In any group G prove that $\displaystyle (a^{-1})^{-1} = a$

Let $\displaystyle f$ be a 1-1 mapping of G onto itself such that $\displaystyle f(a) = a^{-1}$ and let $\displaystyle f^{-1}$ be a 1-1 mapping of G onto itself such that $\displaystyle f^{-1} (a^{-1}) = a$

then $\displaystyle (a^{-1})^{-1} \equiv (f(a))^{-1} = f^{-1} \circ f(a) = f^{-1}(a^{-1}) = a$

Is this a valid argument? I'm just getting started with abstract algebra.