This question already has an answer here:
- Topological manifolds (dimension) 2 answers
For smooth manifolds $A$ and $B$ with respective dimensions $a$ and $b$. If $A$ and $B$ are diffeomorphic, then $a=b$. I guess the same is true for homeomorphic topological ($C^0$, I guess) manifolds (with dimension).
If $A$ and $B$ were instead homeomorphic, then would we still have $a=b$?
- I imagine some weird case how they would have the same dimension viewing them as topological manifolds (with dimension), but they might have different dimensions as smooth manifolds. I actually haven't encountered anything like different dimensions depending on which level of $C^k$ we are since most the manifolds so far I've been studying are smooth, but I think it's similar to the idea of Isomorphic groups but not isomorphic rings.
If yes, then follow-up questions:
Does the "$B$" in this question What does it take for a smooth homeomorphism to be a diffeomorphism? have dimension $k$ because $\dim B = \dim A = \dim \mathbb R^k = k$ ?
Can Corollaries 8.6 and 8.7 in An Introduction to Manifolds by Loring W. Tu be relaxed as follows?
3.1. For 8.6: change "diffeomorphism" to "smooth homeomorphism"
3.2. For 8.7: change "diffeomorphic" to simply "homeomorphic"