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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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    What is $f(uz)$ when u < 1? Note that $uz \not \in S^1$ in this case.2012-12-17

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In this case, looking at the picture clarifies the situation.

You start with a map $f : S^1 \to S^1$ that extends to a map $F : D^2 \to S^1$.

You want a homotopy $H : S^1 \times I \to S^1$ with $H(x, 0) = f(x)$ and $H(x, 1) = \operatorname{const}_e$ for some $e \in S^1$.

Essentially what we do is collapse $S^1 \times \{1\}$ to a point, so that $(S^1 \times I) / (S^1 \times \{1\}) \cong D^2$ (the cone construction).

Collapsing Subspace to a Point.

In this case we define $H := F \circ q$ where $q$ is the quotient map (up to the isomorphism with $D^2$). We can describe this by $q(x, t) = x(1 - t).$

In particular, $H(x, t) = F\big(x(1 -t) \big)$ and so $H(x, 0) = F(x) = f(x)$ and $H(x, 1) = F(0) =: e = \operatorname{const}_e$.

The key part is that for $t \in (0, 1]$, $x(1 - t) \in B^2$, that is, not in $S^1$, so we needed the continuous extension $F$ to tell us what to do on the inside of the disk.