A $C^2$ map $C^1$-close to $F$ has a unique inverse branch for its first component, with $O(\epsilon )$ derivative bounds

Location of the root: fix $(q,q^{\mathrm{\prime }})\in [-R,R{]}^2$ and define $\mathrm{\Phi }(v):=q^{\mathrm{\prime }}+v-A(q,v)$ on $[-R-1,R+1]$. For every such $v$, $\mathrm{\Phi }(v)-q^{\mathrm{\prime }}=v-A(q,v)$ so $\mathrm{\mid }\mathrm{\Phi }(v)-q^{\mathrm{\prime }}\mathrm{\mid }\le \epsilon \le 1$ by [F1]; hence $\mathrm{\Phi }(v)\in [-R-1,R+1]$ by [F4] (because $q^{\mathrm{\prime }}\in [-R,R]$ gives $q^{\mathrm{\prime }}+[-1,1]\subseteq [-R-1,R+1]$). Also $\mathrm{\mid }\mathrm{\Phi }(v)-\mathrm{\Phi }(w)\mathrm{\mid }=\mathrm{\mid }(v-w)-(A(q,v)-A(q,w))\mathrm{\mid }\le {\sup }_{\xi }\mathrm{\mid }1-{\mathrm{\partial }}_{y}A(q,\xi )\mathrm{\mid }\mathrm{\mid }v-w\mathrm{\mid }\le \epsilon \mathrm{\mid }v-w\mathrm{\mid }$ by [F1] and the mean value inequality. Thus $\mathrm{\Phi }$ maps the complete interval into itself and is a contraction with constant $\epsilon <1$.

Unique root: by [F2], $\mathrm{\Phi }$ has exactly one fixed point $v\in [-R-1,R+1]$; the identity $\mathrm{\Phi }(v)=v$ is equivalent to $A(q,v)=q^{\mathrm{\prime }}$. This proves clause 1 and defines $V$.

Regularity: at a root $(q,q^{\mathrm{\prime }})$ with $v=V(q,q^{\mathrm{\prime }})$, the partial derivative ${\mathrm{\partial }}_{y}A(q,v)$ satisfies $\mathrm{\mid }{\mathrm{\partial }}_{y}A(q,v)\mathrm{\mid }\ge 1-\epsilon >0$ by the closeness bound, so $A_y(q,v)\mathrm{\ne }0$. The implicit function theorem [F3] gives a $C^1$ solution $\phi$ near $(q,q^{\mathrm{\prime }})$ with the displayed derivatives, and by clause 1 this local solution coincides with $V$; since $(q,q^{\mathrm{\prime }})\in [-R,R{]}^2$ was arbitrary, $V$ is $C^1$ on the square with the displayed formulas.

Regularity needed by the half-tail construction: choose $0<\delta <1-\epsilon$. For $(q,q^{\mathrm{\prime }})\in (-R-\delta ,R+\delta {)}^2$, both parameters lie in $[-R-1,R+1]$, and every $(q,v)$ with $v\in [-R-1,R+1]$ lies in $S$. Thus the given bounds apply and ${\mathrm{\partial }}_{y}A(q,v)\ge 1-\epsilon >0$. The contraction $\mathrm{\Phi }(v)=q^{\mathrm{\prime }}+v-A(q,v)$ has constant $\epsilon$; moreover $\mathrm{\mid }\mathrm{\Phi }(v)-q^{\mathrm{\prime }}\mathrm{\mid }\le \epsilon$, so $\mathrm{\Phi }(v)\in [-R-1,R+1]$ because $\mathrm{\mid }{q}^{\mathrm{\prime }}\mathrm{\mid }<R+\delta$ and $\delta +\epsilon <1$. Its unique fixed point therefore defines $V$ on this open parameter rectangle, extending the original branch by uniqueness. The implicit function theorem gives a local $C^1$ branch at each point; uniqueness makes these local branches agree with the contraction-defined extension. Since $A$ is $C^2$, the formulas in step 3.1 make its first partial derivatives $C^1$ compositions and quotients with nonvanishing denominator. Consequently this extension is $C^2$ on the open rectangle, as needed by the later root and weighted-tail constructions; clause 2's $C^1$ conclusion remains valid.

Estimates: from $A(q,V(q,q^{\mathrm{\prime }}))=q^{\mathrm{\prime }}$ and the closeness $\mathrm{\mid }A(q,v)-v\mathrm{\mid }\le \epsilon$ evaluated at $v=V(q,q^{\mathrm{\prime }})$, clause 3's first inequality follows: $\mathrm{\mid }V(q,q^{\mathrm{\prime }})-q^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }V-A(q,V)\mathrm{\mid }\le \epsilon$ by [F4]. For the derivatives, by step 3.1 $\mid {\mathrm{\partial }}_{q}V\mid =\frac{\mathrm{\mid }{\mathrm{\partial }}_{x}A(q,V)\mathrm{\mid }}{\mathrm{\mid }{\mathrm{\partial }}_{y}A(q,V)\mathrm{\mid }}\le \frac{\epsilon }{1-\epsilon },\mid {\mathrm{\partial }}_{q^{\mathrm{\prime }}}V-1\mid =\frac{\mathrm{\mid }{\mathrm{\partial }}_{y}A(q,V)-1\mathrm{\mid }}{\mathrm{\mid }{\mathrm{\partial }}_{y}A(q,V)\mathrm{\mid }}\le \frac{\epsilon }{1-\epsilon },$ using $\mathrm{\mid }{\mathrm{\partial }}_{x}A\mathrm{\mid }\le \epsilon$ and $\mathrm{\mid }{\mathrm{\partial }}_{y}A-1\mathrm{\mid }\le \epsilon$ from the hypotheses and $\mathrm{\mid }{\mathrm{\partial }}_{y}A\mathrm{\mid }\ge 1-\epsilon$ from step 3.1.

Conclusion: steps 2.1, 3.1 and 4.2 prove clauses 1, 2 and 3 respectively.