Theorem 4.30.

Let X be an n − 1 - connected space, n ≥ 2. That is, we assume πq(X) = 0 for
q ≤ n − 1. Then Hq(X) = 0 for q ≤ n − 1 and the previously defined “Hurewicz homomorphism”
h : πn(X) → Hn(X)
is an isomorphism.

Proof. We assume the reader is familiar with the analogue of the theorem when n = 1, which
says that for X connected, the first homology group H1(X) is given by the abelianization of the
fundamental group
h : π1(X)/[π1, π1] ∼= H1(X)
where [π1, π1] ⊂ π1(X) is the commutator subgroup. We use this preliminary result to begin
an induction argument to prove this theorem. Namely we assume that the theorem is true for
n − 1 replacing n in the statement of the theorem. We now complete the inductive step. By our
inductive hypotheses, Hi(X) = 0 for i ≤ n − 2 and πn−1(X) ∼= Hn−1(X). But we are assuming that
πn−1(X) = 0. Thus we need only show that h : πn(X) → Hn(X) is an isomorphism.

Consider the path fibration p : P X → X with fiber the loop space ΩX. Now πi(ΩX) ∼= πi+1(X),
and so πi(ΩX) = 0 for i ≤ n − 2. So our inductive assumption applied to the loop space says that
h : πn−1(ΩX) → Hn−1(ΩX)
is an isomorphism. But πn−1(ΩX) = πn(X). Also, by the Serre exact sequence applied to this
fibration, using the facts that
(1) the total space P X is contractible, and
(2) the fiber ΩX is n − 2 - connected and the base space X is (n − 1) - connected.

We then conclude that the transgression,
τ : Hn(X) → Hn−1(ΩX)
is an isomorphism. Hence the Hurewicz map h : πn−1(ΩX) → Hn−1(ΩX) is the same as the
Hurewicz map h : πn(X) → Hn(X), which is therefore an isomorphism.