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Math Help - Graphs

  1. #1
    Senior Member tukeywilliams's Avatar
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    Graphs

    Given  f: X \to Y is a function with graph  G_f \subseteq X \times Y prove that  f is surjective  \Leftrightarrow \forall y \in Y, (X \times \{y \} \cap G_f) \neq \emptyset .

    Proof: '  \Rightarrow ' direction: If  f is surjective, then  \forall y \in Y, \exists x \in X, y = f(x) . Then  X \times \{y \} \cap G_f \Rightarrow x \in X and  y \in \{y \} or  y = y . Then  X \times \{y \} \cap G_f \neq \emptyset because by definition  f is surjective.

    '  \Leftarrow ' direction: If  \forall y \in Y, (X \times \{y \} \cap G_f) \neq \emptyset then  x \in X and  y = y so that  X \times \{y \} \cap G_f \neq \emptyset. Then  \forall y \in Y, \exists x \in X, y = f(x) so that  f is surjective.  \square

    Is this proof correct?

    Thanks
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  2. #2
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    Quote Originally Posted by tukeywilliams View Post
    Is this proof correct?
    Yes it is correct and well done.
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