Let $M=\{(x,x): x \in (1, 1)\}$. Then there is an atlas with only one coordinate chart $(M, (x, x) \mapsto x)$ for $M$. We don't need any coordinate transformation maps to worry about differentiablity. So I thought $M$ is a differentiable manifold. However my teacher says it is not. He says the sharp corner at $x = 0$ is a problem. I can't understand why it is a problem.

3Well, you're right  at least technically :) You do realize that a similar argument will turn any subset of the plane with the cardinality of the continuum into a differentiable manifold, don't you? – Jyrki Lahtonen Jun 16 '11 at 06:40

2@Jyrki: More specifically, into a onedimensional differentiable manifold. This of course includes the plane itself. Also, the same argument can be made with respect to any bijection between any $\mathbb R^m$ and $\mathbb R^n$, so in fact any $\mathbb R^n$ can be equipped with a differentiable structure of arbitrary dimension. – joriki Jun 16 '11 at 06:45

@joriki: That's a good one! – Jyrki Lahtonen Jun 16 '11 at 06:48

4@Jyrki: I don't think it is too similar, because the topology on $M$ is presumably already fixed as the subspace topology in the plane, and the given chart is a homeomorphism of $M$ onto $(1,1)$. – Jonas Meyer Jun 16 '11 at 07:22

1@joriki: As far as I know, $\mathbb{R}^n$ and $\mathbb{R}^m$ cannot be homeomorphic if $m \neq n$. How can a plane itself be a 1manifold? – Lee Jun 16 '11 at 08:36

1@Lee: You're talking about putting a differentiable structure on top of the natural topological structure; I was referring to Jyrki's comment, which ignored the topological structure and talked only about the cardinality. Jonas rightly remarked that this isn't quite analogous to what you're doing, since you're respecting the topological structure, just not the differentiable structure of $\mathbb R^2$. – joriki Jun 16 '11 at 10:00
2 Answers
To add to what joriki wrote and what some people expressed in the comments: the real problem lies in certain implicit information/assumption you are making. The problem is that you have only specified $M$ as a set explicitly, and to ask whether an object is a differentiable manifold requires also specifying the topological and smooth structures! There are generally two definitions of smooth manifolds that I see in textbooks:
 Intrinsic definition A smooth manifold is a topological manifold equipped with an atlas whose transition maps are smooth. In this case the smooth structure is explicitly given by saying that a function defined on the manifold is smooth if it is smooth in each chart belonging to the atlas.
 Extrinsic definition A smooth manifold is defined as a smooth submanifold of some ambient Euclidean space (usually through a defining function). In this case the smooth structure is given by saying that a function defined on the manifold is smooth if it is the restriction of a smooth function on the ambient space.
Your response uses the intrinsic definition: you construct an atlas based on the homeomorphism of your set (as a topological subspace of $\mathbb{R}^2$) to the open unit interval. Your teacher's response implicitly depends on inheriting not only the topological structure from $\mathbb{R}^2$, but also the smooth structure, so is based on the extrinsic definition.
To be really pedantic, as stated the question cannot be answered, since no candidate for a smooth structure was given. In essence you interpreted the question to mean that "Given this topological space $M$, can we give it a smooth structure?" while your teacher interpreted the question to mean "Given this topological subspace $M\subset \mathbb{R}^2$, is it a smooth submanifold?" Therefore two difference answers are expected since two different questions are treated.
To illustrate the problem further: a common exam question is to ask
Is the punctured closed square $[1,1]\times[1,1] \setminus \{(0,0)\} \subset \mathbb{R}^2$ a smooth manifold with boundary?
Observing that if you only induce on the square the topology from the ambient Euclidean space, then you can reason thus: you can explicitly construct a homeomorphism from our punctured closed square to the upper half plane, so we can take that as a single coordinate chart and give the set the structure of a smooth manifold compatible with the induced topology. But if you want the smooth structure to also be inherited....
But unlike joriki, I would say that in context your teacher's response is more reasonable. This is because by using the local homeomorphism of any topological manifold to Euclidean spaces, you can always prescribe local differentiable structures on any topological manifold. The obstruction for a topological manifold to be smooth is very essentially global (and somewhat tangentially related is the fact that the obstruction is on the level of the first derivative only), and the existence of nondifferentiable topological manifolds is actually not that easy to show. Furthermore, it is also known that any topological manifold in dimensions 1, 2 and 3 admits a (unique) compatible smooth structure, so in view of that (also not too easy) result, the question becomes vacuous if you choose to interpret it in the way you did.
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concerning the *extrinsic definition*, how does a topological $d$manifold $M\subseteq\mathbb{R}^n$ inherit (when possible) an atlas from $\mathbb{R}^n$? Why is $\mathbb{S}^1$ a smooth submanifold of $\mathbb{R}^2$, but $\{(x,x): x \in (1, 1)\}$ only a smooth manifold (with a coordinate chart the projection onto the first factor) in it's own right? What does *smooth submanifold* actually mean? – Leo Jun 16 '11 at 11:38

3@Leon: (1) Through the [implicit function theorem](http://en.wikipedia.org/wiki/Implicit_function_theorem) applied to the defining function. Locally the smooth submanifold is the graph of a smooth function, which you you use to build charts by projection. The change of coordinate maps are then rotations (in ambient space), and hence smooth. (2) See next comment (likely to be a bit long). (3) Depends on the textbook. There are several equivalent definitions. One says that it is the image of a smooth immersion of a smooth manifold. Another say that it is the zero set of of a smooth function... – Willie Wong Jun 16 '11 at 12:44

3... with values in $\mathbb{R}^N$ with constant nonzero rank along the zero set. (The case where the submanifold has the same dimension as the manifold requires a separate specification.) ... return to (2): there are several ways to see it. First is that $\mathbb{S}^1$ is the zero set of a smooth function whose derivative is nondegenerate ($f(x,y) = x^2 + y^2  1$), but $\{(x,x)\}$ is not. Also, consider the function $g(x,y) = x+y$ on $\mathbb{R}^2$. This function is smooth. But restricted to $\{(x,x)\}$ using the projection map it is continuous but not differentiable, this shows ... – Willie Wong Jun 16 '11 at 12:51

3... that the constructed smooth structure is not compatible with the "induced" one. – Willie Wong Jun 16 '11 at 12:52
You're right. What your teacher may be trying to say is that the differentiable structure you're defining isn't induced by the differentiable structure of $\mathbb R^2$ in which your manifold is embedded. If your curve didn't have a corner, there would be a natural differentiable structure on it which is compatible with the differentiable structure of $\mathbb R^2$.
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what do you mean by "*induced by the differentiable structure of $\mathbb{R}^2$*"? – Leo Jun 16 '11 at 11:54

1@Leon: Willie has already written a lot in response to your comment. For my purposes, I could have written "compatible" instead of "induced"; I had in mind something like Willie's example $g(x,y)$, which shows that the differentiable structure on $M$ isn't compatible with the one on $\mathbb{R}^2$. – joriki Jun 16 '11 at 21:15