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If g°f is one-to-one, must f and g both be one-to-

  1. Mar 15, 2012 #1
    1. If g°f (g composed with f, two functions) is one-to-one, must f and g both be one-to-one?

    2. The answer is, no, only f has to be, but I just can't see why!
     
  2. jcsd
  3. Mar 15, 2012 #2
    Obviously g needs to be one to one when restricted to Im(f) ...
     
  4. Mar 15, 2012 #3
    Apparently not! I first guessed that both f and g had to be one-to-one, because I could not draw a map otherwise, but the graded work sent back to me said "No! Only f has to be 1-1"
     
  5. Mar 15, 2012 #4
    Also, the next problem is: "Let f:A→B and g:B→C be maps such that g°f is injective. Prove that f must be injective."

    So, twice he's said that it is f that must be injective. I just can't figure out why g doesn't have to be...
     
  6. Mar 15, 2012 #5
    Your grader is right, now try to read my post again and complete the ... part, starting with a 'but'.
     
  7. Mar 15, 2012 #6
    Ok, I am seeing an example and a counterexample.

    f={(1,1),(2,2)} and g={(1,1),(2,2),(3,2)}. Then f is one-to-one, g is not, and g°f={(1,1),(2,2)} is. Right? This works because the domain of g is not restricted to the image of f.

    Also, f={(1,1),(2,2),(3,2)} and g={(1,1),(2,2)}. Then f is not one-to-one, g is, and g°f={(1,1),(2,2),(3,2)} is not.
     
  8. Mar 15, 2012 #7
    Does this apply to the problem?
     
  9. Mar 15, 2012 #8
    yes that is right, but the last part does not prove anything, only the first part is important
     
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