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I am practicing qualifying exam problems and I am having trouble with the following question. Any help is greatly appreciated.

Let $P$ be a polygon with an even number of sides. Suppose that the sides are identified in pairs in any way whatsoever. Prove that the quotient space is a manifold.

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Do you mean polygon as opposed to polynomial? – Michael Albanese Apr 18 '13 at 18:07
Since you have been posting so many questions, it may be a good idea for you to post your ideas/attempts. – user27126 Apr 18 '13 at 18:10
"Any way whatsoever" unfortunately needs a qualifier, or else there are counter-examples. Presumably the identifications on the boundary are 1-1 and linear, and no side is allowed to be identified to two or more other sides, etc. – Ryan Budney Apr 18 '13 at 18:52
Maybe. I may be able to do that problem. But the one posted is word for word. – Ethan Hawver Apr 18 '13 at 18:54
Dear Ethan, I think you should understand the "any way whatsoever" phrase to refer to the way the edges are grouped into pairs (and also about the choice of orientation). The actual gluing should just be the obvious one where the two edges are glued (according to some chosen orientation) so that their vertices match up. (The way you glue the edges of a square to make a torus, or a Klein bottle.) Regards, – Matt E Aug 9 '13 at 12:21

A manifold is, in particular, a locally Euclidean space, meaning that every point has a neighborhood homeomorphic to a disk in $\mathbb{R}^n$. You are working with a surface $(n = 2)$.

Do you see how every point in the interior of the polygon has a neighborhood that is unchanged by the identifications.

What about points on an edge, but not a vertex? They have half-disk neighborhoods in the polygon. Assuming that the identifications are nice (as Ryan Budney points out), two of these half disks are fused along a diameter, so you get a disk.

What happens to the vertices of the polygons?

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