Using retraction for show that: 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} $.
 A: 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.
A: 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 $\lambda$.
