The Sphere from a Polygonal Disk

Collapsing the Boundary of a Disk

A disk is not a sphere: it has a boundary circle, while the sphere has no boundary. But if every point of the disk’s boundary is identified with every other boundary point, the resulting quotient space is a sphere. The point of the construction is that the identification changes the topology, not just the drawing. We will build an explicit homeomorphism and then compare the resulting one-face cell structure with a model made from two polygonal faces.

Let \(D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}\), with its subspace topology from the standard topology on \(\mathbb{R}^2\), and let \(S^2=\{(u,v,w)\in\mathbb{R}^3:u^2+v^2+w^2=1\}\), with its subspace topology from \(\mathbb{R}^3\). Write \(\partial D^2=\{(x,y):x^2+y^2=1\}\) for the boundary circle.

Denote the class of the boundary by \(*\). The quotient topology matters: it is the topology specified by which subsets have open preimages under \(q\). The aim is to identify this topological space, not merely to match its points with points of a sphere.

An Explicit Homeomorphism with the Sphere

For \(r=\sqrt{x^2+y^2}>0\), define \(f(x,y)=\bigl(\frac{\sin(\pi r)}{r}x,\frac{\sin(\pi r)}{r}y,\cos(\pi r)\bigr)\). At \((0,0)\), set \(f(0,0)=(0,0,1)\). This formula moves from the north pole at radius \(0\) to the south pole at radius \(1\), while keeping track of the direction from the disk’s centre.

Proof. First, \(f\) is continuous. Away from the origin this follows from its formula. At the origin, \(\sin(\pi r)/r\to\pi\) as \(r\to0\), so the first two coordinates tend to \(0\); also \(\cos(\pi r)\to1\). Thus the definition at the origin makes \(f\) continuous there. For \(r>0\), the sum of the squares of its coordinates is \((\sin^2(\pi r)/r^2)(x^2+y^2)+\cos^2(\pi r)=\sin^2(\pi r)+\cos^2(\pi r)=1\). At the origin its value also lies in \(S^2\).

If \(r=1\), then \(\sin(\pi r)=0\) and \(\cos(\pi r)=-1\), so every boundary point maps to \((0,0,-1)\). The map is therefore constant on each equivalence class and induces a map \(\widetilde{f}:D^2/\partial D^2\to S^2\) with \(\widetilde{f}\circ q=f\). This induced map is continuous: for any open \(V\subseteq S^2\), the preimage under \(q\) of \(\widetilde{f}^{-1}(V)\) is \(f^{-1}(V)\), which is open; the definition of the quotient topology then makes \(\widetilde{f}^{-1}(V)\) open.

To check bijectivity, the third coordinate of \(f(x,y)\) is \(\cos(\pi r)\), which is strictly decreasing as \(r\) runs from \(0\) to \(1\). Consequently it determines \(r\) uniquely. For \(0<r<1\), the first two coordinates then determine the direction \((x/r,y/r)\), since \(\sin(\pi r)>0\). The north pole comes only from the origin, and the south pole comes exactly from the boundary class. So \(\widetilde{f}\) is both injective and surjective.

Finally, \(D^2\) is compact by the Heine–Borel theorem. Its quotient is compact: an open cover of the quotient pulls back to an open cover of \(D^2\), and a finite subcover of \(D^2\) gives a finite subcover of the quotient. The sphere is Hausdorff because it is a subspace of Euclidean space: two distinct points have positive distance and can be surrounded by disjoint open balls of sufficiently small radius. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. To recall why, every closed subset of the compact domain is compact; its image is compact, hence closed in a Hausdorff space. The bijection is therefore a closed map, so its inverse is continuous. Applying this result to \(\widetilde{f}\) proves the claim. \(\square\)

Reading the One-Face Cell Structure

A \(k\)-cell is a piece of a cell decomposition homeomorphic to an open \(k\)-dimensional disk. In the quotient \(D^2/\partial D^2\), the boundary gives one \(0\)-cell, namely \(*\). There are no \(1\)-cells: every boundary point, including the interior of every boundary edge in a polygonal presentation, belongs to that same single point. The interior of the disk is one open \(2\)-cell, since \(q\) leaves all interior points distinct.

The Sphere as Two Polygonal Faces

There is another useful presentation: take two copies of a disk and glue their boundary circles point by point. One copy becomes the upper hemisphere and the other the lower hemisphere. More generally, each disk can be represented by a polygon, provided corresponding boundary points are glued in the same cyclic order.

Proof. On the first disk define \(g_+(x,y)=(x,y,\sqrt{1-x^2-y^2})\), and on the second define \(g_-(x,y)=(x,y,-\sqrt{1-x^2-y^2})\). These are continuous maps into \(S^2\). On the boundary, where \(x^2+y^2=1\), both maps give \((x,y,0)\). They therefore agree on the points identified by the gluing and induce a continuous map from the quotient to \(S^2\). Every point of \(S^2\) with nonnegative third coordinate is in the image of \(g_+\), and every point with nonpositive third coordinate is in the image of \(g_-\), so the induced map is surjective. The two hemispheres meet exactly along the equator, and the only identifications between the disk copies are the matching boundary points. This makes the induced map injective as well.

The disjoint union of the two disks is compact: restrict any open cover to each disk, take a finite subcover of each compact disk, and combine the two finite collections. Its quotient is compact by the open-cover argument used above. Since \(S^2\) is Hausdorff, the continuous bijection from this compact quotient to \(S^2\) is a homeomorphism by the compact-to-Hausdorff result proved above. \(\square\)

Comparing the Models and Avoiding a Counting Error

The one-face model and the two-face model differ in their cell counts, but both give Euler characteristic \(2\). In the first, the whole boundary is one vertex and the disk interior is one face. In the second, the common boundary is a circle divided into vertices and edges, and the two disk interiors are faces.

Common pitfall. In the collapsed-square model, it is incorrect to count the square’s four boundary sides as four \(1\)-cells after the identification. Every point on every side is part of the same boundary class, so none of the side interiors survives as a separate open cell. The incorrect count \(V=1,E=4,F=1\) would give \(1-4+1=-2\); the error is counting pieces that have all been collapsed to a point. The correct count is \(V=1,E=0,F=1\), giving \(2\). By contrast, in the two-square model the boundary is not collapsed to one point: it remains a circle, so its four edges really are \(1\)-cells.

These models are useful for different purposes. Collapsing the boundary gives a direct quotient description with a particularly small cell structure. Gluing two faces retains a visible common boundary and makes the two hemispheres explicit. In either case, the equivalence relation and the quotient topology determine the space; a cell count is a consistency check, not by itself a proof of homeomorphism.