What is the easiest way to classify all possible smooth orientable closed 2-manifolds? If this has been answered already, please let me know and I'll delete the question. 
ADDED: I'd prefer to assume a smooth structure, rather than a triangulation.
 A: A very simple and low tech solution to this problem is the very first proof, 
given by Möbius in 1863. He assumes that the surface is smoothly embedded
in $\mathbb{R}^3$, and slices it by a family of parallel planes. Assuming that
the orientation of the planes is general and that they are sufficiently close together, this cuts the surface into simple pieces -- either disks, annuli, or pairs of pants.
It is then quite easy to show that the result of assembling such pieces is always a sphere with handles.
A: Count the number of disjoint non-separating embedded circles. 
A: Consider harmonic functions $f$ with exactly 2 log-singularities of weight $\pm 1.$ (locally $f(z)=a log\parallel z\parallel+g,$ $g$ being smooth at $z=0$, $a$ being the weight) on your compact surface equipped with a Riemann metric. They exists by standard elliptic theory ( the two weights $a_1$ and $a_2$ have two add to zero). Consider $\partial f,$ the complex linear part of the differential. This is a meromorphic section of your canonical bundle. Then $deg\partial f=-\frac{1}{2\pi i}\int KdA,$ as the Levi-Civita connection defines a complex linear connection on the canonical bundle.
This shows that if the total curvature is large enough $\geq 4\pi,$ $f$ will not have critical points (only two singularities). Moreover $e^f$ is the real part of a holomorphic bijection onto $CP^1.$
If $f$ has a critical point, then you can easily construct a non-sepreating loop, as in Morse theoretic proofs. You cut your surface, and add two disc (with the right orientation). One, can easily see, that this must increase the total curvature by $4\pi,$ and you end up with the two-sphere after a finite number of steps.
A: Check out this nice paper by Thomassen for a short self-contained proof.
