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Some Compostion Proofs

  1. Mar 14, 2008 #1
    [SOLVED] Some Compostion Proofs

    1. The problem statement, all variables and given/known data

    Prove:
    1.) The composition of subjective functions is subjective
    2.) The composition of injective functions is injective



    2. Relevant equations

    Subjective: A function f: A->B is surjective iff

    For all members of B, there exists a member of A where f(a)=b

    Injective: A funtion f: A->B in injective iff

    f(a)=f(b) -> a=b

    3. The attempt at a solution

    I really don't know how to start this proof, mainly because in the questions, the domain and codomain are not defined in any way. However, both statements seem to be obviously true to me, at least I can't think of any obvioius counter examples.

    I.) Suppose we have functions f an g which are both surjective. The conpostion of these functions can b e written as g(f(x)). If g is subjective, regardless of the values of f(x), then gof will be subjective.

    That proof's really bad I know, and I don't know how to even start 2. Help! It seems like it should be really easy and I feel like an idiot for not knowing how to do these proofs.
     
  2. jcsd
  3. Mar 14, 2008 #2
    Sorry, i meant surjective
     
  4. Mar 14, 2008 #3

    CompuChip

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    For injectivity, suppose that g(f(x)) = g(f(y)). You must prove from this that this implies x = y. (Hint: you know nothing, except that f and g are injective).

    For surjectivity, let z be in the codomain of (g o f). Prove that there is an y in the domain of g such that g(y) = z. Now can you find an x such that g(f(x)) = z, using surjectivity of f?
     
  5. Mar 14, 2008 #4
    Thanks, I think I got the Injective proof spot on now and I think my surjective proof is good to (though my wording is sorta weird, I'll work On it.)

    I have one last problem that I want to check.

    Q: The product of injective functions is injective/


    Can I easily disprove this by letting f(x) = x and g(x) = 2x, which are both injctive and show that their product isnt?
     
  6. Mar 14, 2008 #5

    Dick

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    That will work. Now just give two different x values that map to the same value of f(x)*g(x).
     
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