Given two compact oriented surfaces that have the same number of genus and boundary components. How to construct a homeomorphism that sends one to another?
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1$\begingroup$ I think, as stated, there doesn't have to be one. We have to assume the surface is compact or complicated things can happen. Do you have a specific example in mind or are you looking for the general reason why two connected compact surfaces of the same genus with the same number of boundary components are homeomorphic? $\endgroup$– CallumCommented Mar 9, 2022 at 13:38
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2$\begingroup$ I presume the surfaces are meant to be compact and orientable. $\endgroup$– ThiKuCommented Mar 9, 2022 at 13:41
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1$\begingroup$ Maybe you could be a bit more precise about what you mean by "construct"? Or are you just asking for a proof of the classification of surfaces? (These proofs are "constructive" in some sense. Have you looked at them?) $\endgroup$– NicolastCommented Mar 9, 2022 at 13:52
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A closed surface of genus g can be cut along 2g closed curves (all at one base point) to obtain a 4g-gon. Do this for both surfaces. Then choose a homeomorphism between the 4g-gons which matches the corresponding pairs of curves, so you get a well-defined homeomorphism of surfaces. (A 4g-gon is a cone over some interior cone point. You can first choose an appropriate homeomorphism of boundaries, then fix an interior „cone point“ for both 4g-goes, and then just take the cone map, that extends the chosen homeomorphism of boundaries into the interior, sending one cone point to the other.)
Similarly, surfaces with boundary can be cut into polygons, where now some edges of the polygon correspond to boundary curves.
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$\begingroup$ They should be disjoint except for the common basepoint. And there is always a basepoint-fixing homeomorphism between two closed curves. $\endgroup$– ThiKuCommented Mar 10, 2022 at 12:27
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$\begingroup$ What kind of proof are you looking for, i.e., what do you want to assume? If you assume surfaces are smooth, the easiest proof is via Morse theory. You find it in Morris Hirsch “Differential Topology” or via googling, e.g. uu.diva-portal.org/smash/get/diva2:956989/FULLTEXT02.pdf . If you assume surfaces are triangulated, then the proof gets combinatorial, look at webspace.science.uu.nl/~meier007/ClassificationSurfaces or www3.nd.edu/~andyp/notes/ClassificationSurfaces.pdf $\endgroup$– ThiKuCommented Mar 10, 2022 at 14:40