why $[f_r] =0$ for all $r$?

Question

From Allen Hatcher Book

Theorem $1.8$. Every nonconstant polynomial with coefficients in C has a root in $\mathbb{C}.$

In the proof of the theorem Hatcher say that as $r$ varies, $f_r$ is a homotopy of loops based at $1$. Since $f_0$ is the trivial loop, we deduce that the class $[f_r] \in \pi_1(S^1)$ is zero for all $r$ .

Im not getting that why the class $[f_r] \in \pi_1(S^1)$ is zero for all $r$ ?.

My thinking :

Here $F(s,1)= \frac{p(e^{2\pi is)}/p(1)}{|p(e^{2\pi is)}/p(1)|}=f_1=f(s)$

$F(s,0) = \frac{p(0)/p(0)}{|p(0)/p(0)|}=f_0=1$

By loop homotopy we have $[f(s)] \simeq [1]$

So for any fixed $r$ we have $[f_r(s)]\simeq[1] \implies [f_r] \in \pi_1(S^1)$ is $1$ for all $r$ .

i,e $[f_r]=1\neq 0$ for all $r$

Answer 1

You are confusing a number of things. Under the assumption that $p(z)$ has no roots in $\mathbb C$, Hatcher defines for each real number $r \ge 0$ a loop $$f_r : [0,1] \to S^1, f_r(s) = \frac{p(re^{2\pi is)}/p(r)}{|p(re^{2\pi is)}/p(r)|} .$$ To see that these are loops based at $1 \in S^1$ note that $f_r(0) = f_r(1) = \frac{p(r)/p(r)}{|p(r)/p(r)|} = 1$. Now consider the continous map $$F: [0,1] \times [0,\infty) \to S^1, F(s,r) = \frac{p(re^{2\pi is)}/p(r)}{|p(re^{2\pi is)}/p(r)|} .$$ It has the property $F(s,r) = f_r(s)$. The restriction $F \mid_{[0,1] \times [r,r']}$ describes a homotopy of loops from $f_r$ to $f_{r'}$. Thus all loops $f_r$ are homotopic as loops and therefore give us a single homotopy class of loops based at $1 \in S^1$. In other words, we get a unique element $\phi = [f_r] \in \pi_1(S^1,1)$.

Now we have $f_0(s) = 1 \in S^1$ for all $s$, i.e. $f_0$ is the constant loop at $1$. This means that $\phi = [f_0]$ is the neutral element of the group $\pi_1(S^1,1)$. But $\pi_1(S^1,1) \approx \mathbb Z$ is abelian, and in abelian groups the neutral element is commonly denoted as the zero element. Thus one writes $$\phi = 0 .$$

I think your confusion comes from the fact that $f_0$ is the constant loop based at $1 \in S^1$. Thus you may laxly write $f_0 = 1$ which seems to contradict $[f_0] = 0 \in \pi_1(S^1,1)$. But as we have seen, this is no contradiction.