Dimension of a smooth manifold

Question

I think I have a big confusion in my understanding of smooth manifolds.

Here is the definition of smooth manifolds that I have

Let $M\subset\mathbb{R}^k$, we say that $M$ is a smooth manifold of dimension $m$ if each $x\in M$ has a relative neighborhood in $M$ that is diffeomorphic to some open subset of $\mathbb{R}^m$.

My confusion is the following : I have the feeling that it implies that $m=k$

Here is my reasoning : Let $x\in M$ and denote $W$ a neighborhood of $M$ such that $W\cap M$ is diffeomorphic to some open subset $U\subset\mathbb{R}^m$ through the parametrization $g : U\to g(U)$.

Since $g$ is a diffeomorphism, we can find $V$ an open subset of $\mathbb{R}^k$ such that $g : U\to V$ is a diffeomorphism between open subset of $\mathbb{R}^{m}$ and $\mathbb{R}^k$ but this is well known that this implies that $m=k$

I cannot see where I am wrong but I know that it is the case, if you could provide some helps in order to overcome this misunderstanding please.

One consequence of this is that when I got interested in the tangent space $TM_{x}$, proving that it was a n dimensional vector space was a $2$ lines proof since in my (possible false) setting the derivative at a point of my parametrization $dg_{u}$ is automatically invertible by the chain rule

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I am using the following definition of a diffeomorphism : Let $f:X\to Y$ where $X\subset\mathbb{R}^k$ and $Y\subset\mathbb{R}^m$ are open subsets, $f$ is said to be a diffeomorphism if it is an homeomorphism such that $f$ is $C^1$ and its inverse $f^{-1}$ is also $C^1$.

Answer 1

In such situations it is always a good idea to look at examples. Let $k$ be arbitrary, and consider

$$M=\{(x,0,\ldots,0)\in\mathbb{R}^k\ |\ x\in\mathbb{R}\}$$

You can verify very easily that $M$ is diffeomorphic to $\mathbb{R}$, and thus it is a manifold of dimension $1$. Even though it is a subset of $\mathbb{R}^k$.

through the parametrization $g : U\to g(U)$.

I think you meant $g:W\cap M\to U$? Or vice versa.

Since $g$ is a diffeomorphism, we can find $V$ an open subset of $\mathbb{R}^k$ such that $g : U\to V$ is a diffeomorphism

No. In the example above we have a diffeomorphism

$$g:\mathbb{R}\to M$$ $$g(x)=(x,0,\ldots,0)$$

But there is no open subset $V$ of $\mathbb{R}^k$ that turns this into a diffeomorphism $\mathbb{R}\to V$.

Note that $M=W\cap M$ for $W=\mathbb{R}^k$. But $M$ is not open in $\mathbb{R}^k$, in fact it has empty interior. And thus invariance of domain does not apply.

Here is the definition of smooth manifolds that I have

Let $M\subset\mathbb{R}^k$

Finally note that in the above definition $\mathbb{R}^k$ is not really relevant. As the example above shows our $M$ can be embedded into any $\mathbb{R}^k$. The author gave such definition because it simplifies a lot. But generally one would start with something like: let $M$ be a Hausdorff space, sometimes paracompact, sometimes metrizable, typically second countable. It is a bit more complicated to define differentiability in such scenario, but only a bit.

So the most general definition does not require $M$ to be a subset of $\mathbb{R}^k$. Even though in fact these definitions coincide, due to Whitney embedding theorem. The point is that $\mathbb{R}^k$ is not necessarily important (unless you deal with some specific case in which it is).

Answer 2

I believe you mean "$W$ neighbourhood of $x$", in general it is not possible to find a neighbourhood $W$ of $M$ such that $W \cap M$ is diffeomorphic to an open subset of $\mathbb R^m$. In that case, $g(U)$ is $m$-dimensional. Try to visualize it by taking $M=S^2 \subset \mathbb R^3$, $x$ the North Pole, and $W \cap M$ the upper half sphere without boundary. The latter is diffeomorphic to a disc in $\mathbb R^2$.