open map from a topological space whose connected components aren't open to a connected space Let $X$ be a connected topological space, and $Y$ a space which is not the disjoint union (as topological spaces) of its connected components (ie, the connected components of $Y$ are not all open). Can there exist an open continuous surjective map $Y\rightarrow X$ with finite fibers?
Motivation: I want to argue that any surjective etale map of schemes $Y\rightarrow X$ where $X$ is a connected scheme must have $Y$ be the disjoint union of its connected components.
EDIT: By disjoint union (as topological spaces), I mean the coproduct in the category of topological spaces, sometimes denoted as the "topological sum".
 A: Im not sure if this is what you are looking for (I don't know anything about schemes) but there is at least one Y for which no such map exists for any connected X.
Let $Y = \{1/2,1/3,...,0\}$ equipped with the subspace topology from $\mathbb{R}$. The connected components of this space are the singletons, and all of them are open except for $\{0\}$.
Assume we have an open, continuous, surjective map $f:Y \to X$ with finite fibers. We will show that X is necessarily disconnected.
Since $f$ is an open map with finite fibers, the set $f(Y - \{0\})$ is infinite and all of its points are open. If $f(Y - \{0\}) = X$ then X is disconnected and we are done.
Assume $f(Y - \{0\}) \neq X$. Since f is finite-to-one there must be a largest k such that $f(1/2) = f(1/k)$. Then $f(\{1/(k+1),1/(k+2),...,0\})$ is an open set which contains $f(0)$ but does not contain $f(1/2)$. In fact the complement of $f(\{1/(k+1),1/(k+2),...,0\})$ is a finite number of points, all of which are open in X. Therefore X is disconnected.
On the flip side, there is a space Y and a connected space X for which such a map does exist.
Let $Y = \{(0,1/n) \times \{1-1/n\}: n \geq 1\} \cup \{(0,1)\} \subset \mathbb{R}^2$. Let $X$ be the half open interval $[0,1) \times \{0\}$. Then the projection map from $Y$ to $X$ is an open, continuous, surjective map with finite fibers.
