nLab
mapping cylinder

In Top

Definition

Given a continuous map $f:X\to Y$ of topological spaces, one can define its mapping cylinder as a pushout

$$\begin{array}{ccc}X& \stackrel{f}{\to }& Y\\ {}^{{\sigma }_0}\downarrow & & {\downarrow }^{f_{*}({\sigma }_0)}\\ X\times I& \stackrel{({\sigma }_0{)}_{*}(f)}{\to }& \mathrm{Cyl}(f)\end{array}$$\array{ X &\stackrel{f}\to& Y \\ {}^{\mathllap{\sigma_0}}\downarrow && \downarrow^{\mathrlap{ f_*(\sigma_0)}} \\ X\times I &\stackrel{(\sigma_0)_* (f)}\to & Cyl(f) }

in Top, where $I=[0,1]$ (the unit interval) and ${\sigma }_0:X\to X\times I$ is given by $x\mapsto (x,0)$. By tradition, homotopy theorists sometimes use the inverted (upside-down) mapping cylinder where ${\sigma }_0$ is replaced by ${\sigma }_1:x\mapsto (x,1)$. Of course the two mapping cylinders are homeomorphic so it is matter of convention which one to use, of course, compatibly with other constructions depending on the orientation of $I$.

Set-theoretically, the mapping cylinder is usually represented as the quotient space $(X\times I\coprod Y)/\sim$ where $\sim$ is the smallest equivalence relation identifying $(x,0)\sim f(x)$ for all $x\in X$.

Properties

As any other pushout, the mapping cylinder has a universal property: for any space $Z$ and mapping $g_1:X\times I\to Z$, $g_2:Y\to Z$ such that $g_1(x,0)=g_2(f(x))$ for all $x\in X$, there is a unique $k:\mathrm{Cyl}(f)\to Z$, such that the composition $X\times I\to \mathrm{Cyl}(f)\stackrel{k}{\to }Z$ equals $g_1$ and the composition $Y\to \mathrm{Cyl}(f)\stackrel{k}{\to }Z$ equals $g_2$.

Theorem

Let $f:X\to Y$ be any continuous map. The canonical map $j:=f_{*}({\sigma }_0):Y\to \mathrm{Cyl}(f)$ is a homotopy equivalence. In fact its homotopy inverse can be chosen a deformation retraction.

Proof

We exhibit $j$ as a homotopy equivalence by constructing its homotopy inverse $\tilde{f}$ given by $\tilde{f}:[x,t]\mapsto f(x)$, where $[x,t]$ is a class of $(x,t)\in X\times I$ and $\tilde{f}([y])=[y]$ for $y\in Y$. Clearly this map is well-defined and $\tilde{f}\circ j={id}_Y$. On the other hand, $(j\circ \tilde{f})[x,t]=[f(x)]$. Homotopy $H:\mathrm{Cyl}(f)\times I\to Y$ is given by

$$H([x,t],\tau )=[x,t(1-\tau )],H([y],\tau )=[y].$$H([x,t],\tau) = [x,t(1-\tau)], \,\,\,H([y],\tau)=[y].

It is easy to see that $H(-,0)={id}_{\mathrm{Cyl}(f)}$, $H(-,1)=[-,0]=[f(-)]$ hence $j\circ \tilde{f}\sim {\mathrm{id}}_{\mathrm{Cyl}(f)}$.

Theorem

A continuous map $i:A\to X$ is a Hurewicz cofibration iff there is a retraction $r:X\times I\to \mathrm{Cyl}(f)$ for the canonical map $X\times I\to \mathrm{Cyl}(f)$.

Theorem

A continuous map $f:X\to Y$ is a homotopy equivalence iff $X=X\times \{0\}$ is a deformation retract of the cylinder $\mathrm{Cyl}(f)$.

Theorem

For any $f:X\to Y$, the composition

$$X\stackrel{{\sigma }_1}{\to }X\times I\stackrel{({\sigma }_0{)}_{*}(f)}{\to }\mathrm{Cyl}(f)$$X\stackrel{\sigma_1}\to X\times I\stackrel{(\sigma_0)_* (f)}\to Cyl(f)

is a Hurewicz cofibration. Furthermore, the map $r:\mathrm{Cyl}(f)\to Y$ determined by $r([x,t])=f(x)$ (for all $x\in X$ and $t\in I$) and $r([y])=y$ (for $y\in Y$) is well defined and a homotopy equivalence.

The composition $r\circ ({\sigma }_0{)}_{*}(f)\circ {\sigma }_1=f$, hence this is a decomposition of a continuous map into a cofibration followed by a homotopy equivalence.

Related entries

See also mapping cone.

In homotopy type theory mapping cyclinders can be constructed as higher inductive types. See here.