Let $k$ be an algebraically closed field and $f: \mathbb{A}^2_k \rightarrow \mathbb{A}^2_k$, $(x,y) \mapsto (x,xy)$. I understand what the image looks like, i.e. $\mathbb{A^2_k} - (\mathbb{A}^1-0)$. By observing that: $\mathbb{A}^2_k = \textrm{Im} \cup \mathbb{A}^1_k$ and since $\mathbb{A}^2_k$ is irreducible, by taking the closure we get that $\overline{\textrm{Im}}$ is the whole plane and therefore that the image is not closed. If it was open, the intersection $\textrm{Im} \cap \{X \neq 0\} = 0$ would also be open, which is impossible because $\mathbb{A}^2_k$ is irreducible. Finally, since the image is dense and is not open in the whole set, it can't be locally closed either.
I find this reasoning to be a bit arbitrary. In particular, for the "locally closed" part, I initially wanted to look that at every neighbourhood of $\{(0,0)\}$, which would be the whole plane without some curves, but I feel like that is too tedious.
Is there a more intuitive (understand obvious) way to think about these things ? In general, is there an intuitive way to think about sets being closed or open in the Zariski topology?
