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Thread: 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 $\displaystyle \rightarrow$ B and g : B $\displaystyle \rightarrow$ C be functions.

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

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

    Here is what I have so far:

    $\displaystyle g \circ f : A \rightarrow C. $ Now assume that $\displaystyle g \circ f $ is injective. then $\displaystyle g \circ f $ is one to one and $\displaystyle 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 $\displaystyle 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 $\displaystyle t=s$ then $\displaystyle g(t)=g(s)$ must be true.

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