i an n-cell around it to X, and repeat this process a possibly (countably) infinite number of times. From the first part the homotopy group of the final space Y is isomorphic to that of X, thus to the trivial group. But the space this way obtained is clearly a deformation retract of the complement of {x i } i I, so we are done. Problem 7, p. 53. Let X be the quotient space of S 2 obtained by identifying the north and south poles to a single point. Put a cell complex structure on X and use this to compute π 1 (X). Solution: There exists a cell complex structure on S 2 consisting of two 0- cells (the north and south poles), one 1-cell (a prime meridian γ), and one 2-cell (the remainder of the surface). By identifying the north and south poles to a single point, we get a corresponding cell complex structure of X consisting of: one 0-cell, the north/south pole, one 1-cell, the meridian γ, and one 2-cell, attached along the path γ γ. By Proposition 1.26, we have that π 1 (S 2 ) = π 1 (X)/N, where N = γ = Z. Since π 1 (S 2 ) is trivial, it follows that π 1 (X) = Z.

Problem 8, p. 53. Compute the fundamental group of the space obtained from two tori S 1 S 1 by identifying a circle S 1 {x 0 } in one torus with the corresponding circle S 1 {x 0 } in the other torus. Solution: Let X denote the space in question. Let A and B be the two tori, viewed as subsets of X. Let the basepoint x 0 of X be contained in the intersection of A and B, and make this basepoint implicit in the notation. Let U be the torus A, together with a small neighborhood of A B in B which deformation retracts onto A B. Define V similarly, with the roles of A and B interchanged. Then U and V deformation retract onto A and B, respectively so π 1 (U) = π 1 (A) = Z 2, π 1 (V ) = π 1 (B) = Z 2. The set U V deformation retracts onto the circle A B, so π 1 (U V ) = π 1 (A B) = Z. A generator [γ] of π 1 (A B) corresponds under the inclusion i : A B A to a loop which winds once around one S 1 factor of A and under the inclusion j : A B B to a loop which winds once around one S 1 factor of B. Now, π 1 (A) is free abelian on two generators a and b, corresponding to loops which wind once around its two S 1 factors and are constant in the other factor. Therefore the inclusion i induces the map i : [γ] a. Similarly, if c and d generate π 1 (B) = π 1 (U), then j : [γ] c. By the deformation retractions discussed above, the maps induced by the inclusions of U V into U and V have the same descriptions. Thus, by Van Kampen s theorem, π 1 (X) = π1 (U) π1 (U B) π 1 (V ) = a, b, c, d : ab = ba, cd = dc, a = c = a, b, d : ab = ba, ad = da.

Problem 11, p. 53 The mapping torus T f of a map f : X X is the quotient of X I obtained by identifying each point (x, 0) with (f(x), 1). In the case X = S 1 S 1 with f basepoint-preserving, compute a presentation for π 1 (T f ) in terms of the induced map f : π 1 (X) π 1 (X). Do the same when X = S 1 S 1. Solution: First consider the case when X = S 1 S 1. To form the mapping torus T f, one takes the cross-product of S 1 S 1 with the interval [0, 1], identifying the face S 1 S 1 {0} with S 1 S 1 {1} via the map f, and as f preserves basepoints, the interval [0, 1] with this identification becomes a loop homeomorphic to S 1 attached to S 1 S 1 at the basepoint. This means the space S 1 S 1 S 1 is a subset of T f. Further, to get the rest of T f, we must attach two 2-cells to this wedge product as follows. Suppose a and b are the generators of X. That is, a generates one of the copies of S 1, and b, the other. Let c represent the loop attached to a and b at the basepoint, which we have from the interval [0, 1] with endpoints identified. We attach the first 2-cell wrapping its edges first around a, meeting itself at the basepoint on the face X {0}, but having only outlined one S 1 {0}, then going along c back to the basepoint on X {1}, and gluing itself back to X {0} via the map f a 1, and finally going back along c 1 to where we started. We wrap the second 2-cell to outline the other S 1 {0}, repeating the same process and gluing it back to S 1 {1} exactly as before, only instead going around b, c, f (b) 1 and c 1. Thus by Proposition 1.26 in Hatcher, π 1 (T f ) = < a, b, c acf (a) 1 c 1, bcf (b) 1 c 1 >. The case when X = S 1 S 1 is similar. This time, X is generated by loops a and b, subject to the relation aba 1 b 1 = 1, and again to get the mapping torus, T f, we attach a third generating loop, or 1-cell, c, coming from the interval I with its endpoints identified. To construct T f in full, we now must again add 2-cells as well as one 3-cell. The 3-cell will have no impact on π 1 (T f ), and we attach 2-cells as before. At first glance it seems we must also require f (a)f (b)f (a) 1 f (b) 1 = 1, however, this relation is actually redundant from the relations we already found from attaching 2-cells in the previous discussion. Namely, f (a) = c 1 ac and f (b) = c 1 bc and thus f (a)f (b)f (a) 1 f (b) 1 = c 1 acc 1 bcc 1 a 1 cc 1 b 1 c = c 1 aba 1 b 1 c = 1.

So finally, again by Proposition 1.26, we have π 1 (T f ) = < a, b, c aba 1 b 1, acf (a) 1 c 1, bcf (b) 1 c 1 >. Problem 15, p. 54. Given a space X with base point x 0 X, we may construct a CW complex L(X) having a single 0-cell, a 1-cell e 1 γ for each loop γ in X based at x 0, and a 2-cell e 2 τ for each map τ of a standard triangle P QR into X taking the three vertices P, Q, R of the triangle to x 0. The 2-cell e 2 τ is attached to the three 1-cells that are the loops obtained by restricting τ to the three oriented edges P Q, P R, QR. Show that the natural map L(X) X induces an isomorphism π 1 (L(X)) = π 1 (X, x 0 ). Solution: For every 2-cell e 2 τ, we take U to be the interior of e 2 τ, and take V to be (L(X) e 2 τ) ɛ(e 2 τ) where ɛ(e 2 τ) is a small open part inside of e 2 τ. Then U V = S 1 is path connected, so by Van-Kampen s theorem, π 1 (L(X)) = π 1 (U) π 1 (V )/N τ = π 1 (U)/N τ, where τ is τ restricted to the boundary of the triangle. Then inductively, we can take away all 2-cells, and get that π 1 (L(X)) = e γ 1, γ /N τ, τ, i.e. the free group generated by all 1-cells, quotient out the relation generated by all boundary of 2-cells. On the other hand, recall by definition, π 1 (X, x 0 ) = γ, γ /N h, where N h is the equivalence relation of homotopic paths. Therefore, it suffices to see that the relations in N τ, τ and in N h are the same. Clearly, if γ 1, γ 2, γ 3 corresponds to the three 1-cells attached to τ, then e 1 γ 1 e 1 γ 2 = e 1 γ 3 in π 1 (L(X)), and γ 1 γ 2 = γ 3 in π 1 (X, x 0 ) since there is a homotopy inside the triangle P QR which can be mapped to X through τ to get the homotopy from γ 1 γ 2 to γ 3. This tells use that N τ, τ N h. Also, if γ 1 = γ 2 in π 1 (X, x 0 ), then there is a homotopy from γ 1 to γ 2. Now we can map P Q to be γ 1, P R to be γ 2 and QR to be the constant map, so the homotopy will induced a map τ from a standard triangle to X, which means that e 1 γ 1 = e 2 γ 2 in π 1 (L(X)). This tells use that N τ, τ N h. Hence, we conclude that N τ, τ = N h, and consequently π 1 (L(X)) = π 1 (X, x 0 ).

Problem 1a, Groupoid Handout. Show that there is a category Grpd with objects all groupoids and morphisms the functors between groupoids. Solution: This follows easily from the definition of a category, if we let the composition of morphisms be the typical composition of groupoid functors (composition of functors between groupoids is associative; moreover, if f : G H is a groupoid functor, then 1 H f = f 1 G where 1 H and 1 G are the identity functors on H and G respectively). Problem 1b, Groupoid Handout. Show that the assignment X Π X is a functor from spaces to groupoids. Solution: Let F be the function from topological spaces to groupoids such that F(X) = Π X for a topological space X, and F(f) for a continuous map f : X Y is a groupoid morphism F(f) : Π X Π Y defined by: -on objects (points of X) F(f) equals f -on morphisms (homotopy classes of paths in X with fixed endpoints) F takes the morphism [γ] : x y to the path F([γ]) = [f γ] : f(x) f(y) in Y. To see that F is a functor, note that F(id X ) = id ΠX for every X Top, and that F(g f) is defined by: -on objects (points of X) F(g f) = g f -on morphisms F takes [γ] : x y to [(g f) γ] : (g f)(x) (g f)(y) which is F(g) F(f) by definition, so F respects the composition of morphisms. Problem 2, Groupoid a) Show that if F : G H is a part of an equivalence of groupoids, then F induces an isomorphism of groups F : Aut G (x) Aut H (F x) b) Suppose we have homotopy maps f, g : X Y. Show that a choice of homotopy defines a natural transformation of Π(f) Π(g). c) Suppose spaces X and Y are homotopy equivalent (with no assumption on base points), then Π X and Π Y are equivalent. Solution: a) Let K : H G be the other part of the equivalence, then we have two natural transformations η : 1 G KF and χ : 1 H F K. In other

words, for any x G, we have a morphism η x : 1 G (x) KF (x) such that for all morphism f : x y in Hom G (x, y) the following diagram commutes 1 G (x) 1 G(f) 1 G (y) η x η y KF (x) KF (f) KF (y) Consider the following composition of group homomorphisms Aut G (x) F Aut H (F x) K Aut G (KF x). The composition does not get us back to Aut G (x), but we may use the natural transformation η to correct this. By applying the above commutative diagram about η to the morphism x f x in G, we get η x f = KF (f)η x Since H is a groupoid, η x has a double sided inverse ηx 1, thus we have KF (f) = η x fη 1 x. Clearly, conjugation by η x, henceforth Conj(η x ), is a group isomorphism, hence is bijective. Thus we get K,F x F,x = Conj(η x ) where we used subscript x in F,x to remind ourselves that it maps from Aut G (x), etc. Hence K,F x is surjective and F,x is injective. We may repeat the similar argument for We will get a bijection K,F x : Aut H (F x) Aut G (KF x). F,KF x K,F x = Conj(χ F x ),

hence K,F x is injective as well. Combining with the previous result that K,F x is surjective, we see K,F x is bijective thus has an inverse. Hence F,x = K 1,F x Conj(η x) as a composition of two bijection. Thus F,x is a bijective group homomorphism, i.e. an isomorphism. b) Let H be a homotopy from f to g, i.e., a continuous map H H : X I Y, where H(, 0) = f( ) H(, 1) = g( ). Since Π is a functor from the category of topological spaces to the category of groupoids, f and g as morphism in spaces will be mapped to Π(f) and Π(g) as morphisms in the category of groupoids. However groupoids themselves are categories and morphism between groupoids are functors, hence Π(f), Π(g) are functors from Π X to Π Y. Recall that Π(f) acts on Π X in the following way: x Π X : Π(f)(x) = f(x) γ : x y Hom ΠX (x, y) : Π(f)(γ) = f (γ) where I abused notation to let Π X to denote the set of objects it contains and f is the induced map on the fundamental groupoid. For any x Π X, x is a point in X, we may consider the path γ x : I Y, t H(x, t). Then the homotopy equivalence class [γ x ] =: η x defines a morphism in Π Y from f(x) to g(x). We claim that η is the desired natural transformation. For that we just need to check for any morphism ρ : x y in Π X, the following diagram commutes, Π(f)(x) Π(f)(ρ) Π(f)(y) η x η y Π(g)(x) Π(g)(ρ) Π(g)(y).

Recall that the above morphism are just homotopy equivalence classes of paths in the space Y, we just need to find a representative in each equivalence classes and show that the two ways of composition of paths are equal. Let s denote a path in X representing ρ. We get a map of a square to Y : I I s id X I H Y. We can see that paths along the four edges of the square correspond (modulo the equivalence relation) to the four morphisms of the the above diagram in the following way: (0, t) γ x (t), [γ x ] = η x (1, t) γ y (t), [γ y ] = η y (t, 0) f s(t), [f s] = Π(f)(ρ) (t, 1) g s(t), [g s] = Π(g)(ρ) Then Π(g)(ρ) η x = [(g s) γ x ] corresponds to the straight-line path from (0, 0) to (0, 1) followed by the straight-line path from (0, 1) to (1, 1). Similarly η y Π(f)(ρ) = [γ y (f s)] corresponds to the straight-line path from (0, 0) to (1, 0) followed by the straight-line path from (1, 0) to (1, 1). Since I I is contractible and these two compositions of straight-line paths have the same endpoints, they are homotopic. Composing with H (s id), we get a homotopy from (g s) γ x to γ y (f s). Thus Π(g)(ρ) η x = η y Π(f)(ρ) which proves commutativity of the diagram. c) Since X and Y are homotopy equivalent, we have maps f : X Y and g : Y X, such that 1 X gf and 1 Y fg. More precisely, there is a homotopy H : X I X such that H(, 0) = 1 X and H(, 1) = gf and similarly a homotopy G : Y I Y with similar conditions. By part (b), we see there is a natural transformation η from Π(1 X ) to Π(gf). By the functoriality of Π, we know Π(1 X ) = 1 Π(X), Π(gf) = Π(g)Π(f) where they are viewed as morphism in the category of groupoids. Hence we get a natural transformation η : 1 Π(X) Π(g)Π(f)

and similarly a natural transformation χ : 1 Π(Y ) Π(f)Π(g). Thus, Π(X) and Π(Y ) are equivalent groupoids.