I got confused when calculating the fundamental group of $\Sigma_2$.
We can think of $\Sigma_g$ as the union of two toruses $T_1, T_2$ which intersect in a circle $S^1$.
Using the fundamental polygon, one gets that $\pi_1(T_1)=<a_1,b_1|[a_1,b_1]>$ and $\pi_1(T_2)=<a_2,b_2|[a_2,b_2]>.$ Further the normal subgroup $N$ is generated by $[a_1,b_1]\,[a_2,b_2]=1.$
Thus van Kampen implies that $$\pi_1(\Sigma_2)=\frac{\pi_1(T_1) * \pi_1(T_2)}N=<a_1,b_1,a_2,b_2|\,[a_1,b_1],[a_2,b_2],[a_1b_1]\,[a_2,b_2]>.$$
But the literature says that $\pi_1(\Sigma_2)=<a_1,b_1,a_2,b_2|\,[a_1b_1]\,[a_2,b_2]>.$
Where did I go wrong?
Best Answer
The used partition of the Torus is wrong.
This would give a double Torus united with the two disks which $S^1$ bounds. The proper way to do it is to remove these disks first, and then glue the tori together.
The fundametal group $\pi_1(T_1\setminus D^2)=<a_1,b_1>$, because $T\setminus D^2$ deformation retracts to the wedge sum of the two boundary circles $a_1$ and $b_1$.
Then $$\pi_1(\Sigma_2)=\frac{\pi_1(T_1\setminus D^2) * \pi_1(T_2\setminus D^2)}N=<a_1,b_1,a_2,b_2|\,[a_1b_1]\,[a_2,b_2]>.$$