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Math 442: Final Project
Cup Products and the Cohomology Ring
Due Thursday, March 17
One reason for discussing the cohomology H ∗ (X; R) of a space X with coefficients in a commutative R is that these groups assemble into a graded commutative ring. It is that ring structure that gives cohomology its impact. The
point of this project is define and discuss this product, and to give some basic
examples. The reference for this material is Chapter 3.2 of Hatcher.
In follows, R is a general, but fixed, commutative ring. If X is a topological
space, C• (X) and C • (X; R) will denote the singular chains and cochains on X.
Thus an element in C n (X; R) can be regarded as a function
α : Sn (X) −→ R
on the set Sn (X) of singular n-simplices of X.
If σ : ∆n → X is a singular n-simplex, recall that we have the first p-face
operator fp σ and the last q-face operator `q σ. For example,
fp σ(t0 , . . . , tp ) = σ(t0 , . . . , tp , 0, . . . , 0)
is the induced singular p-simplex.
Definition: Suppose α ∈ C p (X; R) and β ∈ C q (X; R). Define the cup product
α ` β ∈ C p+q (X; R) by the formula
(α ` β)(σ) = α(fp σ)β(`q σ).
Problem 1. Show that the cup product has the following properties:
1. For all α1 , α2 , and β,
(α1 + α2 ) ` β = α1 ` β + α2 ` β.
2. For all α, β, and γ, we have
(α ` β) ` γ = α ` (β ` γ).
3. If f : X → Y is a continuous map and α and β are cochains on Y , then
f ] (α ` β) = f ] α ∪ f ] β. Here f ] : C • (Y ; R) → C • (X; R) is the induced
map on cochains.
4. If e : S0 X → R is the cocycle so that e(σ) = 1 for all singular 0-simplices,
then e ` α = α ` e = α
Problem 2. The cup product descends to cohomology. Prove that if δ is the
coboundary map on cochains, we have
δ(α ` β) = δα ` β + (−1)p α ` δβ
where p is the degree of α. Conclude that there is a natural associative graded
ring structure on H ∗ (X; R). (You may have to define this notion.)
Remark: On cohomology, the cup product has the following commutativity
property: if x ∈ H p (X; R) and y ∈ H q (Y ; R), then
x ` y = (−1)pq y ` x.
This is certainly not true on the chain level and we won’t prove that here. See
Hatcher p. 215.
Problem 3. Here is a formal property of the cup product. Let X and Y be
two spaces and X t Y their disjoint union. Prove that, as rings
H∗ (X t Y ; R) = H∗ (X; R) × H∗ (Y ; R).
The ring structure on the right is
(α1 , α2 ) ` (β1 , β2 ) = (α1 ` β1 , α2 ` β2 ).
Use this to calculate H∗ (X ∨ Y ; R) where X ∨ Y is the one point union of two
locally contractible spaces.
Problem 4. Here are two specific examples. Assuming that you can use a
∆-set structure on a space and the resulting ∆-chains for computations. (You
can.)
1. Let T be the torus. Find generators x, y of H 1 (T, Z) and z of H 2 (T, Z) so
that
x`x=0=y`y
x ` y = z = −y ` x.
2. Let K be the Klein bottle. Find generators x, y of H 1 (T, Z/2Z) and z of
H 2 (T, Z/2Z) so that
x`x=z=x`y
y ` y = 0.
Problem 5. Show that the torus T and the space S 1 ∨ S 1 ∨ S 2 have the same
cohomology groups but different cohomology rings.
Problem 6. Let Mg be the g-holed torus, g ≥ 1. Calculate the cohomology
ring H ∗ (Mg , Z) and show that the cup product
(−) ` (−) : H 1 (Mq ) × H 1 (Mg ) → H 2 (Mg ) ∼
=Z
has the following perfect-pairing property: if x 6= 0, then there exists y so that
x ` y is a generator.
Hint: There is a map Mg → M1 ∨ . . . ∨ M1 , where the g copies of M1 . See
p. 228 of Hatcher.
Problem 7. Consider a continuous map f : Mg → Mk from the g-holed torus
to the k-holed torus and suppose
f∗ : H2 (Mg ) → H2 (Mk ).
is an isomorphism. Prove k ≤ g.
Hint: This problem has two parts. If k ≤ g, you have to construct a map. If
k > g then f ∗ : H 1 (Mk , Z) → H 1 (Mg , Z) must have non-trivial kernel (why?)
and this isn’t possible by the perfect pairing part of the previous problem.
Problem 8. Show that the cup-product extends to relative cohomology in the
following way: if A, B ⊆ X are subspaces, then there is a cup-product pairing
(−) ` (−) : H p (X, A; R) × H q (X, B; R) → H p+q (X, A ∪ B; R).
Conclude that if X is the union of two contractible subspace A and B, then the
cup product of two elements of positive degree in the cohomology of X must be
zero. Show that a suspension of a space Y is an example of such a space.