Transversality of invariant manifolds I am referring to
Kuznetsov's book, p. 230 where in the context of Melnikov's integral, it is said that
$$
T_{(x^0(t_0),0)}\mathcal{W}^u+T_{(x^0(t_0),0)}\mathcal{W}^s=\mathbb{R}^{n+1}
$$
I have problems to see this identity. Why is the sum of the two Tangent spaces on the left Hand side the whole $\mathbb{R}^{n+1}$?
For illustration, let $n=2$. A picture for this situation is shown on page 211 of the linked book: Why is the sum of the two tangent spaces the whole $\mathbb{R}^3$?  
 A: If you are trying to understand why equality on page 211 holds, recall that for linear subspaces always holds equality ${\rm dim}\; ( V + W) = {\rm dim}\; V + {\rm dim}\; W - {\rm dim}\; (V \cap W)$. The same holds for tangent spaces since they are linear subspaces. Both tangent spaces $T_{(x^0(t_0),0)}\mathcal{W}^u$ and $T_{(x^0(t_0),0)}\mathcal{W}^s$ are two-dimensional on this figure, and, according to how it is pictured, at the intersection point they don't coincide, so ${\rm dim}\; (T_{(x^0(t_0),0)}\mathcal{W}^u \cap T_{(x^0(t_0),0)}\mathcal{W}^s) < 2$. Surfaces $\mathcal{W}^u$ and $\mathcal{W}^s$ have a curve $\Gamma_0$ in the intersection, so their tangent spaces at point $(x^0(t_0), 0)$ have common subspace (associated with this curve, of course) and ${\rm dim}\; (T_{(x^0(t_0),0)}\mathcal{W}^u \cap T_{(x^0(t_0),0)}\mathcal{W}^s) = 1$. 
Hence $${\rm dim}\; (T_{(x^0(t_0),0)}\mathcal{W}^u + T_{(x^0(t_0),0)}\mathcal{W}^s) = 3$$
and $T_{(x^0(t_0),0)}\mathcal{W}^u + T_{(x^0(t_0),0)}\mathcal{W}^s = \mathbb{R}^3$.
