# Doubt about the Domain of the chart on a Manifold

I have a doubt about the domain of the chart on a manifold. Suppose $M$ is a smooth manifold and that $(U, \varphi)$ is a chart on $M$, then $\varphi : U \to \mathbb{R}^n$ has $U$ as it's domain. That's fine to me. My doubt has to do with the comparison of this to the case of regular surfaces in $\mathbb{R}^3$. In that case, the surface $S$ is a subset of $\mathbb{R}^3$ and so, if $(V, \psi)$ is a chart on $S$, then I know that $\psi$ takes points in $\mathbb{R}^3$ and takes to $\mathbb{R}^2$. In other, words, I know that I can write:

$$(u,v)=\psi(x_1,x_2,x_3)$$

And so, I know many things, for instance, I know that to build the chart I'll need to find one expression of $x_1, x_2,x_3$ which gives $(u,v)$ respecting the properties required. On general Manifolds, however, I feel a little confused. I mean, the manifold isn't inside another higher dimensional ambient space, so, if $(U, \varphi)$ is a chart, I would have:

$$(x^1, \dots, x^n) = \varphi(\text{what goes here ?})$$

And the $\text{what goes here ?}$ is because if I plug there some $m$-tuple of numbers, I'm supposing that $M$ is a subset of $\mathbb{R}^m$ and that supposition shouldn't be necessary. I've read for instance, that to find charts for the $n$-sphere, we can consider it as a subset of $\mathbb{R}^{n+1}$, but that isn't really necessary.

My point is, I know that elements of the codomain of a map of a chart will be $n$-tuples of numbers, that's fine, I know how to work with such objects. But how will be the elements of the domain, if we do not express the manifold as a subset of a higher dimensional ambient space?

I think I've made clear my point. If I've failed to explain my doubt, please ask and I'll try to explain better. Thanks very much in advance!

EDIT: The definition of Manifold I'm working with is the definition as presented by Manfredo Do Carmo: A smooth manifold of dimension $n$ is a set $M$ with a family of bijective maps $\varphi_\alpha : U_\alpha \to M$ from open sets $U_\alpha\subset \mathbb{R}^n$ to $M$ such that:

1. $\bigcup_\alpha\varphi_\alpha(U_\alpha)=M$
2. For each pair $\alpha, \beta$ with $\varphi_\alpha(U_\alpha)\cap\varphi_\beta(U_\beta)=W\neq\emptyset$ we have $\varphi_\alpha^{-1}(W)$, $\varphi_\beta^{-1}(W)$ open in $\mathbb{R}^n$ and $\varphi_\beta^{-1}\circ\varphi_\alpha$, $\varphi_\alpha^{-1}\circ\varphi_\beta$ are differentiable.
3. The family $\left\{U_\alpha, \varphi_\alpha\right\}$ is maximum with respect to conditions 1 and 2.

The only point is that on my text above, I've decided to change the direction of the maps, but the definition is yet that one.

-
What is your definition of a manifold? I believe that the answer to your question lies essentially in that definition. –  Nils Matthes Apr 3 '13 at 15:02
Hi @NilsMatthes, I've posted the definition of manifold I'm working with. –  user1620696 Apr 3 '13 at 17:29