Is there a suitable definition in categories for a closed continuous function in topology? Working in the category of topological spaces is it possible to give a 'categorical' definition for 'a closed continuous function'? I mean something like: 'a closed continuous function' is an arrow in Top such that ... My intuition tells me that it is not so, but on the other hand this intuition has often turned out to be arrogant and wrong.
 A: Let $\mathcal{C}$ be a category with products and coproducts. Let us denote by $1$ the terminal object. We assume that $\hom(1,-) : \mathcal{C} \to \mathsf{Set}$ is faithful. The motivating example is $\mathcal{C}=\mathsf{Top}$.
Definition. A global element $a$ of an object $A$ is a morphism $a : 1 \to A$.
Definition. An object $A$ is called discrete if every map of global elements $\hom(1,A) \to \hom(1,B)$ is induced by a morphism $A \to B$. An object $A$ is called codiscrete if every map of global elements $\hom(1,B) \to \hom(1,A)$ is induced by a morphism $B \to A$.
Definition. A Sierpinski object $S$ is a an object which has two global elements $x,y$ and which is neither discrete nor codiscrete.
In the case of $\mathcal{C}=\mathsf{Top}$, these elements $x,y$ are distinguished by the property that $\{y\}$ is open, but $\{x\}$ is not. Of course there is also another, isomorphic Sierpinski space with the roles of $x,y$ exchanged. But for us only the first is the correct one.
Now the task is to distinguish these two Sierpinski objects from each other, for general $\mathcal{C}$. The correct $S$ has, for every set $I$, a morphism
$$\cup : \prod_{i \in I} S \to S$$ such that a global element of $\prod_{i \in I} S$, i.e. an $I$-indexed tuple consisting of $x$ and $y$, is mapped to $y$ if and only if there is at least one $y$ in the tuple. And the correct $S$ has a map
$$\cap : \prod_{i \in I} S \to S$$
such that only the constant tuple $y$ is mapped to $y$ only then $I$ is finite.
I don't claim that such an object $S$ exists for every $\mathcal{C}$, but that this is a property in the language of categories which may or not be satisfied by $\mathcal{C}$, and it certainly holds for $\mathcal{C}=\mathsf{Top}$.
After having characterized the correct Sierpinski object $S$ as well its two global elements $x,y$, we are able to define open and closed subobjects, motivated by the fact that the Sierpinski space classifies open subsets in the case of $\mathsf{Top}$.
Definition. An open subobject of an object $A$ is a morphism $U \to A$ which fits into a pullback diagram
$$\begin{array}{c} U & \rightarrow & 1 \\ \downarrow &&~~ \downarrow y \\ A & \rightarrow & S. \end{array}$$
If we take $x : 1 \to S$, we obtain the notion of a closed subobject of $A$. Note that we can define the union of an arbitrary family of open subjects, as well as finite intersections, using the operators $\cup,\cap$ from above.
Definition. Let $f : A \to B$ be a morphism in $\mathcal{C}$. It is called a closed morphism (resp. open morphism) if for every closed (open) subobject $Z \to A$ the composition $Z \to A \to B$ has an image factorization $Z \to Z' \to B$ for some closed (open) subobject $Z' \to B$.
Let me mention that with these definitions it is easy to obtain the rigidity of the category of topological spaces $\mathsf{Top}$. Roughly, this means that the "concrete structure" of the objects can be recovered in the language of category theory. The underlying set of $A$ is just $\hom(1,A)$, and its topology consists of the images of $\hom(1,U) \to \hom(1,A)$ for open subobjects $U \to A$.
The definitions may also be applied to the category of posets $\mathcal{C}=\mathsf{Pos}$, where $x$ and $y$ can not be distinguished from each other. Therefore the automorphism class group of $\mathsf{Pos}$ is $\mathbb{Z}/2$, the automorphism mapping $(X,<)$ to $(X,>)$.
Edit: Freyd had already shown rigidity in his book on abelian categories, Chapter I, Exercise G. He offers a surprisingly simple distinction between the "correct" and the "wrong" Sierpinski object: If every point of a space is open, this space is discrete. This is not the case if every point is closed.
A: Given a continuous function $f: X \to Y$ and letting $\tau(X)$ denote the open sets of $X$. Consider the morphism $f^{-1} : \tau(Y) \to \tau(X)$ -- the fact that this is a morphism is exactly what makes $f$ continuous. Now look at the collection of closed sets of $X$ and $Y$, what does it mean for $f$ to be closed?
A: Topological spaces can be defined as pairs $(X,¯_X)$ where 
$¯_X$ is the closure operator as a function from 
$P_X$ (the power set) to 
$P_X.$ Then a closed continuous map $f:X→Y$ is a map which respects the
closure $¯$, i.e. the compositions 
$$¯_Y∘Pf:PX→PY→PY$$
 and $$Pf∘¯_X:PX→PX→PY$$ are equal.
