Let \(q:A\to B\) be continuous and \(i:I\to I\) be the identity function. Define the function \(q\times i:A\times I\to B\times I\) by \((q\times i)(a,t):=(q(a),i(t))\). Show that \(q\times i\) is continuous. You may assume the fact that if \(q:A\to B\) is a quotient map, then so is \(q\times i\).
Let \(X\) be a path-connected Hausdorff space. The cone \(CX\) of \(X\) is the quotient of the space \(X\times I\) by the smallest equivalence relation such that \((p,0)\sim(q,0)\) for all \(p,q\in X\). Prove that there is a deformation retraction from \(CX\) to the singleton subspace \(\{[(p,0)]\}\).
Solution
1. The map\[ q\times i:A\times I\longrightarrow B\times I, \qquad (a,t)\longmapsto(q(a),t), \]is continuous.
The sets \(U\times V\), with \(U\subseteq B\) open and \(V\subseteq I\) open, form a basis for the product topology on \(B\times I\). For such a basic open set, \[ (q\times i)^{-1}(U\times V) = q^{-1}(U)\times V. \] Since \(q\) is continuous, \(q^{-1}(U)\) is open in \(A\), so the right-hand side is open in \(A\times I\). Thus the inverse image of every basis element is open, hence \(q\times i\) is continuous. This proves part (a).
Let \[ Q:X\times I\longrightarrow CX \] be the quotient map, and denote the cone vertex by \[ v=Q(x,0), \] which is independent of \(x\in X\).
2. Define\[ F:(X\times I)\times I\longrightarrow CX, \qquad F((x,t),s)=Q(x,(1-s)t). \]Then \(F\) is continuous.
The map \[ \Phi:(X\times I)\times I\longrightarrow X\times I, \qquad \Phi((x,t),s)=(x,(1-s)t) \] is continuous: its first coordinate is the projection to \(X\), and its second coordinate is the product of the continuous real-valued functions \((t,s)\mapsto t\) and \((t,s)\mapsto1-s\). Since \(Q\) is continuous by definition of the quotient topology, \[ F=Q\circ\Phi \] is continuous.
3. The map \(F\) is constant on each fiber of\[ Q\times i:(X\times I)\times I\longrightarrow CX\times I. \]
Suppose \[ (Q(x,t),s)=(Q(y,u),r). \] Then \(s=r\) and \(Q(x,t)=Q(y,u)\). By the defining equivalence relation of the cone, either \[ (x,t)=(y,u) \] or \[ t=u=0. \] In the first case the two values of \(F\) are visibly equal. In the second case, \[ F((x,0),s)=Q(x,0)=v=Q(y,0)=F((y,0),s). \] Thus \(F\) is constant on every fiber.
4. There is a unique continuous map\[ H:CX\times I\longrightarrow CX \]satisfying\[ H(Q(x,t),s)=Q(x,(1-s)t). \]
The map \(Q\) is a quotient map. By the fact permitted in the problem, the product \[ Q\times i:(X\times I)\times I\to CX\times I \] is also a quotient map. By <1>2–<1>3, the continuous map \(F\) is constant on its fibers. Therefore the universal property of quotient maps gives a unique map \(H\) with \[ F=H\circ(Q\times i), \] and that induced map \(H\) is continuous. The displayed formula is exactly this factorization.
5. The homotopy \(H\) satisfies\[ H(z,0)=z, \qquad H(z,1)=v \]for every \(z\in CX\).
Write \(z=Q(x,t)\). Using 4, \[ H(z,0)=Q(x,(1-0)t)=Q(x,t)=z, \] and \[ H(z,1)=Q(x,(1-1)t)=Q(x,0)=v. \]
6. The vertex \(v\) is fixed throughout the homotopy:\[ H(v,s)=v \qquad(0\le s\le1). \]
For any \(x\in X\), \[ v=Q(x,0), \] so 4 gives \[ H(v,s)=Q(x,(1-s)0)=Q(x,0)=v. \]
7. Hence \(H\) is a strong deformation retraction of \(CX\) onto the singleton \(\{v\}\).
By 5, \(H(-,0)\) is the identity on \(CX\) and \(H(-,1)\) has image in \(\{v\}\) and restricts to the identity there. By 6, every point of the retracting subspace \(\{v\}\) is fixed at every intermediate time. These are exactly the defining conditions for a strong deformation retraction. This proves part (b).