- Oriented compact 2-surfaces: sphere, torus, 2 holed torus, etc.
- have euler characteristic $V - E + F$ as $2 - 2g$
- Strategy: cut surface into polygonal pieces. Use oriented edges to know cutting. Lay them down on the surface such that the "top part" or "painted surface" will be up [so we retain orientation ].
- Attach all the polygons into one big polygon on the plane.
- For each edge on boundary of the big polygon, it must attach to some other boundary of the big poygon [since the manifold is compact ]. Furthermore, this edge must occur in the
*opposite direction*to make the surface orientable. Otherwise we could pass through the side and flip orientation. Consider:

```
>>>>
| |
>>>>
```

- When I appear from the "other side", my direction wil have flipped. [TODO ]

- So far, we know the edges. What about identifying vertices?

- Next, we need to group vertices together on the big polygon. We can find this by going
*around the edges incident at the vertex*on the*manifold surface*.

- The next step is to reduce the number of vertices to exactly one. We can cut the current polygon and re-paste it as long as we preserve all cutting/pasting relations.

- Suppose I glue all the B vertices to a single vertex. Then, the edges emenating from this B vertex
*must necessarily be the same*. If not, then the edge emenating would need a complementary edge somewhere else, which would give me another "copy" of the B vertex.

- I can imagine such a B vertex as being "pushed inside the polygon" and then "glued over itself", thereby making it part of the
*interior*of the polygon.

- We can repeat this till there is only one type of vertex (possibly multiple copies).
- If we only had two adjacent edges [edges incident against the same vertices ], then we are done, since we get a sphere.
- We can always remove adjacent pairs of edges. What about non-adjacent pairs?
- Take a non adjacent pair. think of these as "left" and "right". We claim that for each edge at the "top", there is a corresponding edge at the "bottom". So we have left and right identified, and top identified with a continugous segment in the bottom. If there wasn't, then we would need another vertex!
- This lets me create a commutator on the boundary, of the form $cdc^{-1}d^{-1}x$. Topologically, this is a handle, since if it were "full" [without the extra $x$], then we would have a torus. Since we do have the $x$, we have a "hole on the torus" which is a handle.
- We keep removing hanldes till we are done.

- If we add a vertex on an edge, we add a vertex and subrtact the (new) edge we have created. Thus $\xi$ is unchanged on adding a vertex on an edge.
- Joining two vertices on a face also does not change $\xi$, since we add an edge and a face.
- Given any two subdivisios, we find a common finer subdivision by these steps. Since the steps we use retain the euler characteristic, finally my original subdiv = common subdiv = friend subdiv.
- Key idea: at each crossing between our subdivsion and the other subdivision, make a new vertex at every crossing. Then "trace over" the other subdivision to make our subdivision agree on the other subdivision on the inside.