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Thread: extension of a surjective map

  1. #1
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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 $\displaystyle \hat f$ be an extension of $\displaystyle f$, i.e. $\displaystyle f\subseteq \hat f$ and $\displaystyle \hat f |_S = f$.
    We are told $\displaystyle f: S\to B$ is surjective.
    And we are given $\displaystyle \hat f : A\to B$.
    Let $\displaystyle b\in B$ then there is $\displaystyle s\in S$ so that $\displaystyle f(s) = b$.
    But since $\displaystyle s\in S\subseteq A$ it means $\displaystyle \hat f (s) = \hat f|_S (s) = f(s) = b$.
    Thus, $\displaystyle \hat f$ 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 $\displaystyle f$ is onto then $\displaystyle B = \left\{ {y:\left( {\exists x} \right)\left[ {\left( {x,y} \right) \in f} \right]} \right\}$.
    So clearly any extension of $\displaystyle f$ looses no pairs.
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