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Math Help - Topological Spaces

  1. #1
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    Topological Spaces

    Let X and Y be a topological spaces.
    Suppose X is the union of two closed subsets A and B of X .
    Let f: A -> Y and g: B-> Y be continuous.
    If f(x)=g(x) for every x in the intersection of A and B, then let h: X -> Y be a function defined by setting h(x)=f(x) for x in A, and h(x)=g(x) for x in B.

    Then could we say that h is continuous ? Why?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lttlbbygurl View Post
    Let X and Y be a topological spaces.
    Suppose X is the union of two closed subsets A and B of X .
    Let f: A -> Y and g: B-> Y be continuous.
    If f(x)=g(x) for every x in the intersection of A and B, then let h: X -> Y be a function defined by setting h(x)=f(x) for x in A, and h(x)=g(x) for x in B.

    Then could we say that h is continuous ? Why?
    This is called the gluing lemma. It applies to finitely many closed sets and arbitrarily many open.

    So, suppose that X=A\cup B where A,B are closed subspaces of X and f:A\to Y and g:B\to Y are continuous functions such that f\mid_{A\cap B}=g\mid_{A\cap B} then f\sqcup g:X\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in A\\g(x)\quad\text{if}\quad x\in B\end{cases} is well-defined since f,g agree on the intersection of their domains. Also, it's continuous since if E\subseteq Y is closed it is easy to see that \left(f\sqcup g\right)^{-1}(E)=f^{-1}(E)\cup g^{-1}(E). But, since f,g are continuous f^{-1}(E),g^{-1}(E) are closed subspaces of A,B respectively but since they themselves are closed subspaces of X it follows that so are f^{-1}(E),g^{-1}(E). Thus, \left(f\sqcup g\right)^{-1}(E) is the finite union of closed sets and thus closed. The conclusion follows.


    This can be generalized, as I said, to a finite partition of closed sets or an arbitary partition of open ones.
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