# Why do all compact connected surfaces have a triangulation?

Does anyone know a reference for a relatively elementary proof that every compact connected surface can be triangulated? By "elementary," I mean that I could present at least a sketch to undergraduates taking a first semester topology course.

When one proves the classification of surfaces, invariably one assumes that the surface is triangulable or smooth, but I'd like to justify this missing step.

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If I remember correctly there is a proof in the book Riemann surfaces by Ahlfors and Sario. Google doesn't let me check, so take this information with a grain of salt, please. Also, it may well be that it doesn't fit the bill elementary at all. –  t.b. Apr 16 '11 at 16:17
Most sources I know point to Moise's 1977 "Geometric Topology in 2 and 3 dimensions", but I don't know if the proof there will count as "relatively elementary", and I'm not sure such a proof exists. You'll likely need to prove, or at least assume, the Jordan separation theorem and probably Schoenflies as well, and even then the argument will be intricate. –  Alon Amit Apr 16 '11 at 17:00
Wikipedia says the proof is hard and cites Seifert & Threlfall, 1934. There are other approaches to classification of surfaces. See for instance the one using Morse functions. –  lhf Apr 16 '11 at 17:05
Oh - there's a relevant MO thread: mathoverflow.net/questions/17578/triangulating-surfaces –  Alon Amit Apr 16 '11 at 17:06
@lhf: initially it may seem obvious that all compact manifolds should be triangulable - just cover them with disks, triangulate each one, and refine things along the boundary so they match up. It then becomes clear that the problem lies in how wildly the boundaries of those disks may intersect. It feels intuitively right that circles can only cause so much trouble, but higher-dimensional spheres may exhibit such wild behavior that it may be impossible to fix. –  Alon Amit Apr 16 '11 at 18:33