Let $M$ be a smooth manifold, $\Delta = \{ (x, x) \in M \times M \}$ and $f : M \to M$ a smooth map. If $$T_{(p,p)} \operatorname{graph}(f) + T_{(p,p)} \Delta \neq T_{(p,p)} (M \times M),$$ why is it true that $$\exists \; 0 \neq v \in T_{(p,p)} \operatorname{graph} (f) \cap T_{(p,p)} \Delta?$$
Thank you!
Note that for finite-dimensional subspaces $W_1, W_2$ of a vector space $V$, we have
$$\dim(W_1\oplus W_2) = \dim W_1 + \dim W_2 - \dim W_1\cap W_2.$$
If $\dim W_1 + \dim W_2 = \dim V$, then we see that $W_1\oplus W_2 = V$ if and only if $W_1\cap W_2 = \{0\}$.
If $M$ is an $n$-dimensional manifold, then $\operatorname{graph}(f)$ and $\Delta$ are $n$-dimensional submanifolds of $M\times M$, so $T_{(p,p)}\operatorname{graph}(f)$ and $T_{(p,p)}\Delta = n$ are $n$-dimensional subspaces of the $2n$-dimensional vector space $T_{(p,p)} M\times M$.