Computing Cones of Familiar Spaces

From a Collapsed Product to a Familiar Shape

A cone has a simple geometric description: take a space, place a copy of it at the bottom, and identify all points of a second copy to a single apex. To identify the resulting topology with a familiar disk or ball, however, it is not enough to draw the shape. We will construct maps on the quotient, show that they are homeomorphisms, and check the coordinates explicitly.

Write \(I=[0,1]\) with its usual subspace topology from \(\mathbb{R}\). Give products the product topology. For \(n\geq 0\), give \(S^n=\{u\in\mathbb{R}^{n+1}:\|u\|=1\}\) and \(D^{n+1}=\{x\in\mathbb{R}^{n+1}:\|x\|\leq 1\}\) their subspace topologies from Euclidean space. In particular, \(S^1\) is the unit circle and \(D^2\) is the closed filled disk.

Recall that the quotient topology is defined by declaring \(V\subseteq CX\) open exactly when \(q^{-1}(V)\) is open in \(X\times I\). Thus, if a continuous map on \(X\times I\) is constant on each equivalence class, it induces a continuous map on \(CX\): the preimage of an open set under the induced map is open precisely when its preimage under the original map is open. We use this quotient-topology test in each construction.

The Cone on an Interval

Represent \(I\) by a coordinate \(x\in[0,1]\), and represent the cone parameter by \(t\in[0,1]\), with \(t=1\) at the apex. The formula \((x,t)\mapsto ((2x-1)(1-t),t)\) sends the product onto the triangle

The triangle has vertices \((-1,0)\), \((1,0)\), and \((0,1)\). Notice that the entire top edge \(I\times\{1\}\) maps to its apex \((0,1)\), as required by the cone identification.

Why This Triangle Is a Disk

We give an explicit radial homeomorphism, so the identification does not depend on a picture. Let \(c=(0,1/3)\), an interior point of \(T\), and write the three inequalities defining \(T\) as \(a_j\cdot p\leq b_j\), where

The slacks \(s_j=b_j-a_j\cdot c\), the distances available in each inequality at the center before moving in a direction, are \(s_1=1/3\), \(s_2=2/3\), and \(s_3=2/3\). For a unit vector \(v\in S^1\), set

At least one \(a_j\cdot v\) is positive. Indeed, if all three were nonpositive, the first would give \(v_y\geq0\), while the second and third, when added, would give \(2v_y\leq0\). Thus \(v_y=0\); the second and third then imply \(v_x\leq0\) and \(-v_x\leq0\), so \(v=(0,0)\), contradicting \(\|v\|=1\). Consequently \(M(v)>0\), and \(\rho\) is defined and continuous.

On the ray \(c+\lambda v\), for \(\lambda\geq0\), the \(j\)-th triangle inequality becomes \(\lambda(a_j\cdot v)\leq s_j\). A nonpositive dot product imposes no upper bound on \(\lambda\); each positive one imposes \(\lambda\leq s_j/(a_j\cdot v)\). The smallest of these upper bounds is \(1/M(v)=\rho(v)\). Therefore the part of this ray in \(T\) is exactly \(c+\lambda v\) for \(0\leq\lambda\leq\rho(v)\).

Proof. Define \(H:D^2\to T\) by \(H(0)=c\), and for \(z=rv\), where \(0<r\leq1\) and \(v\in S^1\), define \(H(rv)=c+r\rho(v)v\). The ray calculation shows that \(H\) maps each radius of \(D^2\) bijectively onto the corresponding segment in \(T\). Distinct rays from \(c\) meet only at \(c\), so \(H\) is bijective. Away from \(0\), its formula is continuous. Also \(T\) is contained in \(\{|x|\leq1,\ 0\leq y\leq1\}\), so \(\rho(v)\leq\sqrt{1+(2/3)^2}=\sqrt{13}/3<2\); this bound shows \(H(rv)\to c\) as \(r\to0\). For the inverse, \(M(v)\leq3\): the first ratio is at most \(3\), and either slanted ratio is at most \(3\sqrt{2}/2<3\). Thus \(\rho(v)\geq1/3\). For \(p\ne c\), put \(v=(p-c)/\|p-c\|\); the inverse is \((\|p-c\|/\rho(v))v\). It is continuous away from \(c\), and its norm is at most \(3\|p-c\|\), so it is continuous at \(c\) as well. Hence \(H\) and its inverse are continuous, proving the proposition. \(\square\)

The Cone on a Sphere Is a Ball

The sphere-to-ball map uses radial coordinates directly. We include the topological justification that a continuous bijection here is a homeomorphism, rather than relying only on the coordinate formula.

Proof. Define \(G:S^n\times I\to D^{n+1}\) by \(G(u,t)=(1-t)u\). This is continuous and constant on \(S^n\times\{1\}\), so it induces a continuous map \(g:CS^n\to D^{n+1}\). The map is onto: if \(x\ne0\), take \(u=x/\|x\|\in S^n\) and \(t=1-\|x\|\in[0,1)\), and then \(G(u,t)=x\); the origin is the image of every \((u,1)\). It is one-to-one on the quotient: a nonzero image \(x\) determines \(t=1-\|x\|\) and \(u=x/\|x\|\), while the only preimages of \(0\) have \(t=1\), which represent the single apex.

It remains to justify continuity of the inverse. The set \(S^n\times I\) is closed and bounded in \(\mathbb{R}^{n+2}\), hence compact by the Heine–Borel theorem. Its continuous image \(CS^n\) under the quotient map is compact: pulling an open cover back to \(S^n\times I\), taking a finite subcover there, and then using surjectivity gives a finite subcover of the quotient. The ball \(D^{n+1}\), as a subspace of Euclidean space, is Hausdorff.

We use the following compact-to-Hausdorff fact. A continuous bijection \(h:K\to Y\) from a compact space to a Hausdorff space is a homeomorphism. To verify it, a closed subset \(C\) of \(K\) is compact, and its continuous image \(h(C)\) is compact. A compact subset \(L\) of a Hausdorff space is closed: for a point \(y\notin L\), separate \(y\) from each \(x\in L\) by disjoint open neighborhoods; finitely many of the neighborhoods of points in \(L\) cover \(L\), and the intersection of the corresponding finitely many neighborhoods of \(y\) misses \(L\). Thus \(h(C)\) is closed. Since \(h\) is bijective, this says its inverse takes closed sets to closed sets, so the inverse is continuous. Applying this fact to \(g\) proves the theorem. \(\square\)

A Sign Error to Avoid

Common pitfall. A sign error in a half-plane inequality can turn a bounded triangle into an unbounded region, invalidating a radial argument based on that triangle. For example, the inequalities \(y\geq0\), \(x-y\leq1\), and \(-x-y\leq1\) do not describe the triangle used above. At the points \((0,s)\), for any \(s\geq0\), they read \(s\geq0\), \(-s\leq1\), and \(-s\leq1\), so the region contains an unbounded vertical ray. The corrected inequalities are \(y\geq0\), \(x+y\leq1\), and \(-x+y\leq1\). For instance, \((0,2)\) satisfies \(y\geq0\) but fails both slanted inequalities, since \(0+2\not\leq1\) and \(-0+2\not\leq1\). For these corrected inequalities and center \(c=(0,1/3)\), the slacks are \(1/3,2/3,2/3\); using the correct signs and slacks is essential to the radial calculation.

The constructions identify the spaces with their quotient topologies, not just as sets of points. Their common pattern is to let \(t\) determine the distance from the origin: \(t=0\) gives the boundary sphere and \(t=1\) gives the ball’s center. For the interval, the same collapse produces a triangle, and the radial homeomorphism identifies that triangle with the disk. These identifications are useful whenever a cone appears as one piece of a larger gluing construction.