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".

• Your condition on $Y$ in the main text is not the same as that listed in the title - the former is never satisfied. – Dustan Levenstein Feb 23 '16 at 20:11
• @DustanLevenstein If $Y$ is an infinite profinite space (a compact hausdorff totally disconnected space), then since it is not a discrete space, it is not the disjoint union of its connected components (which are just the singletons). The key being disjoint union. If $A,B$ are spaces, then each of $A$ and $B$ is open in $A\sqcup B$. – oxeimon Feb 23 '16 at 20:18
• Oh you mean disjoint union as a topology, not as a set? That was not clear at all; you had me quite baffled for a minute there. – Dustan Levenstein Feb 23 '16 at 20:24
• @DustanLevenstein I'm sorry if that was confusing. In general whenever I talk about objects of a certain category, then by default any constructions I mention will take place in that category. (If I wanted a disjoint union of sets, then I would specify that it were a disjoint union as sets) – oxeimon Feb 23 '16 at 20:42
• @oxeimon: Some of us don’t think in category-theoretic terms. I’m a set-theoretic topologist; for me the default is a disjoint union as sets, and every space is the disjoint union of its connected components, since the latter partition the space. I now understand what you’re asking, but I certainly would not have done without the comments. – Brian M. Scott Feb 23 '16 at 22:19

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.

• Nice! Notation's a bit confusing though; I think you want to say $\displaystyle\bigcup_{n \ge 1} \big((0,1/n) \times \{1-1/n\}\big)$ in place of $\{(0,1/n) \times \{1-1/n\}: n \geq 1\}$. It doesn't help that we use the same notation for intervals and for ordered pairs... – Dustan Levenstein Feb 27 '16 at 2:24