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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.

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  • $\begingroup$ 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
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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).

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  • $\begingroup$ Ah, so that's what I was missing... Well-deserved bounty. $\endgroup$ – Najib Idrissi Dec 10 '15 at 8:28
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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.
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  • $\begingroup$ Where can I find a reference on this lemma from #3? Is the proof straightforward? $\endgroup$ – Eric Auld Dec 7 '15 at 8:48
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    $\begingroup$ 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

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