Prove that every continuous map $f: S^1 \to S^1$ is homotopic to a continuous map $g: S^1 \to S^1$ with $g(1) = 1$

Question

I am asked to prove that every continuous map $f: S^1 \to S^1$ is homotopic to a continuous map $g: S^1 \to S^1$ with $g(1) = 1$. Now I thought intuitively I could somehow take each $f(x,y)$ at time $t = 0$ and just go around the circle to get to $g(x,y)$ at time $t = 1$ but apparently I need to use the function $arg(z)$ for that which is not continuous so could someone provide me with a hint on how to approach this problem. Thanks.

Answer 1

Yes, the $\arg$ function is useful here, but there is no need to use $\arg(z)$ as a part of the definition of your homotopy. Suppose that $g(1) = e^{i \theta}$ (equivalently, we could say $\theta = \arg(g(1))$). Consider the function $F: [0,1] \times S^1 \to S^1$ defined by $$ F(t,z) = e^{-i t\theta} g(z). $$ $F$ is indeed continuous.

Answer 2

Although $\text{arg}(z)$ is not continuous, nonetheless many functions which can be written using the expression $\text{arg}(z)$ can also be rewritten without it.

So in your question, suppose that $f(1+0i) = a+bi \in S^1$. Let's write $a+bi = e^{i\theta}$ for some angle $\theta \in [0,2\pi)$. Now let's define a homotopy of rotation maps, namely $h : S^1 \times [0,1] \to S^1$ $$h(z,t) = z e^{-it\theta} $$ For $t=0$ this is the identity map. For $t=1$ this is the rigid rotation of the circle which rotates through the angle $-\theta$. For intermediate values of $t$ this is the rigid rotation which rotates through angle $-t\theta$. Your desired homotopy is therefore given by $$H(z,t) = h(f(z),t) $$