What is the proper definition of a proper map?
Here are a few definitions I've come across:
A continuous map $f: X \to Y$ is proper if preimages of compact subsets are compact.
A continuous map $f: X \to Y$ is proper if it is closed with compact fibres.
Kashiwara and Schapira define a map $f: X \to Y$ to be proper if it is closed and its fibres are compact and relatively Hausdorff (two distinct points in the fibre have disjoint neighbourhoods in $Y$).
Definitions 1. and 2. agree when $X$ is Hausdorff and $Y$ is locally compact Hausdorff. Kashiwara and Schapira state that definition 3. agrees with definition 1. when both $X$ and $Y$ are locally compact (and Hausdorff I assume?).
There are also notions of proper morphisms of schemes and proper geometric morphisms of topoi, though I do not really understand these yet. However, I would assume that they should be attempting to capture the same sort of notion of "properness" as in topological category.
It seems that a proper map in some category should satisfy some property, and I am wondering what that property should be. In other words, what is the correct "abstract" definition of properness? I have a feeling it should have something to do with a proper base change theorem in cohomology.
A continuous map $f:X \to Y$ between topological spaces is proper if it is universally closed, i.e., for all $T \to Y$ the induces map \begin{align} X \times_Y T \to T \end{align} is closed. This definition is equivalent to the one you mentioned when $X,Y$ are locally compact Hausdorff. Here $X \times_Y T$ denotes the fiber product in the category of topological spaces (it is just the subspace of $X \times T$ where the maps to $Y$ agree).
A morphism $f: X \to Y$ between schemes is called proper if it is separated, finite type and universally closed. That means that for all morphisms $T \to Y$ the map \begin{align} X \times_Y T \to T \end{align} is closed. This time, $X \times_Y T$ denotes the fiber product in the category of schemes (or locally ringed spaces).
The connection is as follows: Let $a:X \to \operatorname{Spec} \mathbb{C}$ be a smooth proper variety ($a$ is a proper morphism), then the complex manifold $X^{an}$ is compact. More generally, if $f:X \to Y$ is a proper morphism between smooth proper complex varieties, then the analytification of $f$ is a proper morphism of topological spaces (I don't think the smoothness assumption is at all necessary)