Let $X$ be a compact metrizable space. Let $f$ be a continuous map from $X$ to the cube $[0,1]^m$. I would like to know under which condition of a continuous map $g: X\to [0,1]^n$ there exists a continuous map $h: [0,1]^n \to [0,1]^m$ such that $f=h\circ g$. Is it true that a dense $G_\delta$ set of $C(X, [0,1]^n)$ satisfies the above property? Thanks.
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$\begingroup$ Obviously, this depends on what $f$ is, for example it works for constant $f$. But it won't work in general: if $m=n=1$, $X=[0,1]$ and $f$ is injective and $g(0)=g(1)=0$, $g(1/2)=1$, then any function close to $g$ will fail to be injective. $\endgroup$– Christian RemlingCommented Dec 12, 2023 at 15:20
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