Collapsing One End of a Product
A cone can be built from a space \(X\) by taking a product with an interval and identifying all the points at one end of the product. The points at the other end are left distinct. This construction gives a precise meaning to the idea of moving every point of \(X\) toward one common apex, without requiring a drawing or a particular geometric realization.
Let \((X,\mathcal{T})\) be a topological space, and give \(I=[0,1]\) its usual subspace topology from \(\mathbb{R}\). Give \(X\times I\) the product topology. Define an equivalence relation on \(X\times I\) by identifying all points with interval coordinate \(1\), and making no other identifications.
For \(t<1\), the equivalence class of \((x,t)\) contains only \((x,t)\). There is one class at height \(1\), containing all of \(X\times\{1\}\), provided \(X\) is nonempty. Write \(v\) for this common class, the apex. Thus every point of the cone is either \(q(x,t)\) for a unique \(x\in X\) and \(t<1\), or is \(v\). If \(X=\varnothing\), then \(X\times I=\varnothing\), so the cone is empty and has no apex.
What the Quotient Topology Requires
The quotient topology is part of the construction, not an extra choice. A subset \(W\subseteq C(X)\) is open exactly when its full preimage under \(q\) is open in the product. In particular, the preimage must respect the equivalence classes: if it contains one point of \(X\times\{1\}\), it contains all of \(X\times\{1\}\). Such a preimage is called saturated. For a subset \(O\subseteq X\times I\), being saturated means that whenever \(O\) contains one point of an equivalence class, it contains that entire class.
Proof. If \(g\) is continuous, then \(g\circ q\) is continuous because \(q\) and \(g\) are continuous. Conversely, suppose \(g\circ q\) is continuous. For any open set \(V\subseteq Y\), the set \(q^{-1}(g^{-1}(V))=(g\circ q)^{-1}(V)\) is open in \(Z\). By the definition of the quotient topology, a subset of \(Z/{\sim}\) is open whenever its preimage under \(q\) is open. Therefore \(g^{-1}(V)\) is open, which proves that \(g\) is continuous. \(\square\)
This test is useful whenever a function is first specified on \(X\times I\) and is intended to define a function on the cone. It must take the same value at points identified by \(q\); when it does, the proposition tests continuity by checking the function before passage to the quotient.
Proof. The formula for \(h\) is well-defined: the only class containing more than one point consists of points whose interval coordinate is \(1\), so all representatives of that class give the same value. Let \(p_2:X\times I\to I\) be the projection \(p_2(x,s)=s\). Then \(h\circ q=p_2\), which is continuous in the product topology. The quotient test proves that \(h\) is continuous.
Now fix \(t<1\). The map \(i_t\) is continuous because it is the composition of the continuous map \(x\mapsto(x,t)\) with \(q\). It is injective since no two distinct points at height \(t<1\) are identified, and its image is \(q(X\times\{t\})\). To prove that its inverse onto this image is continuous, take any open \(U\subseteq X\). There is an open set \(J\subseteq I\) containing \(t\) and not containing \(1\), since \(t<1\). The set \(O=U\times J\) is open in \(X\times I\). It is saturated because it contains no point at height \(1\), where the only nonsingleton class occurs. Thus \(q(O)\) is open in \(C(X)\), since \(q^{-1}(q(O))=O\). Its intersection with the level \(q(X\times\{t\})\) is exactly \(i_t(U)\). Therefore \(i_t(U)\) is open in that level with its subspace topology. This proves continuity of the inverse and hence the claimed homeomorphism. \(\square\)
The theorem says that every level below the apex retains the topology of \(X\), while the entire top level has been replaced by one point. It does not assert that the cone is a product: the identification at height \(1\) changes the topology there.
Worked Examples
Worked Example: The Cone on the Empty Space
Let \(X=\varnothing\). Then \(X\times I=\varnothing\times I=\varnothing\). The quotient of the empty set by its (empty) equivalence relation is again empty, so \(C(\varnothing)=\varnothing\). There are no equivalence classes represented by points, and hence no apex. This is why the apex is only defined when \(X\) is nonempty.
Check: The quotient map has domain and codomain both empty. The quotient-topology definition gives just one open subset, \(\varnothing\), as required for the empty space.
Worked Example: The Cone on a One-Point Space
Let \(X=\{a\}\) with its unique topology \(\{\varnothing,\{a\}\}\). The product \(X\times I\) has one point \((a,t)\) for each \(t\in I\). Collapsing \(X\times\{1\}=\{(a,1)\}\) identifies no distinct points, so \(q\) is a bijection from \(X\times I\) to \(C(X)\).
Define \(s:I\to C(X)\) by \(s(t)=q(a,t)\). This is continuous as the composition of \(t\mapsto(a,t)\) with \(q\). It is bijective. The height map \(h:C(X)\to I\) from the theorem is continuous and satisfies \(h(s(t))=t\) for every \(t\in I\). Thus \(h\) is the continuous inverse of \(s\), and \(C(X)\) is homeomorphic to \(I\).
Check: At \(t=1\), \(s(1)\) is the apex. There is only one point at that height, so this endpoint is not duplicated. At every \(t<1\), the point \(s(t)\) is the class of the single product point \((a,t)\).
Worked Example: The Cone on a Two-Point Discrete Space
Let \(X=\{a,b\}\) have the discrete topology \(\{\varnothing,\{a\},\{b\},X\}\). The product \(X\times I\) consists of two copies of \(I\), one for each first coordinate. The quotient keeps the points \(q(a,t)\) and \(q(b,t)\) distinct for every \(t<1\), but identifies \(q(a,1)\) with \(q(b,1)\) as the apex \(v\).
More explicitly, for \(W\subseteq C(X)\), put \(A_a=\{t\in I:q(a,t)\in W\}\) and \(A_b=\{t\in I:q(b,t)\in W\}\). Then \(W\) is open if and only if \(A_a\) and \(A_b\) are open in \(I\). Indeed, the product topology on the two copies of \(I\) makes \(q^{-1}(W)\) open exactly when both coordinate sets are open. Because the apex is one point, \(v\in W\) exactly when \(1\in A_a\) and \(1\in A_b\); this is already required for \(A_a,A_b\) to be the preimages of a subset \(W\).
For example, take \(W=\{v\}\cup\{q(a,t):1/2<t<1\}\). Then \(A_a=(1/2,1]\) and \(A_b=\{1\}\). The first set is open in \(I\), but the second is not: every relative open neighborhood of \(1\) in \(I\) contains points \(t<1\). Therefore \(W\) is not open in the cone.
Check: If instead \(W\) contains both branches for \(1/2<t\leq1\), then \(A_a=A_b=(1/2,1]\), which is open in \(I\). The quotient test therefore makes this \(W\) open. The difference is precisely that a neighborhood of the apex must approach it along both branches.
A Common Mistake: Keeping Several Apexes
A tempting but incorrect construction is to treat the two points \((a,1)\) and \((b,1)\) in the last example as distinct endpoints. That would describe the product \(X\times I\), not the cone: the equivalence relation explicitly puts both points in the same class.
The error also leads to a wrong openness test. In the two-point example, the set \(W=\{v\}\cup\{q(a,t):1/2<t<1\}\) might seem open if one checks only the part of the \(a\)-branch near its endpoint. But its full preimage contains \((b,1)\) and no \((b,t)\) with \(t<1\). The calculation above gives \(A_b=\{1\}\), which is not open in \(I\). Consequently \(q^{-1}(W)\) is not open in \(X\times I\), and \(W\) is not open in the quotient.
The defining principle is to test openness before identifying points: take the entire preimage under \(q\), check that it is open in the product, and remember that all points in each equivalence class must be included together. This principle applies regardless of whether \(X\) is finite or familiar from geometric examples.
Check Your Understanding
Use the quotient definition and the descriptions of the equivalence classes to answer the following questions.
- For nonempty \(X\), which points of \(X\times I\) have singleton equivalence classes, and which points form the apex class?
- Why does the cone on the empty space have no apex?
- In the cone on a two-point discrete space, decide whether the set consisting of the apex and the points \(q(a,t)\) with \(1/2<t<1\) is open. Give its full preimage under \(q\).
- State the quotient test for continuity of a map defined on \(C(X)\), and explain why it is enough to check its composition with \(q\).
- For \(t<1\), what does the level-slice theorem say about \(q(X\times\{t\})\)? Does it make the same claim about the level at \(t=1\)?