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Math Help - composite functions and their injectivity and surjectivity

  1. #1
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    composite functions and their injectivity and surjectivity

    Here is the question:

    let f : A \rightarrow B and g : B \rightarrow C be functions.

    i) show that if g \circ f is injective, then f is injective

    ii) show that if g \circ f is surjective, then g is surjective.

    Here is what I have so far:

    g \circ f : A \rightarrow C. Now assume that  g \circ f is injective. then  g \circ f is one to one and  g(f(x_1)) = g(f(x_2)) \Rightarrow f(x_1) = f(x_2)

    So from here I know the goal is to show that  f(x_1)=f(x_2)  \Rightarrow  x_1=x_2 . But how?? my guess is that inverse functions are involved?? maybe not.

    As far as part b goes, I'm thinking that it is a similar process, almost reversed. but I'm still somewhat confused on this part. thank you for the help!
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  2. #2
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    Realize that if t=s then g(t)=g(s) must be true.

    So suppose that f(p)=f(q). It must be true that g(f(p))=g(f(q)).
    From the given what follows?
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  3. #3
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    if the composition is injective, then (g\circ f)(x_1)=(g\circ f)(x_2)\Rightarrow x_1=x_2, in your first line your implication actually says g is injective, not g composed with f
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