Is open proper map surjective?

Question

I want to know the relationship between "proper" and "surjective", is open proper map surjective? Or it needs more condition to imply surjection? For example, the map is a homogeneous complex map. Thanks in advance.

Answer 1

Here are some sufficient conditions (I assume that sequential compactness is the same as covering compactness. It is correct for metric spaces but I forget under what conditions it is correct for topological spaces): suppose that $ \phi:X \to Y$ is 0) continuous; 1) open; 2) proper; 3) Y is locally compact and connected.

Then $\phi$ is surjective.

Proof. Let $V:= \phi(X)$. By (1) $V$ is an open subset of $Y$. We need to show it is closed.

Let $y_0$ be a cluster point of $V$; since $Y$ is locally compact there is a compact neighbourhood $K$ of $y_0$. The preimage $\phi^{-1}(K)$ is also compact because (2) the map is proper. If $(y_n=\phi(x_n))_{n\in \mathbb N}$ is a sequence in $V\cap K$ converging to $y_0$ then $x_n$ is a sequence in the compact $\phi^{-1}(K)$ and has thus a convergent subsequence $x_{n_k}\to x_0$. By continuity $\phi(x_0)=y_0$ and hence $y_0\in V$. So $V = \overline V$ and it is both open and closed. Since $Y$ is connected, $V=Y$.$\qquad \blacksquare$

Answer 2

Take $Y$ a finite set with discrete topology and $X$ another set with strictly bigger cardinal and again discrete topology, any map from $Y$ to $X$ is proper and open but not surjective! So the answer of the question at the title is negative.