2
$\begingroup$

Let $X$ be a topological space and $$C(X)=X\times [0,1]/X\times \{0\}$$ be the cone on $X$. Call $P$ the apex of the cone. I want to show that $C(X)-P$ deformation retracts onto $X\times \{1\}$.

My try (edited using comments below): The retraction I use is the map $$r:C(X)-P\longrightarrow X\times \{1\}; \;\;[(x,t)]\mapsto (x,1)$$ Now I want to construct a homotopy $$H_t:C(X)-P\rightarrow C(X)-P$$ such that $H_0=id_{C(X)-P}$ and $H_1=i\circ r$ where $$i:X\times \{1\} \rightarrow C(X)-P;\;\;(x,1)\mapsto (x,1)$$ is the inclusion map. The homotopy is given by $$H_t: C(X)-P\longrightarrow C(X)-P;\;\; [x,s]\mapsto [x,(1-t)s+t]$$

$\endgroup$
4
  • $\begingroup$ Hint: the two $t$'s in your question can mean (almost) the same thing. $\endgroup$ – cws Jul 21 '14 at 20:28
  • 4
    $\begingroup$ Isn't $C (X) \setminus \{ P \}$ just homeomorphic to $X \times [0, 1)$? $\endgroup$ – Zhen Lin Jul 21 '14 at 20:29
  • 2
    $\begingroup$ @cws I think the following homotopy $H_t:C(X)\rightarrow C(X)$ defined by $H_t([x,s])=[x,(1-t)s+t]$ verify the conditions. Do you agree ? $\endgroup$ – palio Jul 21 '14 at 20:40
  • 2
    $\begingroup$ @palio Yes, I agree. $\endgroup$ – cws Jul 21 '14 at 20:45
3
$\begingroup$

The quotient map $q:X\times [0,1]\to CX$ which sends $X\times\{0\}$ to a single point restricts to a quotient map $q':X×(0,1]\to CX-\{P\}$ since $X×(0,1]$ is open and saturated in $X×[0,1]$. Since $q'$ is injective, it is a homeomorphism. Now a homotopy $(CX-P)×I\to CX-P$ is determined by a homotopy $X×(0,1]×I→X×(0,1]$. In particular, a retracting homotopy on $X×(0,1]$ to $X×\{1\}$ rel $X×\{1\}$ induces a retracting homotopy on $CX-P$ rel $X×\{1\}$, and the homotopy you just posted in your comment does the job.

$\endgroup$
5
  • $\begingroup$ Thank you Stefan. I edited my post following the comments and your answer. If what I've wrote is correct then i don't see why using the homeomorphism $q'$. Isn't simpler to do it directly as in my edited post ? $\endgroup$ – palio Jul 21 '14 at 21:14
  • 1
    $\begingroup$ You can do it directly, but dealing with equivalence classes as the elements can be a bit cumbersome. Here we have a bijection $[x,s]≘(x,s)$, so it's easy, but in general there is a quotient map $q:X→Z$ and we define the homotopy on $Z×I$ be means of a map $H_t:X×I$ such that for each $t∈I, H_t$ respects the relation. Then we have an induced set map $\tilde H_t:Z×I→W$. There's also a subtlety about the continuity of $\tilde H$. Usually, $q×1_Y:X×Y→Z×Y$ is not a quotient map, but if $Y$ is locally compact, it is! Hence $Z×I$ is a quotient of $X×I$ and this implies the continuity of $\tilde H_t$ $\endgroup$ – Stefan Hamcke Jul 21 '14 at 21:42
  • 1
    $\begingroup$ What prevents the argument in my post to be valid for all $C(X)$ insted of $C(X)-P$ is that the retraction $r$ I gave above is not well defined at $[x,0]$ because there are more than one $x$ and there is no continuous way to choose one of them to send it to $(x,1)$ is that correct ? $\endgroup$ – palio Jul 21 '14 at 21:53
  • 1
    $\begingroup$ @palio: That's correct, yes. $\endgroup$ – Stefan Hamcke Jul 21 '14 at 21:58
  • $\begingroup$ @palio: Well, there may be a continuous map $X×I→X$ which sends $(x,s)$ to $x$ for $s>0$ and $(x,0)$ to a chosen point $x_0$ for all $x\in X$ (and hence factors as $X×I→CX\to X$), but this only works if the only neighborhood around $x_0$ is $X$ itself. $\endgroup$ – Stefan Hamcke Jul 21 '14 at 22:03

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.