Proving a triangle is not a manifold and it's not diffeomorphic to a circle. I struggle with a question I thought was apparently very simple.
Suppose I have a triangle in the plane and a circle in the plane.
How can I prove that the triangle is not a smooth manifold, and how can I prove is not diffeomorphic to a circle?
I looked around and I couldn't find much, I've tried using the definition of smooth manifold but I can't really get anywhere.
The diffeomorphic part I think it will be straightforward once I figure the first part.
Can you help?
 A: If it was a smooth manifold, it would be a submanifold of the plane and hence any point of the triangle, and in particular any vertex, would have a neighbourhood $V$, that, in some basis of the plane could be written $V= \{ (x,f(x)) \ , x \in (-1,1) \}$ where $f:(-1,1) \to \Bbb R$ is a smooth map and $(0,f(0))$ is the coordinates of the vertex in the chosen basis of the plane.
But such an $f$ cannot be smooth because it would also have to be piecewise linear.
Edit: There are many equivalent definitions of a submanifold of $\Bbb R^n$, I used one of them, you can find it  here at (d) in proposition 3.2.1. When working with subsets of $\Bbb R^n$, it is often better to use one of these caracterisations instead of just charts, because it simplifies thing so much. 
So if your triangle was a smooth submanifold of $\Bbb R^2$ it should be locally parametrized by a smooth function, and this is not the case at a vertex, since the only parametrizations are piecewise linear, and henceforth not smooth.
But note that it depends essentially on the fact that your triangle is a subset of $\Bbb R^2$: it inherits the topology of the plane but cannot inherit the differentiable structure because of the "corner". You can find much more details in here (the accepted answer and the comments under it should really help you out).
