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Math Help - Continuous function and compactifications

  1. #1
    CSM
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    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?
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    Quote Originally Posted by CSM View Post
    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.
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    CSM
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    Quote Originally Posted by CSM View Post
    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?
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    Quote Originally Posted by CSM View Post
    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.
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