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Math Help - Inverse Composition of Functions Proof

  1. #1
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    Inverse Composition of Functions Proof

    I'm new here, though I wish I had found this forum long ago. (I'm an applied math major.)

    I've run into trouble on my homework which is, of course, due tomorrow. Here we go:

    If f: A -> B and g: B -> C are one-to-one functions, show that (g o f)^-1 = f^-1 o g^-1 on Range (g o f).

    I'm stuck on where to even start on this proof, and it's driving me mad. For the record, the above notation ^-1 denotes "inverse". Please help?
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  2. #2
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    Quote Originally Posted by Jedi Arashi View Post
    I'm new here, though I wish I had found this forum long ago. (I'm an applied math major.)

    I've run into trouble on my homework which is, of course, due tomorrow. Here we go:

    If f: A -> B and g: B -> C are one-to-one functions, show that (g o f)^-1 = f^-1 o g^-1 on Range (g o f).

    I'm stuck on where to even start on this proof, and it's driving me mad. For the record, the above notation ^-1 denotes "inverse". Please help?
    You need to show that (f^{-1}\circ g^{-1})\circ (g\circ f)= (g\circ f)\circ (f^{-1}\circ g^{-1}) = 1 by 1, it means identity. Because the definition of (g\circ f)^{-1} is a function which composed (either way) with g\circ f gives us the identity.
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  3. #3
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    (f^{-1}\circ g^{-1})\circ (g\circ f)= (g\circ f)\circ (f^{-1}\circ g^{-1}) = 1

    I'm not sure I'm on the same page with you here with the above. Let me see if I can use this math code to produce what I'm saying, because I feel like I'm not quite getting it still.

    (f^{-1}\circ g^{-1}) = f^{-1}\circ g^{-1}

    Okay, that's what I'm supposed to prove. Do I still do the same as above, or does the strategy change?

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  4. #4
    Senior Member tukeywilliams's Avatar
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    Use the following proposition:The functions  f: X \to Y and  g: Y \to X are inverses of each other if and only if  g \circ f = I_X and  f \circ g = I_Y .
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    Quote Originally Posted by tukeywilliams View Post
    Use the following proposition:The functions  f: X \to Y and  g: Y \to X are inverses of each other if and only if  g \circ f = I_X and  f \circ g = I_Y .
    But I'm only given that f is one-to-one, and g is one-to-one (which means  g \circ f is one-to-one as well). They aren't inverses of one another.
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  6. #6
    Senior Member tukeywilliams's Avatar
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    Its an if and only if statement.  f: X \to Y and  g: Y \to X  are inverses of each other  \Leftrightarrow g \circ f = I_X and  f \circ g = I_Y .


    So start by taking  (g \circ f) \circ (f^{-1} \circ g^{-1}) and  (f^{-1} \circ g^{-1}) \circ (g \circ f) .
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  7. #7
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    It's f: A \to B and g: B \to C. Or, I suppose if you prefer, f: X \to Y and g: Y \to Z.

    You'll have to pardon me if I sound rather confused (well, I am, forgive me!). I've spent the past hour or so at least on this problem, and proofs have never been anything near my strong point.

    Crud, I'm sorry! I mixed up the latter equation!

     (g \circ f)^{-1} = f^{-1} \circ g^{-1}

    I'm such an idiot, I'm sorry! That's what I'm supposed to prove.
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  8. #8
    Senior Member tukeywilliams's Avatar
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    First show that  g \circ f is a bijection and then show that its inverse function is  f^{-1} \circ g^{-1} using the above proposition.
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  9. #9
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    I only have to prove it within the range of g \circ f. Is it still required that I prove it to be a bijection, or is there another way to go about it?
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  10. #10
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    Recall that functions are just relations (sets of ordered pairs) with special restrictions.
    In general this property is true for relations:
    \left( {p,q} \right) \in g \circ f\; \Leftrightarrow \;\left( {\exists x} \right)\left[ {\left( {p,x} \right) \in f \wedge \left( {x,q} \right) \in g} \right]
    This means that:
    \left[ {\left( {p,x} \right) \in f \wedge \left( {x,q} \right) \in g} \right]\; \Leftrightarrow \;\left[ {\left( {x,p} \right) \in f^{ - 1}  \wedge \left( {q,x} \right) \in g^{ - 1} } \right]\; \Leftrightarrow \;\left( {q,p} \right) \in f^{ - 1}  \circ g^{ - 1}

    Because you are dealing with only the ranges of the two functions, assume they are bijections.
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