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Math Help - Sets

  1. #1
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    Sets

    Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

    If g of f maps A to C is an injection, then f is an injection.

    Not sure how to do this proof
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by dwsmith View Post
    Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

    If g of f maps A to C is an injection, then f is an injection.

    Not sure how to do this proof
    How about an indirect proof, that is: show that if f:\, A\rightarrow B is not an injection, then g\circ f:\, A\rightarrow C is not an injection either.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Let A,B, and C be nonempty sets and lef f map A to B and g map B to C.

    If g of f maps A to C is an injection, then f is an injection.

    Not sure how to do this proof
    Here are some thoughts on an informal direct proof.
    Let's say you already know (in the given context) the definition of injection.

    For arbitrary x and y in A, suppose that over in B, f(x) = f(y).
    This identity should tell you that (over in C) the image of f(x) under g is identical to the image of f(y) under g, i.e., that g(f(x)) = g(f(y)).
    This can be written, (gof)(x) = (gof)(y).
    Since gof is injective, you now have x = y (by definition).
    So the arbitrariness of x, y and the supposition that f(x) = f(y), leads you to x = y, making f injective (by definition).
    Last edited by PiperAlpha167; April 3rd 2010 at 03:54 AM. Reason: Removed an irrelevant comment.
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