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Math Help - extension of a surjective map

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    extension of a surjective map

    Let A and B be sets. Let S be a subet of A. If f is a surjective map from S to B, then the extension of f from A to B also surjective.

    At first i thought this was false. But then i thought it was true... and now i've really confused myself. Can anyone help?
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    Quote Originally Posted by junebug5389 View Post
    Let A and B be sets. Let S be a subet of A. If f is a surjective map from S to B, then the extension of f from A to B also surjective.

    At first i thought this was false. But then i thought it was true... and now i've really confused myself. Can anyone help?
    Let \hat f be an extension of f, i.e. f\subseteq \hat f and \hat f |_S = f.
    We are told f: S\to B is surjective.
    And we are given \hat f : A\to B.
    Let b\in B then there is s\in S so that f(s) = b.
    But since s\in S\subseteq A it means \hat f (s) = \hat f|_S (s) = f(s) = b.
    Thus, \hat f is surjective.
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    Quote Originally Posted by junebug5389 View Post
    Let A and B be sets. Let S be a subet of A. If f is a surjective map from S to B, then the extension of f from A to B also surjective.

    Here is another way to see it. Every function is a set of ordered pairs.
    If f is onto then B = \left\{ {y:\left( {\exists x} \right)\left[ {\left( {x,y} \right) \in f} \right]} \right\}.
    So clearly any extension of f looses no pairs.
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