# Loops, homotopy, and Z/2Z

I'm in Complex Analysis 2, and I would like to relate loops on the complex plane to $\mathbb{Z}$.

Def. Let $\Omega$ be an open subset of $\mathbb{C}$. A loop is a path $\alpha: [0,1]\rightarrow \Omega$ such that $\alpha(0)=\alpha(1)$.

Def. The composition of paths $\alpha, \beta \in C(\Omega; a,b)$, $t\in[a,b]$ as

$$(\alpha\cdot \beta)[t] = \left\{ \begin{array}{ll} \alpha(2t) & \quad 0\leq t \leq 1/2 \\ \beta(2t-1) & \quad 1/2 \leq t\leq 1 \end{array} \right.$$

Def. Two paths $\gamma_0, \gamma_1$ are homotopic in $C(\Omega; a,b)$, where a and b are fixed endpoints if there exits a continuous map $H: [0,1]\times[0,1]\rightarrow \Omega$ such that $H(t,0)=\gamma_0(t), H(t,1)=\gamma_1(t)$ for $0\leq t\leq 1$ and $H(0,s)=a, H(1,s)=b$ for $0\leq s\leq 1$.

We may show that loops are a homotopic equivalence relations (ie. $\gamma_0\thicksim \gamma_0$; if $\gamma_0\thicksim \gamma_1\Rightarrow \gamma_1\thicksim \gamma_0$; and if $\gamma_0\thicksim \gamma_1$ and $\gamma_1\thicksim \gamma_2 \Rightarrow \gamma_0\thicksim \gamma_2)$ and create a class with them (ie. $[\alpha]:=\{\beta| \alpha\thicksim \beta\}$).

Prop. The composition of paths and the inversion of paths are compatible with homotopies. The composition thus defined on the equivalence classes of paths is associative.

If we have an equivalence class, we can satisfy associativity, along with closure, inverses, and existence of an identity to form a group.

I don't know how to relate this information to analytic continuation of a function like $f(z)=z$ for $z\in \mathbb{C}$ around the origin. I'm told it will get me $\mathbb{Z}$, but I don't know why or how. Similarly for $f(z)=z^{1/2}$ for $z\in \mathbb{C}$, except it supposedly gets me $\mathbb{Z/2Z}$.

Can anyone help?

• It would be related to the concept of winding number, which is the net number of times a loop goes around the origin, taking counterclockwise as positive and clockwise as negative. All the loops which go around the origin the same net number of times are in the same homotopy class, so the fundamental group is isomorphic to the additive group of the integers. – George Law May 7 '16 at 16:42
• Ok, I totally understand now :) Thanks. – User5634 May 7 '16 at 16:47