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The definition of a manifold in my course notes is given as follows:

Let $\mathcal{M}\subseteq\mathbb{R}^n$. A smooth chart on $\mathcal{M}$ consists of a subset $U\subseteq{\mathcal{M}}$, open in $\mathcal{M}$, an open set, $V\subseteq\mathbb{R}^m$, and a diffeomorphism $\phi:U\rightarrow{V}$. A smooth atlas for $\mathcal{M}$ is a collection of charts whose union covers $\mathcal{M}$. We say that $\mathcal{M}$ is a smooth manifold of dimension $m$ (or a smooth $m$-manifold) if it admits a smooth atlas with charts mapping to $\mathbb{R}^m$.

My problem here is that we have a diffeomorphism mapping an open subset $U$ of $\mathbb{R}^n$ to an open set $V$ of $\mathbb{R}^m$. This is then necessarily a homeomorphism, contradicting invariance of dimension (unless $m=n$, but my notes simply say that it is 'easily seen that $m\leq{n}$').

Is there an issue here or am I missing something?

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    $\begingroup$ The set $U$ is not (usually) open in $\mathbb{R}^n$, it is "open in $\mathcal{M}$", also known as "relatively open in $\mathcal{M}$". There exists an open $W\subset \mathbb{R}^n$ such that $U = W \cap \mathcal{M}$. $\endgroup$ – Daniel Fischer Oct 19 '15 at 17:53
  • $\begingroup$ What if $\mathcal{M}$ is open in $\mathbb{R}^n$? Then the intersection $U$ of open sets $W$ and $\mathcal{M}$ is also open in $\mathbb{R}^n$ and we still have a problem? $\endgroup$ – jl2 Oct 19 '15 at 17:57
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    $\begingroup$ Well, then you can't have $m < n$ and $m = n$ follows. But usually, you consider something lower-dimensional, like a hypersurface (e.g. a sphere), or a curve, a surface, and then you typically have $m < n$. $\endgroup$ – Daniel Fischer Oct 19 '15 at 17:59
  • $\begingroup$ So if you have an object of dimension lower than the space it's in, it can't be open in that space? I think I can imagine why that is - any open ball would 'leak' into an extra dimension and therefore not be contained in the object. Correct? $\endgroup$ – jl2 Oct 19 '15 at 18:01
  • $\begingroup$ As long as we talk of manifolds (or similarly nice spaces), and our spaces have the same dimension at each point, yes. If we allow uglier things, we need to add a few conditions to exclude things like a disjoint union of $S^1$ and $S^2$, which globally is a two-dimensional thing, but has an open one-dimensional subspace. [And in full generality, there can be more difficult problems with dimension and openness, but that's not relevant here.] $\endgroup$ – Daniel Fischer Oct 19 '15 at 18:07
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${\cal M}$ is contained in $R^n$ and not always an open subset subset of $R^n$. think of $R^2$, ${\cal M}=R\times\{0\}$, you have a diffeomorphism between ${\cal M}$ and R

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