I wonder how to prove rigorously that the figure eight, or a pair of circles intersecting at one point (looking like OO), is homotopy equivalent to a disjoint pair of circles joined by a straight line segment (looking like O-O)?

On an intuitive level, I can define a map from O-O to OO by contracting the line segment, and I can map OO to O-O by mapping the left-hand circle into itself and the right-hand circle into -O. But I have no idea how to prove that the compositions of those maps are homotopic to constant maps (and it's even not intuitively clear to me). I can trace what the compositions are, but in order to prove it rigorously I guess I have to introduce some kind of coordinates and write down the maps explicitly, which I have a hard time doing.

So what's the easiest way to show that the two spaces are homotopy equivalent? If my maps are not optimal, feel free to suggest other maps for which it's easier to prove that their compositions are homotopic to the identity. Or does the existence of a homotopy equivalence follow from some more general fact?

  • $\begingroup$ Just two possibilities: use the CW-complex formalism, and check out any AT book's discussion of graphs or convince yourself that both spaces are retracts of a disk with two points removed. $\endgroup$ Nov 4, 2018 at 5:23
  • $\begingroup$ This is almost the same question as math.stackexchange.com/q/1191899. $\endgroup$
    – Paul Frost
    Jan 11, 2019 at 14:26

2 Answers 2


The homotopy equivalence follows from the following facts :

(i) Given a Borsuk pair $(X,A)$ such that $A$ is contractible, the canonical projection $X\to X/A$ is a homotopy equivalence

(ii) If $X$ is a CW-complex and $A$ a sub-CW-complex, $(X,A)$ is a Borsuk pair

(iii) Your second space (O-O) is a CW-complex that has - as a sub-CW-complex, and - is contractible

(iv) O-O/- is OO

The hardest part is the point (ii)

Note : since I learned it in a class I don't know whether it's standard terminology : a Borsuk pair $(X,A)$ is a topological pair such that for any continuous $F: X\to Y$ and any homotopy $H: A\times I\to Y$ such that $H(-,0)$ is the restriction of $F$ to $A$, there exists a homotopy $K:X\times I\to Y$ that agrees with $H$ on $A$ and such that $K(-,0)=F$

But of course you can also prove it more directly and concretely with your method if you know how to do the following thing (which is essentially redoing the points (i) through (iv) but only the parts you need) : define a homotopy from the identity of O-O to a map O-O $\to$ O-O in which the - part is sent to a single point of itself (e.g. one of its endpoints or its midpoint).

Once you have such a map and such a homotopy, your construction will essentially work without any trouble if you follow the points (i) through (iv) in your special case.

  • 1
    $\begingroup$ $(X,A)$ is a Borsuk pair iff the inclusion $i : A \hookrightarrow X$ is a cofibration. In my opinion the latter is the standard way to express it. $\endgroup$
    – Paul Frost
    Jan 14, 2019 at 15:38
  • $\begingroup$ @PaulFrost : right, I learnt that now, thank you ! $\endgroup$ Jan 14, 2019 at 15:39
  • $\begingroup$ Some people also use the expression "cofibred pair". $\endgroup$
    – Paul Frost
    Jan 14, 2019 at 15:42

You can also use the methods applied in the answer to Deformation Retraction and Projection/Closest Vector. There you can find an explicit strong deformation retraction $r$ from the doubly punctured plane $P$ to the figure eight OO. Adapting this construction a little, we get a strong deformation retraction $r'$ from $P$ to O-O. Just realize O-O as $S'_{+1} \cup J \cup S'_{-1}$, where $S'_{\pm 1}$ is the circle with radius $1/2$ around $p_{\pm 1} = (\pm 1,0)$ and $J = [-1/2,1/2] \times \{ 0 \}$. On the strips $(-\infty,-1/2] \times \mathbb{R}$ and $[1/2,\infty) \times \mathbb{R}$ proceed as it was done for $r$ on the left/right half plane. On the strip $[-1/2,-1/2] \times \mathbb{R}$ simply define $r'(x,y) = (x,0) \in J$.


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