"Assume that $X = V((z_1 - 1)z_2 - 1) \hookrightarrow \mathbb{C^2}$ and $f(z_1, z_2) = z_1^2(f : X \rightarrow \mathbb{C})$.
Show that f is closed map(in the Zariski topology)."
The book that I read says "It is sufficient to prove $f(X) = \mathbb{C}$".
But I don't know why!
Can you teach me please.
Thank you in advance!