From a Polygon to a Quotient Space
A polygon can serve as a starting space for constructing new topological spaces: select some boundary edges, match each selected pair point by point, and treat matched points as one point. The resulting space is not the polygon itself with a different drawing. It is a set of equivalence classes, equipped with a topology defined from the original polygon. Keeping the points and the classes distinct is essential, especially at vertices, where several identifications can meet.
Let \(P\) be a closed polygon in the plane, with its subspace topology from \(\mathbb{R}^2\). Its interior \(P^\circ\) is its interior in the plane, and its boundary is divided into finitely many closed edges meeting at vertices. We choose pairs of distinct edges, with the requirement that each edge occurs in at most one pair. For each pair, choose a homeomorphism from one edge onto the other. This condition on edge pairs ensures that the construction has the point-matching behavior described below.
Here, “smallest equivalence relation” means that we make the specified identifications and then enforce reflexivity, symmetry, and transitivity—no more and no less. Equivalently, \(x\sim y\) if \(x=y\), or if there is a finite chain from \(x\) to \(y\) in which each step is one of the specified identifications or its reverse.
Which Points Are Identified?
Proof. The generating identifications involve only points on paired boundary edges, so no generating step moves an interior point of \(P\). Thus an interior point has no point other than itself in its class. An interior point of an unpaired edge also occurs in no generating identification, so its class is a singleton. Now let \(x\) be an interior point of a paired edge \(e\), and let \(h:e\to f\) be its chosen homeomorphism to the other edge \(f\). The only generating step involving \(x\) matches it with \(h(x)\). Since \(f\) occurs in no other pair, the only generating step involving \(h(x)\) matches it back with \(x\). A chain starting at \(x\) can therefore visit only \(x\) and \(h(x)\). Finally, a homeomorphism between closed edges maps endpoints to endpoints and interior points to interior points. Thus a generating step involving a vertex only relates it to a vertex, and a chain starting at a vertex cannot reach an edge-interior point or a polygon-interior point. \(\square\)
Vertices require a different description from edge-interior points: a vertex can participate in more than one identification because it is an endpoint of two edges. The equivalence relation includes all consequences of those identifications. In particular, if one vertex is identified with a second, and that second vertex is identified with a third, then all three belong to one class.
Worked Example: No Edges Paired
Take a polygon \(P\) and make no edge pairs. There are no generating identifications, so the only chains in the definition are chains that stay at the same point. Hence \([x]=\{x\}\) for every \(x\in P\).
In this case \(q\) is the bijection that sends each point to its singleton class. By the definition of the quotient topology, a set \(W\subseteq Q\) is open exactly when \(q^{-1}(W)\) is open in \(P\). Since \(q\) merely relabels each point as its singleton class, this is precisely the same collection of open sets as on \(P\). Thus this quotient changes neither the points’ identifications nor the topology, apart from the relabelling by singleton classes.
Check: for any \(x\in P\), \(q^{-1}(\{[x]\})=\{x\}\), as required by the singleton-class calculation.
Worked Example: Pairing Two Opposite Edges of a Square
Let \(P=[0,1]\times[0,1]\) with its usual subspace topology from \(\mathbb{R}^2\). Pair its left edge \(\{(0,t):0\leq t\leq1\}\) with its right edge \(\{(1,t):0\leq t\leq1\}\) using \(h(0,t)=(1,t)\). These are the only paired edges.
For \(0<t<1\), the edge-interior point \((0,t)\) has class \(\{(0,t),(1,t)\}\). For instance, when \(t=1/3\), the matched points are \((0,1/3)\) and \((1,1/3)\). The endpoint pairs are \((0,0)\sim(1,0)\) and \((0,1)\sim(1,1)\). Since the left and right edges are paired only with each other, these endpoint identifications do not connect the bottom corners to the top corners.
Every point with \(0<x<1\) and \(0<y<1\) is in the polygon interior and has a singleton class. The points \((x,0)\) and \((x,1)\), where \(0<x<1\), are on unpaired edges and also have singleton classes. For example, the class of \((1/2,0)\) is just \(\{(1/2,0)\}\).
Check: the chosen map preserves the edge parameter \(t\), so applying the matching from left to right and then back sends \((0,t)\) to \((1,t)\) and then to \((0,t)\). The displayed two-point classes and the endpoint pairs are therefore consistent with the generating identifications and their reverses.
Worked Example: Transitivity at the Vertices
Let \(P\) be a triangle with vertices \(A=(0,0)\), \(B=(1,0)\), and \(C=(0,1)\). Pair edge \(AB\) with edge \(BC\) by the homeomorphism that sends \((1-s)A+sB\) to \((1-s)B+sC\) for \(0\leq s\leq1\). Edge \(AC\) is unpaired.
At \(s=0\), the paired points are \(A\) and \(B\), so \(A\sim B\). At \(s=1\), they are \(B\) and \(C\), so \(B\sim C\). By transitivity, \(A\sim C\); therefore the three vertices lie in one equivalence class. The interior point of \(AB\) at \(s=1/2\) is \((1/2,0)\), and its matched point on \(BC\) is \((1/2,1/2)\). These two edge-interior points form a two-point class, while points in the interior of \(P\) remain singleton classes.
Check: substituting \(s=0\) into the two edge parametrizations gives \(A\) and \(B\); substituting \(s=1\) gives \(B\) and \(C\); substituting \(s=1/2\) gives the two stated midpoints. The vertex conclusion uses transitivity, not an additional edge pairing.
The Quotient Topology and Its Projection
The quotient topology defines openness in terms of the original polygon: a set of classes is open precisely when the union of those classes in \(P\) is open. More explicitly, \(q^{-1}(W)\) is the set of all points whose classes belong to \(W\). It is automatically a union of entire equivalence classes. A subset of \(P\) that is a union of entire classes is called saturated.
Proof. Every class in \(Q\) contains a point of \(P\), so \(q\) is surjective. For any open \(W\subseteq Q\), the quotient-topology definition says that \(q^{-1}(W)\) is open in \(P\); this is exactly the condition for continuity of \(q\). The same definition gives the stated equivalence for every subset \(W\subseteq Q\), so \(q\) is a quotient map. \(\square\)
This definition does not say that \(q\) is an open map. An open map must send every open subset of its domain to an open subset of its codomain; a quotient map is instead characterized by the openness test using preimages. Nor does the definition by itself establish that \(Q\) is a surface. It constructs a topological space; whether its neighborhoods have the local form required of a surface is a further question.
Proof. Since \(f\) is constant on each class, define \(\bar f([x])=f(x)\). If \([x]=[y]\), then \(f(x)=f(y)\), so this definition is independent of the representative. It also gives \(f=\bar f\circ q\). If another map has this factorization, its value at \([x]\) must be \(f(x)\), so it equals \(\bar f\); this proves uniqueness. To prove continuity, let \(V\subseteq Y\) be open. Then \(q^{-1}(\bar f^{-1}(V))=f^{-1}(V)\), which is open in \(P\) because \(f\) is continuous. The quotient-topology definition now implies that \(\bar f^{-1}(V)\) is open in \(Q\). Therefore \(\bar f\) is continuous. \(\square\)
A Common Pitfall: Pairwise Disjointness and Transitivity
A tempting but incorrect setup is to pair one edge with two different edges and then claim that every interior point of the first edge has just one partner. For example, take a triangle and pair edge \(AB\) with \(AC\), and also pair \(AB\) with \(BC\), choosing the edge homeomorphisms so that the midpoint of \(AB\) is sent to the midpoint of each other edge. The midpoint of \(AB\) is then identified with two distinct points. It does not have a unique partner. This is why the construction here requires that each edge occur in at most one pair.
A separate error is to record only the listed point-to-point matches and forget that they generate an equivalence relation. In the triangle example, the endpoint matches give \(A\sim B\) and \(B\sim C\). Writing down two separate pairs and leaving \(A\) unrelated to \(C\) would fail transitivity. The actual equivalence class contains all three vertices. More generally, to describe a quotient point correctly, one must collect every point reachable by a finite chain of matches and reversed matches.
| Location in the polygon | Equivalence class under the stated pairing rule | Reason |
|---|---|---|
| Interior of \(P\) | Singleton | No generating identification involves an interior point |
| Interior of an unpaired edge | Singleton | No generating identification involves that edge |
| Interior of a paired edge | The point and its matched point | Each edge belongs to at most one pair |
| Vertex | A class of vertices, possibly containing several vertices | Endpoint matches combine by transitivity |
Polygon quotients provide a precise language for gluing boundary pieces: the equivalence relation specifies which points become one, while the quotient topology specifies which sets count as open after gluing. The distinction between edge interiors and vertices is useful because their classes arise differently. It is also a reminder that a polygon diagram is not, on its own, a proof of the topology of the resulting space; the classes and their neighborhoods are determined by the definitions.
Check Your Understanding
Use the definitions in this tutorial to answer the following questions.
- Why does the construction require that each edge occur in at most one chosen pair if paired edge-interior points are to have unique partners?
- In a quotient of a triangle, suppose the endpoint identifications give \(A\sim B\) and \(B\sim C\). Which vertices must lie in the same equivalence class, and which property of an equivalence relation justifies the conclusion?
- For the square example, describe the class of \((0,1/4)\) and the class of \((1/2,0)\).
- State the condition on \(W\subseteq Q\) that defines openness in the quotient topology.
- Why does the factorization property require that \(f\) be constant on each equivalence class?