Let $f: X \rightarrow Y$ be a morphism of schemes: $X$, $Y$ are regular schemes, let $Z_1, Z_2$ be two closed regular subschemes of $X$, let $x \in X \times_Y Z_2$ such that $y = f(x) \in Z_1 \cap Z_2$. Then I would like to know how I can deduce that the map $$ T_xX \rightarrow (T_yY/T_yZ_2) \otimes_{k(y)} k(x) $$ is surjective knowing that

1. the image of $T_{x} (X \times_Y Z_1)$ generates $T_yY/T_yZ_2$

2. $T_x(X \times_Y Z_1) \rightarrow T_y Z_1 \otimes_{k(y)} k(x)$ is surjective.

It is supposed to follow from these things but not seeing how... I would appreciate any explanation on this. Thank you!

Let $f: X \rightarrow Y$ be a morphism of schemes: $X$, $Y$ are regular schemes, let $Z_1, Z_2$ be two closed regular subschemes of $Y$, let $x \in X \times_Y Z_2$ such that $y = f(x) \in Z_1 \cap Z_2$. Then I would like to know how I can deduce that the map $$ T_xX \rightarrow (T_yY/T_yZ_2) \otimes_{k(y)} k(x) $$ is surjective knowing that

1. the image of $T_{x} (X \times_Y Z_1)$ generates $T_yY/T_yZ_2$

2. $T_x(X \times_Y Z_1) \rightarrow T_y Z_1 \otimes_{k(y)} k(x)$ is surjective.

It is supposed to follow from these things but not seeing how... I would appreciate any explanation on this. Thank you!