For $n$-degree complex polynomial $p(z) = a_nz^n + \cdots + a_1z + a_0$ defined on $S^2$ is $p$ homotopic to $z^n$? If I have an $n$-degree complex polynomial $p : S^2 \to S^2$, can I just construct the straight-line homotopy on all non-leading coefficients from $a_i$ to $0$? Is this continuous?
Full disclosure, this is for a homework problem (Hatcher's Algebraic Topology 2.2.8) but this fact is not the primary focus of the problem.
Edit: added that this is $S^2 \to S^2$ instead of $\mathbb C \to \mathbb C$ (which is kind of important...)
 A: This is related to this other question, though it's not strictly speaking a duplicate.
Let $p(z) = a_n z^n + \dots + a_1 z + a_0$ with $a_n \neq 0$. It's easier (for me at least) to do this in two steps. First you can define
\begin{align}
H : S^2 \times [0,1] & \to S^2 \\
(z,t) & \mapsto a_n z^n + t (a_{n-1} z^{n-1} + \dots + a_1 z + a_0)
\end{align}
This satisfies $H(z,0) = a_n z^n$ and $H(z,1) = p(z)$.
The main fact to prove is that $H$ is continuous. It is clearly continuous on $\mathbb{C} \times [0,1]$ so it remains to prove that it's continuous at each point of $\{\infty\} \times [0,1]$. Note that for all $t \in [0,1]$, $H(\infty,t) = \infty$, because $a_n \neq 0$. If you think about the $\epsilon-\delta$ definition of continuity for this case, you see that for a given $t_0$, you need to prove that for all $A > 0$, there exists $B > 0$ and $\epsilon > 0$ such that
$$|z| > B \text{ and } |t - t_0| < \epsilon \implies |H(z,t)| > A.$$
But as $a_n z^n$ dominates the rest of the expression, it's easy to see that it's true. You can take $\epsilon = 1$ (i.e. impose no condition on $t$), and $B = A |a_{n-1} / a_n|$ or something like that.
Second, you want to get rid of $a_n$. Write $a_n = r e^{i\theta}$ for $r > 0$ and $\theta \in \mathbb{R}/2\pi\mathbb{Z}$. The homotopy between $z \mapsto a_n z^n$ and $z \mapsto z^n$ is given by:
$$J(z,t) = r^t e^{it\theta} z^n.$$
This satisfies $J(z,0) = z^n$ and $J(z,1) = a_n z^n$. The proof that $J$ is continuous is similar (even simpler).
