(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?

  • $\begingroup$ The extension is unique, are you just asking if the extension is or not an open surjection? $\endgroup$ – André Caldas Oct 11 '11 at 19:54
  • $\begingroup$ @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? $\endgroup$ – Asaf Karagila Oct 11 '11 at 20:10

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.

  • $\begingroup$ 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 $\beta Y$. $\endgroup$ – Asaf Karagila Oct 12 '11 at 9:23
  • 1
    $\begingroup$ @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 $0$. $\endgroup$ – Emil Jeřábek Oct 12 '11 at 10:27
  • $\begingroup$ 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? $\endgroup$ – Asaf Karagila Oct 12 '11 at 10:42
  • $\begingroup$ Sure, post it on math.SE, Asaf. I don't have an account there. $\endgroup$ – Bill Johnson Oct 12 '11 at 15:55

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