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Math 527 - Homotopy Theory Spring 2013 Homework 3 Solutions Problem 1. An H-space (named after Hopf) is a pointed space (X, e) equipped with a “multiplication” map µ : X × X → X such that the basepoint e is a two-sided unit up to pointed homotopy. In other words, both maps µ(e, −) : X → X µ(−, e) : X → X and pointed-homotopic to the identity map idX . Note that µ is not assumed to be associative, not even up to homotopy. Show that the fundamental group π1 (X, e) of an H-space is abelian. Solution. Let us show a stronger result: The π1 (X, e)-action on πn (X, e) is trivial for all n ≥ 1. Let [γ] ∈ π1 (X, e) and [θ] ∈ πn (X, e) be represented by pointed maps γ : S 1 → X and θ : S n → X (by abuse of notation), or equivalently, maps of pairs γ : (I, ∂I) → (X, e) and θ : (Dn , ∂Dn ) → (X, e). Consider the continuous map H : Dn × I → X defined by H(z, s) = µ (θ(z), γ(s)) . Then H restricted to the bottom face is H|Dn ×{0} = θe whereas H restricted to the remaining faces is H|∂Dn ×I∪Dn ×{1} = (eγ) · (θe) so that H provides a pointed homotopy between θe and (eγ) · (θe). Because right multiplication by e is pointed-homotopic to the identity of X, the composite S n θ / µ(−,e) X / 7 X θe is pointed-homotopic to θ, yielding the equality [θe] = [θ] in πn (X, e). Likewise, the equality [eγ] = [γ] holds in π1 (X, e). We obtain the equality [γ] · [θ] = [eγ] · [θe] = [θe] = [θ] in πn (X, e). 1 Problem 2. Let f : X → Y be a map of spaces, and x ∈ X any basepoint. Show that the induced map πn f : πn (X, x) → πn (Y, f (x)) for n ≥ 1 is a map of π1 -modules, in the sense that it is π1 f -equivariant. More precisely, for any γ ∈ π1 (X, x) and θ ∈ πn (X, x) the equation (πn f )(γ · θ) = (π1 f )(γ) · (πn f )(θ) holds in πn (Y, f (x)). Solution. The equation to be proved can be written as the commutative diagram • / π1 (X, x) × πn (X, x) π1 f ×πn f • π1 (Y, f (x)) × πn (Y, f (x)) πn (X, x) / (1) πn f πn (Y, f (x)) • − πn (X, x) is obtained by applying the functor Recall that the action map π1 (X, x) × πn (X, x) → op [−, X]∗ : Top∗ → Set∗ to the coaction map c : S n → S 1 ∨ S n . The map f : (X, x) → (Y, f (x)) in Top∗ yields the postcomposition natural transformation f∗ : [−, X]∗ → [−, Y ]∗ . Applying f∗ to the coaction map c yields the commutative right-hand square of the diagram [S 1 , (X, x)]∗ × [S n , (X, x)]∗ o ∼ = [S 1 ∨ S n , (X, x)]∗ f∗ ×f∗ [S 1 , (Y, f (x))]∗ × [S n , (Y, f (x))]∗ o ∼ = c∗ f∗ / [S 1 ∨ S n , (Y, f (x))]∗ c∗ [S n , (X, x)]∗ / (2) f∗ [S n , (Y, f (x))]∗ where the left-hand square also commutes, since the wedge is the coproduct in Top∗ and in Ho(Top∗ ). But the outer diagram in (2) is precisely the diagram (1). 2 Problem 3. Let X be the topologist’s sine curve: 1 X = {0} × [−1, 1] ∪ {(x, sin ) | 0 < x ≤ 1}. x Consider the map f : S 0 → X which picks out the points (0, 1) and (1, sin 1). Show that this map f is a weak homotopy equivalence but not a homotopy equivalence. Solution. Write A = {0} × [−1, 1] and B = {(x, sin x1 ) | 0 < x ≤ 1} with X = A ∪ B where the union is disjoint. Recall that X is connected (being the closure of the connected subset B ⊂ R2 ), but not path-connected. The two path components of X are A and B. f is a weak homotopy equivalence. Write a := (0, 1) ∈ A and b := (1, sin 1) ∈ B, and S 0 = {∗a , ∗b }, with f (∗a ) = a and f (∗b ) = b. The map f : S 0 → X induces a bijection on the ' sets of path components π0 f : π0 (S 0 ) − → π0 (X) = {[a], [b]}. Since S n is path-connected for all n ≥ 1, any pointed map α : S n → (X, a) lands inside the path component A ⊆ X. But A is contractible, so that πn (A, a) = 0 and α : S n → (A, a) is ' pointed-null-homotopic. This proves πn (X, a) = 0. Therefore πn f : πn (S 0 , ∗a ) − → πn (X, a) is an isomorphism (between trivial groups!) for all n ≥ 1. ' Likewise, B is contractible, so that πn f : πn (S 0 , ∗b ) − → πn (X, b) is also an isomorphism (between trivial groups) for all n ≥ 1. f is not a homotopy equivalence. Let g : X → S 0 be any continuous map. Since X is connected, the image g(X) is connected and is therefore a singleton {∗a } or {∗b }. Hence g cannot induce a bijection on π0 , and thus f has no homotopy inverse. 3