Zero discrete force does not select a unique continuum path

For each $c\in \mathbb{R}$, $R_c(t)=(t,c\sin (\pi t))$ on $[0,1]$ is regular because its first coordinate derivative is $1$, and it has endpoints $A,B$. Since $\mathrm{\nabla }V=0$, every normal gradient component is zero, so [F1] makes every $R_c$ an MEP. Distinct $c$ give distinct paths.

For $M=2$, each listed interior point is equidistant from $A$ and $B$: the two distances are $1\mathrm{/}2$ when $R_1=(1\mathrm{/}2,0)$ and both $\sqrt{5}\mathrm{/}2$ when $R_1=(1\mathrm{/}2,1)$. Thus the adjacent lengths agree. Since the gradient is zero, the string projected force is zero by [F2], and the ordinary NEB spring term is also zero by [F3]. Both chains therefore have exactly zero force and spacing residuals.

The two chains differ in their interior image, and the continuum family in step 1.1 is nonunique. In this example the string residual map has no locally invertible Jacobian that can isolate a path: its force components vanish identically and its spacing equation alone cannot determine the two coordinates of the interior image. Thus the inverse/stability premise [F4] is absent. Independently of that missing premise, the two distinct MEPs and two zero-residual chains explicitly prove nonuniqueness. Each chain's midpoint lies on its corresponding exact MEP $R_c$, so this is not an assertion of inaccuracy relative to every MEP. Relative to the prescribed reference $R_0([0,1])$, however, the second chain's midpoint has Euclidean distance $1$ while its residual is zero. Therefore residual alone cannot select a unique continuum MEP or bound distance to that reference by an error bound that vanishes with the residual.