# Composition of a differentiable map with a homemorphism.

Suppose I have a $C^{\infty}$ map $f:X\to\mathbb{R}^{n}$, for some differential manifold $X$.

Then I also have a homeomorphic coordinate map $h$ from a subset of $\mathbb{R}^{n}$ to a subset of $X$.

My question is, since $f$ is $C^{\infty}$, then are necessarily $f\circ h$ going to be $C^{\infty}$?

I would think the answer is yes since homeomorphisms preserve convergence, but I'm hesitant to conclude so because I don't know if homeomorphisms are necessarily differentiable.

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Homeomorphisms aren't necessarily differentiable. Consider for example $f(x) = x^{\frac{1}{3}}$ from $\mathbb{R}$ to itself with the regular smooth structure. If they are, they are called diffeomorphisms. There's no reason a composition of something that isn't necessarily smooth with a smooth map will be smooth. – levap Oct 27 '12 at 0:50

I found out the answer is yes, but only because the definition of a Euclidean valued function defined on a manifold being $C^{\infty}$ requires it to be true, assuming that $h$ is one of the coordinate maps. For any other homeomorphism, the answer is not in general yes.