Let $\mathbb{C}_3$ denote the configuration space of 3 distinct ordered points in $\mathbb{C}$. This space carries a natural free action of the affine group $A := Aff(\mathbb{C})$, which consists of maps $$\mathbb{C}\rightarrow\mathbb{C} \quad\text{of the form}\;\;z\mapsto az+b\quad\text{for $a,b\in\mathbb{C}$, $a\ne 0$}$$ Let $F\subset\mathbb{C}_3$ be the orbit of $(0,1,2)$ under $A$, then we have a sequence of maps: $$F\hookrightarrow \mathbb{C}_3\rightarrow\mathbb{C}_3/A$$
Is this a fiber bundle? (how can we see this?)
In any case, the map $\mathbb{C}_3\rightarrow\mathbb{C}_3/A$ admits a section $\mathbb{C}_3/A\rightarrow \mathbb{C}_3$ given by sending $A\cdot (a,b,c)\in\mathbb{C}_3/A$ to $(0,1,\lambda)\in\mathbb{C}_3$ where $\lambda = (b-a)c+a\in\mathbb{C}$ is image of $c$ under the unique transformation $f\in A$ sending $(a,b)\mapsto (0,1)$.
The fiber $F$ is a principal homogeneous space under $A$, and hence is $\mathbb{C}\times\mathbb{C}^\times$, and has fundamental group $\mathbb{Z}$.
Also, we have $\pi_1(\mathbb{C}_3) = PBr_3$ (the pure braid group on 3 strands).
Let $\Gamma := \pi_1(\mathbb{C}/A)$, then assuming the sequence $F\hookrightarrow \mathbb{C}_3\rightarrow\mathbb{C}_3/A$ is a fibration, we get an exact sequence $$1\rightarrow \mathbb{Z}\rightarrow PBr_3\rightarrow \Gamma\rightarrow 1$$ which is right split, and hence we find $PBr_3\cong \mathbb{Z}\rtimes\Gamma$.
Is this semi-direct product actually a direct product? (How can we compute the action of $\Gamma$ on $\mathbb{Z} = \pi_1(A)$?)