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When we introduce a smooth structure on $m$-manifold we want to have an atlas which covers all the points of the manifold such that both of its charts are smoothly related. Does it mean that nay space $C$ homeomorphic to $D\subseteq \mathbb R^m$, $D$ - an open set, is a smooth $m$-manifold?

My reasoning: we can take an atlas $\mathcal A = \{(C,h)\}$ where $h:C\to D$ is a homeomorphism.

Edited: Since the first version of the question is unclear, I'd like to clarify it. If the space $M = \{(y,x)\mid y=|x|,x\in (a,b)\}$ with an induced topology is a smooth manifold? If yes, then what is the tangent space $T_{(0,0)}$? Finally, if it is a smooth manifold, what is the advantage of raising assumptions on the smooth relations between charts if a non-smooth curve $M$ is a smooth manifold?

The atlas was asked: $(M,\pi)$ for $\pi(x,y) =x$.

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Yes. Smoothness is not relevant to the question, since you are only asking for homeomorphisms. –  Zhen Lin Aug 4 '11 at 14:21
    
@Zhen: I've edited a question. I wonder if it is a smooth $m$-manifold. –  Ilya Aug 4 '11 at 14:26
    
That's absolutely true (and the first example in any reasonable expositions of manifold theory). A large part of the construction is then to extend to manifolds all the differential calculus you know on open sets in $\mathbb R^n$... –  PseudoNeo Aug 4 '11 at 14:30
    
@PseudoNeo: I thought that once we have a smooth manifold, we have also a differential calculus independently of the atlas. While here we can take the graph of $|x|,x\in[0,1]$ which will be a smooth $1$-manifold. –  Ilya Aug 4 '11 at 14:35
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I think the question now is about the notion of submanifold. –  Pierre-Yves Gaillard Aug 4 '11 at 15:01

3 Answers 3

up vote 3 down vote accepted

This is the difference between intrinsic and extrinsic geometries: given a continuous embedding of a topological manifold $X$ into another topological manifold $Y$, it is possible that $X$ can be endowed with a smooth structure, and $Y$ endowed with a smooth structure, but that the embedding of $X$ into $Y$ is non-smooth.

For more details, please see my answer to Is $M=\{(x,|x|): x \in (-1, 1)\}$ not a differentiable manifold? .

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Here's my interpretation of your question: Suppose $X$ is a topological space homeomorphic to an open subset of $\mathbb R^n$. Can $X$ be given a smooth structure? Phrased this way, the answer is clearly yes because you can just take the open subset as the single chart in the atlas, as you say.

Here's a different question. Suppose $X$ is a smooth manifold which is homeomorphic to an open subset of $\mathbb R^n$. Is $X$ diffeomorphic to an open subset of $\mathbb R^n$? This is false in dimension $4$! There are things called large exotic $\mathbb R^4$'s which are homeomorphic to $\mathbb R^4$ but not diffeomorphic to any open subset of $\mathbb R^4$.

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Thank you for the explanation. Unfortunately, that's not exactly what I asked about, I've made the question clear now. –  Ilya Aug 4 '11 at 14:46

Just to point out that the statement of whether a space M 'is a manifold' can be ambiguous, in that a manifold involves a choice of charts and chart maps, and the choice of these that you are making/assuming may not always be clear. So , I am assuming you're asking about the "induced charts".

If by induced charts you refer to the subspace charts, then the answer is no, meaning that, with the induced (subspace) topology, f(x)=|x| is not a submanifold of $\mathbb R^2$; the definition of submanifold I know of needs for the manifold to look locally like the inclusion of $\mathbb R^n$ into $\mathbb R^m$ ; in your case , you want a neighborhood U of every point (let's choose the problem point 0, and an open set $U_0$), and a chart map $\phi$ with $\phi(U_0)$=$(x_k,0)$

BTW: if your tangent space existed at (0,0) , then, by dimension reasons, it would have to be 1-dimensional, i.e., a copy of $\mathbb R$. Is that what you were asking?

A point that may be worth making is that f(x)=|x| can be made into a manifold, using this fact: If two topological spaces X,Y are homeomorphic, and one of them, say X, is a manifold, then the other one is (can be made into) a manifold too: just compose each chart map by the homeomorphism, to "pullback" the charts. In your case, you can homeomorph f(x)=|x|, so that its corner disappears, and you get a manifold.

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