1-1 and onto functions

  • Thread starter Lee33
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  • #1
Lee33
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Homework Statement



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

Homework Equations





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?
 

Answers and Replies

  • #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.
 
  • #3
Lee33
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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?
 
  • #4
Office_Shredder
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That looks better. Any ideas on what to do for g?
 
  • #5
Lee33
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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?
 
  • #6
Office_Shredder
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That looks good to me.
 
  • #7
Lee33
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Thank you very much!
 

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