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Topology Midterm 3 Solutions (1) Let Xi (i = 1, 2) be topological spaces and let xi ∈ Xi be an arbitrary point. Let X1 × X2 be the product, with the product topology. Let πi : X1 × X2 → Xi be the projections. Prove that π1∗ × π2∗ : π1 (X1 × X2 , (x1 , x2 )) → π1 (X1 , x1 ) × π1 (X2 , x2 ) is an isomorphism of groups. Solution: We saw in class that each πi∗ is a well-defined group homomorphism, so the map in question is certainly a well-defined group homomorphism (by the universal property of the product group, if you like). For injectivity, suppose γ and γ 0 are loops in X1 × X2 based at (x1 , x2 ) with ([π1 γ], [π2 γ]) = ([π1 γ 0 ], [π2 γ 0 ]). Then we have a homotopy (of loops based at xi ) Hi : I × I → Xi from πi γi to πi γ 0 (i = 1, 2). By the universal property of the product X1 × X2 , the map H := H1 × H2 : I × I → X1 × X2 is continuous and is clearly a homotopy (of loops based at (x1 , x2 )) from γ to γ 0 . Surjectivity is similar: ([γ1 ], [γ2 ]) is the image of γ1 × γ2 . (2) Prove that π1 (S 2 ) = {1}. You can use, without proof, the fact that any loop in S 2 is homotopic (as a loop) to a loop which is not surjective. (This fact is not so obvious! There are loops in S 2 which are surjective.) Solution: By the hint, any loop is homotopic to a loop missing some point x ∈ S 2 and S 2 \ {x} ∼ = R2 . Since every loop in R2 is contractible (by straight-line homotopy), we conclude that every loop in S 2 is contractible. (3) Prove that S 1 × S 1 is not homeomorphic to S 2 . Solution: By the result of Problem 1 and the computation π1 (S 1 ) ∼ = Z in 1 1 ∼ 2 2 class, we have π1 (S × S ) = Z , while π1 (S ) = 0, so the spaces in question have non-isomorphic fundamental groups. Choose any two of these last four problems and do only those two! P 2 (4) The group {±1} acts on S n = {x ∈ Rn+1 : n+1 i=1 xi = 1} by ±1 · x = ±x. 1 Prove that S /{±1} (with the quotient topology) is homeomorphic to S 1 . We noted in class that the projection map S 2 → S 2 /{±1} =: RP2 is a covering space. Use your knowledge of covering spaces and the fundamental group to prove that π1 (RP2 ) = {±1} is the group with two elements. Conclude that S 2 is not homeomorphic to RP2 . Solution: Let q : S 1 → S 1 /{±1} be the quotient map. Making the standard identification R2 = C, we can view S 1 as the set of complex numbers of magnitude 1. Then f (z) := z 2 is a continuous map f : S 1 → S 1 constant on the fibers of q, so, since q is a descent map, there is a unique continuous map g : S 1 /{±1} → S 1 such that gq = f . This map g is bijective since z 2 = w2 iff z = ±w, hence it is a homeomorphism because its domain is compact (it is the image of the compact space S 1 under q) and its codomain is Hausdorff. 2 (5) For n ∈ Z>0 , let Sn1 ⊆ R2 be the circle of radius 1/(2n) centered at (1/(2n), 0). Let X ⊆ R2 be the Hawaiian earring (the union of the Sn1 ). Prove that there does not exist a covering space f : X̃ → X with X̃ simply connected. Solution: No one did this right. Let f : X̃ → X be a covering space. Take some neighborhood` U of x0 := (0, 0) in X such that f −1 (U ) is a (nonempty!) disjoint union i Ui of open subsets of X̃ with f |Ui : Ui → U a homeomorphism for each i. Pick n large enough that Sn1 ⊆ U . Define r : U → Sn1 by setting r(x) := x if x ∈ Sn1 ⊆ U , r(x) := x0 otherwise. As we noted in class, this r is a continuous retract of the inclusion Sn1 ,→ U . Now pick any i and let x̃0 ∈ Ui be the unique lift of x0 in Ui . Let S̃n1 be the lift (preimage) of Sn1 in Ui . Let r̃ : Ui → S̃n1 be the retract of S̃n1 ,→ Ui corresponding to r under the homeomorphisms Ui ∼ = U , S̃n1 ∼ = Sn1 . The compact space S̃n1 ∼ = S 1 is closed in X̃ since X̃ is Hausdorff (it is easy to see that any covering space of a Hausdorff space is Hausdorff). Let g : X̃ \ S̃n1 → S̃n1 denote the constant map onto the point r̃(x̃0 ) ∈ S 1 . The maps g and r̃ are defined on open subspaces of X̃ covering X̃ and they agree on the overlap, so they glue to a continuous map s : X̃ → S̃n1 ∼ = S 1 which is clearly a retract of the inclusion S 1 ∼ = S̃n1 ,→ X̃, hence the fundamental group of X̃ must surject onto π1 (S 1 ) ∼ = Z, so X̃ can be simply connected. (6) A topological group is a set G equipped with a group structure and a topology which are compatible in the sense that the multiplication map G × G → G and the inverse map G → G are continuous. Given two loops γ, ν based at 1 ∈ G, define a new loop γ ⊗ ν based at 1 by (γ ⊗ ν)(t) := γ(t) · ν(t). (This is continuous because multiplication is continuous.) (a) Show that ⊗ is well-defined on homotopy classes (of loops based at 1) and that (π1 (G, 1), ⊗) is a group. (b) Show that, in fact, the group operation ⊗ is the same as the usual “concatenation of loops” group operation on π1 (G, 1). (c) Show that π1 (G, 1) is abelian. Solution: Let 1 : I → G denote the constant loop based at 1 ∈ G. (a) If H is a homotopy (of loops in G based at 1) from γ to γ 0 and K is a homotopy from ν to ν 0 , then H · K is a homotopy from γ ⊗ ν to γ 0 ⊗ ν 0 . Evidenlty 1 is the identity for ⊗. Associativity of ⊗ follows from associativity of · and the existence of inverses for ⊗ follows from the existence of a continuous inverse map for G (γ −1 (t) := γ(t)−1 ). (b) It is clear from the definitions that (γ ∗ 1) ⊗ (1 ∗ ν) = γ ∗ ν (on the nose, not just up to homotopy). Since γ ∼ γ ∗ 1 and ν ∼ 1 ∗ ν, the equality [γ ∗ ν] = [γ ⊗ ν] now follows from the invariance established in (a). 3 (c) Just compute [ν] ∗ [γ] = [1 ∗ ν] ∗ [γ ∗ 1] = [1 ∗ ν] ⊗ [γ ∗ 1] = [γ] ∗ [ν]. (7) Suppose f : X̃ → X is a covering space with X̃ simply connected. Fix a base point x0 ∈ X and a point x̃0 ∈ f −1 (x0 ). Prove that [γ] 7→ γ̃(1) (c.f. the Path Lifting Lemma) is a well-defined bijection from the set of homotopy classes of paths in X based at x0 to X̃. (This is supposed to motivate the general “universal cover” construction we discussed in class.) Solution: Well-definedness follows from the Homotopy Lifting Lemma (using uniqueness in the Path Lifting Lemma) and the fact that each fiber f −1 (x) is totally disconnected, so there are no non-constant paths in any such fiber. Injectivity follows from simple-connectedness of X̃: If γ̃(1) = ν̃(1), then there must be a homotopy (of paths) H from γ̃ to ν̃, else γ ∗ ν would be a nontrivial loop in X̃, then f H is a homotopy of path from γ to ν. Surjectivity follows from path-connectedness of X̃.