Collapsing a Disk Boundary to Form a Sphere

The Quotient and Its Equivalence Classes

A quotient space records which points of a space are to be regarded as the same. For the closed disk, identifying every point of its boundary while leaving its interior points distinct produces a space with one special point and an open-disk-like remainder. We will construct a map from this quotient to the unit sphere and verify that it is a homeomorphism.

Let \(D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}\), with the subspace topology from \(\mathbb{R}^2\), and let \(\partial D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}\). Define an equivalence relation on \(D^2\) by declaring \(p\sim p'\) if either \(p=p'\), or both \(p\) and \(p'\) belong to \(\partial D^2\). Write \(Q=D^2/{\sim}\) for the set of equivalence classes, with the quotient topology, and \(q:D^2\to Q\) for the quotient map. Thus \(U\subseteq Q\) is open exactly when \(q^{-1}(U)\) is open in \(D^2\).

The sphere in the construction is \(S^2=\{(u,v,w)\in\mathbb{R}^3:u^2+v^2+w^2=1\}\), with the subspace topology from \(\mathbb{R}^3\). We will send the boundary class to the south pole \((0,0,-1)\). An interior point at distance \(r\) from the origin will map to a point at polar angle \(\pi r\) from the north pole.

An Explicit Map from the Disk

For \(0<r\leq 1\), put \(h(r)=\sin(\pi r)/r\), and set \(h(0)=\pi\). For \((x,y)\in D^2\), let \(r=\sqrt{x^2+y^2}\), and define

At the origin this definition gives \(F(0,0)=(0,0,1)\). If \(r>0\), the squared length of the image is \((h(r)x)^2+(h(r)y)^2+\cos^2(\pi r) =h(r)^2r^2+\cos^2(\pi r) =\sin^2(\pi r)+\cos^2(\pi r)=1\). At \(r=0\), the image also has length one. Thus \(F\) takes values in \(S^2\).

The function \(h\) is continuous for \(r>0\), and \(\lim_{r\to0}\sin(\pi r)/r=\pi=h(0)\), using the standard calculus limit \(\lim_{t\to0}\sin(t)/t=1\). Hence \(h\) is continuous on \([0,1]\), so the displayed coordinate functions show that \(F\) is continuous on \(D^2\). If \(r=1\), then \(h(1)=\sin(\pi)=0\) and \(\cos(\pi)=-1\). Consequently \(F(x,y)=(0,0,-1)\) at every boundary point.

The Induced Map Is a Bijection

Since \(F\) is constant on each equivalence class, it defines a function \(\overline{F}:Q\to S^2\) by \(\overline{F}([p])=F(p)\). This is well-defined: if \([p]=[p']\), then either \(p=p'\), or both are boundary points, and in the latter case \(F(p)=F(p')=(0,0,-1)\).

Proof. First, \(\overline{F}\) is continuous. For an open set \(V\subseteq S^2\), the defining identity \(\overline{F}\circ q=F\) gives \(q^{-1}(\overline{F}^{-1}(V))=F^{-1}(V)\). The right-hand side is open because \(F\) is continuous. By the definition of the quotient topology, \(\overline{F}^{-1}(V)\) is open in \(Q\).

We next prove bijectivity. Let \(p=(x,y)\) have radius \(r>0\). The distance of \(F(p)\) from the north pole is encoded by its third coordinate \(w=\cos(\pi r)\). As \(r\) ranges from \(0\) to \(1\), \(\pi r\) ranges from \(0\) to \(\pi\), where cosine is strictly decreasing. Thus different radii in \([0,1]\) give different third coordinates. For a fixed radius \(0<r<1\), the first two coordinates of \(F(p)\) are \(\sin(\pi r)(x/r,y/r)\). Since \(\sin(\pi r)>0\), these coordinates determine the direction \((x/r,y/r)\). Therefore \(F\) is one-to-one on the interior, including the origin: only \(r=0\) maps to the north pole. No interior point maps to the south pole, because \(\cos(\pi r)=-1\) requires \(r=1\). The boundary maps to that pole, so \(\overline{F}\) is one-to-one on quotient classes.

For surjectivity, take any \((u,v,w)\in S^2\). There is a unique \(\phi\in[0,\pi]\) with \(w=\cos\phi\). If \(\phi=\pi\), then \(u^2+v^2=0\), so the point is the south pole, which is the image of the boundary class. If \(\phi=0\), the point is the north pole, which is \(F(0,0)\). If \(0<\phi<\pi\), define \(r=\phi/\pi\), so \(0<r<1\). The sphere equation gives \(u^2+v^2=1-w^2=\sin^2\phi\), hence \(\sqrt{u^2+v^2}=\sin\phi\). Set \((x,y)=\frac{r}{\sin\phi}(u,v)\). Then \(x^2+y^2=r^2<1\), and \(h(r)x=\frac{\sin\phi}{r}\frac{r u}{\sin\phi}=u\), \(h(r)y=v\), and \(\cos(\pi r)=\cos\phi=w\). Thus \(F(x,y)=(u,v,w)\), proving surjectivity.

It remains to prove that this continuous bijection is a homeomorphism. By the Heine–Borel theorem, \(D^2\) is compact. The map \(q\) is continuous and surjective, so \(Q\) is compact: given an open cover of \(Q\), its inverse images under \(q\) form an open cover of \(D^2\); a finite subcover of \(D^2\) therefore gives a finite subcover of \(Q\). The space \(S^2\) is Hausdorff because it has the subspace topology from Euclidean \(\mathbb{R}^3\): for distinct points at distance \(\delta>0\), the intersections with \(S^2\) of the open balls of radius \(\delta/3\) about them are disjoint open neighborhoods.

We use the compact-to-Hausdorff theorem and give its argument here. If \(C\) is closed in a compact space \(Q\), then \(C\) is compact: an open cover of \(C\), together with \(Q\setminus C\), covers \(Q\), so a finite subcover of \(Q\) restricts to a finite subcover of \(C\). A continuous image of a compact space is compact, since any open cover of the image pulls back to an open cover of the domain. Finally, a compact subset \(K\) of a Hausdorff space \(Y\) is closed: for \(y\notin K\), separate \(y\) from each \(x\in K\) by disjoint open neighborhoods; compactness gives finitely many neighborhoods covering \(K\), and the intersection of the corresponding finitely many neighborhoods of \(y\) is an open neighborhood disjoint from \(K\). Therefore a continuous map from compact \(Q\) to Hausdorff \(S^2\) sends closed sets to closed sets. Applying this to the continuous bijection \(\overline{F}\), its inverse is continuous because the preimage under \(\overline{F}^{-1}\) of a closed set \(C\subseteq Q\) is \(\overline{F}(C)\), which is closed. Hence \(\overline{F}\) is a homeomorphism. \(\square\)

A Common Mistake and Why the Construction Matters

Common pitfall. The formula \(G(x,y)=(x,y,\sqrt{1-x^2-y^2})\) does take values on the sphere, but it does not define a map on this quotient: it is not constant on the boundary class. For example, \((1,0)\) and \((0,1)\) are equivalent boundary points, yet \(G(1,0)=(1,0,0)\) and \(G(0,1)=(0,1,0)\), which are different. In fact, \(G\) maps only to the upper hemisphere. The correction is to use a formula whose boundary values agree; the map \(F\) above sends every boundary point to the south pole and covers the entire sphere.

The quotient topology is essential to the argument. A continuous map on the disk that is constant on equivalence classes gives a continuous map on the quotient precisely because a subset of \(Q\) is open when its inverse image under \(q\) is open. The explicit formula identifies which sphere point each class represents, while compactness and the Hausdorff property ensure that the resulting continuous bijection has a continuous inverse.