Cone Maps and Contractibility

Maps from a Cone and the Quotient Test

A cone adds a single apex to a space by collapsing one end of its product with an interval. The quotient construction makes the apex easy to describe, but it also raises a practical question: how can we define a map from the cone and know that it is continuous? The quotient topology gives an exact test. It also lets us build a homotopy that moves every point of the cone to its apex.

Let \((X,\mathcal{T})\) be a topological space, and give \(I=[0,1]\) its standard subspace topology from \(\mathbb{R}\). Give \(X\times I\) the product topology. Define the cone \(CX\) to be the quotient of \(X\times I\) that identifies all points of \(X\times\{1\}\) with one another and makes no other identifications. Write \(q:X\times I\to CX\) for the quotient map, and write \([x,t]=q(x,t)\). The common equivalence class of the points \((x,1)\) is the apex, denoted \(v\). In particular, \([x,1]=v\) for every \(x\in X\).

Proof. Suppose first that \(f\) is continuous. The quotient map \(q\) is continuous by the definition of the quotient topology, so the composite \(f\circ q\) is continuous. Conversely, suppose \(f\circ q\) is continuous. For any open set \(V\subseteq Y\), we have \(q^{-1}(f^{-1}(V))=(f\circ q)^{-1}(V)\), which is open in \(X\times I\). The defining property of the quotient topology now implies that \(f^{-1}(V)\) is open in \(CX\). Thus \(f\) is continuous. For the equivalent formulation, a function \(g\) defines a function on equivalence classes precisely when it takes equal values on equivalent points. The only distinct points identified by \(q\) are those at height \(1\), so this condition is exactly that \(g(x,1)\) be independent of \(x\). The first part of the theorem then gives the continuity assertion in both directions. \(\square\)

This criterion is the cone’s universal property in concrete form: to define a continuous map out of \(CX\), it is enough to give a continuous map on \(X\times I\) whose values agree wherever the quotient identifies points. No separation assumption on \(Y\), or compactness assumption on \(X\), is required.

Using the Criterion

An Explicit Contraction

A space \(Z\) is contractible if there are a point \(z_0\in Z\) and a continuous map \(H:Z\times I\to Z\) such that \(H(z,0)=z\) and \(H(z,1)=z_0\) for every \(z\in Z\). Such a map is a homotopy from the identity map to the constant map with value \(z_0\). We use the convention that a contractible space must contain its contraction point.

For the continuity check, we use the standard product theorem for quotient maps: if \(q:E\to Z\) is a quotient map and \(K\) is locally compact Hausdorff, then \(q\times\operatorname{id}_K:E\times K\to Z\times K\) is a quotient map. The interval \(I\) is compact Hausdorff and hence locally compact Hausdorff, so \(q\times\operatorname{id}_I\) is a quotient map. This product theorem is a standard result about quotient and product topologies.

Proof. Define \(G:X\times I\times I\to CX\) by \(G(x,t,s)=[x,t+s(1-t)]\). The expression \(t+s(1-t)\) lies in \(I\): it equals \((1-s)t+s\), a sum of nonnegative terms, and \(1-((1-s)t+s)=(1-s)(1-t)\geq0\). The map \(G\) is continuous, as it is the composite of the continuous height function \((x,t,s)\mapsto (1-s)t+s\) with the quotient map \(q\).

For any fixed \(s\), if \((x,t)\) and \((y,u)\) represent the same point of \(CX\), then either they are the same point of \(X\times I\), or \(t=u=1\). In the latter case, both transformed heights are \((1-s)1+s=1\), so both images under \(G\) are the apex. Therefore \(G\) is constant on the fibers of \(q\times\operatorname{id}_I\). The quotient property gives a unique continuous map \(H:CX\times I\to CX\) satisfying \(H\circ(q\times\operatorname{id}_I)=G\). Its formula is the one in the theorem.

At \(s=0\), \(t+0(1-t)=t\), so \(H([x,t],0)=[x,t]\). At \(s=1\), \(t+1(1-t)=1\), so \(H([x,t],1)=[x,1]=v\), independently of \(x\) and \(t\). Hence \(H\) is a homotopy from \(\operatorname{id}_{CX}\) to the constant map with value \(v\). Since \(X\) is nonempty, \(CX\) has this apex and is nonempty, so this is a contraction. \(\square\)

Why the Hypotheses Matter

The construction explains contractibility without requiring any special property of \(X\): every point moves along its cone segment toward the common apex. The map criterion is useful whenever a formula is easiest to write before passing to a quotient. First check continuity on the product, then check agreement on identified points; both checks are necessary.

The nonempty hypothesis in the theorem ensures that the cone has an apex. If \(X=\varnothing\), then \(X\times I=\varnothing\), and its quotient \(CX\) is empty. Under the stated convention, the empty space is not contractible because it has no point to serve as the contraction value. For a nonempty \(X\), the cone’s apex always supplies that value.

Common pitfall. A continuous formula on \(X\times I\) need not define any map on \(CX\). In the two-point example, the two constant values \(0\) and \(1\) are continuous on their respective pieces, but they assign different values to representatives of the same apex. The correction is to test the equivalence relation explicitly: every pair identified by the quotient must receive equal values. Once that condition holds, the quotient criterion—not an assumption that the resulting map is continuous—supplies the conclusion.