The tangent space to the intersection is the intersection of the tangent spaces.

Solution 1:

First, note that since $X \cap Z \subseteq X$ and $X \cap Z \subseteq Z$, we have that for $y \in X \cap Z$, $T_y(X \cap Z) \subseteq T_y(X)$ and $T_y(X \cap Z) \subseteq T_y(Z)$, or $$T_y(X \cap Z) \subseteq T_y(X) \cap T_y(Z),$$ by the definition of intersection.

Since $X \pitchfork Z$, we have that $$\mathrm{codim}(X \cap Z) = \mathrm{codim}(X) + \mathrm{codim}(Z),$$ or $$\dim(X \cap Z) = \dim(X) + \dim(Z) - n,$$ which tells us that $$\dim(T_y(X \cap Z)) = \dim(T_y(X)) + \dim(T_y(Z)) - n.$$ On the other hand, from linear algebra we know the formula for the dimension of an intersection of vector spaces: \begin{align*} \dim(T_y(X) \cap T_y(Z)) & = \dim(T_y(X)) + \dim(T_y(Z)) - \dim(T_y(X) + T_y(Z)) \\ & = \dim(T_y(X)) + \dim(T_y(Z)) - n, \end{align*} where $\dim(T_y(X) + T_y(Z)) = n$ since $X \pitchfork Z$. Then since $T_y(X \cap Z) \subseteq T_y(X) \cap T_y(Z)$ and both vector spaces have the same dimension, it must be that they are equal. We conclude that for all $y \in X \cap Z$, $$T_y(X \cap Z) = T_y(X) \cap T_y(Z).$$

Note that we can't automatically conclude the second part. Here transversality is important. For example if $$X = \{(x, 0) : x \in \Bbb R\} \subset \Bbb R^2, \quad Z = \{(x,x^2) : x \in \Bbb R\} \subset \Bbb R^2,$$ we have that $X \cap Z = \{(0,0)\}$, so that $T_{(0,0)}(X \cap Z) = \{0\}$, a zero-dimensional vector space. But $T_{(0,0)} (X) = T_{(0,0)} (Z)$ is a $1$-dimensional vector space, so that we necessarily have a strict inclusion $$T_{(0,0)}(X \cap Z) \subset T_{(0,0)}(X) \cap T_{(0,0)} (Z).$$ The problem here is that $X$ and $Z$ do not intersect transversally.