Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(Cross posted from this math.SE 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?

share|improve this question
    
The extension is unique, are you just asking if the extension is or not an open surjection? –  André Caldas Oct 11 '11 at 19:54
    
@Andre: Exactly. Under the assumptions that $Y$ is paracompact, and possibly we may add that $X$ is Cech-complete, is the extension remains an open map? –  Asaf Karagila Oct 11 '11 at 20:10

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
Bill, the extension is onto $\beta Y$, I also fail to see why the closure of $2B$ is clopen, and why the closure of $\frac{1}{2}Y$ is not clopen in $\beta Y$. –  Asaf Karagila Oct 12 '11 at 9:23
1  
@Asaf: The indicator function of $2B$ is continuous, and its extension to $\beta X$ gives a partition of $\beta X$ into two clopen subsets that are easily seen to be the closures of $2B$ and its complement, respectively. (In general, if $C$ is a clopen subset of $W$, then the closure of $C$ in $\beta W$ is clopen.) By a similar argument, the closure of $\frac12Y$ omits all points $-1/(2n+1)$, hence it does not contain any neighbourhood of its element $0$. –  Emil Jeřábek Oct 12 '11 at 10:27
    
@Emil: Thanks!! –  Asaf Karagila Oct 12 '11 at 10:41
    
Bill: If you have a math.SE, could you post this on the linked question as well? If not, may I have your permission to post it as community wiki? –  Asaf Karagila Oct 12 '11 at 10:42
    
Sure, post it on math.SE, Asaf. I don't have an account there. –  Bill Johnson Oct 12 '11 at 15:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.