"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 $f(X) = \mathbb{A}^{1}_{\mathbb{C}}$.
