Example of a non extendable rational map

Question

A rational map is a class of equivalence on $$F=\{f:U \rightarrow Y \mid \mbox{morphisms with } U\subset X \mbox{ open subvariety}\}$$ given by $$f\sim g \iff f_{| (\operatorname{Dom}f \cap \operatorname{Dom}g)}= g_{|(\operatorname{Dom}f \cap \operatorname{Dom}g)}$$ In this way, William Fulton says in his book Algebraic Curves that "a rational map can be defined as a function $f$ from an open subvariety $U$ of $X$ to $Y$ such that $f$ cannot be extended to a morphism from any larger open subset of $X$ to $Y$." I really think that a I get this, but I can't find an example of a rational map $f$ defined on a proper subvariety $U\subsetneq X$.

Answer 1

The example of Eric Wofsey is kind of "artificial" because one could extend this map using the projective space as target variety. More generally, if $Y$ is projective and $X$ is a smooth curve, then any rational map $f : X \to Y$ will extends.

On the other hand, if $P = [0:0:1]$ and $X = \Bbb P^2 \backslash \{P\}$ the map $$ \pi : X \to \Bbb P^4, [x:y:z] \mapsto [x^2:y^2:xy:xz:yz] $$ can't be extended. So this is a "true" rational map.

Edit : let met add a counter-example of a morphism $X \to Y$ which can't be extended where $C$ is projective but not smooth. We take $X = \{(x,y,z) \in \Bbb P^2 : zx^2 = y^3\}$ and $Y = \Bbb P^1$. We have a bijective map $f : \Bbb P^1 \to C, (t,s) \mapsto (t^3, t^2s, s^3)$. On the other hand, we have a well-defined inverse defined on the smooth locus of $C$ (i.e $C \backslash \{(0:0:1)\}$) defined as $g(x,y,z) = (x,y)$. If one could extend $g$ into a regular map, then $f,g$ would be inverse and we would have an isomorphism $C \cong \Bbb P^1$ but $C$ is not smooth.

Answer 2

Take $X=Y=\mathbb{A}^1$ and define $f(x)=1/x$. This is defined on $U=\mathbb{A}^1\setminus\{0\}$ but cannot be extended to $0$.