Can a birational morphism surject from an affine to a projective variety? Let $X$ be an affine variety and $Y$ a projective variety, both integral (reduced and irreducible). Assume that $\phi:X\to Y$ is a birational morphism. I would venture to say that $\phi$ can not be surjective, but I have no proof for the statement. So, my question: Can $\phi$ be surjective? If no, why?
 A: Here's a thought to show that this cannot happen if $\dim Y>0$. It's 5:30 AM though, and so I can't in good conscience post this not as community wiki. :) Anyone should feel free to point out a silly mistake, or finish the proof in some way. It proceeds as classic dévissage.
Indeed, suppose first that $X$ and $Y$ are regular curves. Then, this certainly cannot happen. Indeed, open embed $X$ into a regular projective curve $C$. Suppose that the birational map gives an isomorphism $X\supseteq U\xrightarrow{\approx} V\subseteq Y$ and let $f:V\to U$ be the inverse isomorphism. Then, consider the composition $V\to U\to C$. By the curve-to-projective theorem this extends to a map $X\to C$ which must be an isomorphism since it's degree $1$. But, in particular, by the surjectivity of $X\to Y$, this implies that $C=X$ which contradicts that $X$ is affine.
Suppose now that $X\to Y$ is as in the problem statement, but now with no assumptions on $\dim Y$. Take a curve $C\subseteq Y$. Then, consider the fiber product 
$$\begin{matrix}X_C & \to & X\\ \downarrow & & \downarrow\\ C & \to & Y\end{matrix}$$
By assumption, the map $X_C\to C$ is surjective and birational (edit: this may need to be tweaked, but I think it's fixably tweakable. see comments below edit v 2.0: this is possible by a simple application of Bertini's lemma, when $Y$ is normal. To show that it suffices that $Y$ is normal, take its normalization and show, similarly to below, that everything works out OK)
. Moreover, since $C$ closed embeds into $Y$ it's projective, and since $X_C$ closed embeds into $X$ it's affine.
Now, by checking component by component we may (hopefully I'm not making a silly error here--I haven't checked through all the details) assume that $C$ and $X_C$ are irreducible. Moreover, by passing to their reduced subschemes, we may assume they are reduced, and thus integral. 
Now, consider the normalization $C'\to C$ and form, once again, the fibered diagram
$$\begin{matrix}X_{C'} & \to & X_C\\ \downarrow & & \downarrow\\ C' & \to & C\end{matrix}$$
Using similar ideas to above, we see that $X_{C'}$ is affine, $C'$ is projective, and we may assume that $X_{C'}$ is integral. Then, finally, by passing to the normalization of $X_{C'}$, call it $X'_{C'}$ and looking at the composition
$$X'_{C'}\to X_{C'}\to C'$$
we obtain a surjective birational map with $X'_{C'}$ affine, and $C'$ projective. But, since both of these objects are a curve, this contradicts our original case. 
