Superposition dimension a la Kolmogorov/Arnold

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(Now asked at MO.)

Given a topological space $\mathcal{X}$, let $$\mathsf{Cl_C}(\mathcal{X})=\bigcup_{n\in\mathbb{N}}C(\mathcal{X}^n,\mathcal{X})$$ be the clone of finite-arity continuous functions on $\mathcal{X}$ (where $\mathcal{X}^n$ is given the product topology). Motivated by the Kolmogorov-Arnold superposition theorem, let the superposition dimension of $\mathcal{X}$ be the minimal cardinality of a set of continuous functions $\mathfrak{F}\subseteq\mathsf{Cl_C}(\mathcal{X})$ such that the clone generated by $\mathfrak{F}\cup C(\mathcal{X},\mathcal{X})$ is all of $\mathsf{Cl_C}(\mathcal{X})$.

Some quick observations:

  • Any space homeomorphic to its square trivially has superposition dimension $1$: there is no real distinction between single- and multi-variable continuous functions over such a space once we bring in an appropriate pairing function.

  • The K/A theorem says that the superposition dimension of $[0,1]$ is $1$, witnessed by $\{+\}$.

    • Admittedly the failure to distinguish between Cantor space and the unit interval suggests that the word "dimension" is being wildly misused, but meh.
  • It is not clear to me that shifting attention to maps with codomain $[0,1]$ results in the same notion (I strongly suspect it doesn't), so e.g. Ostrand's extension of the K/A theorem doesn't seem relevant here.

I'm generally curious for any relevant information, but the following particular question seems interesting: is there, for each finite $n$, a "nice" topological space with superposition dimension $n$? Of course this is really a whole family of questions, one for each meaning of "nice." Tentatively I think the most interesting case is likely to be the following:

Question: Is there, for each finite $n$, a connected topological manifold (with or without boundary, not necessarily compact) with superposition dimension $n$?

I suspect that every topological manifold has superposition dimension $1$, by some not-too-complicated extension of K/A, but I don't immediately see how to show that.