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First time encountering manifolds in general.

A smooth manifold $M$ is said to be of dimension $n$ if its (complete) atlas is of dimension $n$.

Now, what if we replace smooth by continuous ? Does the dimension is even well defined ?

I know from this that in case of smooth manifolds, there exists invariance of domain which means that there does not exist two U,V open sets of $M$ such that $U$ is homeomorphic to $\mathbb{R}^n$ and $V$ is homeomorphic to $\mathbb{R}^m$ with $m<n$. It all boils down to the fact that there does not exist an homeomorphism between an open subset of $\mathbb{R}^n$ and an open subset of $\mathbb{R}^m$.

I do not know if this holds in the case of continuous maps.

Flewer47
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1 Answers1

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The proof is exactly the same as in the smooth case. Invariance of domain doesn't use smooth maps, but continuous ones, and smooth maps happen to be continuous (in fact, smooth charts are defined to be homeomorphisms), which is why they also work. The real reason why invariance of domain can be applied is that smooth charts are homeomorphisms. But continuous charts are too, by definition, so the same reasoning applies.

Vercassivelaunos
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