The algebraic concordance map is a surjective homomorphism

For a smooth oriented knot $K$, choose a surface and basis from [F1] and define $a(K)=[V_F]\in G_{\mathbb{Z}}$. The class is independent of this choice: [F4] relates any two such matrices by finitely many congruences and elementary enlargements or their inverses. A congruence is an isomorphism of the integral lattice, and each enlargement preserves the class. This also handles every integral basis choice. Applied to a transported surface under an oriented ambient isotopy, the same surface-independence statement and the pairing convention give a function on oriented knot types. This part is proved without a slice-disk premise.

Put representatives of $K,J$ and their chosen surfaces in separated balls, and join the surfaces by the connected-sum band away from the old basis curves. Their old first-homology bases together form a basis of the boundary-connected sum: collapsing the connecting band gives the wedge of the two surfaces, so first homology is their direct sum. Positive push-offs and the curves from opposite balls have linking number zero, since any curve in one ball bounds a rational, indeed integral, singular two-chain there, disjoint from the other ball. Thus the matrix is $V_K\oplus V_J$, and [F2] gives $a(K\mathrm{\#}J)=a(K)+a(J)$. The unknot has its disk surface with empty matrix and hence $a(U)=0$. By step 1.1 these calculations are independent of the separated representatives and chosen surfaces.

Work first in the smooth category. A knot smoothly concordant to $U$ bounds a smooth proper disk by [F6], so [F8] gives $a(K)=0$. The inverse of $[J]$ is represented by $J^{†}$, and therefore $J\mathrm{\#}{J}^{†}$ is concordant to $U$. Step 2.1 gives $a(J^{†})=-a(J)$. If $K$ and $J$ are smoothly concordant, their group difference $K\mathrm{\#}{J}^{†}$ is concordant to $U$, so $0=a(K\mathrm{\#}{J}^{†})=a(K)-a(J)$. Hence $a$ descends to a well-defined map ${\mathrm{\Phi }}_{\mathrm{s}\mathrm{m}}$ on smooth concordance classes. Step 2.1 proves that this map is a group homomorphism. Here [F8] applies the proved locally flat metabolizer argument to the smooth slice disk.

The smooth homomorphism is onto. Each positive generator $[V]$ of $G_{\mathbb{Z}}$ is the image of the realizing knot from [F3]. The image of a group homomorphism is a subgroup, so it also contains the negative of every such generator. Every element of the presented integral group is a finite sum of signed generators, hence is in the image. This proves the entire smooth surjective-homomorphism assertion.

In the locally flat category, [F10] supplies the group law and inverse needed for descent. If $K$ is TOP concordant to $U$, cap the unknot end of its collared annulus with the standard disk in a standard ball. Gluing that ball to $S^3\times I$ adds a radial collar to the ball, and the matching end product charts make the result a locally flat proper slice disk. By [F9] its Seifert form is metabolic, so $a(K)=0$ by [F2]. For TOP-concordant $K,J$, their group difference $K\mathrm{\#}{J}^{†}$ is TOP concordant to $U$. The smooth inverse calculation in step 3.1 already gives $a(J^{†})=-a(J)$, and step 2.1 therefore yields $0=a(K\mathrm{\#}{J}^{†})=a(K)-a(J)$. Hence $a$ descends to ${\mathrm{\Phi }}_{\mathrm{t}\mathrm{o}\mathrm{p}}$. Step 2.1 proves additivity, and the same realization argument as in step 4.1 makes its image contain every signed generator, hence all of $G_{\mathbb{Z}}$. This proves the TOP surjective-homomorphism assertion under the unchanged full-AC hypothesis, using the explicitly external group input and the exact TOP metabolizer interface rather than the former conditional hold.