TITLE: Image of a morphism of varieties
QUESTION [5 upvotes]: Suppose $A$ and $B$ are two algebraic varieties, and $f:A\to B$ is a morphism of algebraic varieties. I guess it is true that $\text{im}(f)$ is itself an algebraic variety. But how to prove it?

REPLY [13 votes]: You can't prove it because it is not true!  
Consider the (dominant)  morphism $f\colon \mathbb C^2\to \mathbb C^2\colon (x,y)\mapsto (x,xy)$.
Its image is the subset $Im(f)=\{(u,v)\in \mathbb C^2 |  u\neq 0  \} \cup \{(0,0)\}$.
 This set is not locally closed in $\mathbb C^2$ and so $Im(f)$ is not a subvariety of $\mathbb C^2$.
Feel free to soup up the example by introducing arbitrary fields, schemes,... :-)