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Let $K$ be a compact Hausdorff space. I'm wondering: Does there always exist a subset $J \subseteq C(K, K)$ such that:

  • $J$ is closed under composition,
  • there is an element $f \in C(K)$ such that the map $J\rightarrow C(K)$, $g \mapsto f\circ g$ is bijective.

If not, how does a possible counterexample look like?

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  • $\begingroup$ What about $J=C(K,K)$ and $f$ the identity? $\endgroup$
    – Benjamin Steinberg
    Commented May 3, 2018 at 15:23
  • $\begingroup$ I want the function $f$ to be an element of $C(K)$. Since $K$ is not necessarily a subset of $\mathbb{C}$ there might be no identity in $C(K)$ $\endgroup$
    – worldreporter
    Commented May 3, 2018 at 15:34
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    $\begingroup$ So to clarify, $C(K)$ means $C(K, \mathbb{C})$? $\endgroup$
    – Nate Eldredge
    Commented May 3, 2018 at 18:05

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Fix any $f \in C(K)$. Then the range $f(K)$ is compact, so there is some $z_0 \in \mathbb{C}$ which is not in the range of $f$. The constant function $z_0$ is thus not of the form $f \circ g$ for any $g \in C(K,K)$, so your map $J \to C(K)$ cannot be surjective, no matter what $J$ is.

Is there something missing from the statement of the problem?

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