Let \(X=D^2\), \(Y=\{y\}\) a singleton, and \(f:X\to Y\) be the constant function. Construct and describe the mapping cylinder.
Solution
For a map \(f\colon X\to Y\), the mapping cylinder is \[ M_f=\bigl((X\times[0,1])\sqcup Y\bigr)\big/\bigl((x,1)\sim f(x)\text{ for all }x\in X\bigr). \]
1. For \(f\colon D^2\to\{y\}\), \(M_f\) is the (unreduced) cone \(CD^2=(D^2\times[0,1])/(D^2\times\{1\})\), which is also the join \(D^2\ast\{y\}\).
Since \(Y=\{y\}\), the relation identifies all of \(D^2\times\{1\}\) with the single point \(y\) and nothing else, which is the defining quotient of the cone. The base \(D^2\times\{0\}\) is a copy of \(D^2\) in \(M_f\).
2. \(M_f\) is homeomorphic to the closed \(3\)-ball.
Regard \(D^2\subset\mathbb R^2\times\{0\}\subset\mathbb R^3\) and let \(a=(0,0,1)\). The map \(D^2\times[0,1]\to\mathbb R^3\), \((x,t)\mapsto(1-t)x+ta\), is continuous, collapses exactly \(D^2\times\{1\}\) to \(a\), and is otherwise injective, so it induces a continuous bijection from the compact space \(M_f\) onto the solid cone \(\{(1-t)x+ta\}\), which is Hausdorff; hence a homeomorphism. The solid cone is a compact convex subset of \(\mathbb R^3\) with nonempty interior, hence homeomorphic to \(D^3\).