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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 $\mathbb{R}^3$?

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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 $T_{(x^0(t_0),0)}\mathcal{W}^u + T_{(x^0(t_0),0)}\mathcal{W}^s = \mathbb{R}^3$.

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    The very last conclusion is not clear to me yet. Why does dimension 3 imply that it equals $\mathbb{R}^3$? – 2017-02-06
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    Oh, that's simple. Since this picture lives in $\mathcal{M} = \mathbb{R}^3$, the overall tangent space $T_{(x^0(t_0), 0)} \mathcal{M}$ is $\mathbb{R}^3$. The sum $T_{(x^0(t_0),0)}\mathcal{W}^u + T_{(x^0(t_0),0)}\mathcal{W}^s$ is a part of $T_{(x^0(t_0), 0)} \mathcal{M}$ and has the same dimension as $T_{(x^0(t_0), 0)} \mathcal{M}$. So, it has to be the whole $T_{(x^0(t_0), 0)} \mathcal{M}$ -- the only subspace of $\mathbb{R}^m$ with dimension $m$ is $\mathbb{R}^m$ itself. – 2017-02-07