Is $M=\{(x,|x|): x \in (-1, 1)\}$ not a differentiable manifold? 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.
 A: 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. 
A: 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 $\mathbb R^2$.
