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I have $f:S^3 \rightarrow S^2$ given by $$f(x_0,x_1,x_2,x_3) = (x_0^2+x_1^2-x_2^2-x_3^2, 2x_0x_3+2x_1x_2,2x_1x_3-2x_0x_2)$$

I define $\tilde{f}:S^3\subset \mathbb{R^4} \rightarrow S^2 \subseteq \mathbb{R}^3$ so that I have $$d\tilde{f}_{(x_0,x_1,x_2,x_3)} = \begin{pmatrix} 2x_0 & 2x_1& -2x_2 &-2x_3\\ 2x_3& 2x_2& 2x_1&2x_3 \\ -2x_3 & 2x_3 & -2x_0 & 2x_1 \end{pmatrix}$$ I argued that since the differential has maximum rank, equal to 3 and equal to the dimension of the image, then the differential is surjective and therefore $f$ is a submersion. Is it correct? Or, for example, do I have to argue something more regarding the differential ? Like $d\tilde{f}_x: T_xS^3 \subseteq T_x\mathbb{R^4} \rightarrow T_xS^2 \subseteq T_x\mathbb{R^3} $ ?

Son Gohan
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m120p
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    The last equality follows from the fact that the image of $\tilde f$, when restricted to $S^3$ lands in $S^2$. So you are done once you have done that. – Thomas Rot Apr 20 '21 at 09:13
  • @ThomasRot Thank you for your reply. Do you mean that $\tilde{f}|_{S^3}=f $? – m120p Apr 20 '21 at 09:34
  • Yeah basically that f is well defined – Thomas Rot Apr 20 '21 at 09:47
  • @Paul Frost I looked at the question, although is similar (or maybe equal) the answer is a bit odd for my knowledge. I am more familiar with the approach I used and with the comments and the answer here I should be fine. – m120p Apr 20 '21 at 16:59
  • You consider a map $\tilde f : \mathbb R^4 \to \mathbb R^3$ and show that $d\tilde f_x$ has rank $3$ in all points. However, as you say, you have to check additionally that $d\tilde f_x(T_x S^3) = T_{f(x)}S^2$ for all $x \in S^3$. See point 1. of the linked answer. – Paul Frost Apr 20 '21 at 16:59

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You reasoning is perfectly fine. Just as a remark: you need to show it has maximal rank for every point $x=(x_0, x_1, x_2, x_3) \in S^3$, and indeed this is the case, just use Gauss reduction algorithm to see it has maximal rank.

Recall indeed the definition of submersion : a submersion is a differentiable map between differentiable manifolds whose differential is everywhere surjective. It is exactly what you proved.

Different would be to ask if it is a Riemannian submersion, but this is a different notion.

m120p
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Son Gohan
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