Finite-generation cohomological UCT gives integral cone comparison

Cohomology UCT surjects the displayed degree-i cohomology onto Hom(H_i(C),Q), respectively Hom(H_i(C),F_p). In the finite abelian-group decomposition, a nonzero free summand has a nonzero map to Q; any nonzero p-primary summand has a nonzero map to F_p. Thus the vanishing of all these Hom groups forces H_i(C)=0. No H^{i+1} vanishing is used, and the Ext term is not mistaken for the Hom term. In fact all prime fields alone detect a finitely generated nonzero abelian group; Q is included to match the topological coefficient comparisons.

Apply this to C_f. The degreewise split cone sequence, dualized to any coefficient field, gives H^{i-1}(Y;F)→H^{i-1}(X;F)→H^i(Hom(C_f,F)) →H^i(Y;F)→H^i(X;F). This follows from the inspected long exact sequence of complexes after reindexing cochains; degreewise splitting ensures Hom remains exact. Therefore field isomorphisms f*:H^i(Y;F)→H^i(X;F) for 0≤i≤D imply cone cohomology vanishing for 0≤i≤D, including degree zero with the negative groups zero. The previous lemma and finite generation then give H_i(C_f;Z)=0 for i≤D. The integral cone long exact sequence yields the stated isomorphism below $D$ and surjection at $D$.

f_:H_i(X;Z)→H_i(Y;Z) is an isomorphism for i<D, f_:H_D(X;Z)→H_D(Y;Z) is surjective. Claiming injectivity in degree D would require H_{D+1}(C_f)=0 and is not a consequence of these hypotheses. For the actual Thom comparison take D=2r−1. Items 2 and 4 give the rational and odd-prime field isomorphisms through D, since both reduced cohomologies vanish there. The separate mod-two metastable comparison must give f* isomorphisms through D. If it does, the argument proves integral homology isomorphisms through 2r−2 and surjectivity at 2r−1. Neither odd-primary vanishing nor a mod-two comparison at 2r is required. For a later degree n+r in the isomorphism range choose r≥n+2.