Let $f:X\to Y$ be a surjective morphism of smooth irreducible varieties over $\mathbb{C}$. Assume further that $Y$ is complete and that every fiber $f^{-1}(y)$ for $y\in Y$ is complete and irreducible. Does it necessarily follow that $X$ is complete as well? If no, what additional assumptions can we put so that this follows?


Edit: According to the remark of vrz, I added the assumption that each fiber is irreducible (which implies connectedness). If I don't require equidimensionality, is there also a counterexample?



Complete target and complete fibers imply complete source?
Let $k$ be a field. Let $Y$ be a separated, finite type $k$-scheme that is geometrically connected and normal. Let $f:X\to Y$ be a separated, finite type morphism from a geometrically connected and reduced $k$-scheme to $Y$ such that the fiber over every geometric point of $Y$ is connected and proper.


**Proposition.** The morphism is proper.


**Proof.** By Nagata compactification, there exists a dense open immersion $i:X\hookrightarrow \overline{X}'$ into a proper $k$-scheme. The product morphism, $$(i,f):X\to \overline{X}'\times\_{\text{Spec}\ k}Y,$$ is also an open immersion between separated, finite type $k$-schemes. Thus, the closure of the image is also a separated, finite type $k$-scheme. Denote this closure by $\overline{X}$. Also denote the restriction to $\overline{X}$ of the second projection by
$$\overline{f}:\overline{X}\to Y.$$


By construction $\overline{f}$ is a proper, surjective morphism. Thus, there exists a Stein factorization, $$\overline{X}\xrightarrow{h} Z\xrightarrow{g} Y,$$ where $h$ has geometrically connected fibers and where $g$ is a finite surjective morphism. Since $X$ is a dense open subscheme of $\overline{X}$ that is connected, also $\overline{X}$ is connected. Therefore $Z$ is also connected. Similarly, since $X$ is reduced, also $Z$ is reduced. Since the geometric generic fiber of $f$ is connected and dense in the geometric generic fiber of $\overline{f}$, also the geometric generic fiber of $\overline{f}$ is connected. Thus, $g$ is birational. Since $Y$ is normal, by Zariski's Main Theorem, the morphism $g$ is an isomorphism. In other words, the fiber of $\overline{f}$ over every geometric point is connected and proper.


The fiber of $f$ over that same geometric point is an open subscheme of the fiber of $\overline{f}$. By hypothesis, it is also proper, and thus it is a closed subscheme of the fiber of $\overline{f}$. Since the fiber of $\overline{f}$ is connected, the fiber of $\overline{f}$ equals the fiber of $f$ for every geometric point of $Y$. For every geometric point of $\overline{X}$, for the image geometric point of $Y$, the geometric point of $\overline{X}$ is a point of the fiber of $\overline{f}$ over that geometric point of $Y$. Thus, it is also a point of $X$. In other words, the open subscheme $X$ of $\overline{X}$ equals all of $\overline{X}$. Therefore, also $f$ equals $\overline{f}$, so that the morphism $f$ is proper. **QED**


