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Let $f:\mathbb{S^2} \rightarrow \mathbb{R^2} \diagdown \{(0,0)\}$ a continuous application. Proof that there is $(x_0,y_0,z_0)\in \mathbb{S^2}$ such that $f(x_0,y_0,z_0)=\lambda(x_0,y_0)$ for some $\lambda \in \mathbb{R} $.

Henfe
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  • there is something quite unclear for me here. $\mathbb{R}^2 \setminus (0,0)$ has as retract $\mathbb{S}^1$, and $\mathbb{S}^2 \setminus (0,0) \cong \mathbb{R}^2$. And $\mathbb{S}^2$ is compact in $\mathbb{R}^3$ but $\mathbb{R}^2$ not so there isn't surjective continuous map between this two spaces – Riccardo May 02 '13 at 20:39

2 Answers2

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Suppose the statement does not hold, we can then define $$f_t(x)=\frac{tf(x)+(1-t)x}{|tf(x)+(1-t)x|}, S^2\to S^2$$ Since $f_0(x)=x, \quad f_1=f(x)/|f(x)|$, consider the degree you will get a contradiction.

Ma Ming
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  • can you explain wich kind of contradiction you get considering the degree? Thank you in advance :) – Riccardo May 02 '13 at 21:09
  • @RicPed The degree of $f_0$ is $1$, while the degree of $f_1$ is zero since $f_1:S^2\to S^2, \operatorname{im}f_1\subset{S^1}$, pick up $p\in S^2- S^1$, then $\operatorname{degree}(f_1)=#f_1^{-1}(p)=0$. – Ma Ming May 02 '13 at 21:14
  • I do not quite understand why the degree of $f_1$ is $0$, and also do not know what it is$#f_1^{-1}(p)$. Sorry – Henfe May 02 '13 at 21:34
  • Recall the definition of degree, it is the (signed) number of preimages for regular point (this works for smooth map, why also for continuous map is yours.) – Ma Ming May 02 '13 at 21:42
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Let $r\colon\mathbb{R}^2\setminus\{0\}\rightarrow S^1$ be given by $r(x,y)=(x,y)/||(x,y)||$. We fix some $z_0\in(-1,1)$ and let $S(z_0)=\{(x,y,z_0)\in S^2\}$ and note that $r\circ f|_{S(z_0)}$ is a map from the circle to the circle and, because $f$ is homotopic to the constant map, so is $r\circ f|_{S(z_0)}$. It is a general fact that a map from the circle to itself which is not surjective has a fixed point.

You then only require to show that a $z_0$ exists such that $r\circ f|_{S(z_0)}$ is not surjective. This is easy though because $S(1)$ is a point and so you can choose some $z_0$ sufficiently close to $1$ so that $r\circ f|_{S(z_0)}$ is not surjective (this is guarenteed to exist by continuity of $f$). The fixed point $(x_0,y_0)$ of this map, together with the chosen $z_0$ is then your point which is mapped to $\lambda(x_0,y_0)$ for some $\lambda$.

Dan Rust
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