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Domain and range can't be negative

  1. Jul 31, 2010 #1
    1. The problem statement, all variables and given/known data
    Given f:A→B and g:B→C, let h=g(f(a))

    If h is injective, then g is injective.

    Give a counter example.

    2. Relevant equations
    Injection: Let f:A→B

    For all f(x1)=f(x2) implies x1=x2

    3. The attempt at a solution

    f(a)=[tex]\sqrt{a}[/tex] from [0, infinity]→[0,infinity]

    g(b)=b2 from [all real numbers]→[0,infinity]

    h(a)=a from [0, infinity]→[0,infinity]

    Assuming h is injective and g is not, then this is a counter example. My problem is I am not sure if the square root messes things up here. I know that the square root of a squared is plus or minus a, but because the domain and range can't be negative ( I think), then this works.

    So is this correct?
  2. jcsd
  3. Jul 31, 2010 #2
    Re: Injection

    I believe you are correct.
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