On the notion of 'winding numbers' of maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$

Question

In complex analysis, the winding number (around the origin) of a continuous loop $\gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ is the number of times the loops "winds" around zero, which is given by the integral

$$\frac{1}{2 \pi i}\int_\gamma \frac{dz}{z}$$

One of the basic results of algebraic topology is that loops with the same winding numbers are homotopic.

I think it is pretty clear that any continuous map $f: \mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ should also carry a similar notion of 'winding number'. It should be defined by the integral $\frac{1}{2 \pi i} \int_\gamma \frac{dz}{z}$ (where $ \gamma: [0,1] \to \mathbb{C} \setminus \{0\}$ is given by $\gamma(t) = f(e^{2 \pi it})$).

In this scenario, does the result above still hold, i.e. that continuous maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\}$ with the same winding number are homotopic (through continuous maps $\mathbb{C} \setminus \{0\} \to \mathbb{C} \setminus \{0\})$? How can I see this? Is it possible to use the result above (the equivalent one for loops) to construct this homotopy?

Answer 1

Suppose $f : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \}$ is any continuous map such that the loop $$\gamma : S^1 \to \mathbb C \setminus \{ 0 \}, \ \ \ \gamma (t) = f(e^{2\pi it})$$ has winding number $n$.

I claim that $f$ is homotopic (via maps $\mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \}$) to the map $$ h_n : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \} , \ \ \ h_n (re^{2\pi it})=e^{2\pi int}.$$ If I can prove this, then this would be sufficient for me to conclude that any two maps $f_1, f_2 : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \}$ such that $t \mapsto f_1(e^{2\pi i t})$ and $t \mapsto f_2 (e^{2\pi i t})$ have the same winding number must be homotopic to each other.

To construct the homotopy between $f$ and $h_n$, I proceed in two steps.

First, I define the map $$ g : \mathbb C \setminus \{ 0 \} \to \mathbb C \setminus \{ 0 \} , \ \ \ g (re^{2\pi it})=f(e^{2\pi it}),$$ and note that $f$ is homotopic to $g$ via the homotopy, $$ F : \mathbb C \setminus \{ 0 \} \times [0,1] \to \mathbb C \setminus \{ 0 \} , \ \ \ F(re^{2\pi it}, s) = f(r^{1-s} e^{2\pi it}).$$

The next step is to use the fact that, since $\gamma (t) = f(e^{2\pi it})$ has winding number $n$, $\gamma$ must be homotopic to the map $$w_n : S^1 \to \mathbb C \setminus \{ 0 \}, \ \ \ w_n (e^{2\pi it}) = e^{2\pi int}.$$

Suppose that $Z : S^1 \times [0, 1] \to \mathbb C \setminus \{ 0 \}$ is a homotopy between $\gamma$ and $w_n$. Then

$$ G : \mathbb C \setminus \{ 0 \} \times [ 0, 1] \to \mathbb C \setminus \{0 \}, \ \ \ G(re^{2\pi it}, s) = Z(e^{2\pi it}, s)$$ is a homotopy between $g$ and $h_n$.

Having shown that $f$ is homotopic to $g$ and $g$ is homotopic to $h_n$, we conclude that $f$ is homotopic to $h_n$, and we're done.

Answer 2

Kenny Wong's answer is instructive because it does it "by hand", but here's a much more condensed version:

A very basic result is that the winding number of $\gamma$ only depends on the path homotopy class of $\gamma$.

Let $r:\mathbb{C}\setminus\{0\} \to S^1, z\mapsto \frac{z}{|z|}$, and $i:S^1\to \mathbb{C}\setminus\{0\}$. It is then a classical result that $r$ is a retraction by deformation onto $S^1$.

Let $f,g:\mathbb{C}\setminus\{0\}\to \mathbb{C}\setminus\{0\}$ be two maps with the same winding number. Then compare $i\circ r\circ f\circ i$ and $i\circ r\circ g\circ i$: they are two maps $S^1\to \mathbb{C}\setminus\{0\}$ and since $i\circ r \simeq id_{\mathbb{C}\setminus\{0\}}$, $i\circ r \circ f\circ i \simeq f\circ i$ and similarly with $g$, thus they have respectively the winding number of $f$ and $g$. Thus they are homotopic (using the result for $S^1$). Thus $f\circ i$ and $g\circ i$ are. Thus $f\circ i \circ r$ and $g\circ i \circ r$ are, and thus $f$ and $g$ are.