The fundamental path-fibration class has the normalized relative lift

Put $P=P{K}_{n+1}$ and $B=K_{n+1}$. Choose a based map $f:S^n\to F$ representing the marked nonzero element of ${\pi }_n(F)=\mathbb{Z}\mathrm{/}2$. Its absolute Hurewicz image $h_F[f]=f_{\ast }[S^n]$ is the nonzero element of $H_n(F;\mathbb{Z})$: for $n>1$, use the marked weak equivalence $h:K_n\to F$ of the path-loop input lemma. Absolute Hurewicz is an isomorphism on the CW source $K_n$; the weak-equivalence definition gives an isomorphism on homotopy, and the integral weak-equivalence homology theorem in [F1] gives the isomorphism on integral homology, and Hurewicz naturality therefore makes $h_F$ an isomorphism too. For $n=1$, absolute Hurewicz says abelianization for every path-connected space, and ${\pi }_1(F)=\mathbb{Z}\mathrm{/}2$ is already abelian.

Since $P$ is contractible, $f$, considered as a map into $P$, has a nullhomotopy. Cone that nullhomotopy to obtain a continuous map $A:D^{n+1}\to P$ restricting to $f$ on the boundary; identify the disk with the cone on the sphere and choose its boundary orientation accordingly. Let $c_D$ be a finite triangulated integral fundamental chain of the disk, whose boundary $c_S$ is the corresponding fundamental cycle of its sphere. The chain $A_{\ast }{c}_D$ is a relative cycle for $(P,F)$ since its boundary is $f_{\ast }{c}_S$ in $F$. Write $z=[A_{\ast }{c}_D]\in H_{n+1}(P,F;\mathbb{Z})\mathrm{.}$ The homology pair-LES computes its boundary directly as $\mathrm{\partial }z=[f_{\ast }{c}_S]=h_F[f]$. Because $P$ is contractible and $n\ge 1$, that boundary map is an isomorphism $H_{n+1}(P,F;\mathbb{Z})\to H_n(F;\mathbb{Z})$. Thus $z$ is its nonzero generator. No relative Hurewicz map or theorem has been invoked.

The composite $pA$ maps the whole boundary sphere to the marked basepoint. Hence it factors through the collapsed disk, giving $g:D^{n+1}\mathrm{/}\mathrm{\partial }{D}^{n+1}\cong S^{n+1}⟶B\mathrm{.}$ In the fibration connecting construction, $A$ is a lift of this disk representative and its boundary lift is $f$; therefore the connecting homomorphism sends $[g]$ to $[f]$. It is an isomorphism because the path-space total is contractible. The fiber marking was defined by precisely this connecting isomorphism, so $[g]$ is the marked nonzero base element of ${\pi }_{n+1}(B)$.

On chains, $p_{\ast }z$ is the relative class of $(pA{)}_{\ast }{c}_D$. Collapsing the disk boundary sends its oriented relative fundamental class to the fundamental sphere class. The sphere boundary is a nonempty closed subspace with a radial collar retracting onto it, so cor-homology-of-good-pairs-is-reduced-homology-of-the-quotient identifies relative disk homology with the reduced quotient homology. In the one-top-cell description of $D^{n+1}\mathrm{/}\mathrm{\partial }{D}^{n+1}$ the quotient sends the characteristic disk generator to the same top-cell generator; naturality of thm-cellular-homology-computes-singular-homology makes this the asserted comparison of actual singular classes. Choose the quotient sphere orientation accordingly. Thus, under $H_{n+1}(B,\ast ;\mathbb{Z})\cong H_{n+1}(B;\mathbb{Z})$, $p_{\ast }z=g_{\ast }[S^{n+1}]=h_B[g]\mathrm{.}$ Absolute Hurewicz of the $n$-connected base identifies this with the nonzero generator of $H_{n+1}(B;\mathbb{Z})=\mathbb{Z}\mathrm{/}2$, including $n=1$, where the base is simply connected and its first nonzero homotopy degree is two. This is the required generator comparison entirely through the pair-LES and the two absolute Hurewicz maps.

Check the UCT obstruction exactly in degree $n+1$. For the relative pair it is ${\mathrm{Ext}}_{\mathbb{Z}}^1(H_n(P,F;\mathbb{Z}),{\mathbb{F}}_2)\mathrm{.}$ For $n>1$, the homology pair-LES identifies $H_n(P,F)$ with $H_{n-1}(F)$, which is zero by the integral homology comparison with the $(n-1)$-connected CW model $K_n$ obtained by applying the integral weak-equivalence homology theorem in [F1] to its marked weak equivalence. For $n=1$, the relevant LES segment is $0=H_1(P)⟶H_1(P,F)⟶H_0(F)\stackrel{\cong }{\to }{H}_0(P),$ so again $H_1(P,F)=0$. Thus the relative Ext term is zero in every case. The base UCT Ext term is ${\mathrm{Ext}}^1(H_n(B),{\mathbb{F}}_2)=0$, since $H_n(B)=0$ for the $n$-connected base. For the fiber in degree $n$, its Ext term is zero because $H_{n-1}(F)=0$ if $n>1$, while $H_0(F)=\mathbb{Z}$ is free if $n=1$. The three evaluations are therefore the actual normalized fundamental-class evaluations, with no unidentified Ext summand.

Finally compute the two target evaluations. If $u$ represents ${\iota }_n$ on $F$ and $b$ extends $u$ to $P$, the cohomology pair connector is represented by $db$. Therefore $⟨\delta {\iota }_n,z⟩=(db)(A_{\ast }{c}_D)=b(f_{\ast }{c}_S)=⟨{\iota }_n,h_F[f]⟩=1.$ Naturality of the Kronecker pairing likewise gives $⟨p^{\ast }{\iota }_{n+1},z⟩=⟨{\iota }_{n+1},p_{\ast }z⟩=⟨{\iota }_{n+1},h_B[g]⟩=1.$ The relative UCT, with its now-verified zero Ext term, says evaluation is an isomorphism onto $\mathrm{Hom}(\mathbb{Z}\mathrm{/}2,{\mathbb{F}}_2)$. Equality of these evaluations proves the claimed equality of classes. Mod-two coefficients eliminate the possible boundary-orientation sign.