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I'm trying to understand the proof of the Fundamental Theorem of Algebra in Theorem 3.7 here.

I can't get my head around this sentence though

If $p(z)$ has no roots at all, the map $p|_{S1(R)}$ factors through the complex plane $\mathbb{C}$ and is therefore nullhomotopic (as $\mathbb{C}$ is contractible).

I know that $\mathbb{C}$ is convex so if $p$ is a map into $\mathbb{C}$ then it is homotopic to any other map into $\mathbb{C}$. Does this help? Also surely any map into $\mathbb{C}\setminus\{0\}$ is a map into $\mathbb{C}$. Doesn't this make the whole thing vacuous?!

Many thanks in advance.

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If $p(z)$ has no roots at all then it defines a map $\mathbb C \to \mathbb C - \{0\}$. Therefore, in this case we can write $p\vert_{S^1(R)}$ as a composition $ S^1(R) \stackrel{i}{\to} \mathbb C \stackrel{p(z)}{\to} \mathbb C - \{0\} $ where the first map is the inclusion. Since $\mathbb C$ is contractible, the map $p(z)$ is nullhomotopic so that the composition, which is $p\vert_{S^1(R)}$, is also nullhomotopic.

The reasoning it's called factoring is that we've decomposed the original map $p\vert_{S^1(R)}$ into a composition $p\circ i$.

Note that even if $p$ has roots, $p\vert_{S^1(R)}$ is still $p \circ i$ but now the functions in the homotopy may have 0 as a value.

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    You're quite right - sorry wasn't thinking straight! Many thanks.2012-04-05