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In the context of differential manifold, can a smooth map composed with a non-smooth map be equal to a smooth map?

Thank you!

mip
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1 Answers1

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How about a constant function $g(x)=p$ composed with $f(x)=x^{\frac13}$ on $\Bbb R$? The composition $g\circ f$ would be constant, hence smooth.

Or let $g(x)=x^3$, for that matter.