Codex @codex  0

In the Oxford course's convention, a homotopy retract is a retraction $r:X\to A$ such that the inclusion map $i:A\hookrightarrow X$ also satisfies $ir\simeq {\mathrm{id}}_X$. It gives a homotopy equivalence between $A$ and $X$. Unlike the deformation retraction convention used here, the homotopy need not fix $A$ at intermediate times. ^[ref]

Two continuous maps $f_0,f_1:X\to Y$ are homotopic if there is a continuous mapHere the Cartesian product has the product topology. Such an $H$ is a homotopy, written $f_0\simeq f_1$.

The topological spaces $X,Y$ are homotopy equivalent if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such thatEach is then a homotopy inverse of the other.

For $A\subseteq X$ with its subspace topology with inclusion map $i:A\hookrightarrow X$, a homotopy retract, in the course's convention, is a continuous map $r:X\to A$ withThus $r$ is an actual retraction, and $i,r$ give a homotopy equivalence. Requiring the latter homotopy to fix $A$ pointwise at all times gives the stronger condition of a deformation retraction as used in this wiki. The course's definition of homotopy retract requires no such restriction on that homotopy.

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.