Suppose $X$ is a topological space which is contractible. I want to show that the cone on $X$ deformation retracts onto $X$.

My retraction $r: CX \to X$ is just the homotopy which contracts $X$ to a point. Now I need a homotopy $H: CX \times I \to CX$ between $1_{CX}$ and $ir$ rel $X$.

If I visualize such a homotopy when $X$ is $D^2$, I can see intuitively how it should behave. The explicit homotopy is eluding me, though.

  • $\begingroup$ Where are you getting this from? Is it an exercise from a book? $\endgroup$ – Pedro Tamaroff Oct 29 '15 at 3:46
  • $\begingroup$ @PedroTamaroff No. I thought it should be true, and asked an instructor, who said it was true, and gave me a quick explanation as to why that I didn't totally understand. $\endgroup$ – Eric Auld Oct 29 '15 at 4:10

Let $H : X \times I \to X$ be a deformation retraction of $X$ onto a point $* \in X$, so $H(x,0) = x$ and $H(x,1) = *$ for all $x$. Then an explicit deformation retraction of $CX = (X \times I) / (X \times \{1\})$ onto $X = X \times \{0\}$ is given by $$G([x,s],t) = \begin{cases} [H(x,2st), s] &\text{ for $0\leq t\leq 1/2$} \\ [H(x,s), s(2-2t)] &\text{ for $1/2\leq t\leq 1$} \end{cases}$$

First, this is well-defined (and hence continuous): when $s=1$, the right-hand side does not depend on $x$, and for $t=1/2$, the two definitions agree. The fact that $G$ is a deformation retraction onto $X\times\{0\}$ then follows from the following computations: $$G([x,s],0)=[H(x,0),s]=[x,s]$$ $$G([x,s],1)=[H(x,s),0]\in X\times\{0\}$$ $$G([x,0],t)=[H(x,0),0]=[x,0]\text{ for $0\leq t\leq 1/2$}$$ $$G([x,0],t)=[H(x,0),0]=[x,0]\text{ for $1/2\leq t\leq 1$}$$

The intuition behind this is as follows. First, homotope the $X$-coordinate from $x=H(x,0)$ to $H(x,s)$, keeping the cone coordinate constant. Second, move the cone coordinate down to $0$ while keeping the $X$-coordinate constant. Note that if you try to do these steps at the same time instead of one after the other (as Najib Idrissi and Nitrogen did in their now-deleted answers), the map fails to be well-defined at the cone point, because the first step is only well-defined at the cone point if you keep the height constant (so the cone point stays at the cone point).

  • $\begingroup$ Ah, so that's what I was missing... Well-deserved bounty. $\endgroup$ – Najib Idrissi Dec 10 '15 at 8:28

Another approach:

  1. Any cone $CY$ on a space $Y$ is contractible as it deformation retracts to the apex.
  2. If $X$ is contractible, then the inclusion $i:X\hookrightarrow CX$ of $X$ as the base of $CX$ is a homotopy equivalence. This follows for example since contractible spaces are terminal objects in the category $h\mathbf{Top}$ and a morphism between terminal objects is necessarily an isomorphism.
  3. Now, this inclusion $i$ is also a cofibration, and there is a lemma saying that a cofibration which is at the same time a homotopy equivalence (an acyclic cofibration) is the inclusion of a deformation retract.
  • $\begingroup$ Where can I find a reference on this lemma from #3? Is the proof straightforward? $\endgroup$ – Eric Auld Dec 7 '15 at 8:48
  • 2
    $\begingroup$ A rather ad hoc proof is given in Hatcher's Algebraic Topology in chapter 0. It is also a corollary of the theory of the track groupoid $\pi X^A$ operating on the set of homotopy classes, see Brown's Topology and Groupoids, chapter 7.2 @EricAuld $\endgroup$ – Stefan Hamcke Dec 7 '15 at 10:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.