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Homework Help: Functions - is this proof satisfactory?

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f be a function from A to B and g a function from B to C. Show that if the composite function g°f is one-one, then f is one-one.

    2. Relevant equations

    3. The attempt at a solution

    By definition of a function every value of A has to be maped to a value in B. Likewise every value of B has to be maped to a value in C. So by definition of 1-1... since g°f satisfies this rule and g°f is a composite function... Therefore f must be 1-1.

    Does this make a valid argument? and justify the answer?
  2. jcsd
  3. Mar 16, 2012 #2
    Its not so clear to me.
    If f(x) = f(y) for some x,y in A you have to show x=y.
  4. Mar 16, 2012 #3
    hmm I think I dont understand.
  5. Mar 16, 2012 #4
    You have to prove that if f(x) is not one one
    then function g(f(x) is not one one.

    Suppose that f(x) is not one one and for two values of x lets say x1 and x2(both lying in the domain) give a value y1.

    Now this y1 lies on domain of function g.

    What happens when you calculate g(y1)?

    What does it tell us about our original assumption the f(x) is not one one.


    Your proof is not right.(even if you understand why f should be one one, the proof is not satisfactory)
  6. Mar 16, 2012 #5
    But I have to prove f is one-one.

    But from the assumption say I have 3 sets A,B,C the same as the original question.
    Now x1 & x2 lie in A. when we plug in the values x1 & x2 into f wwe have they both = y1 which lies in B. So from this picture alone the function f :A→B is not one-one. But we have to prove the opposite!

    And to this statement.

    "What happens when you calculate g(y1)?"

    if the value of f at x1 and x2 = y1.... then the above statement is just saying g(f(x)).

    I'm confused?
  7. Mar 16, 2012 #6
    You are given g(f(x)) is one one.now you are supposed to prove f(x) is one one.

    To prove this we say let f(x) be many-one(i.e not one one).

    In that case we observe that g(f(x)) comes out to be many one and not one one(How?).However we are told that g(f(x)) is one one.
    This contradiction signifies that our assumption of f(x) being many one is wrong and therefore f(x) is one one.

    This is the way you can proceed.Solving the indivudual bits is now upto you
  8. Mar 16, 2012 #7


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    What is the definition of a one-to-one function ?
  9. Mar 16, 2012 #8
    No 2 elements in X can be maped to the same element in Y. for f:X→Y
    ie, f(x1) & f(x2) cannot both = y1.
  10. Mar 16, 2012 #9


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    As I read your proof, there is nothing in it which explicitly uses this fact.
  11. Mar 16, 2012 #10
    Can I prove it using a venn diagram?
    Im not so good putting what I want to say in words.
  12. Mar 16, 2012 #11
    Hello charmed beauty.

    The most elegant propf will be by proving that f(x) cant be many one(which i mentioned a while ago).

    Drawing Venn-Diagrams is also a v useful approach.
    However, as the no. of functions forming the composite increase, proving using Ven Diagrams will get v big.

    But they are good way to start as they convey a lot of information
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