A noncompact continuation datum can lose its trajectories at infinity

The two ends are valid: by [F1] and [F2] both pairs are complete Morse--Smale pairs with exactly one nondegenerate critical point of index $1$, so both Morse chain complexes are $\mathrm{\Lambda }$ concentrated in degree $1$ with zero differential.

Rigidity of the upper end. Let $x$ solve the continuation equation with ${\lim }_{s\to +\mathrm{\infty }}x(s)=x_0$. On $s\ge 1+\delta$ the equation is $\dot{x}=F(x)$, and every solution with $x(1+\delta )>x_0$ increases and converges to $+\mathrm{\infty }$, while every solution with $x(1+\delta )<x_0$ decreases, crosses $0$ (where $F(0)=-1$) and converges to $-\mathrm{\infty }$; hence necessarily $x(1+\delta )=x_0$, and then $x(s)=x_0$ for all $s\ge 1+\delta$.

Backward blow-up. Extend the solution of step 2.1 backward from $x(1+\delta )=x_0$. On the interpolation region the field is $G_s(x)=-\frac{d}{dx}((1-\theta (s))\frac{1}{3}{x}^3+\theta (s)f^{+}(x))=(1-\theta (s))(-x^2)+\theta (s)F(x)$ with $0\le \theta \le 1$, and at $x=x_0$ this equals $-(1-\theta (s))x_0^2<0$ whenever $\theta (s)<1$, In backward time the vector field at $x_0$ is nonnegative, so uniqueness (or the scalar differential inequality for the negative part of $x-x_0$) makes $x_0$ a lower barrier. On $[x_0,2x_0]$ all fields $G_s$ are bounded in absolute value by a common $C>0$. Choosing $\delta <x_0\mathrm{/}C$, the integral bound $\mathrm{\mid }x(s)-x_0\mathrm{\mid }\le C\delta$ precludes a first exit through $2x_0$; the lower barrier precludes a first exit below $x_0$. Thus the backward trajectory exists throughout this short interpolation and stays in $[x_0,2x_0]$. Entering the plateau at $s=1$ with $x(1)\in [x_0,2x_0]$, its backward evolution there obeys $\frac{dx}{d\tau }=x^2$ in the backward time $\tau =1-s$, with solution $x(\tau )=x(1)\mathrm{/}(1-x(1)\tau )$, which blows up at $\tau =1\mathrm{/}x(1)$; since $x(1)\ge x_0>\frac{1}{2}$ we have $1\mathrm{/}x(1)<2$, so the blow-up occurs strictly inside the plateau $[-1,1]$ and the backward trajectory is not defined beyond it.

Consequently no solution of the continuation equation has both the lower limit $p=0$ and the upper limit $q=x_0$: a solution with upper limit $x_0$ must satisfy $x(1+\delta )=x_0$ by step 2.1, and its backward continuation blows up in finite time by step 3.1, so it has no lower limit at all. Hence $\mathcal{C}(p,q)=\mathrm{\varnothing }$ although $\mathrm{ind}(p)=\mathrm{ind}(q)=1$, and the continuation count is the zero map $\mathrm{\Lambda }\to \mathrm{\Lambda }$, which is not an isomorphism.

The claim is therefore false, and the failure is exactly the mechanism recorded in Flow and compactness hypotheses for noncompact Morse homology: the end pairs have a single critical point each, but the interpolating field $\dot{x}=-x^2$ on the plateau is not complete, so the connecting trajectory escapes to spatial infinity in finite time, the energy identity has no finite-energy solution to apply to, and no compactness-up-to-breaking statement is available.