ProofComplete

Proof of the Hurewicz Theorem

We prove the Hurewicz theorem: for an (n1)(n-1)-connected space XX with n2n \geq 2, the Hurewicz homomorphism h:πn(X)Hn(X)h : \pi_n(X) \to H_n(X) is an isomorphism.


Proof

Theorem (Hurewicz): Let XX be an (n1)(n-1)-connected CW complex with n2n \geq 2. Then H~k(X)=0\tilde{H}_k(X) = 0 for k<nk < n and h:πn(X)Hn(X)h : \pi_n(X) \to H_n(X) is an isomorphism.

Step 1: Reduction to a CW complex with trivial (n1)(n-1)-skeleton.

Since XX is (n1)(n-1)-connected, by cellular approximation and CW approximation, we may assume XX is a CW complex with X(n1)={x0}X^{(n-1)} = \{x_0\} (a single point). This is because πk(X)=0\pi_k(X) = 0 for k<nk < n allows us to collapse the (n1)(n-1)-skeleton to a point without changing homotopy groups (by the Whitehead theorem applied to the quotient map).

With this CW structure, XX has no cells in dimensions 11 through n1n-1, so the cellular chain complex has CkCW(X)=0C_k^{CW}(X) = 0 for 1kn11 \leq k \leq n-1, giving Hk(X)=0H_k(X) = 0 for 1k<n1 \leq k < n.

Step 2: Identification of πn(X)\pi_n(X) and Hn(X)H_n(X) via the attaching maps.

The nn-cells of XX are attached by maps φα:Sn1X(n1)={x0}\varphi_\alpha : S^{n-1} \to X^{(n-1)} = \{x_0\}, which are all constant. Therefore each nn-cell gives a map Φα:SnX(n)=αSαn\Phi_\alpha : S^n \to X^{(n)} = \bigvee_\alpha S^n_\alpha (the wedge of nn-spheres), representing an element [Φα]πn(X(n))[\Phi_\alpha] \in \pi_n(X^{(n)}).

For the wedge of spheres, we have: πn(αSαn)αZHn(αSαn)\pi_n\left(\bigvee_\alpha S^n_\alpha\right) \cong \bigoplus_\alpha \mathbb{Z} \cong H_n\left(\bigvee_\alpha S^n_\alpha\right)

The first isomorphism holds because the wedge of nn-spheres is (n1)(n-1)-connected and the inclusion SαnSαnS^n_\alpha \hookrightarrow \bigvee S^n_\alpha generates a free summand. The second is immediate from cellular homology.

Step 3: The Hurewicz map commutes with attaching.

Consider the exact sequence from the pair (X,X(n))(X, X^{(n)}). Since XX is obtained from X(n)X^{(n)} by attaching cells of dimension >n> n, and these higher-dimensional cells do not affect πn\pi_n or HnH_n (by cellular approximation and the cellular chain complex), we have: πn(X)πn(X(n))/N\pi_n(X) \cong \pi_n(X^{(n)}) / N Hn(X)Hn(X(n))/BH_n(X) \cong H_n(X^{(n)}) / B where NN and BB are the subgroups generated by the attaching maps of the (n+1)(n+1)-cells.

Step 4: The subgroups NN and BB correspond under the Hurewicz map.

Each (n+1)(n+1)-cell has attaching map ψβ:SnX(n)\psi_\beta : S^n \to X^{(n)}. The element [ψβ]πn(X(n))[\psi_\beta] \in \pi_n(X^{(n)}) maps to the cellular boundary n+1(eβn+1)CnCW(X(n))\partial_{n+1}(e^{n+1}_\beta) \in C_n^{CW}(X^{(n)}) under the Hurewicz map. Since hh maps NN isomorphically onto BB, it descends to an isomorphism on the quotients: h:πn(X)=πn(X(n))/N    Hn(X(n))/B=Hn(X)h : \pi_n(X) = \pi_n(X^{(n)}) / N \xrightarrow{\;\cong\;} H_n(X^{(n)}) / B = H_n(X)

This completes the proof. \square


RemarkThe case $n = 1$

For n=1n = 1, the Hurewicz theorem states that h:π1(X)H1(X)h : \pi_1(X) \to H_1(X) is the abelianization map, with H1(X)π1(X)ab=π1(X)/[π1(X),π1(X)]H_1(X) \cong \pi_1(X)^{\text{ab}} = \pi_1(X) / [\pi_1(X), \pi_1(X)]. The proof in this case is different, relying on the fact that H1H_1 is the abelianization of the fundamental group for path-connected spaces.