Questions: The Hurewicz Theorem

Question 1

A simply connected space X has π_2(X) ≅ Z^3. What does the Hurewicz theorem tell us about H_2(X)?

• A. H_2(X) = 0 because X is simply connected
• B. H_2(X) ≅ Z^3, since X is simply connected (π_1 = 0) and π_2 is the first nontrivial homotopy group
• C. H_2(X) ≅ Z^3/(some torsion subgroup)
• D. Nothing — the Hurewicz theorem only applies to spheres

Answer: (B)

X is path-connected (being simply connected) with π_1(X) = 0. The Hurewicz theorem (n = 2 case) says that if π_1(X) = 0, then h: π_2(X) → H_2(X) is an isomorphism. Since π_2(X) ≅ Z^3, we get H_2(X) ≅ Z^3. The theorem also tells us H_1(X) = 0 (the abelianization of the trivial group is trivial). This is the power of Hurewicz: it converts the first nontrivial homotopy group into a computable homology group.

Question 2

The Hurewicz theorem for n = 1 says H_1(X) is the abelianization of π_1(X). This means H_1 of a space with non-abelian fundamental group carries less information than π_1.

• T. True
• F. False

Answer: True

For n = 1, the Hurewicz homomorphism h: π_1(X) → H_1(X) is surjective with kernel equal to the commutator subgroup [π_1, π_1]. So H_1(X) = π_1(X)/[π_1(X), π_1(X)] — the abelianization. If π_1 is non-abelian (like the free group on two generators for the figure-eight), the abelianization loses the non-commutativity: F_2 has abelianization Z^2, which cannot distinguish it from Z × Z. This information loss is why the fundamental group is a finer invariant than H_1.

Question 3

If a CW complex X satisfies H_k(X) = 0 for all k ≥ 1, what can you conclude about X?

• A. X is contractible
• B. X has the same homology as a point, but may not be contractible
• C. π_n(X) = 0 for all n ≥ 1, so X is contractible (using Whitehead's theorem)
• D. Both A and C, since they say the same thing

Answer: (A)

For a CW complex with trivial reduced homology: H_1(X) = 0 implies π_1(X) is a perfect group (equal to its own commutator subgroup). But actually, if X is simply connected AND has all homology vanishing, then by the Hurewicz theorem, π_n(X) = 0 for all n (by induction: each π_n is the first to be nontrivial, so it equals H_n = 0). Then Whitehead's theorem gives contractibility. If X is not simply connected, there exist acyclic spaces that are not contractible (like certain quotients of trees). So the full answer requires assuming simple connectivity or being more careful.

Question 4

Apply the Hurewicz theorem to compute π_n(S^n) for n ≥ 2.

Model answer: S^n is (n-1)-connected: π_k(S^n) = 0 for k < n (by cellular approximation, since S^n has cells only in dimensions 0 and n). The Hurewicz theorem then says the Hurewicz homomorphism h: π_n(S^n) → H_n(S^n) is an isomorphism. Since H_n(S^n) ≅ Z, we get π_n(S^n) ≅ Z. The generator is the homotopy class of the identity map id: S^n → S^n.

This is one of the most important applications of the Hurewicz theorem. It confirms that the degree of a map S^n → S^n (defined homologically as the integer d with f_*[S^n] = d[S^n]) coincides with its homotopy class — degree completely classifies self-maps of S^n up to homotopy. The Hurewicz theorem thus provides the foundation for degree theory.

Question 5

Does the Hurewicz theorem help compute π_3(S^2)?

Model answer: No, not directly. The Hurewicz theorem says π_2(S^2) ≅ H_2(S^2) ≅ Z (the first nontrivial homotopy group matches the homology). But for π_3(S^2), we are above the first nontrivial dimension, and the Hurewicz map h: π_3(S^2) → H_3(S^2) = 0 is the zero map — it provides no information. Computing π_3(S^2) ≅ Z requires different tools, such as the long exact sequence of the Hopf fibration S^1 → S^3 → S^2.

This illustrates the limitation of the Hurewicz theorem: it gives an isomorphism only in the first nontrivial dimension and only a surjection one dimension higher (by the relative Hurewicz theorem). Beyond that, the relationship between homotopy and homology breaks down, and the two invariants diverge dramatically.