= Solution
{c}
Two <continuous maps> $f_0,f_1:X\to Y$ are <homotopic> if there is a <continuous map>
$$
H:X\times I\longrightarrow Y,\qquad H(x,0)=f_0(x),\quad H(x,1)=f_1(x).
$$
Here 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 that
$$
\boxed{gf\simeq\operatorname{id}_X,\qquad fg\simeq\operatorname{id}_Y.}
$$
Each 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> $r:X\to A$ with
$$
\boxed{ri=\operatorname{id}_A,\qquad ir\simeq\operatorname{id}_X.}
$$
Thus $r$ 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>.
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