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Consider the space $X:=\{(z_1, ..., z_n) \in \Bbb C^n : \sum z_k^2 \neq 0 \} $ and its closed subspace $Y:=\{(z_1, ..., z_n) \in X : \sum z_k^2=1 \}$. Is there a retraction of $X$ onto $Y$? Or, more generally, is there a continuous surjection $X \to Y$? Intuitively it seems true, but I got stuck constructing such a map. Any helps will be appreciated. (Both $X$ and $Y$ have the Euclidean topology.)

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    $\begingroup$ Won't $ z \mapsto z/||z|| $ work? $\endgroup$ – P-addict Sep 29 '19 at 6:14
  • $\begingroup$ @P-addict It doesn't. It is not well defined, for example in $ \Bbb C^2$, consider the image of $( \sqrt{2}, i)$ $\endgroup$ – Quadr Sep 29 '19 at 7:06
  • $\begingroup$ correct, its not $\mathbb{R^{n}}$. $\endgroup$ – P-addict Sep 29 '19 at 11:40
  • $\begingroup$ $f(z) = \sum_j z_j^2$ then image by $f$ of the intermediate sets must be contained in $f(X) = \Bbb{C}^*$ which is not simply connected so it cannot retract to a point $\endgroup$ – reuns Sep 30 '19 at 1:43
  • $\begingroup$ @reuns What are the "intermediate sets"? And $\mathbb C^*$ retracts to a point (although it doesn't deformation retract). $\endgroup$ – Paul Frost Sep 30 '19 at 9:05
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No for $n=1$. We have $X = \{ z \in \mathbb C \mid z^2 \ne 0\} =\{ z \in \mathbb C \mid z \ne 0\} = \mathbb C \setminus \{ 0 \}$ and $Y = \{ z \in X \mid z^2 =1 \} = \{-1,1\}$. Since $X$ is connected whereas $Y$ is not, there does not exist a continuous surjecton $X \to Y$.

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