Let $(M,g)$ be a compact smooth orientable riemannian manifold, and let $f: M \to \mathbb R$ be a Morse function. All functions here are assumed to be smooth. We will be considering the integral curves and the flow $\varphi$ of the gradient vector field of $f$. That is \begin{align} & \varphi : \mathbb R \times M \to M, \\ & (t,x) \mapsto \varphi^t(x), \end{align} where $\varphi^t(x)$ is the solution of the ODE $\dot{u} = -\nabla(f(u))$ at time $t$, with initial value $x$. It is known that $\varphi^t(x)$ converges to critical points of $f$, as $t \to \pm \infty$. Given $p,q$ critical points of $f$, we set \begin{align} \mathcal M(p,q) := \{u: \mathbb R \to M : \dot{u} = -\nabla(f(u)), \lim_{t \to -\infty}u(t) = p, \lim_{t \to +\infty}u(t) = q \}. \end{align} Next, we assume that $f$ is Morse-Smale, so that $\mathcal M(p,q)$ is a smooth manifold (it essentially corresponds to the transverse intersection of a stable manifold with an unstable manifold). We consider on $\mathcal M(p,q)$ the $C^\infty_{loc}$-convergence, that is, the uniform convergence on every compacta of the derivatives of all orders. There is a $\mathbb R$-action of $\mathcal M(p,q)$, defined by translation: \begin{equation} (\tau \cdot u)(s) = \varphi^\tau(u(s)) = u(s+\tau). \end{equation} The quotient $\widehat{\mathcal M}(p,q) = \mathcal M(p,q) / \mathbb R$ is the space of gradient flow lines from $p$ to $q$.
I'd like an answer or a reference for the following question:
How do you prove that the above action is proper and free, so that the quotient $\widehat{\mathcal M}(p,q) = \mathcal M(p,q) / \mathbb R$ is indeed a smooth manifold?
This question is possibly related to this one.