*(Cross posted from [this math.SE](https://math.stackexchange.com/questions/66375/) question)* 


Let $X$ be a Cech-complete space, and $Y$ a paracompact space. Suppose $f\colon X\to Y$ is a continuous and open surjection.


Since $Y$ is completely regular we have that $\beta(Y)$ is homeomorphic to $Y$ as a dense subset of $\beta Y$ (the Stone-Cech compactification).


We can, if so, take $\hat f\colon X\to\beta Y$ defined as $\beta\circ f$, as a continuous function from $X$ into a compact Hausdorff space.


By the universal property of $\beta X$ we can uniquely extend $\hat f$ to a continuous $\tilde f\colon\beta X\to\beta Y$ such that $\tilde f|\_{\beta(X)} = \hat f\circ\beta$. In particular $\tilde f$ is onto $\beta Y$ due to two reasons:


1. $\tilde f$ is continuous from a compact domain, therefore its image is closed; and
2. $\tilde f$ is onto a dense subset of $\beta Y$.


Therefore it is onto its closure which is $\beta Y$.


**My question** is whether or not the fact $Y$ is paracompact implies that the extend is also an open surjection?



Extending open maps to Stone-Cech compactifications
Let $Y=(-1/n)\_{n=1}^\infty \cup \{0\}$, $B$ the positive integers, $X=Y\cup B$ with the topology they inherit from the real line. Define $f:X\to Y$ to be the identity on $Y$ and $f(n)=-1/n$ for $n$ in $B$. The closure of $2B$ in $\beta X$ is open and onto $\{0\} \cup (1/2n)\_{n=1}^\infty$ in $Y$, which is not open. 


Thanks to Todd Elsworth for pointing out that my first example was wrong.


