One can show that for any given homology theory $H$ with non-trivial coefficient group $G$ there does not exist a retract $\partial B^n \subset B^n$. Brower's fix point theorem states that any continuous map $f:B^n\to B^n$ has at least one fixpoint. If we now assume by contradiction that $f(x) \neq x \space (\forall x \in B^n) $ then one can define $r: B^n \to \partial B^n$ by taking the ray $r_{f(x),x}$ from $f(x)$ to $x$ and defining $r(x)=r_{f(x),x} \cap \partial B^n$. Obvisouly we have $r\mid_{\partial B^n}=id_{\partial B^n}$.
Question: Why is r continuous?
Edit: I'm not seeking an intuitive answer as I already have an intuitive understanding on what is going on, but rather an explicit proof. If possible not an $\epsilon - \delta$ one.
You can calculate explicitly the intersection between the line defined by $x$, $f(x)$ and the border of the ball.
The line in parametric form is $$t\longmapsto f(x)+(x-f(x))t.$$ The condition "the point is in the border" is $$\|x+(f(x)-x)t\| = 1$$ (or any other radius $R$).