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Composite and one to one function

  1. Sep 16, 2006 #1
    Hi,

    I have this question:
    if f and [tex] f \circ g [/tex] are 1 to 1 functions, does it follow that g is 1-1 function?

    My answer is NO g doesn't have to be 1-1 function.

    for example ,

    if g(3)=4 and g(7)=4,
    then f( g(3) )= f( G(7) )=f(4) will produce the 1-1 function property.

    Is my reasonning OK??

    B
     
  2. jcsd
  3. Sep 16, 2006 #2
    The best way to look at "true/false" questions like this is often to try to prove whatever it's asking, and if you can't, then see if you can build a counterexample by paying attention to what "breaks" your proof.

    Your counterexample doesn't work, since then you'll have, as you wrote, [itex](f\circ g)(3) = (f\circ g)(7)[/itex], which obviously means [itex]f \circ g[/itex] is not 1-1.

    So try to see if you can prove it. I'd suggest doing it by contradiction.

    (And just to remind you, a function [itex]h[/itex] is 1-1 iff h(x) = g(y) implies x=y).
     
    Last edited: Sep 16, 2006
  4. Sep 17, 2006 #3
    Ok thank I found the problem
     
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