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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!

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This question has an open bounty worth +50 reputation from Johnny T. ending in 7 days.

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  • $\begingroup$ $f(x)\in Y$, but $Z_1\cap Z_2\subset X$, so did you mean something else? $\endgroup$ – Mohan 5 hours ago

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