Questions: Homology with Coefficients

Question 1

RP^2 has H_1(RP^2; Z) = Z/2Z. What is H_1(RP^2; Z/3Z)?

• A. Z/2Z
• B. Z/3Z
• C. 0
• D. Z/6Z

Answer: (C)

By the universal coefficient theorem: H_1(X; G) ≅ (H_1(X; Z) ⊗ G) ⊕ Tor(H_0(X; Z), G). H_1(RP^2; Z) ⊗ Z/3Z = Z/2Z ⊗ Z/3Z = Z/gcd(2,3)Z = 0. Tor(H_0(RP^2; Z), Z/3Z) = Tor(Z, Z/3Z) = 0 (Z is free). So H_1(RP^2; Z/3Z) = 0. The 2-torsion in H_1(RP^2; Z) becomes invisible with Z/3Z coefficients because gcd(2,3) = 1. This illustrates that different coefficient groups detect different torsion phenomena.

Question 2

With field coefficients k, all homology groups H_n(X; k) are vector spaces over k, and the universal coefficient theorem simplifies to H_n(X; k) ≅ H_n(X; Z) ⊗ k.

• T. True
• F. False

Answer: True

When G = k is a field, Tor(−, k) = 0 (every module over a field is free, hence flat, so the tensor product is exact). The universal coefficient theorem gives H_n(X; k) = H_n(X; Z) ⊗_Z k ⊕ Tor(H_{n-1}(X; Z), k) = H_n(X; Z) ⊗ k. Concretely: Z ⊗ k = k (free summands survive) and Z/mZ ⊗ k depends on whether m is invertible in k. Over Q: Z/mZ ⊗ Q = 0 for all m (torsion vanishes). Over Z/pZ: Z/mZ ⊗ Z/pZ = Z/gcd(m,p)Z (only p-torsion survives).

Question 3

Why is Z/2Z a natural choice of coefficients for studying non-orientable manifolds?

• A. Because all non-orientable manifolds have H_1 = Z/2Z
• B. Because Z/2Z coefficients eliminate all sign issues, making orientation irrelevant — Poincaré duality holds for ALL closed manifolds with Z/2Z coefficients
• C. Because Z/2Z is the simplest field
• D. Because non-orientable manifolds only have 2-torsion

Answer: (B)

Poincaré duality with integer coefficients requires orientability (a Z-fundamental class). But every closed manifold — orientable or not — has a Z/2Z-fundamental class, because Z/2Z ignores orientation (there is only one 'sign'). So H^k(M; Z/2Z) ≅ H_{n-k}(M; Z/2Z) for ALL closed n-manifolds. This makes Z/2Z the natural choice for studying topological properties that do not depend on orientation, including Stiefel-Whitney classes and mod-2 intersection theory.

Question 4

Explain the relationship between H_n(X; Z) and H_n(X; Z/pZ) for a prime p. What information about the integer homology does the mod-p homology detect?

Model answer: By the universal coefficient theorem: H_n(X; Z/pZ) ≅ (H_n(X; Z) ⊗ Z/pZ) ⊕ Tor(H_{n-1}(X; Z), Z/pZ). The tensor part detects the free rank of H_n (each Z summand contributes Z/pZ) and the p-primary torsion (Z/p^kZ ⊗ Z/pZ = Z/pZ). The Tor part detects p-primary torsion in H_{n-1} (Tor(Z/p^kZ, Z/pZ) = Z/pZ). So mod-p homology sees: (1) the free rank, and (2) the p-primary torsion of the integer homology in dimensions n and n-1. By computing H_n(X; Z/pZ) for all primes p, one can recover the full torsion structure of H_n(X; Z).

This is the algebraic topology version of 'localization at p.' Computing mod-p homology for various primes is often the most efficient route to understanding the integer homology: first determine the free rank (from rational homology), then determine the p-primary torsion for each prime (from mod-p homology). The universal coefficient theorem reassembles these local computations into the global integer answer.