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This exercise $2.1.10$ in page $131$ of Hatcher's book Algebraic topology.

(a) 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 $\mathbb{R^2}$.

(b) Show the edges can always be oriented so as to define a $\Delta$-complex structure on the quotient surface. [author: This is more dificult.]

I've done the 1st one, but I got stuck while solving the 2nd part. Currently I've no clue for this.

I am not able to find any kind of algorithm which fits for it for any arbitrary $n$. For example, if we take this operation using two 2-simplex as a result of various quotien,t we can get different spaces like torus, Klein bottle, projective plane etc...and for all this spaces the ordering will be different.

So as a result I cannot guess the algorithm and I need some serious help, a way of thinking. Thank you.

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    $\begingroup$ Hint: Start by choosing a linear ordering of the set of vertices of the complex. $\endgroup$ – Jim Belk May 29 '15 at 0:49
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I can think of the following algorithm:

First, choose an ordering of all the vertices of the quotient surface, and then orient all the all the edges that aren't loops along this ordering. This way, all the 2-simplices (triangles) are oriented properly.

So now you only have to consider, for every vertex $v$, the edges that go from $v$ to $v$. You use the local homeomorphism to $\mathbb R^2$ at $v$ to be able to talk about "clockwiseness". Then, you iterate over the edges at $v$ in a clockwise fashion, so that every loop will appear exactly twice. I will elaborate later which edge you start this process at. At each edge, you do the following:

  • If that edge goes to a vertex other than $v$, it is already oriented, so you do nothing.
  • Otherwise, if the edge is not oriented yet, you orient it so that it points away from $v$ in the region you're considering.
  • Otherwise, you already oriented that edge earlier, so now it points towards $v$.

Now, to prove that this results in a proper $\Delta$-complex structure, we can consider that a $2$-simplex is properly oriented if, in the original space, there is one vertex of the $2$-simplex that has two edges pointing away from it.

So let us consider an arbitrary $2$-simplex (in the original space), whose three vertices map to $v$ under the canonical map. Let us call the edge that was oriented first $a$. Then let us call the vertex that $a$ points away from $x$. Now, let us call the other edge at $x$ $b$. Then there are two cases:

  • $b$ is oriented in the next step after $a$ was oriented: In this case, $b$ points away from $x$ as well, so we are done!
  • $b$ is the last in the sequence of edges, while $a$ is the first. In this case, $b$ will point towards $x$, so the algorithm only works if the third edge points from the other end of $b$ towards the other end of $a$. We will choose the starting edge of our iteration to ensure that.

So at which edge should the process start? If one edge is not a loop, we can start at this edge, since then the $2$-simplices adjacent to it will be properly oriented already. Otherwise, we need some definitions:

A half-edge will be the part of an edge (in the quotient space) that touches $v$. The twin of a half-edge $\alpha$ (denoted $T(\alpha)$) will the other half-edge that touches $v$. The successor of a half-edge $\alpha$ (denoted $S(\alpha)$) will be the next half-edge in the clockwise ordering. Let $X(\alpha) = T(S(\alpha))$.

The next $2$-simplex of a half-edge will be the the $2$-simplex between that half-edge and its successor. For any half-edge $\alpha$, the $2$-simplex next to $\alpha$ will also be the $2$-simplex next to $X(\alpha)$, so $X^3(\alpha) = \alpha$.

So if you go from $\alpha$ to $X(\alpha)$ to $X(X(\alpha))$ and back to $\alpha$ while always going clockwise, you will take either one or two turns around $v$. If you took one turn, we will call the next $2$-simplex of $\alpha$ bad, otherwise we will call it good.

Now, there has to be at least one good $2$-simplex at $v$. Otherwise, let $\alpha$ be some half-edge. Then, we have $T(X(\alpha) = S(\alpha)$, so in the clockwise ordering, $X(S(\alpha))$ will be strictly between $S(\alpha)$ and $X(\alpha)$. Similarly, $X(S(S(\alpha)))$ will be strictly between $S(S(\alpha))$ and $X(S(\alpha))$, and so on, which is a contradiction since there are only finitely many edges.

So let $\beta$ be some half-edge so that the next $2$-simplex is good. Then we start the process at $S(\beta)$ and end it at $\beta$. To see that this $2$-simplex will be properly oriented, observe the sequence in which its half-edges are visited:

  • $S(\beta)$ is oriented to point away from $v$, so $X(\beta)$ points towards $v$.
  • $X(X(\beta))$ is oriented to point away from $v$, so $S(X(\beta))$ points towards $v$.
  • $S(X(X(\beta))$ is oriented to point away from $v$, so $\beta$ points towards $v$.

So since both $X(X(\beta))$ and $S(X(\beta))$ point away from $v$, this $2$-simplex is properly oriented.

I feel like I probably reinvented the wheel pretty badly here, so does anybody know some concepts that I used, or some useful notation, or some way of simplifying this?

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For a single 2-simplex, use the axiom of choice to assign the ordinals $\{1,2,3\}$ to the vertices. Then $1 \rightarrow 2$, $1 \rightarrow 3$, and $2 \rightarrow 3$ is an ordering.

Do you see how to proceed if there are more vertices?

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  • $\begingroup$ Hi Eric, a single 2-simplex should not be allowed in this case. For we are identifying each edge of a simplex with exactly one other edge. $\endgroup$ – user135520 Feb 9 '16 at 4:34
  • $\begingroup$ @user135520 : I agree. And yet, if you understand the construction on a single simplex, you understand the construction on an arbitrary complex. $\endgroup$ – Eric Towers Feb 9 '16 at 4:39
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    $\begingroup$ But aren't we trying to determined an orientation based on the gluing. For each edge there two ways to glue it to some other edge and so forth. I'm having trouble seeing the connection here $\endgroup$ – user135520 Feb 9 '16 at 5:25
  • $\begingroup$ There is a new solution, have a look at that.. Thanks. $\endgroup$ – Anubhav Mukherjee Jun 8 '16 at 17:10

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