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

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)