If $h:[0,1]\to [0,1]$ is continuous and surjective, then does there exist a continuous function $f$ such that $ff=h?$ If so, is $f$ unique?

Question

Let $h:[0,1]\to [0,1]$ be a surjective continuous real function. There surely exist many functions $f:[0,1]\to [0,1]$ such that $f(f(x)) = h(x)\ $ on $x\in [0,1].$

Does there exist a continuous function $f:[0,1]\to [0,1]$ such that $f(f(x)) \equiv h(x) ?$ If so, is this function unique?

If the answer is yes, a further question could be: would the answer still be yes if we did not require $h$ to be continuous?

If the answer is no, a further question could be: would the answer be yes if in the question we required $h$ be bijective (and therefore monotone) (rather than just surjective)?

Answer 1

We have the following theorems:

• If $X$ is any set and $h : X \to X$ is injective (resp. surjective) and $f : X \to X$ satisfies $h = f\circ f$, then $f$ is injective (resp. surjective).
• If $f : [0,1] \to [0,1]$ is continuous and injective, then $f$ is strictly monotonic.
• If $X$ is an ordered space and $f : X \to X$ is (strictly) monotonic, then $f\circ f$ is (strictly) increasing.

Together, this means that when $h$ is assumed to be injective, the existence of continuous $f$ such that $h = f\circ f$ is sufficient to prove that $h$ is strictly increasing.

Since the assumptions in your question allow for $h(x) = 1-x$ (or $h(x) = 1-x^3$, or $h(x) = \cos(\pi x/2)$, etc.), all your answers are "no".

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To me, the natural follow-up question would be "what if we assume $h$ is continuous, bijective and strictly increasing?" to which the answer is "Yes, then $f$ exists, but it's not unique unless we restrict to $f$ being increasing."