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Function counterexample

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]f: A\to B[/tex]. I'm trying to find a function [tex]g: B\to C[/tex] such that [tex]g[/tex] is not [tex]1-1[/tex] but [tex]g\circ f[/tex] is.

    The original assignment (which I've completed) was to prove that for all functions [tex]f: A\to B[/tex] and [tex]g: B\to C[/tex], if [tex]g\circ f[/tex] is [tex]1-1[/tex], then so is [tex]f[/tex]. However, in the process of completing the assignment, I tried (out of curiosity) to find g's that weren't 1-1. But I couldn't.
     
  2. jcsd
  3. Jan 31, 2009 #2
    f: n -> (n,n)
    g: (m,n) -> m
    gf = identity, neither f nor g is one-to-one
     
  4. Jan 31, 2009 #3
    Have you thought about why it may not be possible? If the composite is defined then [tex]g\circ f[/tex] = g(f(x)) where x is in A. If f(x) is bijective then g(f(x)) = g(B)
     
  5. Jan 31, 2009 #4
    First of all, when specifying a function (and worrying about its properties), without the domain and codomain, you can't really tell much.

    But assuming you have f: R -> R^2, and g: R^2 -> R, then f actually is one to one.

    edit:
    But since g isn't one-to-one, you have answered my question. So, thanks!
     
  6. Jan 31, 2009 #5
    The domain is any arbitrary set A with more than 1 element, the codomain is A x A.
     
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