Solution
= Solution
The inverse factors are generally in the wrong order. For a counterexample, take the <bijections> $f(x)=x+1$ and $g(x)=2x$, with <inverse functions> $f^{-1}(x)=x-1$ and $g^{-1}(x)=x/2$. Then
$$
((f\circ g)\circ(f^{-1}\circ g^{-1}))(x)=2(x/2-1)+1=x-1.
$$
Thus
$$
\boxed{\text{False in general; the displayed composition need not be }\operatorname{id}.}
$$
For particular <bijections> that commute the equality can hold; the claim fails as a universal assertion.