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No, not really. In dimension 4, for example, an open subset of R^4$\mathbb{R}^4$ can be homeomorphic to R^4$\mathbb{R}^4$ but not diffeomorphic, as there are exotic smooth R^4's$\mathbb{R}^4$'s that embed smoothly in R^4$\mathbb{R}^4$.

But in dimensions different from 4, R^n$\mathbb{R}^n$ admits a unique smooth structure. So your neccessary and sufficient condition can be that the open subset is homeomorphic to R^n$\mathbb{R}^n$. That's probably not what you want to hear?

No, not really. In dimension 4, for example, an open subset of R^4 can be homeomorphic to R^4 but not diffeomorphic, as there are exotic smooth R^4's that embed smoothly in R^4.

But in dimensions different from 4, R^n admits a unique smooth structure. So your neccessary and sufficient condition can be that the open subset is homeomorphic to R^n. That's probably not what you want to hear?

No, not really. In dimension 4, for example, an open subset of $\mathbb{R}^4$ can be homeomorphic to $\mathbb{R}^4$ but not diffeomorphic, as there are exotic smooth $\mathbb{R}^4$'s that embed smoothly in $\mathbb{R}^4$.

But in dimensions different from 4, $\mathbb{R}^n$ admits a unique smooth structure. So your neccessary and sufficient condition can be that the open subset is homeomorphic to $\mathbb{R}^n$. That's probably not what you want to hear?

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Ryan Budney
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No, not really. In dimension 4, for example, an open subset of R^4 can be homeomorphic to R^4 but not diffeomorphic, as there are exotic smooth R^4's that embed smoothly in R^4.

But in dimensions different from 4, R^n admits a unique smooth structure. So your neccessary and sufficient condition can be that the open subset is homeomorphic to R^n. That's probably not what you want to hear?