Solution
= Solution
The proposed <inverse function> cancels on both sides:
$$
\begin{aligned}
(f\circ g)\circ(g^{-1}\circ f^{-1})&=\operatorname{id},\\
(g^{-1}\circ f^{-1})\circ(f\circ g)&=g^{-1}\circ(f^{-1}\circ f)\circ g=\operatorname{id}.
\end{aligned}
$$
The uniqueness of an <inverse function> therefore gives
$$
\boxed{\text{True: }(f\circ g)^{-1}=g^{-1}\circ f^{-1}.}
$$
The operation applied last must be undone first.