Relative Hurewicz comparison through a choice-free weak model

Apply the first clause of [F1] to obtain $Q:(Z,A)\to (X,A)$ equal to the identity on $A$ and weak on total spaces, with only relative cells of dimensions at least $n$. This uses the actual all-extension-data construction, not a selection of representatives or its later homotopy-inverse clause. The restriction to $A$ is the identity, hence weak. Thus both hypotheses of each comparison in [F2] hold. They give isomorphisms $Q_{\pi }:{\pi }_n(Z,A,a)⟶{\pi }_n(X,A,a),Q_H:H_i(Z,A)⟶H_i(X,A)$ for every $i\ge 0$ in homology. The relative basepoint remains literally $a$.

Write $Z_k=A\cup \{\text{relative cells of dimension at most }k\}$. We now use only the calculations of [F3] on this supplied model. Its layer quotient is a wedge of $k$-spheres, so each layer's relative homology is zero except for the free group on its $k$-cells in degree $k$. The homology triple sequences and finite support of each test chain yield $H_i(Z,A)=0$ for $i<n$ and $H_n(Z_{n+1},A)\cong H_n(Z,A)$, exactly as in the homology computation of [F3]. Its high-cell connectivity and homotopy triple sequence give ${\pi }_n(Z_{n+1},A,a)\cong {\pi }_n(Z,A,a)$, as in its homotopy stability computation. None of these arguments asks for an equivalence of $Z$ with a second replacement space.

Let $F_n$ and $F_{n+1}$ be the free abelian groups on the relative cells in these two dimensions. Since $A$ is simply connected, the single-layer basis calculation in [F3] applies to $(Z_n,A)$, and $Z_n$ is simply connected, so it also applies to $(Z_{n+1},Z_n)$. The homotopy and homology triple boundary maps have the identical matrix $D_{\alpha \beta }=\mathrm{deg}(p_{\alpha }{q}_n{\phi }_{\beta }),$ where $q_n:Z_n\to Z_n\mathrm{/}A$, $p_{\alpha }$ is the sphere projection and ${\phi }_{\beta }$ has its disk-boundary orientation. The two incidence computations in [F3] identifies these coefficients for each actual characteristic disk, including their sign; its final cokernel computation and the stability in step 2.1 identify both full degree-$n$ groups with $F_n\mathrm{/}\mathrm{im}D$ and their actual Hurewicz map with the identity of that quotient. Consequently $h_Z$ is surjective because each finite cell vector represents a homotopy class with that homology image, and injective because a vector mapping to zero belongs to precisely the same relation subgroup on both sides. This includes $n=2$, where the surjection from the free abelian cell group proves abelianness of the full relative group.

Naturality [F4] gives $h_X{Q}_{\pi }=Q_H{h}_Z$. All three maps $Q_{\pi },Q_H,h_Z$ on the right of $h_X=Q_H{h}_Z{Q}_{\pi }^{-1}$ are isomorphisms by steps 1.1 and 3.1. This equality proves that the isomorphism on $(X,A)$ is its actual oriented-disk Hurewicz homomorphism. In particular a target homology class can be pulled back through $Q_H$, lifted through $h_Z$, and pushed through $Q_{\pi }$, proving surjectivity. If $h_{X}u=0$, the displayed commuting square and injectivity of $Q_H$ and $h_Z$ show $Q_{\pi }^{-1}u=0$, proving injectivity. The homology comparisons in step 1.1 also transfer every lower vanishing in step 2.1.

No inverse map of spaces was chosen or asserted: $Q_{\pi }^{-1}$ and $Q_H^{-1}$ are inverses of bijections and hence unique functions. The model construction is choice-free by [F1]; its comparisons test only finite domains by [F2]; and [F3] explicitly makes its supplied-model computations choice-free. Thus this proof removes precisely the inverse-of-spaces use of AC. Empty relative cell sets give zero free groups, one cell gives the ordinary one-generator presentation, and zero vectors and zero incidence columns are retained. Equal pairs have zero relative groups; a point subspace is allowed. An empty $A$ has no specified $a$ and is excluded, while degrees zero and one occur only in the lower homology assertion, not as a relative Hurewicz isomorphism here. The first admissible degree and both isomorphism directions were checked in steps 3.1–4.1.