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Show the quotient space of a finite collection of disjoint 2-simplices obtained by identifying pairs of edges is always a surface, locally homeomorphic to $\mathbb{R}^2$.

I have thought about doing the following: I think we have to consider several cases

To prove that this space is a surface, we must take a point and prove that there is an open that contains it that is homeomorphic to the plane, if the point belongs to the interior of a 2-simplex that this space includes, we are ready the open is 2-simplex itself, the problem is if the point in question belongs to the intersection of two or more 2-simplices, how can I do in this case to be well defined? Thank you!

Edit: This question is part of the exercises in Hatcher's book, in particular, exercise $10.(a)$ (pag 131), the complete exercise is:

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Note that: Each edge is identified with exactly one other edge.

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  • $\begingroup$ You must make precise what "obtained by identifying pairs of edges" means. Are only some edges identified? Or is any edge identified with another edge? Can three edges be identified? I voted to close your question. After clarification it can be reopened. $\endgroup$
    – Paul Frost
    Commented Jul 17, 2019 at 10:07
  • $\begingroup$ @PaulFrost "obtained by identifying pairs of edges" means that two faces of two different 2-simplices are stuck together. $\endgroup$
    – Nash
    Commented Jul 17, 2019 at 22:32
  • $\begingroup$ @PaulFrost you still do not understand what I mean? $\endgroup$
    – Nash
    Commented Jul 18, 2019 at 1:15
  • $\begingroup$ It seems that your question is an exercise (from a book or from homework). Check whether there are more details in the context (or ask if it is homework). In the present form it is too vague. Here are some questions that must be answered. 1) Can one identify two edges of one simplex? (I guess no.)2) Is it allowed that more than $2$ edges are identified? I guess no.) 3) Is it allowed than some edges remain unidentified? 4) Say you have a number $k$ of $2$-simplices. Then you have $3k$ edges. What if $k$ is odd? (Consider 3.) If you can make it precise, please edit your question. $\endgroup$
    – Paul Frost
    Commented Jul 18, 2019 at 7:32
  • $\begingroup$ @PaulFrost I have already edited the question. $\endgroup$
    – Nash
    Commented Jul 19, 2019 at 2:01

2 Answers 2

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One must demonstrate an atlas of coordinate charts with continuous transitions. For any point on the interior of a polygon, this is trivial. For other points, the strategy is to draw circles around the point and traverse through identifications.

For a point on an edge of the polygon that is not a vertex, you can capture a neighborhood by drawing a semicircle that does not go around any vertices, then following the gluing and drawing another semicircle. The enclosed region is evidently homeomorphic to a neighborhood of $\mathbb{R}^2$.

Around verticies, you do essentially the same thing: you draw segments of circles, following identifications around until you're back to where you started. Note that this process is well-defined as every edge is glued to exactly one other edge, and must terminate in finitely many steps since each edge only gets crossed once. Mapping the enclosed neighborhood into a disk in the plane (reparametrizing angles so as to match up at the end) then gives a chart around the vertex. A bit of thought shows that the charts these describe can be made to have continuous transitions.

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See John Lee's Introduction to Topological Manifold (e2) proposition 6.4 (b) for local homeo, and 5.23 and 6.4 (a) for hausdorff and 2nd countable.

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