Let $f_i\in C[x_1,\dots, x_n]$ with $i\leq m$. Suppose $f_i$ are algebraically independent functions. Consider the map $\phi:A^n\to A^m$ by $\phi(x_1,\dots, x_n)=(f_i(x_1,\dots, x_n))_i$. Suppose $\phi$ is non surjection. Then $Im(\phi)$ is closed algebraic subset of $A^n$.
$\textbf{Q:}$ Why is $Im(\phi)$ a closed subset if $\phi$ is not surjection? I could imagine $A^2\to A^2$ by $(x,y)\to (x,xy)$. Clearly $x,xy$ are algebraically independent. Furhtermore, the map is not surjection as part of $x=0$ axis is missing. It is worse that the image is not closed algebraic set. Have I misunderstood the statement?
Context: The book tries to explain $f_i$ algebraic independence/essential parameters is equivalent to surjectivity of the map $\phi$ above.
Ref. Beltrametti M.C., et al. - Lectures on curves, surfaces and projective varieties Sec 3.2 of Chpt 3, Pg 62
This is just an error in the text. You are correct that the image of $\varphi$ need not be closed. The result being proved (that $\varphi$ is surjective iff the $f_i$ are algebraically independent) is simply wrong, as your example also shows. The correct statement is that $\varphi$ is dominant (has dense image) iff the $f_i$ are algebraically independent.