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Let $X=\{(k,0) \mid k =0,1,\dots\} \subset \Bbb R^2$ and consider the cone $CX$ over $X$. Let $C'X$ be the subspace of $\Bbb R^2$ obtained by taking the union of the line segments from $(k,0)$ to $(0,1)$ for all $k=0,1,\dots$ Let $q:X \times I \to CX, ((k,0),t) \mapsto[((k,0),t)]$ be the quotient map and $h: X \times I \to C'X$ the map sending $((k,0),t) \mapsto t(0,1)+(1-t)(k,0)$. Show that the inverse $\tilde{h}^{-1}$ of the induced map $\tilde h:CX \to C'X$ sending $[((k,0),t)] \mapsto t(0,1)+(1-t)(k,0)$ is not continuous.

I've tried to figure out what is the problem coming up with continuity for the inverse, but can't figure it out. The spaces look very similar except that the "vertex" of $C'X$ is the point $(0,1)$ which is different from the vertex of $CX$. Is this what's causing the discontinuity here or am I missing something fundamental?

Walker
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  • Consider the sequence $[((k,0),1-1/\sqrt{k})]\in CX$. These tend to the vertex of $CX$, but are sent by the map to $(\sqrt{k},1-1/\sqrt{k})$. – plop Oct 04 '22 at 21:55
  • How did you come up with this sequence? @user85667 – Walker Oct 04 '22 at 22:04
  • I think on Wikipedia it's the other way around, there the vertex is at $t=0$ or am I mistaken? However, in this exercise the vertex should clearly be at $[(k,0),1]$. The important insight is that in $CX$ we only need $t\rightarrow 1$, no matter the $k$, while in $C'X$ we need both coordinates to tend to $(0,1)$, jointly. There are many suitable sequences $(t_n,k_n)$, as long as $t_n\rightarrow 1$ and $\limsup_{n\rightarrow\infty}(1-t_n)k_n>0$. For example, you can also take $(t_n,k_n)=(1-\frac{1}{n},n)$ or $(1-\frac{1}{\ln(n)},n)$... – Matija Oct 04 '22 at 22:23
  • $\tilde h^{-1}$ is continuous iff $\tilde h$ is a homeomorphism. This was treated in https://math.stackexchange.com/q/2973093. Note that that $X$ is not compact. – Paul Frost Oct 04 '22 at 23:01

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