A finite local trace misses a disconnected solution branch

The equation factors as $(x-1)(x+1)=0$, so its complete solution set is $S=(\{1\}\times \mathbb{R})\cup (\{-1\}\times \mathbb{R})\mathrm{.}$ Each set is connected and open-and-closed relative to $S$, and the two sets are disjoint; therefore they are the two connected components of $S$.

Choose $h>0$ and $0<R_0<\min \{h\mathrm{/}2,1\mathrm{/}4\}$. The seed section is $G_{-1}(x,\lambda )=(x^2-1,\lambda )$ with center $(1,0)$ and radius $R_0$. Its residual is zero, its Jacobian inverse there is $\mathrm{diag}(1\mathrm{/}2,1)$ with norm $1$, and its Jacobian is $2$-Lipschitz. Thus the exact-data certificate has $q=2R_0<1$ and zero residual term, and certifies the seed ball by [F2]. In $V=\{(x,\lambda ):\mathrm{\mid }x-1\mathrm{\mid }<1,-h<\lambda <2h\}$ every zero lies on $c(s)=(1,s)$ for $s\in [-h,2h]$. Rank is one there.

At $(1,0)$, $DF=[20]$ has rank $1$ and unit tangent $t=(0,1)$. Choose previous orientation and candidate tangent $\tau =\widehat{t}=t=(0,1)$. The chart has $a=\kappa =1$, $r_{\mathrm{c}\mathrm{h}}^{+}=2h$, and $h>R_0$, $h+R_0<2h$, so the previous-normal offset and chart tests in [F1] pass. The augmented equation is $(x^2-1,\lambda -h)=0$, whose unique root in any ball of radius less than $1$ centered at $(1,h)$ is $(1,h)$. Its bordered derivative is $\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]$, so the local corrector is nonsingular.

At the exact center $(1,h)$ the augmented residual is zero and the Jacobian is invertible. The augmented Jacobian is $2$-Lipschitz; with exact data, choose $0<R<\min \{1\mathrm{/}4,h\mathrm{/}4\}$, so the whole corrected ball lies in $V$, set $\widehat{\beta }=1$, and use zero residual and derivative errors. Then $q=2R<1$ and $\beta \rho =0$, so [F2] certifies the unique local section root. With $N_{\mathrm{r}\mathrm{e}\mathrm{q}}=N_s=1$ and $N_c=0$, the exact predictor already solves this section and both the seed and next ball are certified. The ball of radius $R<1$ also excludes the other root $(-1,h)$.

The other component $\{-1\}\times \mathbb{R}$ contains no point of that local chart around $(1,0)$, and the accepted one-step trace contains no point with $x=-1$. Nevertheless both components satisfy $F=0$. By [F1], the finite trace correctly certifies its local step but cannot certify that the second component is absent or that all branches have been found.