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Let $X\subset\mathbb{R}^n$ open. Let $f,g:X\rightarrow X$ be functions such that $f$ is a diffeomorphism and $g$ is a homeomorphism. Is $g\circ f \circ g^{-1}$ a diffeomorphism?

I'm currently working through Lee's Intro to Smooth Manifolds, trying to show that any non-degenerate manifold that admits a smooth structure admits uncountably many distinct smooth structures. If the above question is true, it would enable me to complete my proof.

My intuition for the special case in $\mathbb{R}$ is as follows: Let $x\in X\subset \mathbb{R}$, and $(x_n)$ a sequence converging to $x$. Choose $(y_n)$ so that $y_n = g(x_n)$ for each $n$, and let $y = \lim_{n\rightarrow\infty} y_n$. Then \begin{align} &\lim_{n\rightarrow\infty}\frac{(g\circ f\circ g^{-1})(x_n) - (g\circ f\circ g^{-1})(x)}{x_n - x}\\ = &\lim_{n\rightarrow\infty}\frac{g(f(g^{-1}(x_n))) - g(f(g^{-1}(x)))}{g(g^{-1}(x_n)) - g(g^{-1}(x))}\\ = &\lim_{n\rightarrow\infty}\frac{g(f(y_n)) - g(f(y))}{g(y_n) - g(y)} \end{align} But here I reach a problem because I don't see how I can unwrap this any further. Moreover, there is no guarantee that $y$ exists. ($y$ may not be in X.)

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  • $\begingroup$ Since $g^{-1}:X\to X$ and $y=g^{-1}(x)$ you can conclude that $y\in X$. Note that $g^{-1}$ is a well defined continuous function. $\endgroup$ Commented Nov 20, 2017 at 8:35

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No: Here is a counterexample in one dimension with $X={\mathbb R}$: $$ g(x) = x^3\qquad f(x) = x+1\qquad g\circ f \circ g^{-1}(x) = (x^{1/3}+ 1)^3 $$

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