Is it possible to have two distinct one-way functions (called, say, $h$ and $g$) such that their composition $h \circ g = [\, x \mapsto h(g(x)) \,]$ is not one-way?
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The function $f$ introduced by Maeher in this answer to a related question should also do the job here (as both $g$ and $h$). For convenience, let me quote that answer here:
Assume that a one-way function $h$ exists where in- and output length are the same. We call this length $n/2$. I.e. we have a one-way function $$h : \{0,1\}^{n/2} \to \{0,1\}^{n/2}.$$
From this function, we now construct a new function $$f : \{0,1\}^{n} \to \{0,1\}^{n}$$ as follows: $$f(x_1\Vert x_2) = 0^{n/2}\Vert h(x_1),$$ where $|x_1|=|x_2|=n/2$.
As shown in Maeher's answer, $f$ is one-way if $h$ is. However, $f(f(x)) = 0^{n/2}\Vert h(0^{n/2})$ is constant, independently of $x$ (except for its length $n$), and thus finding pre-images is trivial.
Ilmari Karonen
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Ilmari's answer is incorrect.
$f(f(x))$ is not equal to $0^{n/2}\| h(0^{n/2})$. Instead it equals $0^{n/2}\| h(h(x))$ which is not necessarily easy to invert. In particular, if $h \circ h$ is also one-way then $f \circ f$ would also be one way.
Anon
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