# What does 'factors through' mean in this context?

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?!

-

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.
I don't understand why $p$ is then null-homotopic. Is it not the case that if $X$ contractible and $f:Y\rightarrow X$ a map then $f$ null-homotopic, not necessarily the other way round...? –  Edward Hughes Apr 5 '12 at 13:47
@hgbreton: We may define a nullhomotopy $p_t : \mathbb C \to \mathbb C - \{0\}$ by $p_t(z) = p(tz)$. –  Eric O. Korman Apr 5 '12 at 13:48
Ah right so this isn't a general result, just a property of $p$ then? Thanks! –  Edward Hughes Apr 5 '12 at 13:51
@hgbreton: I think it is general: let $X$ be contractible and $H_t$ be a homotopy from the identity function to a constant function. Then given any $f: X \to Y$, $f_t := f \circ H_t$ is a nullhomotopy of $f$. –  Eric O. Korman Apr 5 '12 at 13:58