About triviality of a path in Residue Theorem Reading the Residue theorem in the following form :

Theorem : Let $D \subseteq \mathbb{C}$ open, $f: D - S \longmapsto \mathbb{C}$ holomorphic, $S$ closed and discret in $D$.
Let $R \subseteq D$ compact with $C^{1}$ boundary. $R \cap S = \left\lbrace z_{1},\cdots, z_{k}\right\rbrace$ is finite and $\partial R \cap S = \varnothing$
Then we have $\int_{\partial R}f(z)dz = 2\pi i \sum\limits_{1 \leq i \leq k} Res(f,z_{i})$.

I've stumble across the following problem :
(Adding a picture to clarity)

Let $\beta = \gamma \ast l_{k} \ast \bar{\alpha_{k}} \ast \bar{l_{k}} \ast l_{k-1} \ast \cdots \ast l_{1} \ast \alpha_{1} \ast \bar{l_{1}}$
Where for example $\bar{\alpha_{k}}$ denote the inverse path of $\alpha_{k}$, i.e $\alpha_{k}(1-t)$, and each $\alpha_{i}$ denotes a little circumference around $z_{i}$ traveled counterclockwise.
What I don't get is why $\beta$ is homotopically trivial in $R - \left\lbrace z_{1},\cdots, z_{k}\right\rbrace$. I have the same problem with the proof of Laurent expansion of homolorphic function on Annulus.
I think the reasoning are the same, which should be that $\beta$ runs through a disk, so it's hotopically trivial in $R - \left\lbrace z_{1},\cdots, z_{k}\right\rbrace$.
There is a topological way to see this fact ? I would like to avoid proofs that require any stronger  characterization.
The assert to prove is very clear from the picture I attached, but it doesn't seem trivial or obvious to me prooving it.
 A: First we want to prove this lemma:

Lemma: Consider two concentric circumferences in the plane centered in the origin parametrized by $A(t)=ae^{2\pi it}$ and $B(t)=be^{2\pi i (1-t)}$ with $a,b\in \mathbb R^+$, $a>b$ and $t\in [0,1]$. Consider the segment $l(t)=bt + (1-t)a$ with $t\in [0,1]$. Then $\gamma(t) = A(t) * l(t) * B(t) * l(1-t)$ is homotopy trivial in $\mathbb R^2  \ \backslash \  {(0,0)}$.

Proof: The idea is to rotate around the origin. At each moment $s$ you don't want all the Annulus, but just a piece of it. For, define the following path:
\begin{gather}
A_s(t)=ae^{2\pi i [(1-s)t+s]}\\
B_s(t)=be^{2\pi i [(1-s)(1-t)+s]}\\
l_s(t)=tae^{2\pi s} + (1-t)be^{2\pi i s}
\end{gather}
They are respectively the parametrization of a part of the largest circumference (from the point $ae^{2\pi i s}$ to $a$), the parametrization of a part of the smallest circumference (from the point $b$ to $be^{2\pi i s}$) and the segment that joins $be^{2\pi s}$ and $ae^{2\pi s}$.
Consider now the path $\gamma_s = A_s(t) * l(t) * B_s(t) * l_s(t)$. and define the map:
$$
F:I^2\rightarrow \mathbb R^2 \quad F(t,s) = \gamma_s(t)
$$
This is an homotopy between $\gamma_0 = \gamma$ in our hypotesis, and the path $\gamma_1(t)$ that is:
$$
\gamma_1(t) = ae^{2\pi i} * l(t) * be^{2\pi i} * l(1-t)
$$
and this is homotopy trivial since $ae^{2\pi i}$ and $be^{2\pi i}$ are constant and $l(1-t)$ is the inverse of $l(t)$.

Observe that you can choose $\gamma$ as a bunch of $k$ path where the path $\eta_i$ goes only around the point $z_i$ and it is a simple curve (without intersection). This is clearly possible because the points $z_i$ are a discrete set. 
Take now the space that $\eta_i$ bounds and its boundary $\eta_i$: this portion of space is omeomorphic to a circle minus a point. Using the lemma and returning back thanks to the omeomorphism you have the statement. 
