# The set of 1-1 Mapping of S Onto itself

1. Jan 23, 2012

### kmikias

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. Jan 23, 2012

### micromass

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

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

Then $fg(x)=x^3+1$ and $gf(x)=(x+1)^3$. These are not equal.

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

$$g^{-1}gf=g^{-1}f^{-1}g$$

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