We want to show that $\mathbb C^{\infty}\setminus\{0\}$ retracts on $S^{\infty}$.
$C^{\infty}\setminus\{0\}$ should be the same as $\mathbb R^{\infty}\setminus\{0\}$. Maybe the retraction could be $(1-t)x + t \frac{x}{|x|}$?
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With which topology do you view $\mathbb C^\infty$? – Bart Michels Aug 11 '18 at 19:02
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@barto the one given by limit induction on $\mathbb C^n / 0$, which has the one used for $\mathbb C^n /0 \subset \mathbb P^n C$. – Maffred Aug 11 '18 at 19:14
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I think you need to specify which $\mathbb{R}^\infty$ you mean. If it has a norm, then your retraction works. If it just has the product topology, then the usual norm is not continuous, so your retraction won't work. But then it's not clear how you even define $S^\infty$, etc. I think actual definitions for your spaces would make this question easier to parse (and answer). – Steve D Aug 12 '18 at 02:34
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@SteveD consider $\mathbb C^{n} / {0}$ as an euclidean space, and it's limit for $n \to \infty$. Then consider the $S^{\infty}$ of radius $1$ it contains. – Maffred Aug 12 '18 at 04:45
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@SteveD I want to show that $\mathbb C^{\infty} / {0}$ is contractible, that's why I retract it to the sphere and then use that it is contractible. – Maffred Aug 12 '18 at 04:46
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Oh, that is much easier! you can do a straight-line homotopy $$ (1-t)(x_1,x_2,\ldots) + t(0, x_1, x_2,\ldots)$$ and then another straight-line homotopy $$ (1-t)(0, y_1, y_2, \ldots) + t(1, 0, 0, \ldots) $$ the product is a deformation retraction to a point. – Steve D Aug 12 '18 at 04:52
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@SteveD Wow! Nice moves! – Maffred Aug 12 '18 at 05:03
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@SteveD maybe you have some clues on how to perform another contraction? It also has 200 points bounty but no one seems to know the answer! https://math.stackexchange.com/questions/2876889/mat-infty-n-mathbb-c-of-max-rank-is-contractible – Maffred Aug 12 '18 at 05:48
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You have to define properly what $\mathbb C^\infty$ is. In its present form it is impossible to give an answer. – Paul Frost Jul 21 '19 at 11:12