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Could we always locally represent a continuous function $F(x,y,z)$ in the form of $g\left(f(x,y),z\right)$ for suitable continuous functions $f$, $g$ of two variables? I am aware of Vladimir Arnold's work on this problem, but it seems that in that context $F(x,y,z)$ is written as a sum of several expressions of this form. Can one reduce it to just a single superposition $g\left(f(x,y),z\right)$; or does anyone know a counter example?

For related posts see:

Is there any continuous ternary function which can not be represented by composition of continuous binary functions?

Kolmogorov superposition for smooth functions

Kolmogorov-Arnold theorem for (just-)functions

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Proposition. The function $F(x,y,z)=x(1-z)+yz$ cannot be represented as $F(x,y,z)=g(f(x,y),z)$, where $f,g:\mathbb{R}^2\to\mathbb{R}$ are continuous.

Proof. Suppose to the contrary that we have such a representation. Let $g_1(t)=g(t,0)$. Then $$ g_1(f(x,y))=g(f(x,y),0)=F(x,y,0)=x. $$ Let $g_2(t)=g(t,1)$. Then $$ g_2(f(x,y))=g(f(x,y),1)=F(x,y,1)=y. $$ Let $h=(g_1,g_2):\mathbb{R}\to\mathbb{R}^2$, $h(t)=(g_1(t),g_2(t))$. Then $$ (h\circ f)(x,y)=(g_1(f(x,y)),g_2(f(x,y)))=(x,y). $$ In particular, it implies that $f:\mathbb{R}^2\to\mathbb{R}$ is one-to-one.

It remains to show there there are no one-to-one continuous functions from $\mathbb{R}^2$ to $\mathbb{R}$. Suppose to the contrary that such a function $f$ exists. Let $\mathbb{S}^1\subset\mathbb{R}^2$ be the unit circle. Then $$ f|_{\mathbb{S}^1}:\mathbb{S}^1\to\mathbb{R} $$ is continuous and one-to-one. Since $\mathbb{S}^1$ is compact, $f|_{\mathbb{S}^1}$ is a homeomorphism of $\mathbb{S}^1$ onto $f(\mathbb{S}^1)$.

Since $\mathbb{S}^1$ is connected and compact $f(\mathbb{S}^1)\subset\mathbb{R}$ is connected and compact. Therefore it is a closed interval, $f(\mathbb{S}^1)=[a,b]$ and we arrive to a contradiction, because $[a,b]$ is not homeomorphic to $\mathbb{S}^1$ (to see this observe that removing an interior point from $[a,b]$ makes the space disconnected while if we remove a point form $\mathbb{S}^1$, the space will remain connected).

Remark. For a statement and a proof of the Kolmogorov theorem, see for example:

G. G. Lorentz, Approximation of functions, 1966.

Edit. I modified my original answer that was not fully correct. A mistake was pointed out by Aleksei Kulikov in his comment. The correction is related to the arguments in the comments of KhashF and user44191.

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    $\begingroup$ I’m sorry but where you used continuity of $f$ and $g$? Because without it answer to question in OP is that every function can be represented in such a manner (via bijection between $\mathbb{R}$ and $\mathbb{R}^2$). And also what is $g_1^{-1}(x)$ really? I see no reasons for $g_1$ to be injective. $\endgroup$ Feb 1 '19 at 2:51
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    $\begingroup$ @AlekseiKulikov Very good point. However, under the continuity assumption my proof works. The continuity assumption is used implicitly and I will explain how. I will add details tomorrow. $\endgroup$ Feb 1 '19 at 4:44
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    $\begingroup$ P.S. I think even the following works: $g_1(f(x,y))=x$, $g_2(f(x,y))=y$ imply that the continuous $f:\Bbb{R}^2\rightarrow\Bbb{R}$ is injective, a contradiction. $\endgroup$
    – KhashF
    Feb 1 '19 at 20:53
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    $\begingroup$ As an alternative approach, $h(t) := (g_1(t), g_2(t))$ is a continuous inverse to the continuous function $f(x, y)$, and therefore a homeomorphism. This approach should work category-theoretically: you can construct such functions $f, g$ on $X^2$ only if $X \simeq X^2$ in whichever category you've chosen - and since $\mathbb{R} \not \simeq \mathbb{R}^2$ topologically, we're done. $\endgroup$
    – user44191
    Feb 1 '19 at 21:08
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    $\begingroup$ It should be noted that there are non continuous functions f and g for this example, which depend on f "encoding" two reals into one and g using the decoding inverses to f to represent F. Gerhard "Enhancing The Continuity Of Explication" Paseman, 2019.02.13. $\endgroup$ Feb 13 '19 at 18:10

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