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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 $$H_t: C(X)-P\longrightarrow C(X)-P;\;\; [x,s]\mapsto [x,(1-t)s+t]$$

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  • $\begingroup$ Hint: the two $t$'s in your question can mean (almost) the same thing. $\endgroup$
    – cws
    Jul 21, 2014 at 20:28
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    $\begingroup$ Isn't $C (X) \setminus \{ P \}$ just homeomorphic to $X \times [0, 1)$? $\endgroup$
    – Zhen Lin
    Jul 21, 2014 at 20:29
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    $\begingroup$ @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, 2014 at 20:40
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    $\begingroup$ @palio Yes, I agree. $\endgroup$
    – cws
    Jul 21, 2014 at 20:45

1 Answer 1

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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.

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  • $\begingroup$ 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, 2014 at 21:14
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    $\begingroup$ 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, 2014 at 21:42
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    $\begingroup$ 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, 2014 at 21:53
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    $\begingroup$ @palio: That's correct, yes. $\endgroup$
    – Stefan Hamcke
    Jul 21, 2014 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, 2014 at 22:03

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