Deformation Retractions of Mapping Cylinders

The Mapping Cylinder and Its Target

Let \(f:A\to Y\) be a continuous map. The mapping cylinder keeps a copy of \(Y\) and attaches the bottom of \(A\times I\) to it using \(f\). Its geometry suggests a simple motion: slide each point of the cylinder down to the attached end, while leaving the target alone. The main issue is to make that motion a continuous map on the quotient, including where the two pieces have been identified.

Give \(A\times I\) the product topology, give \(A\times I\bigsqcup Y\) the disjoint union topology, and let \(q:A\times I\bigsqcup Y\to M_f\) be the quotient map for the equivalence relation generated by \((a,0)\sim f(a)\). Thus \(M_f=(A\times I\bigsqcup Y)/{\sim}\). Write \(q(a,t)\) for \(q((a,t))\), and \(q(y)\) for the image of a point in the \(Y\) component. The mapping cylinder has the quotient topology.

The map \(q|_Y:Y\to q(Y)\) is a homeomorphism, so we can regard the target subspace as a copy of \(Y\). Here is a direct check. It is continuous and injective, since the equivalence relation identifies a point in the \(Y\) component only with points of the form \((a,0)\), and does not identify distinct points of \(Y\). If \(V\subseteq Y\) is open, then \(V\bigsqcup\bigl(f^{-1}(V)\times[0,1)\bigr)\) is an open saturated subset of \(A\times I\bigsqcup Y\): \(f^{-1}(V)\) is open, and \([0,1)\) is open in \(I\). Its image under \(q\) is open by the definition of quotient topology, and its intersection with \(q(Y)\) is \(q(V)\). Thus \(q(V)\) is open in \(q(Y)\), proving that the inverse of \(q|_Y\) is continuous.

The Candidate Homotopy

For a point represented by \((a,t)\), the natural motion replaces its height \(t\) by \((1-s)t\). A point in the target is to stay fixed. These prescriptions agree at the identified end: when \(t=0\), the cylinder representative and its target representative are equivalent.

The formulas specify the right motion, but continuity does not follow just from checking them separately on the two pieces. In general, a quotient map \(q\) need not remain a quotient map after taking its product with another space. We therefore justify continuity using maps into a function space, and check explicitly that the formulas match on the identified end.

Proof. To prove continuity of \(\widehat G\), it is enough to check the preimage of a subbasic set \([K,U]\). Suppose \(\widehat G(p)\in[K,U]\), so \(G(p,s)\in U\) for every \(s\in K\). For each \(s\in K\), continuity of \(G\) gives open sets \(N_s\subseteq P\) and \(J_s\subseteq I\) with \(p\in N_s\), \(s\in J_s\), and \(G(N_s\times J_s)\subseteq U\). The sets \(J_s\) cover the compact set \(K\), so finitely many, say \(J_{s_1},\ldots,J_{s_n}\), cover it. The open neighborhood \(N_{s_1}\cap\cdots\cap N_{s_n}\) of \(p\) is then mapped by \(\widehat G\) into \([K,U]\). This proves that \(\widehat G^{-1}([K,U])\) is open. If \(K\) is empty, \([K,U]\) is the whole function space, so its preimage is open as well.

For evaluation, suppose \(g(s)\in U\), where \(U\) is open. The set \(g^{-1}(U)\) is an open neighborhood of \(s\) in \(I\). There is a compact interval \(K\subseteq I\) such that \(s\) lies in its relative interior and \(K\subseteq g^{-1}(U)\); at an endpoint of \(I\), use a closed interval with that endpoint. Then \([K,U]\) is an open neighborhood of \(g\), and the relative interior of \(K\) is an open neighborhood of \(s\). Every pair in their product evaluates into \(U\). This proves continuity of \(\operatorname{ev}\). \(\square\)

Defining the Motion Before Passing to the Quotient

Set \(P=A\times I\bigsqcup Y\), so the quotient map above is \(q:P\to M_f\). Define \(F:P\times I\to M_f\) by the following two formulas, one on each component of \(P\):

Each formula is continuous on its component. On \(A\times I\times I\), the coordinate function \((a,t,s)\mapsto(a,(1-s)t)\) is continuous, and composing it with \(q\) gives the first formula. On \(Y\times I\), the second formula is the continuous map \(q|_Y\) composed with projection onto \(Y\). Since \(P\) has the disjoint union topology, continuity on both components proves continuity of \(F\).

At the identified end, the two values agree for every \(a\in A\) and \(s\in I\):

Thus equivalent points of \(P\) give the same entire path in \(M_f\), not just the same point at one chosen time. The adjoint \(\widehat F:P\to C(I,M_f)\) is continuous by the lemma. Since it is constant on equivalence classes, the quotient property gives a continuous map \(\overline F:M_f\to C(I,M_f)\) satisfying \(\overline F(q(p))=\widehat F(p)\). Consequently \(H(m,s)=\operatorname{ev}(\overline F(m),s)\) is continuous, because it is the composite of the continuous map \((m,s)\mapsto(\overline F(m),s)\) and the continuous evaluation map.

This construction also shows precisely why continuity must be addressed across the identified end. Defining continuous formulas on the two disjoint pieces is not, by itself, enough: they must agree on equivalent representatives, and the induced homotopy on \(M_f\times I\) must be shown continuous.

Verifying the Deformation Retraction

Proof of the theorem. The preceding construction proves that \(H\) is a well-defined continuous map on \(M_f\times I\). For a cylinder representative, at the initial time, \(H(q(a,t),0)=q(a,(1-0)t)=q(a,t)\); for a target representative, \(H(q(y),0)=q(y)\). Thus \(H(m,0)=m\) for every \(m\in M_f\).

At the final time, \(H(q(a,t),1)=q(a,0)=q(f(a))\), which lies in \(q(Y)\); also \(H(q(y),1)=q(y)\in q(Y)\). Hence the whole image of \(H(-,1)\) lies in the target subspace. For every \(y\in Y\) and every \(s\in I\), the formula gives \(H(q(y),s)=q(y)\), so the target is fixed pointwise throughout the homotopy. These are exactly the conditions for a strong deformation retraction. \(\square\)

The final map, regarded as a retraction \(r:M_f\to q(Y)\), has the explicit formulas \(r(q(a,t))=q(f(a))\) and \(r(q(y))=q(y)\). In particular, points at every height in the same cylinder segment end at the target point prescribed by \(f\). The value at the bottom is consistent with the identification, because \(q(a,0)=q(f(a))\).

A Common Pitfall

A formula for \(F\) must use inputs from the component on which it is being defined. The domain of \(F\) is \(P\times I\), where the \(Y\) component of \(P\) has points \(y\in Y\). Writing \(F(q(y),s)=q(y)\) on that component is ill-typed: \(q(y)\) is already a point of \(M_f\), not an input in the \(Y\) component of \(P\). The correct formula is \(F(y,s)=q(y)\). The matching condition is then \(F((a,0),s)=q(a,0)=q(f(a))=F(f(a),s)\). This check matters: without a well-defined map on \(P\times I\), there is no induced homotopy on the mapping cylinder.

The construction also explains why the target must be fixed. A homotopy that merely sends every point into \(q(Y)\) at its final time need not be a deformation retraction: it could move points of \(q(Y)\) during the homotopy. Here the separate formula \(H(q(y),s)=q(y)\) verifies the stronger, pointwise-fixed condition.