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.

  • $\begingroup$ That is very kind of you thanks, I will be waiting for it. It's great to have some expansion on the original solution $\endgroup$
    – Melba1993
    Commented May 7, 2016 at 22:32

1 Answer 1


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 $h$.

  • $\begingroup$ That made things clearer, thanks, so one of the homotopies actually apparently works as a deformation retract then. And just as a side note, looking at the solution, my first impression was "how in heavens name could I possibly come up with such a map alone" but do you have any approaches you take when you have these problems? Finding some map, homotopy, homeomorphism etc... Or is it really up to seeing many many many problems and getting used to it? $\endgroup$
    – Melba1993
    Commented May 7, 2016 at 22:49
  • $\begingroup$ 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. $\endgroup$
    – Arthur
    Commented May 7, 2016 at 22:56

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