I have a polynomial $p(z) = z^n + a_{n-1} z^{n-1} + \ldots + a_0$, and I extend it continuously from $p : \mathbb{C} \to \mathbb{C}$ to a map from $S^2 \to S^2$. $(n > 0)$.
In the plane, I have the homotopy $tp(z)$ between $p$ and the constant zero function. Why does this not extend to a homotopy on the sphere? Why does $(t -1)p(z) + tz$ not extend to a homotopy iff $n > 1$. (These results have to be true in order for the degree of the polynomial to determine the degree of the map, since degree is a homotopy invariant, but I'm not seeing why they hold independently of a degree argument.)
(I am trying to prove the fundamental theorem of algebra using homology and degree of maps. The outline of my argument is to show that $p$ is homotopic to $z^n$, whose degree I can compute locally, in particular by examining $z^n : S^1 \to S^1$ and using the algebraically or geometrically evident fact that there is an nth root of unity. The degree will be n, since locally each map is a dilation + rotation – which is homotopic to the identity. Since the degree is thus nonzero, the map must be surjective.)
(This is mostly inspired by Hatcher problem 2.2.8)
Best Answer
For each $t$, the function $f_t(z) = (t-1)p(z)+tz$ does determine a map $\mathbf P^1 \to \mathbf P^1$, but the resulting map $I \times \mathbf P^1 \to \mathbf P^1$ is not continuous.
As $t$ approaches $1$, the zeroes of $f_t(z)$ are unbounded; indeed, these zeroes are also the zeroes of $p(z)+\frac{t}{t-1}z$. Assuming WLOG that $p(z)$ is monic, the (negative of) the sum of the zeroes is the coefficient of $z$, namely $p'(0) + \frac{t}{t-1}$, which $\to \infty$ as $t\to 1$. Therefore, there is a sequence $(t_n, z_n)$ such that $t_n \to 1$, $z_n \to \infty$, and $f_{t_n}(z_n)=0$ for each $n$. But in the limit, $f_1(\infty) = \infty$; this limit would be $0$ if $(t, z)\mapsto f_t(z)$ were continuous.