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Let $X$ be a Čech-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-Čech 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 lets us extend the map such that $\tilde f$ is also an open surjection.

(The motivation is to write a proof for the theorem mentioned in my previous question, and a result as above would give a quick solution to the problem. Regardless, this question is interesting on its own accord)

Edit (Oct. 3rd): If in a few more days there won't be an answer, I'll try cross-posting this on MathOverflow as well.

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Posted on MathOverflow. –  Asaf Karagila Oct 11 '11 at 19:50

1 Answer 1

up vote 2 down vote accepted

The question was answered on MathOverflow by user Bill Johnson. With his permission, I post it here as well.

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.

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