Let $f: X \rightarrow Y$ be an invertible function. Show that the inverse of $f^{-1}$ is $f$,i.e.,$\left(f^{-1}\right)^{-1}=f$.
(A) Let $f : X \rightarrow Y$ be an invertible function.
By definition,a function $f$ is invertible if there exists a function $g : Y \rightarrow X$ such that $g \circ f = I_X$ and $f \circ g = I_Y$,where $I_X$ and $I_Y$ are identity functions on $X$ and $Y$ respectively.
In this case,$g = f^{-1}$.
Substituting $g = f^{-1}$ into the conditions,we have:
$f^{-1} \circ f = I_X$ and $f \circ f^{-1} = I_Y$.
Now,consider the function $f^{-1} : Y \rightarrow X$. For $f^{-1}$ to be invertible,there must exist a function $h : X \rightarrow Y$ such that $h \circ f^{-1} = I_Y$ and $f^{-1} \circ h = I_X$.
From the conditions $f \circ f^{-1} = I_Y$ and $f^{-1} \circ f = I_X$,we can see that $f$ acts as the function $h$.
Therefore,$f$ is the inverse of $f^{-1}$,which means $\left(f^{-1}\right)^{-1} = f$.
By definition,a function $f$ is invertible if there exists a function $g : Y \rightarrow X$ such that $g \circ f = I_X$ and $f \circ g = I_Y$,where $I_X$ and $I_Y$ are identity functions on $X$ and $Y$ respectively.
In this case,$g = f^{-1}$.
Substituting $g = f^{-1}$ into the conditions,we have:
$f^{-1} \circ f = I_X$ and $f \circ f^{-1} = I_Y$.
Now,consider the function $f^{-1} : Y \rightarrow X$. For $f^{-1}$ to be invertible,there must exist a function $h : X \rightarrow Y$ such that $h \circ f^{-1} = I_Y$ and $f^{-1} \circ h = I_X$.
From the conditions $f \circ f^{-1} = I_Y$ and $f^{-1} \circ f = I_X$,we can see that $f$ acts as the function $h$.
Therefore,$f$ is the inverse of $f^{-1}$,which means $\left(f^{-1}\right)^{-1} = f$.
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