Past exam of the topology and groups course of the University of Oxford / 2023 / 1 / b

Writefor the quotient map defining the mapping cylinder, and put $j(x)=q(x,0)$. Since the equivalence relation only attaches points at height $1$ to $Y$, distinct points at height $0$ remain distinct. Thus $j$ is an injective continuous map.

Define the homotopyIt is a continuous map, with $H(x,0)=q(x,1)=i(f(x))$ and $H(x,1)=j(x)$. HenceEven if $f$ identifies many points, the free end of the mapping cylinder retains a separate copy of every point of $X$.

DefineThe two formulas agree on the attached end, so the universal property of the quotient topology makes $r$ a continuous map. Also $ri={\mathrm{id}}_Y$.

Slide each cylinder point towards its attached end usingAt $t=1$ these formulas both give $i(f(x))$, so they define a homotopy on the mapping cylinder. For continuity, the product of a quotient map with the unit interval is a quotient map; the formulas before taking the quotient topology are continuous maps and agree on every identified pair. ThusMoreover $Y$ stays fixed throughout. We have obtained the stronger conclusionIn particular $r$ is the requested homotopy retract.

Identify the bottom $j(X)=q(X\times \{0\})$ with $X$; its natural map $j$ is a topological embedding. To verify this directly, an open subset $O\subseteq X$ gives the saturated open subset $O\times [0,1/2)$ of the space before taking the quotient topology; its image is open in $M(f)$ and meets the bottom in $j(O)$.

Suppose first that there is a retraction $R:M(f)\to j(X)$. Define the continuous mapThenis a homotopy with $L(x,0)=x$ and $L(x,1)=g(f(x))$. Hence $gf\simeq {\mathrm{id}}_X$.

Conversely, suppose a continuous map $g:Y\to X$ has $gf\simeq {\mathrm{id}}_X$. Reverse the parameter if necessary to choose a homotopy $L$ with $L(x,0)=x$ and $L(x,1)=gf(x)$. SetAt the glued end, $j(L(x,1))=j(gf(x))$, exactly the value prescribed at $i(f(x))$. The universal property of the quotient topology therefore gives a continuous map $R:M(f)\to j(X)$. At the bottom, $R(j(x))=j(x)$, proving that $R$ is a retraction. We have established the mapping-cylinder bottom retraction criterion: