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Can a cube (meaning $g(x) = f(x)^3 = f(x) \cdot f(x) \cdot f(x)$) of discontinuous function $f: D \to \mathbb{R}$ ($D$ is subset of $\mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.

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    $\begingroup$ The point is not that $x \longmapsto x^3$ is injective, as much as $x \longmapsto x^{1/3}$ is continuous. $\endgroup$ – Mindlack Jan 8 at 1:03
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    $\begingroup$ What is your domain? It matters really quite a lot. $\endgroup$ – user3482749 Jan 8 at 1:04
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    $\begingroup$ I think the injectivity is very much to the point. $f(x) =\sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)\cdot g(x)$ continuous. $\endgroup$ – MJD Jan 8 at 1:29
  • $\begingroup$ No. But the cube of a non-differentiable function can be differentiable : $|x|^3$ $\endgroup$ – Eric Duminil Jan 8 at 7:53
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    $\begingroup$ I misspoke: it's the range that matters. If it's $\mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity). $\endgroup$ – user3482749 Jan 8 at 15:19
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If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.

So the contrapositive is also true, which is:

If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.

(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)

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Since $\phi : \mathbb{R} \to \mathbb{R}, \phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $\phi \circ f$ is continuous.

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    $\begingroup$ To lower the level of this answer, note that $\phi^{-1}(x) = \sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $\phi\circ f$ is continuous, so is $f = \phi^{-1}\circ \phi \circ f$. $\endgroup$ – Paul Sinclair Jan 8 at 18:09

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