Mapping Cylinders as Map Factorizations

The Mapping Cylinder Separates a Map into Two Stages

A continuous map \(f:X\to Y\) may fail to be injective, surjective, or a homeomorphism. Its mapping cylinder provides a space in which the domain and codomain occupy different ends of a cylinder: the domain is represented at the bottom, while the top is attached to the codomain using \(f\). This construction turns \(f\) into a composite of an embedding and a retraction. The factorization is useful because it separates the way points enter the construction from the way the construction returns to \(Y\).

Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, let \(I=[0,1]\) have its standard topology, and let \(f:X\to Y\) be continuous. Give \(X\times I\) the product topology and \(X\times I\bigsqcup Y\) the disjoint union topology. Write \(q\) for the quotient map in the definition below.

Thus \([x,1]=k(f(x))\). Points of the cylinder with height less than \(1\) are not identified with points of \(Y\), and two top points \((x,1)\) and \((x',1)\) are identified precisely when \(f(x)=f(x')\). The map \(k\) names the copy of \(Y\) in the quotient; the theorem below verifies that this copy has the original topology of \(Y\).

The Canonical Factorization

Here an embedding means a continuous injective map that is a homeomorphism onto its image with the subspace topology. A retraction onto \(k(Y)\) means a continuous map \(r:M_f\to Y\) satisfying \(r\circ k=\operatorname{id}_Y\); its image, viewed in \(M_f\), is \(k(Y)\). We prove the claims, including that \(k\) is an embedding.

We will use the quotient test for continuity. If \(g:M_f\to Z\) is a map into a topological space \(Z\), then \(g\) is continuous if and only if \(g\circ q\) is continuous. Indeed, for every open \(W\subseteq Z\), \(q^{-1}(g^{-1}(W))=(g\circ q)^{-1}(W)\). If \(g\circ q\) is continuous, this preimage is open, so the definition of the quotient topology makes \(g^{-1}(W)\) open. The other implication follows by composition of continuous maps.

Proof. Define \(j(x)=[x,0]\). The map \(x\mapsto(x,0)\) from \(X\) to \(X\times I\) is continuous, so \(j\), its composite with \(q\), is continuous. No bottom points are identified with one another, so \(j\) is injective. To check its image topology, take any open \(U\subseteq X\). The set \(U\times[0,1/2)\) is open in \(X\times I\) and, regarded as a subset of the disjoint union, is open and saturated for the quotient: none of its points lies at height \(1\), where identifications occur. Therefore its image under \(q\) is open in \(M_f\), and that image intersects \(j(X)\) in exactly \(j(U)\). Thus every \(j(U)\) is open in \(j(X)\). Together with continuity and injectivity of \(j\), this proves that \(j\) is an embedding.

Next, \(k\) is continuous as the composite of the inclusion of the \(Y\)-summand into the disjoint union with \(q\), and is injective because no two points of \(Y\) are identified. For an open \(V\subseteq Y\), the set \(\bigl(f^{-1}(V)\times(1/2,1]\bigr)\bigsqcup V\) is open and saturated: its cylinder points at height \(1\) are identified only with points of \(V\), and it includes every such identified point. Its image under \(q\) is open and intersects \(k(Y)\) in \(k(V)\). Hence \(k\) is an embedding.

It remains to prove that \(r\) is well-defined and continuous. On the disjoint union, define a map to \(Y\) by \((x,t)\mapsto f(x)\) on \(X\times I\) and \(y\mapsto y\) on \(Y\). Both restrictions are continuous, so the map on the disjoint union is continuous. Its values agree on each identified pair because \((x,1)\) is sent to \(f(x)\), as is the point \(f(x)\) in the \(Y\)-summand. It therefore induces a well-defined map \(r:M_f\to Y\); the quotient test shows that \(r\) is continuous. By its definition, \(r(k(y))=y\), so \(r\circ k=\operatorname{id}_Y\). Finally, for every \(x\in X\), \((r\circ j)(x)=r([x,0])=f(x)\). Thus \(f=r\circ j\), as required. If \(X\) is empty, these same definitions give \(M_f=Y\), \(j\) the empty embedding and \(r=\operatorname{id}_Y\); if a map to \(Y=\varnothing\) exists, then \(X=\varnothing\), so this also covers that case. \(\square\)

The map \(j\) places the domain at height \(0\), not at the attached end. Meanwhile, \(r\) sends every point on the cylinder over \(x\) to \(f(x)\), and fixes the copy of \(Y\). The construction therefore preserves the original map when the domain first enters the cylinder and then retracts to the codomain.

Worked Factorizations

A Common Mistake: Choosing the Attached End

One might try to represent the domain by the top of the cylinder, defining \(j_{\mathrm{top}}(x)=[x,1]\). This can produce the correct composite with the retraction, but it need not be an inclusion. For the constant map in the previous example, \(j_{\mathrm{top}}(0)=[0,1]=k(1/3)\) and \(j_{\mathrm{top}}(1)=[1,1]=k(1/3)\). The two distinct points \(0\) and \(1\) therefore have the same image. The calculation \(r(j_{\mathrm{top}}(x))=1/3=f(x)\) is correct, but it does not give the required factorization through an embedding: \(j_{\mathrm{top}}\) is not injective. Using the bottom instead gives \(j(0)=[0,0]\) and \(j(1)=[1,0]\), distinct points, while still satisfying \(r(j(x))=f(x)\).

The distinction between the two ends is structural. The top is where points are identified according to the values of \(f\), so it records the identifications made by the map. The bottom retains a separate point for each element of \(X\), which is why it gives an embedded copy of the domain. The retraction then records the value of \(f\) along each cylinder segment while leaving the codomain copy fixed.