Proposition: Descent on the span of one null presentation

Let ${\mathcal{P}}_{\mathrm{int}}$, $\mathsf{Rec}$, $O$, $m^{\circ }$, $Y$ be as in XV-B's linear interface, let $n\in \ker \mathsf{Rec}$, and let $F_0$ be any preliminary lift. Then $F_0(n)\in Y$, and for a correction class $\mathcal{A}\subseteq \mathrm{Hom}({\mathcal{P}}_{\mathrm{int}},Y)$ the component of the obstruction class on $\mathrm{span}\{n\}$ is the class of $F_0(n)$ in $Y/\{a(n):a\in \mathcal{A}\}$. If every $a\in \mathcal{A}$ takes values in $V_C$ for a closed $C\subseteq W\setminus O$, then $\mathrm{supp}{F}_0(n)\subseteq C$ is necessary for descent in $F_0+\mathcal{A}$. For $n=(qi\theta +1)M_{\mathrm{R}}$ the descent equation is

Proof. $r_O{F}_0(n)=m^{\circ }\mathsf{Rec}(n)=0$, so $F_0(n)\in \ker r_O=Y$. A candidate $F_0+a$ annihilates $n$ iff $a(n)=-F_0(n)$, which is the quotient statement. If $a(n)\in V_C$ for all $a$, then $F_0(n)=-a(n)\in V_C$ is necessary. The last display is linearity.