Is a closed morphism with proper fibres proper?



morphism closed + fibres proper => proper?
The answer is no. Consider an integral nodal curve $Y$ over an algebraically closed field, normalize the node and remove one of the two points lying over the node. Then you get a morphisme $f : X\to Y$ which is bijective (hence homeomorphic), separated and of finite type, and the fibers are just (even reduced) points. But $f$ is not proper (otherwise it would be finite and birational hence coincides with the normalization map). 


In the positive direction, you can look at EGA, IV.15.7.10. 


[**Add**] There is an elementary way to see that $f$ is not proper just using the definition. Let $Y'\to Y$ be the normalization of $Y$. So $X$ is $Y'$ minus one closed point $y\_0$. It is enough to show that the base change of $f$ to $X\times Y' \to Y \times Y'$ is not closed. Consider the closed subset 
$$\Delta=\left\lbrace (x, x) \mid x\in X \right\rbrace \subset X\times Y'.$$ 
Its image by $f\_{Y'} : X\times Y' \to Y\times Y'$ is $\left\lbrace (f(x), x) \mid x\in X\right\rbrace$ which is the graph of $Y'\to Y$ minus one point $(f(y\_0), y\_0)$. So $f$ is not universally closed, thus not proper.


