Irreducible Components and Surjectivity I guess this is very easy, but after I thought some time about it, I still
didn't find the idea.
If I have a continuous map $f$ from topological spaces $X$ to $Y$ where $Y$ is irreducible
and $X$ has finitely many irreducible components and $f$ is surjective, does then already
exist an irreducible component of $X$ such that the restriction of $f$ to this component is still surjective to $Y$?
 A: I don't see why this should hold. Let $X=\{0,1\}=\{0\}\cup \{1\}$ with the discrete topology and let $Y=\{0,1\}$ with the trivial topology (so that its open sets are $\emptyset$ and $Y$). The function $f:X\to Y$ given by $f(0)=0, f(1)=1$ is continuous, surjective, but $X$ is not irreducible.
A: No, take the identity $Id:\{0,1\}, discrete \to \{0,1\}, indiscrete$.   
However the answer is yes if your map $f$ is closed.  
In Algebraic Geometry, the main example of closed maps is proper maps. This may sound like a tautology, but actually it is often easy to check properness. For example closed immersions and projective maps are proper . You also have a valuative criterion for properness. Properness has very pleasant stability properties: under base-change, composition... Also, properness of a morphism can be checked on coverings of the target.
So properness is what you have to watch for if you want your property to hold.
Exercise: Is there a scheme with underlying topological space $( \{ 0,1\}, indiscrete)$?
