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Inverse Functions

  1. Aug 12, 2009 #1

    jgens

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    Gold Member

    1. The problem statement, all variables and given/known data

    Suppose [itex]g[/itex] is a function with the property that [itex]g(x) \neq g(y)[/itex] if [itex]x \neq y[/itex]. Prove that there is a function [itex]f[/itex] such that [itex]f \circ g = I[/itex]

    2. Relevant equations

    A function is collection of ordered pairs with the property that if [itex](a,b)[/itex] and [itex](a,c)[/itex] are in the collection, then [itex]b = c[/itex].

    3. The attempt at a solution

    Here are my thoughts so far (this isn't a proof):

    From the restrictions on [itex]g[/itex] we have that [itex]g(a) = g(b)[/itex] if and only if [itex]a = b[/itex]; hence, [itex]g[/itex] is the collection of ordered pairs [itex](x,g(x))[/itex] with the property that [itex](a,c)[/itex] and [itex](b,c)[/itex] are in the collection if and only if [itex]a = b[/itex].

    If the composition of [itex]f[/itex] and [itex]g[/itex] is the identity function then [itex]f \circ g[/itex] is the collection of ordered pairs [itex]((x,g(x)),(f \circ g)(x)) = ((x,g(x)),x) = (g(x),x)[/itex] such that if [itex](a,b)[/itex] and [itex](a,c)[/itex] are in the collection, then [itex]b = c[/itex]. Since [itex]g[/itex] has this property, it seems like there should be some function [itex]f[/itex] such that [itex]f \circ g = I[/itex].

    Now, I'm sure that what I've shown so far is an abuse of/improper notation and probably is not valid. I would just like some help to string everything together. Thanks!
     
  2. jcsd
  3. Aug 12, 2009 #2
    Since you are defining a function as a collection of ordered pairs, why not define f to be the collection of ordered pairs that corresponds to reversing the order of the pairs in the collection g(x). Then the fact that fog = I is by definition, and you have only to prove that f is a function, which should be automatic given the definition of g.
     
  4. Aug 12, 2009 #3

    jgens

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    I actually thought about that about a half-hour after my last post. Thanks!
     
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