= Solution
Write $f:U=\mathbb A_\mathbb C^1\setminus\{1\}\to\mathbb A_\mathbb C^1$ for $f(t)=t^2$. For any closed point $a\in\mathbb C$, the fiber is the zero-dimensional closed subscheme of $U$ cut out by $t^2-a$. It is finite over $\mathbb C$, and every <finite morphism> is <proper morphism>[proper]. This includes $a=1$, whose fiber in $U$ consists only of $t=-1$.
The morphism itself fails the <valuative criterion for properness>. Take the <discrete valuation ring> $R=\mathbb C[[s]]$ with <fraction field> $K=\mathbb C((s))$. The $K$-point $t=1+s$ lies in $U(K)$ and lies over the $R$-point of the target defined by
$$
y=(1+s)^2.
$$
Any extension $\operatorname{Spec}R\to\mathbb A^1$ of that $K$-point must still be given by $t=1+s$, whose closed point maps to the deleted point $t=1$. It therefore cannot factor through $U$. The required lift does not exist, so $f$ is not proper. This is an instance of the fact that <proper closed-point fibers do not imply properness>.
Back to article page
