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If g o f = idA and f o g = idB, then g = f^-1 proof?

  1. Mar 28, 2012 #1
    Hey guys,

    I need to prove: Suppose A and B are sets, and f and g are functions with f: A -> B and g: B -> A. If g o f = idA and f o g = idB, then f is invertible and g = f^-1.

    So far I have understood why g must be the inverse of f, but I do not know how to prove it.

  2. jcsd
  3. Mar 28, 2012 #2
    Take an arbitrary element a in A and b in B and show that they relate via composition of both functions.
  4. Mar 28, 2012 #3
    Hint: f(a) = b, g(b) = a
  5. Mar 28, 2012 #4
    Okay, not really sure how to do that but here's my attempt:

    Suppose g o f = idA,
    then dom(g o f) = dom(idA) = A
    and (g o f)(a) = g(f(a)) = idA(a) = a.

    Suppose f o g = idB,
    then dom(f o g) = dom(idB) = B
    and (f o g)(b) = f(g(b)) = idB(b) = b.
  6. Mar 28, 2012 #5
    A function is only invertible if it's 1 - 1 and onto. Is this the case? You didn't specify.
  7. Mar 28, 2012 #6
    So what I have is sufficient for g=f^-1? But first I need to show that f^-1 exists?
  8. Mar 28, 2012 #7
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