If $g(X)$ and $h(X)$ are both non-zero functions in $X$ such that $g^2(X) \perp h(X)$, does it follow that $g(X) \perp h(X)$?
The context is an answer given on this site that uses this assumption in a proof. But I'm not sure this is actually true, as information may be lost in the transformation $g(X) \to g^2 (X)$ if $g(X)$ can take negative values.
Consider $X$ a random variable with Rademacher distribution. Then $X^2=1$ is a constant, thus $X^2 \perp X$. However, $X$ is not independent of itself.
Besides if $g^2(X)\perp h(X)$, then $\sqrt{g^2(X)} = |g(X)|\perp h(X)$. So if $g\geq 0$, then indeed $g(X)\perp h(X)$.