Is a morphism of schemes which is proper at every fiber proper? Let $f\colon X \rightarrow Y$ be a separated morphism of finite type of schemes.
Suppose $f^{-1}(y)$ is proper over $Spec(k(y))$ for every $y \in Y$, where $k(y)$ is the residue field of $y$.
Is $f$ proper?
If not, what conditions (in addition to the above one) should $Y$ (or $X$) satisfy to make $f$ proper?
 A: No, the morphism $f$ needn't be proper.  
For a counterexample,   take $ X=\mathbb A^1_\mathbb C \setminus \lbrace 0\rbrace , Y=\mathbb A^1_\mathbb C$ and let $f:\mathbb A^1_\mathbb C \setminus \lbrace 0\rbrace  \hookrightarrow \mathbb A^1_\mathbb C $ be the inclusion.
All fibers $f^{-1}(y)$ are proper over $Spec(k(y))$ but $f$ is not proper since it is not  closed.
A: I think, it is instructive to consider the case that $f$ has only finitely many points in each fiber, i.e. it is quasi-finite. It is not so difficult to show that every finite morphism is both proper and quasi-finite. But the converse is also true: At least if $X$ and $Y$ are noetherian, every proper quasi-finite morphism is automatically finite (Lemma 3.1.4 in Jonathan Wang's notes on the Zariski Main Theorem). Thus, in this case you would be asking whether every quasi-finite morphism is also finite. As in Georges example, open immersion are usually counter-examples to this. 
The surprising thing is that these are more or less the only counter-examples. More precisely, a version of Zariski's main theorem states that under mild assumptions every quasi-finite morphism can be factored into an open embedding and a finite morphism. These mild assumptions could, e.g., be that $Y$ is quasi-compact and (quasi-)separated and $f$ is separated and of finite presentation (see Section 4 in Wang's notes). 
