Twisted Edge Identification for the Möbius Band

The Rectangle and Its Twist

A rectangle can be turned into a band by joining two opposite sides. If points at the same height are joined, the result is a cylinder. If the height coordinate is reversed as the sides are joined, the result is the Möbius band. The reversal is not a visual detail: it determines the equivalence classes and changes which edges meet along the boundary.

Let \(R=[0,1]\times[-1,1]\) have the subspace topology inherited from the Euclidean plane. Write points as \((x,t)\), where \(x\) is the coordinate along the rectangle and \(t\) is the transverse coordinate. Define an equivalence relation by identifying \((0,t)\) with \((1,-t)\) for every \(t\in[-1,1]\), and making no other identifications between distinct points. Let \(M=R/{\sim}\) have the quotient topology, and let \(q:R\to M\) be the quotient map.

More explicitly, the equivalence relation consists of equality together with the pairs \((0,t)\sim(1,-t)\) and their reversals. No point with \(0<x<1\) is identified with another point. On the two vertical edges, each point is paired with exactly one point on the opposite edge. The horizontal edges are not identified with other points along those edges, although their endpoints participate in the vertical-edge identifications.

Proof. By the definition of the relation, the only generating pairs involve a point whose first coordinate is \(0\) and the corresponding point whose first coordinate is \(1\). The map \(t\mapsto -t\) is its own inverse, since \(-(-t)=t\). Thus \((0,t)\) is paired with \((1,-t)\), and \((1,-t)\) is paired back with \((0,t)\). A point with first coordinate strictly between \(0\) and \(1\) occurs in no generating pair, so its class contains only itself. The listed pairs and singletons account for every point of \(R\). \(\square\)

What Happens to the Edges?

Let \(L_0=\{0\}\times[-1,1]\) and \(L_1=\{1\}\times[-1,1]\) be the vertical edges, and let \(A=([0,1]\times\{-1\})\cup([0,1]\times\{1\})\) be the union of the horizontal edges. The full preimage of the image of a set is called its saturation: for \(E\subseteq R\), this is \(q^{-1}(q(E))\). A set is saturated when it contains every point equivalent to each of its points.

The vertical edges are joined to each other, so they do not remain as boundary edges of the resulting band. The horizontal edges do remain, but their endpoints are paired across the two edges. In particular, the top-left corner \((0,1)\) is identified with the bottom-right corner \((1,-1)\), while the bottom-left corner \((0,-1)\) is identified with the top-right corner \((1,1)\).

Proof. We first verify that the quotient topology on \(q(A)\) is the quotient topology induced by restricting \(q\) to \(A\). The set \(A\) is closed in \(R\), and it is saturated: the only identifications involving its points pair the four corners, all of which belong to \(A\).

We will use that \(q\) is a closed map. If \(F\) is closed in \(R\), then its saturation is \(q^{-1}(q(F))=F\cup\{(1,-t):(0,t)\in F\}\cup\{(0,-t):(1,t)\in F\}\). Each added set is closed in its vertical edge, and each vertical edge is closed in \(R\); hence the saturation is closed in \(R\). By the definition of the quotient topology, \(q(F)\) is closed in \(M\): its complement has open preimage \(R\setminus q^{-1}(q(F))\). Now if \(C\) is closed in \(A\), then \(C\) is closed in \(R\), since \(A\) is closed in \(R\). Thus \(q(C)\) is closed in \(M\), and in particular in \(q(A)\). Therefore the surjection \(q|_A:A\to q(A)\) is a closed map. A closed surjection is a quotient map: if the preimage of a subset is open, the preimage of its complement is closed, so its image, which is that complement in the target, is closed. This proves the required quotient-topology claim.

Next, consider the interval \([0,2]\) with its endpoints identified. Map the first half to the top edge by \(x\mapsto(x,1)\), and the second half to the bottom edge by \(x\mapsto(x-1,-1)\). After applying \(q\), the two definitions meet at the same quotient point because \(q(1,1)=q(0,-1)\); the endpoints also agree in the quotient because \(q(0,1)=q(1,-1)\). This gives a continuous map from the interval with endpoints identified onto \(q(A)\). It is one-to-one on the resulting quotient: the only identifications among points of \(A\) are precisely those two cross-edge endpoint pairings. It is onto because every point of \(A\) lies on one of its two horizontal edges. The inverse correspondence is continuous as well: on the top edge assign the interval coordinate \(x\), and on the bottom edge assign \(1+x\), with \(0\) and \(2\) identified. These assignments respect the corner pairings, so they induce a continuous map from the quotient \(q(A)\). Thus the two quotient spaces are homeomorphic. The interval with its endpoints identified is homeomorphic to \(S^1\), as in the endpoint-identification construction of the circle. Finally, locally at a point of \(q(A)\), the rectangle gives a half-rectangle neighborhood, including at the paired corners; points not in \(q(A)\) have rectangle neighborhoods without an edge. Hence \(q(A)\) is exactly the manifold boundary. \(\square\)

Local Coordinates and the Common Mistake

Points away from the joined vertical edges inherit ordinary rectangle neighborhoods. At a point on a joined edge with transverse coordinate strictly between \(-1\) and \(1\), the two half-neighborhoods on opposite sides of the seam fit together. To see the reversal in coordinates, use \(u=x\), \(v=t\) on the left side of the seam, and \(u=x-1\), \(v=-t\) on the right side. The paired points receive the same coordinates, and the two sides together provide a neighborhood with \(u\) both positive and negative. Such a point is an interior point of the band, not a boundary point.

At a horizontal edge point the local rectangle neighborhood is cut in half by that edge. At a corner on a joined vertical edge, the corresponding corner on the opposite side supplies the other half of the seam neighborhood, but the horizontal edge remains: in the coordinates just given, the local region lies on one side of \(v=1\) or \(v=-1\). These are boundary neighborhoods. This local description agrees with the global boundary calculation \(q(A)\cong S^1\).

Common pitfall. Replacing the rule \((0,t)\sim(1,-t)\) with \((0,t)\sim(1,t)\) changes the construction. For example, the actual class of \((0,\tfrac{1}{2})\) is \(\{(0,\tfrac{1}{2}),(1,-\tfrac{1}{2})\}\), whereas the unreversed rule would pair it with \((1,\tfrac{1}{2})\). The latter is a different edge identification and gives a cylinder, not the Möbius band. Another error is to say that the horizontal edges are paired point by point: only their endpoints are paired, with the top-left corner joined to the bottom-right corner and the bottom-left to the top-right.

The quotient description is useful because it replaces an informal instruction to “glue with a twist” by precise data: a space, an equivalence relation, and a quotient topology. The classes specify exactly which points represent the same point of the band, while the saturation test identifies which subsets of the rectangle genuinely come from subsets of the quotient.