Math 527 - Homotopy Theory Spring 2013 Homework 3 Solutions ...

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 : (D n , ∂D n ) →
(X, e).
Consider the continuous map H : D n × I → X defined by
Then H restricted to the bottom face is
whereas H restricted to the remaining faces is
H(z, s) = µ (θ(z), γ(s)) .
H|D n ×{0} = θe
H|∂D n ×I∪D n ×{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
X
θe
µ(−,e)
X
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
in πn(X, e).
[γ] · [θ] = [eγ] · [θe]
= [θe]
= [θ]
1

Problem 2. Let f : X → Y be a map of spaces, and x ∈ X any basepoint. Show that the
induced map
πnf : πn(X, x) → πn(Y, f(x))
for n ≥ 1 is a map of π1-modules, in the sense that it is π1f-equivariant. More precisely, for
any γ ∈ π1(X, x) and θ ∈ πn(X, x) the equation
holds in πn(Y, f(x)).
(πnf)(γ · θ) = (π1f)(γ) · (πnf)(θ)
Solution. The equation to be proved can be written as the commutative diagram
π1(X, x) × πn(X, x)
π1f×πnf
π1(Y, f(x)) × πn(Y, f(x))
•
•
πn(X, x)
πnf
πn(Y, f(x))
Recall that the action map π1(X, x) × πn(X, x) • −→ πn(X, x) is obtained by applying the functor
[−, X]∗ : Top op
∗ → 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)]∗
f∗×f∗
[S 1 , (Y, f(x))]∗ × [S n , (Y, f(x))]∗
∼=
∼=
[S 1 ∨ S n , (X, x)]∗
f∗
[S 1 ∨ S n , (Y, f(x))]∗
c ∗
c ∗
[S n , (X, x)]∗
f∗
[Sn , (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
(1)
(2)

Problem 3. Let X be the topologist’s sine curve:
X = {0} × [−1, 1] ∪ {(x, sin 1
) | 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 1)
| 0 < x ≤ 1} with X = A ∪ B where
x
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 π0f : 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 πnf : πn(S 0 , ∗a) −→ πn(X, a) is an
isomorphism (between trivial groups!) for all n ≥ 1.
Likewise, B is contractible, so that πnf : 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