From a Polygon Diagram to Cells
A polygon diagram records how boundary points are identified, but its sides and corners do not automatically give the cells of a decomposition. A paired pair of sides usually contributes one edge, not two; several corners may become a single vertex; and a self-pairing can identify interior points of a side, so that side may need subdivision. The aim here is to turn the diagram into a precise list of vertices, edges, and faces.
Let \(P\) be a closed polygon, and let \(X\) be its quotient under the stated boundary identifications, with the quotient topology. The open polygon means the interior of \(P\). A polygon diagram may also leave some sides unpaired; their interiors remain boundary edges in \(X\). Unless otherwise stated, assume each paired side pair consists of two distinct sides identified by a homeomorphism, and that no interior point of the polygon is identified with another point.
In a polygon diagram, a vertex class is an equivalence class of polygon corners under the endpoint identifications generated by the side pairings. An edge class is a side, or a collection of side pieces, whose interiors are identified with one another by the gluing. Count each such class once. A side left unpaired contributes its own edge class.
The Counting Principle
Proof. For each endpoint class, take a point of the quotient as a \(0\)-cell. Each unpaired edge piece has interior homeomorphic to an open interval. For a pair of distinct pieces identified by a homeomorphism, their interiors have a common image homeomorphic to an open interval: the quotient identifies one interval with the other, and either interval gives a continuous parametrization with continuous inverse on that image. The endpoints of each interval map to endpoint classes, so the closed interval gives a characteristic map for its \(1\)-cell.
The interior of the polygon is not identified with any other point, so its quotient image is homeomorphic to an open disk. The quotient map restricted to the closed polygon is a characteristic map for this \(2\)-cell: it is continuous, is a homeomorphism on the interior, and sends the boundary into the union of the vertices and edges. For a subdivision into polygonal faces, use the closed pieces as the characteristic domains. Their interiors are disjoint and map homeomorphically to the respective open \(2\)-cells; the two copies of each internal edge are identified, giving one edge cell. Finally, the quotient topology is the topology obtained by attaching these finitely many closed cells along their boundaries: the polygonal pieces cover the original polygon, and the only identifications are the specified boundary and subdivision identifications. This gives the asserted finite cell decomposition. \(\square\)
The theorem is a counting rule, not a test that the quotient is a surface. Whether every point has a neighborhood homeomorphic to a disk or half-disk is a separate question, addressed by the link tests for polygonal surface quotients. The counting rule still describes the cells when the quotient fails to be a surface.
Worked Example: One Paired Pair and Two Unpaired Sides
Let a square have cyclically ordered vertices \(v_0,v_1,v_2,v_3\) and sides \(e_i\) running from \(v_i\) to \(v_{i+1}\), with indices read modulo \(4\). Pair \(e_0\) with \(e_2\) by a homeomorphism that reverses their boundary orientations, and leave \(e_1,e_3\) unpaired. Find the cells.
The endpoint identifications are \(v_0\sim v_3\) and \(v_1\sim v_2\): reversing the boundary orientations sends the initial endpoint of \(e_0\) to the terminal endpoint of \(e_2\), and its terminal endpoint to the initial endpoint of \(e_2\). Thus the vertex classes are \(\{v_0,v_3\}\) and \(\{v_1,v_2\}\), so there are two \(0\)-cells.
The interiors of \(e_0,e_2\) form one paired edge class. Each unpaired side contributes one more edge class. Hence there are \(1+1+1=3\) edge cells. Their endpoints are also a useful check: the paired edge joins the two distinct vertex classes; \(e_1\) has both endpoints in \(\{v_1,v_2\}\); and \(e_3\) has both endpoints in \(\{v_0,v_3\}\). The square interior contributes one \(2\)-cell. The resulting counts are two vertices, three edges, and one face.
Check: the two paired sides were counted once, not twice; the two unpaired sides were each counted once. The endpoint identifications listed above account for all four corners.
Worked Example: Subdividing a Self-Paired Edge
Use a square with the same cyclic vertices and sides. Identify \(e_0\) with itself by reflection across its midpoint, leave \(e_2\) unpaired, and pair \(e_1\) with \(e_3\). For the latter pairing, use the map matching the initial endpoint of \(e_1\) to the initial endpoint of \(e_3\) and the terminal endpoint of \(e_1\) to the terminal endpoint of \(e_3\). Determine a cell decomposition.
Write points of \(e_0\) as parameters \(t\in[0,1]\), from \(v_0\) to \(v_1\). The self-pairing sends \(t\) to \(1-t\). In particular, \(v_0\sim v_1\), while the midpoint \(m\), with parameter \(1/2\), is fixed. The \(e_1,e_3\) pairing identifies \(v_1\sim v_3\) and \(v_2\sim v_0\). Combining these relations puts all four corners in one class, denoted \(V\). The midpoint \(m\) is not identified with a corner, so it gives a second vertex class.
Subdivide \(e_0\) at \(m\). Its two half-edges are exchanged by reflection, so their interiors make one edge cell, with endpoints \(V\) and \(m\). The paired sides \(e_1,e_3\) make one edge cell. The unpaired side \(e_2\) makes one edge cell. The square interior is one face. Thus this decomposition has two vertices, three edges, and one face.
Check: the self-pairing sends the first half of \(e_0\) onto the second half and fixes only their shared midpoint. The midpoint must therefore be included among the vertices; without it, the image of the side interior would not be a single open interval cell.
Subdivision is not always a finite remedy for a self-pairing. The reflection above exchanges the two halves after a single cut. For a different self-pairing \(h(t)=t^2\) on \([0,1]\), a finite subdivision into pieces paired with distinct pieces cannot be obtained by choosing finitely many subdivision points invariant under \(h\). Indeed, if an interior point \(t\) belongs to such an invariant finite set, its iterates \(t,t^2,t^4,t^8,\ldots\) must also belong to that finite set. They are distinct because \(0<t<1\), a contradiction. Thus subdivision advice for self-pairings requires a suitable hypothesis: not every side self-pairing can be resolved in this way.
Worked Example: Subdividing a Face Along a Diagonal
Take a square with no boundary identifications, and draw a diagonal from \(v_0\) to \(v_2\). Use the two resulting triangles as faces. Count the cells of this decomposition and compare its quotient space with the unsplit square.
The four corners remain four distinct vertices. There are four boundary edge cells and one diagonal edge cell, for a total of five. There are two triangular face cells, one on each side of the diagonal. Each triangle’s interior maps homeomorphically to its open face, and the two copies of the diagonal are identified point-for-point as the shared internal edge.
The space is still the original square, with its original topology. More precisely, the two closed triangles, glued along their common diagonal, form a space canonically homeomorphic to the square: the map that sends each triangle to its original position is continuous and bijective, and its inverse is continuous because it is the same piecewise identification on the two triangles. The quotient of these pieces with the internal edges glued is therefore homeomorphic to the original square. Subdivision changes the chosen cell decomposition, not the underlying space.
Check: the diagonal occurs in the boundaries of both triangles but is one edge cell, not two. The four boundary sides plus that diagonal give five edges; the two triangle interiors give two faces.
Reading Counts Reliably
A useful procedure is to separate the three dimensions rather than count directly from the picture.
Write every endpoint identification induced by the side maps, then take the equivalence classes generated by those relations. Do not assume that different corners remain distinct.
Count each paired pair once and each unpaired piece once. If a side is paired with itself, check whether its interior quotient is actually an open interval; subdivide only when an appropriate finite subdivision exists.
A polygon with no face subdivision contributes one face. A subdivision contributes one face per polygonal piece, with internal edges glued and counted once.
Check that every edge has the stated endpoint classes and that each proposed open cell is homeomorphic to an open ball of its dimension.
Common pitfall. Counting every original side as a separate edge is wrong whenever two side interiors are identified. In the self-paired-edge example, counting \(e_0\) as one ordinary edge and ignoring its fixed midpoint would also miss a vertex: the interior point \(m\) is a separate equivalence class, and the quotient of the side interior is not a single open interval. The correction is to split at \(m\), count the two paired half-edges as one edge cell, and count \(m\) as a vertex. A second common error is to count the two copies of a subdivided diagonal as two edges; they are identified along the internal seam and form one edge cell.
These cell counts provide a finite description of a polygon quotient. They are useful because later calculations can be made from vertices, edges, and faces rather than from the entire quotient space. The description is only as reliable as the cell checks: endpoint classes must be computed from the actual gluing maps, and each proposed open cell must have the required homeomorphism type.
Check Your Understanding
For each question, state the relevant vertex, edge, and face counts and explain which identifications produce them.
- A square has \(e_0\) paired with \(e_2\) by an endpoint-reversing map, while \(e_1,e_3\) are unpaired. What are the vertex classes and cell counts?
- In a polygon diagram, two distinct sides are paired and all other sides are unpaired. How many edge cells arise from those two paired sides?
- A square is divided by both diagonals into four triangles, with no boundary identifications. Count the vertices, edges, and faces, counting each internal edge once.
- For the self-pairing \(t\mapsto 1-t\) of one side, which point must be added as a vertex when subdividing into paired edge pieces?
- Why does a face subdivision change the cell counts without changing the quotient space?