MATH 547 ALGEBRAIC TOPOLOGY HOMEWORK ASSIGNMENT 4

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1 MATH 547 ALGEBRAIC TOPOLOGY HOMEWORK ASSIGNMENT 4 ROI DOCAMPO ÁLVAREZ Chapter 0 Exercise We think of the torus T as the quotient of X = I I by the equivalence relation generated by the conditions (, s) (, s) (s, ) (s, ) s I (note that in this case I denotes the interval [, ]). If X denotes the boundary of X in R 2, then X/ is homeomorphic to S S (where is as above). The torus with a point removed T p} is homeomorphic to (X (0, 0)})/. The boundary X is a deformation retract of X (0, 0)}. The retraction is given by x f t (x) = ( t)x + t, x where (a, b) = max a, b }, as usual. The retraction f t is compatible with the equivalence relation (in the sense that f t (x) f t (y) whenever x y). Therefore it induces a deformation retraction g t from (X (0, 0)})/ to X/, in other words from T p} to S S (note that g t is given by g t ([x] ) = [f t (x)] ). Chapter 0 Exercise 9 Let X be a contractible space and A be a retract of X. Denote by f t the contraction of X (the homotopy between id X and a constant c x0 ), by r the retraction of X onto A, and by i the inclusion of A into X. Consider g t = r f t i. Then G(a, t) = g t (a) is continuous, since it can be factored as a composition of continuous maps: A I i idi where F (x, t) = f t (x). Moreover: G X I F X r A g 0 = r f 0 i = r id X i = r i = id A g = r f 0 i = r c x0 i = c r(x0) Therefore g t defines a homotopy between id A and a constant, and we conclude that A is contractible.

2 Chapter 0 Exercise 6 We think S n as the one point compactification of R n. From this point of view, the inclusion S n S n+ is just the inclusion of R n } inside R n+ } given by the hyperplane x n+ = 0 (and that identifies the points at infinity). Therefore S = n S n is homeomorphic to R }, where R is the union n R n equipped with the weak topology. Although S n is not contractible, it is possible to deform it into a point if we allow to move the space inside S n+. The equations for this deformation are very easy if we use the notation of R n }: R n } is defined inside R n+ } by the condition x n+ = 0; the contraction is obtained by making the coordinate x n+ go to infinity. The idea is to combine all these maps to get the contraction of S. First we define the inclusion of S into S + : ϕ : R } R } (x, x 2,...) (0, x, x 2,...) ϕ is continuous, since R } has the weak topology an the restriction maps ϕ R n } : R n } R n+ } are continuous. Moreover, using the convexity of R, we can easily find a homotopy between ϕ and the identity: f t ( ) = f t (x) = ( t)x + tϕ(x) if x R Again, if we restrict to R n } we can check that f t (x) is continuous (in both x and t). After shifting the first coordinate, we can make everything go to infinity: ϕ t : R } R } (x, x 2,...) t ( t, x, x 2,...) and ϕ = c. These maps are continuous (to check restrict again to the finite level), and they define a homotopy between ϕ = ϕ 0 and ϕ = c. Since the identity of S is homotopic to ϕ, and ϕ is nullhomotopic, we can conclude that S is contractible. Chapter Section. Exercise 6 Assume that f and g are paths in X such that f g is a loop. Then we claim that f g and g f define the same homotopy class in [S, X]. To see this, denote by h 0 and h the maps from S to X corresponding to f g and g f respectively. h 0 (e 2πis f(2(s k)) if s k + [0, / 2 ] for some k Z ) = g(2(s k) ) if s k + [ / 2, ] for some k Z h (e 2πis g(2(s k)) if s k + [0, / 2 ] for some k Z ) = f(2(s k) ) if s k + [ / 2, ] for some k Z Note that h (z) = h 0 (ze πi ) = h 0 ( z). Then the homotopy between h 0 and h is given by h t (z) = h 0 (ze πit ). 2

3 Assume that X is path-connected and let [f ] be a class in [S, X]. Since X is path-connected, we can find a path h joining y = f() and x 0. Then h f h is a loop based at x 0, and [[h f h] = Φ([h f h]) is in the image of Φ. But, by the previous result, [f ] = [f h h] = [h f h], and we see that Φ is surjective. If [f] and [g] are conjugate in π (X, x 0 ), then [f] = [h g h] for some loop h, and we can use the claim to show that: Φ([f]) = Φ([h g h]) = [h g h] = [g h h] = [g ] = Φ([g]) Conversely, assume that Φ([f]) = Φ([g]). Then [f ] = [g ] and we can find a homotopy h t between f and g. Define γ t (s) = h ts (). Then γ = γ is a loop that joins f() = x 0 with g() = x 0, and f is homotopic (as a loop) to γ g γ. The homotopy of loops is given by f t = γ t h t γ t. Therefore [f] = [γ g γ] = [γ][g][γ], that is, [f] is conjugate to [g] in π (X, x 0 ). Chapter Section. Exercise 3 Assume that the homomorphism i induced by the inclusion i : A X is surjective, and let f be a path in X with endpoints a 0 and a in A. Since A is path-connected, we can find a paths h 0, h inside A from a 0 and a to x 0. The composition (i h 0 ) f (i h ) is a loop in X centered at x 0. Since i is surjective, there is a path g in A such that [i g] = i ([g]) = [(i h 0 ) f (i h )]. In other words, f is homotopic to (i h 0 ) (i g) (i h ) = i (h 0 g h ), which is a path in A. Conversely, if every path in X with endpoints in A is homotopic to a path in A, every loop f in X based at x 0 A is homotopic to a loop of the form i (g) = i g for some loop g in A. Hence every class [f] of π (X, x 0 ) has a preimage for the map i, and i is surjective. Chapter Section. Exercise 6 d. A retraction from D 2 D 2 to S S would define a surjective homomorphism from π (D 2 D 2 ) = 0 to π (S S ) = Z Z, which is impossible (Z Z 0). e. Let X = D 2 /x, x} be a disk with two points on its boundary identified. Let Y be the diameter joining x and x. Then Y is a deformation retract of the disk, and this deformation is compatible with the identification of the two points. Therefore X is homotopy equivalent to Y/x, x} S. A retraction from X to its boundary S S would define a surjection from π (X) = π (S ) = Z to π (S S ) = Z Z, which is impossible (Z Z is nonabelian, and the image of an abelian group is always abelian). f. Define Y = [0, ] [, ] and consider the equivalence relation generated by (0, s) (, s) s [, ]. Then the Möbius band X can be obtained as the quotient space Y/. The maps f t (a, b) = (a, ( t)b) define [0, ] 0} as a deformation retract of Y, and induce a homotpy equivalence between X and the central circle S ([0, ] 0})/ 3

4 (given by f : [(a, b)] [(a, 0)] = e 2πia ). The map i : A X e 2πit [(2t, )] if t [0, / 2 ] [(2t, )] if t [ / 2, ] defines the inclusion of the boundary circle A S in X. The induced map i : Z = π (A) Z = π (X) is determined by the image of the generator of π (A). This generator is [ω ], where ω (t) = e 2πit. Then: [(2t, 0)] = e 2πi2t if t [0, / 2 ] f (i(ω (t))) = ω 2 (t) = [(2t, 0)] = e 2πi(2t ) = e 2πi2t if t [ / 2, ] and we see that i ([ω ]) is twice the generator of π (X). In other words, i is just multiplication by 2 : i (n) = 2n. Assume that there is a retraction r : X A. Then the map r : Z Z verifies which is impossible. 2 r () = r (2) = r (i ()) = (r i) () = id () = id() =, Chapter Section. Exercise 7 Let Y and Y 2 be two copies of S and X = Y Y 2. Points of Y i are written in the form e 2πis i, and we assume that e 0 is identified with e 0 2 in X. The fundamental group of X (based at e 0 = e 0 2) is isomorphic to Z Z. The generators are a = [f] and b = [g] where f(t) = e 2πit and g(t) = e 2πit 2. Consider the maps: ϕ n : X Y e 2πis e 2πis e 2πis 2 e 2πins Then each ϕ n is a retraction of Y, and the induced maps are given by: (ϕ n ) : Z Z Z a b n Since the induced maps are different, the retractions ϕ n are nonhomotopic. Chapter Section.2 Exercise 3 Let A = p,..., p k } be a finite set in R n (with n 3), and X its complement. Clearly X is path-connected, so to prove that it is simply-connected we only have to show that it has trivial fundamental group. Let B be a closed ball in R n and x and y any two points in the interior of B. Then the boundary B of B is a deformation retract of both B x} and B y}, and hence there is a homotopy equivalence between B x} and B y} that fixes B. This can be understood in the following way: we can move the points p,..., p k in small neighborhoods without changing the homotopy type of X. Therefore we will assume that p i = (i, 0,..., 0). 4

5 Let Y i be the closed ball of center p i and radius / 2 with respect to the norm (x,..., x n ) = max x,..., x n }. Let Z be the closure of the complement of i Y i. Then the boundary Y i is a deformation retract of Y i p i } for i =,..., k, and Z is a deformation retract of Z. We can combine all these deformations and we obtain that i Y i = ( i Y i ) Z is a deformation retract of X = ( i Y i p i }) Z. But i Y i is homotopy equivalent to the wedge sum of k copies of S n, which is simply-connected since n 3. Therefore the fundamental group of X is trivial. Chapter Section.2 Exercise 5 Let X be a finite graph in R 2 whose edges are straight segments. Consider R 2 embedded in S 2 = R 2 }, and denote by Y 0, Y,..., Y k the connected components of S 2 X. Since X R 2, one of these components contains ; assume it is Y 0. Therefore Y,..., Y k are the bounded complementary regions of X and Y 0 } is the unbounded one. One version of the Jordan curve theorem states that any closed curve in S 2 separates S 2 into a collection of disjoint regions, each one isomorphic to an open disk (and if the curve is simple closed, the number of such regions is exactly two). Y i is one of the regions determined by the curve Y i, therefore each Y i is homeomorphic to an open disk. Set p 0 = and choose points p i Y i for i =... k. Since Y i is homeomorphic to an open disk, each boundary Y i is a deformation retract of Y i p i }. Combining all these retractions, we see that X = i Y i is a deformation retract of S 2 p 0,..., p k } = R 2 p,..., p k }. Denote by r : R 2 p,..., p k } X this retraction. Choose any basepoint z 0 in Z = R 2 p,..., p k }. Since Z has the homotopy type of the wedge sum of k copies of S, the fundamental group π (Z, z 0 ) is free of rank k. A collection of generators is obtained in the following way: choose small disks centered at p,..., p k and consider loops g,..., g k parameterizing the boundaries of these disks; choose paths h 0,..., h k in Z from z 0 to g (0),..., g k (0); consider f i = h i g i h i ; then [f ],..., [f k ] generate the fundamental group. Set x 0 = r(z 0 ). Since r : Z X is a deformation retraction, π (X, x 0 ) has generators [r f ],..., [r f k ], and r f i = (r h i ) (r g i ) (r h i ). Recall that g i is the boundary of a small disk centered in p i, and that r maps the punctured disk Y i p i } to its boundary Y i. If the radius of the disk of g i is chosen small enough, clearly r g i is a path parameterizing the boundary Y i. Since we have a lot of freedom in the choice of h 0,..., h k, we can view r h 0,..., r h k as arbitrary paths joining the basepoint x 0 to the boundaries Y,..., Y k. To sum up, if X is a finite plane graph whose edges are straight segments, and Y,..., Y k are the bounded complementary regions of X, then π (X, x 0 ) is free of rank k with generators [γ 0 Y 0 γ 0 ],..., [γ k Y k γ k ], where γ 0,..., γ k are paths in X joining x 0 to Y,..., Y k. 5