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1-1 and onto functions

  1. Sep 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Let f: A→B be a surjection and g:B→C be such that g∘f is an injection. Prove that both f and g are injections

    2. Relevant equations



    3. The attempt at a solution

    Suppose a = a' then g(f(a)) = g(f(a')) since g∘f is 1-1 we have that f(a) = f(a') hence f is 1-1.

    Is this correct?
     
  2. jcsd
  3. Sep 25, 2013 #2

    Office_Shredder

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    You assumed that a=a'and proved that f(a) = f(a'), which is simply showing that f is a function, not anything to do with whether f is 1-1. Furthermore, you went from g(f(a)) = g(f(a')) to f(a) = f(a') which seems to assume that g is injective already.

    You never used the fact that f is a surjection, which should also tip you off that you missed something.
     
  4. Sep 25, 2013 #3
    Let me redo.

    Suppose a,a' ∈ A and that f(a) = f(a') then g(f(a)) = g(f(a')) since g∘f is an injection then we have a = a' therefore f is injective.

    Is this correct?
     
  5. Sep 25, 2013 #4

    Office_Shredder

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    That looks better. Any ideas on what to do for g?
     
  6. Sep 25, 2013 #5
    Thanks! For g:

    Let b, b' ∈ B. Then since f is onto there exists an a, a' ∈ A such that f(a) = b and f(a') = b'. Suppose g(b)=g(b') then we get g(b) = g(f(a)) = g(f(a'))= g(b') since g∘f is 1-1 then a = a' hence f(a) = f(a') therefore b = b' which means g is 1-1.

    Is this correct?
     
  7. Sep 25, 2013 #6

    Office_Shredder

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    That looks good to me.
     
  8. Sep 25, 2013 #7
    Thank you very much!
     
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