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This is a theorem from my lecture notes:

If the continuous map $f: S^1 \to S^1$ extends to a continuous map $F: B(0,1) \to S^1$ the $f$ is homotopic to a constant map.

The proof just defines a homotopy $G$ by $G(z,u) = F(uz)$ so $G(z,1) = F(z) = f(z)$ for $Z\in S^1$ and $G(z,0) = F(0)$.

I don't understand why the criterion that $f$ ends to a continuous map $F: B(0,1) \to S^1$ is necessary? Surely you can still just define $G(z,u) = f(uz)$ to get the same result?

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If you don't, the map $G(z,u)=f(uz)$ is not necessarily continuous at $0$. It is stated as a hypothesis that this map is continuous, but without the hypothesis, you can't make that claim. – Mario Carneiro Dec 17 '12 at 18:26
What is $f(uz)$ when $u < 1$? Note that $uz \not \in S^1$ in this case. – WimC Dec 17 '12 at 18:30

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