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why $(X\cup CA)/CA=X/A$ and $(X\cup CA)/X=\Sigma A$ where $CA$ is the cone on $A$ and $\Sigma A$ is the suspension of $A$.

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In the first one, there's an obvious bijection, and this bijection is easily seen to be a homeomorphism. For the second one, I think you have a typo: it should be $(X\cup CA)/X \cong \Sigma A$. This has an obvious bijection, too.

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In other words: make a picture! – Mariano Suárez-Alvarez May 7 '11 at 22:18
this does not seem to be obvious to me. $X\cup CA=(A\times I )\sqcup X /\sim$ such that $(a,0)\sim a$ and $(a,1)\sim (a',1)$ for all $a,a'\in A$ and we have a map $f:(A\times I )\sqcup X /\sim\rightarrow X/A$ sending $x \mapsto x$ and $(a,t)\mapsto a$ then we factor by $CA$ to have the homeomorphism is that correct? – studento May 8 '11 at 11:48
Well, that doesn't quite work: the map $X\cup CA \rightarrow X/A$ does not factor through $X$ (which is what you imply) since there's nowhere to put the cone point. Rather, we simply take $X\cup CA$ and crush the entire cone $CA$ to a point. The points in the quotient are all of the form $x\in X-A$ except for the equivalence class $\overline{(a,t)}$ (for every $(a,t)\in CA$). Observe that there is a bijection between this and $X/A$, whose points are all of the form $x\in X-A$ except for the equivalence class $\overline{a}$ (for every $a\in A$). – Aaron Mazel-Gee May 8 '11 at 23:49

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