# 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.

• That is very kind of you thanks, I will be waiting for it. It's great to have some expansion on the original solution Commented May 7, 2016 at 22:32

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 $h$.
• Seing many problems definitely helps. As for the "straight line" homotopy, that's a standard one, and perhaps one that is worth remembering: If you have a map $f(x)$ and a map $g(x)$, the straight line homotopy between them is $$h(x, t) = (1-t)f(x) + tg(x)$$That does require a few things, though. First of all that the operations makes sense: in this case $f(x)$ and $g(x)$ are points in the plane, so you have vector addition and the usual multiplication by $t$. Second: you have to make sure that it's a valid homotopy, i.e. that all $h(x, t)$ are actually within the relevant domain. Commented May 7, 2016 at 22:56