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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 $\pi_1(T_1)=$ and $\pi_1(T_2)=.$ Further the normal subgroup $N$ is generated by $[a_1,b_1]\,[a_2,b_2]=1.$

Thus van Kampen implies that $$\pi_1(\Sigma_2)=\frac{\pi_1(T_1) * \pi_1(T_2)}N=.$$

But the literature says that $\pi_1(\Sigma_2)=.$

Where did I go wrong?

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    Since $a_1$ commutes with $b_1$ and $a_2$ commutes with $b_2$ then obviously always $[a_1, b_1][a_2, b_2]=1$. I.e. these two groups are isomorphic since the last relation is a consequence of previous 2. – 2017-02-02
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    @freakish this is clear, but how can we show that $[a_1,b_1][a_2,b_2]=1$ implies that $a_1$ commutes with $b_1$ – 2017-02-02
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    No, the implication is other way around. $[a_1, b_1]=1$ and $[a_2, b_2]=1$ so $[a_1, b_1][a_2,b_2]=1$. – 2017-02-02
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    Sorry, I really don't know much about algebra, so i need to ask these dumb questions... @freakish: If we only have the implication in one direction, then why can we conclude more than that one group is a subgroup of the other? – 2017-02-02
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    Oh, wait, there's no `,` in the other group. My mistake. But the group should be $\pi_1(\sum_2)=$, shouldn't it? Isn't that the one the literature says? That's unlikely that the one you've written is correct. – 2017-02-02

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The used partition of the Torus is wrong.

We can think of $\Sigma_g$ as the union of two toruses $T_1, T_2$ which intersect in a circle $S^1$.

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 $\pi_1(T_1\setminus D^2)=$, because $T\setminus D^2$ deformation retracts to the wedge sum of the two boundary circles $a_1$ and $b_1$.

Then $$\pi_1(\Sigma_2)=\frac{\pi_1(T_1\setminus D^2) * \pi_1(T_2\setminus D^2)}N=.$$