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Math Help - Problem involving Indentity, Coposition functoin

  1. #1
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    Problem involving Indentity, Coposition functoin

    Hey, I don't know how to prove this.
    So... help me

    Let f: X -> Y and g: Y -> X be functions so that g o f = 1x(which means Identity function). Prove that f is injective and g is surjective. Need either be bijective?

    On hints sheet, This has 3 parts. Justify your answer to last part w/ a proof or counterexample.

    Thank you
    Last edited by 892king; April 29th 2008 at 02:27 PM. Reason: sr
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  2. #2
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    Quote Originally Posted by 892king View Post
    Hey, I don't know how to prove this.
    So... help me

    Let f: X -> Y and g: Y -> be functions so that g o f = 1x(which means Identity function). Prove that f is injective and g is surjective. Need either be bijective?

    On hints sheet, This has 3 parts. Justify your answer to last part w/ a proof or counterexample.

    Thank you
    For Part 1: Suppose f is not injective. Then there exist a, b \in X such that a is not equal to b and f(a) = f(b) = c \in Y. By definition, g(c) = a and g(c) = b. But in order for g to be a function, a = b, which is a contradiction.
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  3. #3
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    Quote Originally Posted by icemanfan View Post
    For Part 1: Suppose f is not injective. Then there exist a, b \in X such that a is not equal to b and f(a) = f(b) = c \in Y. By definition, g(c) = a and g(c) = b. But in order for g to be a function, a = b, which is a contradiction.
    Thank you for your help.
    Can I prvoe g's being surjective by same way?
    Like,

    Suppose g is not surjective. Then there exists a \in Y such that a is not contained in the image of set Y. By definition, g(a) doesn't exist. But in order for g to be a function, g(a) should be defined(not sure about this part).

    892King
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