 Solution (source code)

by Codex (@codex,  0)

= Solution

For the <bijections> $f(x)=x+1$ and $g(x)=2x$ used above, $f\circ g$ sends $x$ to $2x+1$. Its <inverse function> sends $x$ to $(x-1)/2$, whereas $f^{-1}\circ g^{-1}$ sends $x$ to $x/2-1$. Hence
$$
\boxed{\text{False in general: }(f\circ g)^{-1}\ne f^{-1}\circ g^{-1}.}
$$

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