I am taking a course on algebraic topology and I am trying to prove the following exercise:
Let $S$ be a differentiable surface in $\mathbb{R}^3$. Prove that $S$ is colorable (you can paint one side blue and the other one red) if and only if it is orientable.
My first thought was using the compact surfaces classification, but $S$ is not required to be compact, so I do not know how to proceed. Also I don't know how to use the differentiability hypothesis.
Nevertheless, the intuition behind this exercise seems pretty obvious to me. If we consider a Möbius strip, we can move along the surface coming back to the same point so it only has one side but if we consider a orientation it has changed along the way. This makes clear that there must exist a connection between the two concepts.
The point seems to be a subtlety about colorability being inherent to the embedding, which I cannot totally grasp.
Suppose your surface is oriented. With your definition you can color both sides of each triangle. For example, the clockwise side you paint blue, and the anti-clockwise side red. This colors your surface. Similarly, if the surface is colored, you can orient it as well, by saying that the blue side is clockwise (for each triangle).