Vanishing of the primary obstruction is equivalent to extension over the next skeleton

Suppose the prior-stage restriction extends to $F:X^{n+1}\cup A\to Y$. Put $f_1=F{\mathrm{\mid }}_{X^n\cup A}$. Every attaching sphere then bounds its characteristic-disk restriction, so $\theta (f_1)=0$ by [F3]. The independence theorem, applied to the relevant prior-stage homotopy, gives $[\theta (f)]=[\theta (f_1)]=0$.

Let $d\in C^n(X,A;\mathcal{P})$ and consider one oriented relative $n$-cell with characteristic disk $D^n$ and cell map $f_e$. Choose a based sphere map $u_e:S^n\to Y$ representing the sign-adjusted value $d(e)$ in the stalk fixed by the cell's whisker. There is a relative pinch map $q:D^n\to D^n\vee S^n$: choose a small closed ball in the interior, collapse its boundary to the wedge point, map the outside quotient to the first disk by a radial homeomorphism fixed on $\mathrm{\partial }{D}^n$, and map the collapsed inner ball with degree $+1$ to the sphere summand. Define $f_{d,e}=(f_e\vee u_e)q$. It agrees with $f_e$ on $\mathrm{\partial }{D}^n$. In the prism-boundary sphere of [F2], collapse the stationary side and the region on which the two disk maps agree. What remains is exactly the degree-one sphere carrying $u_e$; choosing the sign of $u_e$ according to [F2]'s convention therefore gives $d(f,\mathrm{const},f_d)(e)=d(e)$.

Apply Step 1.2 to every relative $n$-cell. AC in [A1] selects the sphere representatives for an arbitrary family; only finitely many choices occur for a finite pair. The modified cell maps agree with the unchanged map on $X^{n-1}\cup A$, so the CW pushout and weak topology glue them to $f_d:X^n\cup A\to Y$, with $d(f,\mathrm{const},f_d)=d$. No homotopy from $f$ to $f_d$ on the $n$-cells is constructed or needed.

Conversely, assume $[\theta (f)]=0$ and choose $d$ with $\theta (f)=\delta d$. Apply Step 2.1 and put $f^{\mathrm{\prime }}=f_d$. By [F1], $\theta (f)-\theta (f^{\mathrm{\prime }})=\delta d=\theta (f)$, hence $\theta (f^{\mathrm{\prime }})=0$. By [F3], $f^{\mathrm{\prime }}$ extends over every relative $(n+1)$-cell. AC selects all fillers for an arbitrary cell family, and the pushout glues them to an extension on $X^{n+1}\cup A$. This extension restricts to the original map on $X^{n-1}\cup A$.

Steps 1.2–2.1 also prove the more precise realization assertion, and Steps 1.1 and 3.1 prove both implications. Zero cochains, absent cells, and the case $A=X$ are included. The proof does not say that the original fixed map $f$ extends when merely its cohomology class vanishes: it may first be changed, generally nonhomotopically rel boundary, on the $n$-cells while staying fixed on the prior skeleton. $\mathrm{\square }$