# Proof that map is closed(in Zariski topology)

"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!

The map $$X \to \mathbb{A}_{\mathbb{C}}^{1} - \{0\}$$ defined as $$(z_{1}, z_{2}) \to z_{1} - 1$$ is an isomorphism, and $$\mathbb{A}^{1}_{\mathbb{C}}- \{0\} \subset \mathbb{A}_{\mathbb{C}}^{1}$$ is an open subset. Since the only closed subets of $$\mathbb{A}^{1}_{\mathbb{C}}$$ are finite set of points or $$\mathbb{A}_{\mathbb{C}}^{1}$$ itself, so is $$\mathbb{A}_{\mathbb{C}}^{1}- \{0\}$$ and $$X$$, i.e. closed subsets of $$X$$ are also finite subsets or itself. Since an image of a point, which is also a point, is clearly a closed subet, we only need to show that $$f(X)$$ is closed, which is true if $$f(X) = \mathbb{A}^{1}_{\mathbb{C}}$$.