# Retraction of $\{(z_1, …, z_n) \in \Bbb C^n : \sum z_k^2 \neq 0 \}$ onto $\{(z_1, …, z_n) \in \Bbb C^n : \sum z_k^2=1 \}$

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.)

• Won't $z \mapsto z/||z||$ work? – P-addict Sep 29 '19 at 6:14
• @P-addict It doesn't. It is not well defined, for example in $\Bbb C^2$, consider the image of $( \sqrt{2}, i)$ – Quadr Sep 29 '19 at 7:06
• correct, its not $\mathbb{R^{n}}$. – P-addict Sep 29 '19 at 11:40
• $f(z) = \sum_j z_j^2$ 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 – reuns Sep 30 '19 at 1:43
• @reuns What are the "intermediate sets"? And $\mathbb C^*$ retracts to a point (although it doesn't deformation retract). – Paul Frost Sep 30 '19 at 9:05

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 $$X \to Y$$.