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The set of 1-1 Mapping of S Onto itself

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I was reading my text book and i encountered this....--->>

    " For instance if f,g,h are in A(S) and fg = fh then g=h " I understand this part... because we can take the the inverse of f both sides and say g=h.

    then it says--->> " If gf = f^(-1)g but since f ≠ f^(-1) we cannot cancel the g here"

    SO MY QUESTION IS
    IF gf = f^(-1)g then ....why can't we take inverse of g both sides and show f = f^(-1) and say f = f^(-1) and g = g.
     
  2. jcsd
  3. Jan 23, 2012 #2

    micromass

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    Note that for functions it is NOT true that fg=gf in general.

    This is very easy to see. Let [itex]f(x)=x+1[/itex] and let [itex]g(x)=x^3[/itex].

    Then [itex]fg(x)=x^3+1[/itex] and [itex]gf(x)=(x+1)^3[/itex]. These are not equal.

    Now, back to your question. Let's assume that [itex]gf=f^{-1}g[/itex]. If we take [itex]g^{-1}[/itex] of both sides, then we end up with

    [tex]g^{-1}gf=g^{-1}f^{-1}g[/tex]

    The left side is equal to f, no problem. But we can do nothing with the right side. Indeed, we can't just switch [itex]g^{-1}[/itex] and [itex]f^{-1}[/itex]!!
     
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