Two continuous functions taking circles into circles are not in general homotopic

Question

Assume $f:S^1 \rightarrow S^1$, $g:S^1 \rightarrow S^1$ are two continuous functions. Then define the function $F: S^1 \times [0,1] \rightarrow S^1$ as $$F(z, t) = e^{i((1-t)arg(f(z)) + (t)arg(g(z)))}$$ where $arg(z)$ is defined to be between $0$ and $2\pi$. Then at what point does this fail to be continuous in general? I ask this because according to this post:What are two continuous maps from $S^1$ to $S^1$ which are not homotopic? there are continuous functions from $S^1$ into $S^1$ which are not homotopic.

Note: I have just started learning algebraic topology and so I do not know anything about homology or fundamental group or anything advanced so please keep the explanation as simple as possible, thanks.

Answer 1

So what properties $arg: S^1\to\mathbb{R}$ has to satisfy in order for $F$ to be a homotopy? The following:

1. it is continuous
2. for any $z\in S^1$ we have $e^{iarg(f(z))}=f(z)$ due to homotopy condition $F(z,0)=f(z)$
3. for any $z\in S^1$ we have $e^{iarg(g(z))}=g(z)$ due to $F(z,1)=g(z)$

If one of $f,g$ functions is "onto" (note that any continuous function $S^1\to S^1$ that is not "onto" is nullhomotopic, and so this is a reasonable assumption) then this simply reduces to

$$e^{iarg(z)}=z$$

In other words, your $arg(z)$ is actually the complex argument. And that function is not continuous on $S^1$.