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Let \(f : X \to Y\) be an invertible function. Show that \(f\) has unique inverse.

(Hint: suppose  \( g_1 \)and  \(g_2\) are two inverses of \(f\). Then for all \( y \in Y \), \( fog_1(y) = 1 _Y (y) = fog_2 (y). \)   Use ono-one ness of \(f\))          
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  • To show that $f:X \to Y $ has unique inverse function we take two functions $g_1\; and \;g_2$ as two inverse functions of f then show $g_1=g_2$
Given $f$ is invertible. We need to show that $f$ has a unique inverse.
If a function $f$ has two inverses $g_1$ and $,g_2 $ then $fog_1 (y)=I_y\;and\;fog_2(y)=I_y$ for $y \in Y$
$\Rightarrow fog_1 (y)=I_y(y)=fog_2(y)$
$\Rightarrow f(g_1(y))=f(g_2(y))$
Since $f $ is invertible $f$ is one-one, $\Rightarrow g_1(y)=g_2(y)$
There if $f$ has two inverses, then $g_1=g_2$. In otherwords, $f$ will only have a unique inverse.
answered Feb 26, 2013 by meena.p
edited Mar 19, 2013 by balaji.thirumalai

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