Consider the ODE \begin{align} \dot x = f(x) \tag{1} \end{align} Let $x_0$ and $y_0$ be hyperbolic critical elements (fixed points or periodic solutions) of $(1)$, and let $W^u$ and $W^s$ denote the unstable and stable manifolds. Assume that the intersection of stable and unstable manifolds of $x_0$ and $y_0$ satisfy the transversality condition and $\{W^u(x_0)- x_0 \} \cap \{W^s(y_0)- y_0 \} \ne \emptyset $. Then according to Smale we have $$\text{dim} \: W^u(x_0) \ge \text{dim} \: W^u(y_0)$$
where the equality sign holds when $x_0$ is an equilibrium point and $y_0$ is a closed orbit.
I appreciate it if anyone can provide an intuition for this result. I cannot justify this for the pendulum heteroclinic orbit where both $x_0$ and $y_0$ are equilibrium points, but their unstable manifolds have the same dimension.