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Note the followin theorem

Theorem : any continuous map from a unit two dimensional disk $E^2$ into itself has a fixed point.

To prove this theorem, Harper and Greenberg's book use the following argument :

If $f : E^2 \to E^2$ has no fixed point then we have $F(x) = \frac{f(x)-x}{|f(x) - x|}$.

So from the pertubation of $F$ we have a contraction $r$.

Here I cannot understand why $r$ is a contraction. If $F|_{S^1}$ has degree $1$, then it is plausible. But $F|_{S^1}$ may have degree $>1 $.

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Show that $F|_{S^1}$ has no fixed point. Then it is homotopic to the antipodal map $x\to-x$ via the homotopy $H:\ S^1\times I\to S^1,\ \ (x,t)\mapsto{-tx+(1-t)F(x)}/{||-tx+(1-t)F(x)||}$. This implies that the degree of $F|_{S^1}$ equals that of the antipodal map, which is $(-1)^{n+1}$ for $S^n$, as it is a composition of $n+1$ reflections.
On the other hand, you don't need to know the degree, as it is obvious that the antipodal map (at least on $S^1$), and thus also $F|_{S^1}$ is homotopic to the identity. This homotopy can then be extended to the entire disk since $(E^2,\partial E2)$ has the homotopy extension property. Hence you obtain a homotopy between $F$ and a retraction.