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Let $f: X \rightarrow Y$ 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 $y \in Y, f_1(y)=I_y(y)=f_1(y)$. Use one-one ness of $f$.
Hint: Suppose $g_1$ and $g_2$ are two inverses of $f$. Then for all $y \in Y, f_1(y)=I_y(y)=f_1(y)$. Use one-one ness of $f$.
Solution:
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Verified Answer
If $\mathrm{f}$ is invertible $g \circ f(\mathrm{x})=\mathrm{I}_{\mathrm{x}}$ and fog $(\mathrm{y})=\mathrm{I}_{\mathrm{y}}$ $\therefore f$ is one-one and onto.
Let there be two inverse $g_1$ and $g_2$ $\operatorname{fog}_1(\mathrm{y})=\mathrm{I}_{\mathrm{y}}$, fog $_2(\mathrm{y})=\mathrm{I}_{\mathrm{y}}$
$\mathrm{I}_{\mathrm{y}}$ being unique for a given function $\mathrm{f}$
$$
\Rightarrow \mathrm{g}_1(\mathrm{y})=\mathrm{g}_2(\mathrm{y})
$$
$\therefore \mathrm{f}$ is one-one and onto
$\therefore$ fhas a unique inverse.
Let there be two inverse $g_1$ and $g_2$ $\operatorname{fog}_1(\mathrm{y})=\mathrm{I}_{\mathrm{y}}$, fog $_2(\mathrm{y})=\mathrm{I}_{\mathrm{y}}$
$\mathrm{I}_{\mathrm{y}}$ being unique for a given function $\mathrm{f}$
$$
\Rightarrow \mathrm{g}_1(\mathrm{y})=\mathrm{g}_2(\mathrm{y})
$$
$\therefore \mathrm{f}$ is one-one and onto
$\therefore$ fhas a unique inverse.
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