Contradictory; Homotopy equivalence and deformation retract problem The question

Show that $S^1$ is a deformation retract og $D^2\setminus\{(0,0)\}$ the unit punctured disc.

The solution

the inclusion map $i:S^2 \to D^2\setminus\{(0,0)\}$ and
$$j:D^2\setminus\{(0,0)\} \to S^1: (x,y) \to \frac{a}{\sqrt{x^2+y^2}}(x,y)$$
are inverse homotopy equivalences via the straight line homotopy.

Question marks on this one. I recall the definition of homotopy equivalences

$X,Y$ are homotopy equivalent if there exists $f:X \to Y$ and $g:Y \to X$ such that we can construct homotopies (note that it is plural)
$$h:gf \cong id_X, k:fg \cong id_Y$$

What troubles me with the solution; simply, So, what is the deformation retract anyway?
A deformation retract is a map, a single map $h$. But the solution is talking about $i,j$ being inverse homotopy equivalences, so this means we have two homotopies $h,k$ for $ij$ and $ji$,
SO what, is the solution $h$?$k$? What is it saying? Very unclear to me.
 A: Constructing a deformation retract is the standard way of showing that a topological space is homotopy equivalent to a given subspace. The reason is that a deformation retract may be used explicitly to construct both the homotopies you seek in order to fulfill the definition of homotopy equivalence.
Setting $X = D^2\setminus\{(0,0\}$ for brevity, the deformation retract in question is the map $h:X\times I \to X$ given by
$$
h((x, y), t) = (1-t)(x, y) + \frac{t}{\sqrt{x^2 + y^2}}(x, y)
$$
which if we fix $t = 0$ is the identity map on $X$, and if we fix $t = 1$ is the map $j$ (for $a = 1$, which is just saying that $S^1$ is embedded in the plane as the unit circle). Intuitively, the map $h$ "smears out" the points of $X$, shoving them away form the center and towards the edge of $D^2$ along straight lines as $t$ increases.
As for the homotopy equivalence, we have $j \circ i = Id_{S^1}$ already, no homotopy needed, and $i \circ j$ is homotopy equivalent to $Id_X$ via (the reverse of) $h$.
This works for any deformation retract $h$ of a space $X$ onto a subspace $Y$ the following way: the composition $$Y\overset{\text{inclusion}}\hookrightarrow X \overset{h(\cdot, 1)}{\longrightarrow}Y$$ is the identity map on $Y$, and $$X \overset{h(\cdot, 1)}{\longrightarrow}Y\overset{\text{inclusion}}\hookrightarrow X$$ is homotopic to the identity on $X$ via $h$.
