In order to solve the fundamental group of $\mathbb{R}^{2}$ minus $n$ distinct points i stumbled across the following problem :
I'd have liked to create the following homopoty equivalence between $S^{1}$ and just $\mathbb{R}^{2}$ minus a point $w$ and then iterate the process :
$$\mathbb{R}^{2}-w \overset{r}{\longmapsto} S^{1} \overset{i}{\longmapsto} \mathbb{R}^{2}-w$$
Such that $r(v) =\frac{v-w}{\|v-w\|_{2}}$ and $i(v) =\frac{v-w}{\|v-w\|_{2}} +w$ for any $v \in \mathbb{R}^{2}-w$.
And then define $H : (\mathbb{R}^{2}-w) \times [0,1] \longmapsto (\mathbb{R}^{2}-w)$ such that $H(v,t) = (1-t)(\frac{v-w}{\|v-w\|_{2}} + w)+tv$.
The problem however is that the map is not well defined since there are values of $t$ for which $(1-t)(\frac{v-w}{\|v-w\|_{2}} + w)+tv = w$.
What am I missing here ? Another approach I tried is : given the $n$ points define $n$ disjoint balls around them and try something as suggested here but I was unable to show that $\mathbb{R}^{2}$ minus these points retracts onto the boundary of these balls.
The problem has been asked here as well without direct proof, I think the more accurate proof and similar to my idea is this. However, instead of simply understand the solution given I'd like to understand the problem in mine in order to comprehend what is wrong, which is why opened the question.
You did a little mistake in the definition of $H(v,t)$.
Consider the stadard retraction from $\mathbb R^2 \ \backslash \{(0,0)\}$ to $S^1$ given by: $$ F:\mathbb R^2 \ \backslash \{(0,0)\}\times [0,1] \rightarrow S^1\subset \mathbb R^2 \ \backslash \{(0,0)\}, \quad F(v,t) = \frac{v}{||v||}(1-t) + tv $$
The map you are looking for is $$H:\mathbb R^2 \ \backslash \{w\}\times [0,1] \rightarrow \mathbb R^2 \ \backslash \{w\}, \quad H(v,t) = \tau_w \circ F_t \circ \tau_{-w}(v)$$ where $\tau_w$ is the traslation by the vector $w$: $\tau_w(v) = v+w$. The idea is to translate the problem in the origin, resolve it and come back.
If you expand the formula you obtain: $$ H(v,t) = \frac{v-w}{||v-w||}(1-t) + (v-w)t + w $$