# Continuous function and compactifications

• May 26th 2011, 02:05 AM
CSM
Continuous function and compactifications
Can someone help me on this one?

Given two compact Hausdorff Space $X,Y$ and a continuous function $f:X\rightarrow Y$ which conditions are necessary and sufficient to garantuee the existence of a continuous function $\hat{f}:\hat{X}\rightarrow \hat{Y}$ such that $\hat{f}|X=f$ ( $\hat{f}$ restricted to $X$) where $\hat{X},\hat{Y}$ are the one-point compactifications?
• May 26th 2011, 03:34 AM
Opalg
Quote:

Originally Posted by CSM
Can someone help me on this one?

Given two compact Hausdorff Space $X,Y$ and a continuous function $f:X\rightarrow Y$ which conditions are necessary and sufficient to garantuee the existence of a continuous function $\hat{f}:\hat{X}\rightarrow \hat{Y}$ such that $\hat{f}|X=f$ ( $\hat{f}$ restricted to $X$) where $\hat{X},\hat{Y}$ are the one-point compactifications?

If X and Y are already compact, then their one-point compactifications either don't exist at all, or they just consist of an isolated point at infinity (depending on whose definition you want to use). In the latter case, $\hat{f}$ can always be defined by taking $\hat{f}(\infty_X) = \infty_Y.$

If you meant to say that X and Y are locally compact spaces, then the condition for $\hat{f}$ to exist is that f should have a limit at $\infty_X$ in the space $\hat{Y}$. In other words, there should exist $y_0\in \hat{Y}$ such that, given any neighbourhood U of $y_0$, there is a compact set $K\subset X$ such that $f(x)\in U$ for all $x\in X\setminus K$.
• May 26th 2011, 04:29 AM
CSM
Quote:

Originally Posted by CSM
Can someone help me on this one?

Given two compact Hausdorff Space $X,Y$ and a continuous function $f:X\rightarrow Y$ which conditions are necessary and sufficient to garantuee the existence of a continuous function $\hat{f}:\hat{X}\rightarrow \hat{Y}$ such that $\hat{f}|X=f$ ( $\hat{f}$ restricted to $X$) where $\hat{X},\hat{Y}$ are the one-point compactifications?

I am sorry. I meant locally compact indeed.
Is that condition you give sufficient? How do I see that $\hat{f}|X=f$ is guaranteed?
• May 26th 2011, 06:48 AM
Opalg
Quote:

Originally Posted by CSM
Is that condition you give sufficient? How do I see that $\hat{f}|X=f$ is guaranteed?

The way to define $\hat{f}$ is to take $\hat{f}(x) = f(x)$ for all $x\in X$. Then we just have to define $\hat{f}$ at the point at infinity (which I denote by $\infty_X\in\hat{X}$), say $y_0 = \hat{f}(\infty_X).$ The function $\hat{f}:\hat{X}\to\hat{Y}$, defined in that way, will be continuous at all points of X (because f is continuous on X), so we just have to check that $\hat{f}$ is continuous at $\infty_X$. Since a base of neighbourhoods of $\infty_X$ is given by complements of compact subsets of X, the condition for that continuity is the one that I gave in my previous comment.