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I have heard that one can prove the classical classification of surfaces theorem using Morse theory. I am planning on learning this approach as a way to motivate and get comfortable with Morse theoretic ideas, but I am curious if there is some special insight that this approach has for this specific classification effort. On Wikipedia they describe Morse homology theory and say that it is an "easy" way to understand the homology of smooth manifolds. Why? Can this been seen in the classification of surfaces, Morse-style?

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Maybe this should be a comment, but it was a bit too long. First of, let me state that I have never studied the classification of surfaces through Morse theory. A quick google search gave me the following document (which you probably already found), which seems to describe it in detail. .

Classical Morse theory describes how sublevel sets $f^a=\{x\in M|f(x)\leq a\}$ change topology when one passes through critical values of a sufficiently nice function $f$. By studying these topology changes, one in the end get a full description of the homotopy type of the manifold $M$. This can be done using the gradient flow of $f$. By working harder you can improve this with handlebodies to get the diffeomorphism type.

The following is just my speculation on the classification. In two dimensions, only critical points of index $0,1,2$ are possible of course. Not all possible combinations of these types of critical points can occur. For example on a compact surface you always have at least a critical point of index 0 and 2, coming from the maximum and minimum of the Morse function. These of course exist because $M$ is compact. By thinking on how these critical points can occur, you can classify the manifold $M$ using $f$.

As a side remark, I don't think that Morse homology is the easiest way to do homology theory. To get the full Morse theorem, you need quite a bit of work. However, the Betti numbers can be extremely easy to compute using Morse theory. The Morse Lacunary principle says for example that if you can construct a Morse function with no odd index critical points on $M$, then the Betti numbers are equal to the number of critical points with that index.

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