TITLE: Do we always get the same dimension regardless of the manifold/topological structure?
QUESTION [5 upvotes]: Note: I ask this motivated by this other question: Are manifold subsets that are immersed submanifolds (regular/embedded) submanifolds?

Maybe a weird question, but:
Let $A$ be a set s.t. it is possible to endow $A$ with 2 different smooth (or topological or $C^k$ or possibly even holomorphic/complex/Kähler or whatever) manifold structures. Endow $A$ with 2 different smooth manifold structures $\mathscr F$ and $\mathscr G$ to get, resp, $(A,\mathscr F)$ and $(A,\mathscr G)$.
If $(A,\mathscr F)$ and $(A,\mathscr G)$ have respective dimensions $f$ and $g$, then is $f=g$?

Edit: Note: Oh right so I really mean that $A$ is a set and not a topological manifold, so the way $(A,\mathscr F)$ and $(A,\mathscr G)$ are topological manifolds in the 1st place are that they are based on the same topological structure i.e. they are based on the same topology that makes the set $A$ into a topological space (and then this topological space is indeed a topological manifold).

What I have in mind:

So, like, there are many smooth (or whatever) manifold structures on $\mathbb R^n$, but I'm wondering if they all make $\mathbb R^n$ a smooth $n$-manifold. Perhaps there's some wild smooth manifold structure to make $\mathbb R^n$ locally $\mathbb R^{n-1}$ (i guess in the 1st place, such structure would be s.t. the topological structure makes $\mathbb R^n$ locally $\mathbb R^{n-1}$). I think of something like $\mathbb R^n$ and $\mathbb R^{n-1}$ as diffeomorphic/homeomorphic, but then it's not under the standard manifold or even topological structures.

I believe there's a rule that says for $(A,\mathscr F)$ to be a smooth $f$-manifold, we must have $f$ equal to the same dimension $h$ that allows us to say in the 1st place $(A,\mathscr F)$ is a topological $h$-manifold. But here...I'm not sure but I think I recall that the creation of a smooth manifold begins with a topological manifold and then this creation relies on same dimension.

$f=g$ if $(A,\mathscr F)$ and $(A,\mathscr G)$ are diffeomorphic/homeomorphic, but I recall $(A,\mathscr F)$ and $(A,\mathscr G)$ need not be diffeomorphic/homeomorphic.

REPLY [2 votes]: Dimension needs to be consistent over all smooth structures.
Indeed, you can prove that if $U \subset \mathbb{R}^n$ and $V \subset \mathbb{R}^m$ are homeomorphic, then $n = m$.
The proof (if you're familiar with homology theory), is by looking at the relative homology of the pair $(U, U\backslash\{0\})$. In particular, you can show that it is isomorphic to the reduced homology of the $n$-sphere, while the homology of $(V, V\backslash\{0\})$ is isomorphic to the reduced homology of the $m$-sphere. So asking for $U \cong V$ enforces $m = n$. (Let me know if you want me to get into more details on that)
In particular, assume that have two different smooth structures on your manifold, making it locally homeomorphic to respectively $\mathbb{R}^m$ and $\mathbb{R^n}$. Pick a point $x$, two charts $U$ and $V$ around it. Then, using the result above, $m $ and $n$ have to be the same.