# Thread: Continuous function and compactifications

1. ## Continuous function and compactifications

Can someone help me on this one?

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

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

Given two compact Hausdorff Space $\displaystyle X,Y$ and a continuous function $\displaystyle f:X\rightarrow Y$ which conditions are necessary and sufficient to garantuee the existence of a continuous function $\displaystyle \hat{f}:\hat{X}\rightarrow \hat{Y}$ such that $\displaystyle \hat{f}|X=f$ ($\displaystyle \hat{f}$ restricted to $\displaystyle X$) where $\displaystyle \hat{X},\hat{Y}$ 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, $\displaystyle \hat{f}$ can always be defined by taking $\displaystyle \hat{f}(\infty_X) = \infty_Y.$

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

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

Given two compact Hausdorff Space $\displaystyle X,Y$ and a continuous function $\displaystyle f:X\rightarrow Y$ which conditions are necessary and sufficient to garantuee the existence of a continuous function $\displaystyle \hat{f}:\hat{X}\rightarrow \hat{Y}$ such that $\displaystyle \hat{f}|X=f$ ($\displaystyle \hat{f}$ restricted to $\displaystyle X$) where $\displaystyle \hat{X},\hat{Y}$ are the one-point compactifications?
I am sorry. I meant locally compact indeed.
Is that condition you give sufficient? How do I see that $\displaystyle \hat{f}|X=f$ is guaranteed?

4. Originally Posted by CSM
Is that condition you give sufficient? How do I see that $\displaystyle \hat{f}|X=f$ is guaranteed?
The way to define $\displaystyle \hat{f}$ is to take $\displaystyle \hat{f}(x) = f(x)$ for all $\displaystyle x\in X$. Then we just have to define $\displaystyle \hat{f}$ at the point at infinity (which I denote by $\displaystyle \infty_X\in\hat{X}$), say $\displaystyle y_0 = \hat{f}(\infty_X).$ The function $\displaystyle \hat{f}:\hat{X}\to\hat{Y}$, 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 $\displaystyle \hat{f}$ is continuous at $\displaystyle \infty_X$. Since a base of neighbourhoods of $\displaystyle \infty_X$ is given by complements of compact subsets of X, the condition for that continuity is the one that I gave in my previous comment.