# A “Trivial” Smooth Structure

Let M be a topological manifold of dimension $n$. Suppose that there exists a homeomorphism $$\phi \colon M \longrightarrow \mathbb{R}^n$$ defined for all $x \in M$. That is, $\phi$ is a continuous bijection with a continuous inverse. Thus, it would be appropriate to call $\phi$ a chart on $M$ and, it seems to me, we could define an atlas $\mathcal{A}$ on $M$ consisting of the single chart $\phi$, $$\mathcal{A} := \{(\phi, M)\}$$ Using this atlas, if my reasoning is correct, one can consider $M$ to be a smooth manifold. Therefore, any topological manifold $M$ for which there exists a homeomorphism $\phi \colon M \longrightarrow \mathbb{R}^n$ can actually be considered as a smooth manifold.

So, my question is, Is this line of reasoning correct?

All of these statements seem to follow directly from the definitions but it is somewhat counterintuitive in the sense that clearly not all homeomorphisms have smooth inverses. e.g., if $\phi(x) = x^3$, $\phi^{-1}(x) = x^{1/3}$ and it is easily seen that $\phi$ is a homeomorphism but the derivative of $\phi^{-1}$ does not exist at the origin.

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Regarding your question about $x \mapsto x^3$, you might want to read this part of Wikipedia. – kahen Jan 16 '12 at 19:49

As for your example on $\mathbb R$, there is a smooth manifold structure on $\mathbb R$ coming from the chart $x\mapsto x^3$, and for the reason you gave, this is not compatible with the standard smooth manifold structure coming from the chart $x\mapsto x$.
However, the smooth manifolds obtained in this way are diffeomorphic. Let $\mathbb R_{\text{cube}}$ be $\mathbb R$ with the cubing chart, and let $\mathbb R_{\text{st}}$ be $\mathbb R$ with the standard atlas. Then $f:\mathbb R_{\text{cube}}\to\mathbb R_{\text{st}}$ defined by $f(x)=x^3$ is a diffeomorphism.
It is generally the case that the structure you put on $M$ in this way is diffeomorphic to $\mathbb R^n$. Somewhat analogous to turning a homeomorphism into a diffeomorphism in this way would be to induce a topology on a set using a bijection. E.g., you could set up a rather unnatual topology on $[0,1]$ by choosing a bijection of $[0,1]$ with $\mathbb R$ and making it a homeomorphism. Also somewhat analogous, you can put a complete metric on $(0,1)$ by choosing a homeomorphism of $(0,1)$ with $\mathbb R$ and making it an isometry. In this case, the sequence $(1/n)$ will not be a Cauchy sequence.
I guess I'm missing something, but I don't see how $x \mapsto x^3$ is a diffeomorphism (meaning that it's a smooth homeomorphism with smooth inverse) since the derivative of the inverse map $x \mapsto x^{1/3}$ does not exist at the origin. – ItsNotObvious Jan 16 '12 at 20:06
@3Sphere: To see that $f^{-1}$ is differentiable, you only need to check that for all charts $\phi:\mathbb R_{\text{cube}}\to\mathbb R$ and $\psi:\mathbb R_{\text{st}}\to\mathbb R$, the map $\phi\circ f^{-1}\circ \psi^{-1}:\mathbb R\to\mathbb R$ is differentiable. Note for example that if $\phi(x)=x^3$, this removes the singularity at the origin of $x\mapsto x^{1/3}$. – Jonas Meyer Jan 16 '12 at 20:11