I'm currently self-studying some introduction to algebraic topology. In the book Topology an Invitation from K. Parthasarathy, the author presents the attached proof.
However I can't see why, in the last section, he asserts that $(1-s)z^n+sh_t(z) \neq 0$ for all 'large' $t$. From the previous limit I can deduce that for some $z \in S^1$ then given a large $t$ the equation is true, but why is it true for all $z \in S^1$?
Thank you very much.
Here is the proof:
If we prove the result for monic polynomials, we are done. Consider a polynomial $p(z) = z^n + a_1z^{n-1} + \ldots + a_0$ and suppose it has no zeros in $\mathbb{C}$.
Then we can define $h_t(z) := \frac{p(tz)}{|p(tz)|}$ to obtain a morphism $h_t : S^1 \to S^1$ for each $t > 0, t \in \mathbb{R}$. Note also that $h_0$ is a constant and $F(z, s) = h_{ts}(z), 0 \leq s \leq 1,$ gives a continuous morphism $F = F_t : S^1 \times [0, 1] \to S^1$ for each $t > 0, t \in \mathbb{R}$.
Therefore, $F(z, 0) = h_0(z)$ and $F(z, 1) = h_t(z)$. Thus, $F$ is a homotopy from $h_0$ to $h_t$. Now note that since $h_0$ is a constant, it induces the trivial homomorphism $h_* = e_*$ in $\pi_1(S^1, 1)$, and therefore, $h_t$ also induces the trivial homomorphism for all $t > 0$.
On the other hand, we will now show that for sufficiently large $t$, $h_t$ is homotopic to the morphism $z \mapsto z^n$. We know that this latter morphism induces the non-trivial homomorphism $k \mapsto nk$, as seen in a previous example, and therefore $h_t$ would also induce a non-trivial homomorphism, leading to a contradiction.
To see the homotopy to $z^n$, first observe that $\lim_{t \to \infty} h_t(z) = z^n$ for each $z \in S^1$. In particular, $(1 – s)z^n + sh_t(z) \neq 0$, $0 \leq s \leq 1$, for sufficiently large $t$. If we now define, with a sufficiently large $t$,
$$
H(z, s) := \frac{(1 – s)z^n + sh_t(z)}{|(1 – s)z^n + sh_t(z)|},
$$
then $H : S^1 \times [0, 1] \to S^1$ is continuous, $H(z, 0) = z^n$ and $H(z, 1) = h_t(z)$, and we have a homotopy $H = H_t$ from the function $z^n$ to $h_t$ for each sufficiently large $t$.
Best Answer
Indeed, pointwise convergence $h_t(z) \to z^n$ isn't enough. What would suffice is uniform convergence: for any $\epsilon>0$ there is some $t_0$ such that $|h_t(z) - z^n| < \epsilon$ for all $z \in S^1$ and $t > t_0$. (Then there exists $t$ with $|h_t(z)-z^n|<2$ for all $z \in S^1$, which implies $(1 - s)z^n + sh_t(z) \ne 0$.) We could appeal to the Arzelà –Ascoli theorem, if we can show that $h_t'(z)$ is uniformly bounded for large enough $t$. However, it's probably easiest to prove uniform convergence directly:
For $|z| = 1$ we have $$ \left| \frac{p(tz)}{(tz)^n} - 1 \right| \le t^{-1} |a_{n-1}| + \cdots + t^{-n} |a_0| $$ so that $p(tz)/(tz)^n \to 1$ uniformly as $t \to\infty$. Then $|p(tz)|/t^n = |p(tz)/(tz)^n| \to 1$ uniformly, so $$\frac{h_t(z)}{z^n} = \frac{p(tz)}{|p(tz)|z^n} = \frac{p(tz)/(tz)^n}{|p(tz)|/t^n} \to 1 $$ uniformly. Thus $h_t(z) \to z^n$ uniformly (since $z^n$ is uniformly bounded).