Questions: The Universal Coefficient Theorem

Question 1

If H_1(X; Z) ≅ Z/6Z and H_0(X; Z) ≅ Z, what is H^1(X; Z)?

• A. Z/6Z
• B. Z
• C. Z/6Z ⊕ Z
• D. 0

Answer: (D)

By the universal coefficient theorem: H^1(X; Z) ≅ Hom(H_1(X; Z), Z) ⊕ Ext^1(H_0(X; Z), Z). Now Hom(Z/6Z, Z) = 0 (the only homomorphism from a finite group to Z is zero), and Ext^1(Z, Z) = 0 (Z is free, so Ext vanishes for free groups). Therefore H^1(X; Z) = 0. The torsion Z/6Z in H_1 does NOT appear in H^1 — it appears in H^2 via the Ext term. This shift is a key feature of the universal coefficient theorem: torsion in H_{n-1} contributes to H^n.

Question 2

For a space with free (torsion-free) homology groups, the universal coefficient theorem simplifies to H^n(X; Z) ≅ Hom(H_n(X; Z), Z).

• T. True
• F. False

Answer: True

When all homology groups are free abelian, Ext^1(H_{n-1}, Z) = 0 for all n (since Ext vanishes for free groups). The short exact sequence collapses to H^n(X; Z) ≅ Hom(H_n(X; Z), Z), which is the ordinary algebraic dual. In this case, cohomology and homology carry exactly the same information as abelian groups. Spaces with free homology include all spheres, complex projective spaces, and products of spheres.

Question 3

The splitting in the universal coefficient theorem is natural (functorial with respect to continuous maps).

• T. True
• F. False

Answer: False

The universal coefficient theorem guarantees that the short exact sequence splits, but the splitting is NOT natural — there is no functorial way to choose the splitting. This means H^n(X; G) ≅ Hom(H_n, G) ⊕ Ext(H_{n-1}, G) as abstract groups, but the isomorphism is not compatible with maps between spaces. A continuous map f: X → Y induces compatible maps on the short exact sequences, but the splitting cannot be chosen to commute with f*. The non-naturality is a genuine subtlety, not a mere technicality.

Question 4

Compute H^2(RP^2; Z) using the universal coefficient theorem, given that H_0(RP^2) = Z, H_1(RP^2) = Z/2Z, H_2(RP^2) = 0.

Model answer: H^2(RP^2; Z) ≅ Hom(H_2(RP^2), Z) ⊕ Ext^1(H_1(RP^2), Z) = Hom(0, Z) ⊕ Ext^1(Z/2Z, Z) = 0 ⊕ Z/2Z = Z/2Z. The torsion Z/2Z in H_1 reappears in H^2 via the Ext term. This is a manifestation of the general principle: torsion in homology dimension n-1 contributes torsion to cohomology dimension n.

Computing Ext^1(Z/2Z, Z): take a free resolution 0 → Z →^2 Z → Z/2Z → 0, apply Hom(−, Z) to get 0 → Hom(Z, Z) →^2 Hom(Z, Z) → 0, which is 0 → Z →^2 Z → 0. The cokernel is Z/2Z, so Ext^1(Z/2Z, Z) = Z/2Z. This computation exemplifies the general formula Ext^1(Z/nZ, Z) = Z/nZ.