I want to show that $\pi_1(P_g)=\langle c_1,\ldots, c_{2g+1}\ |\ c_1^2\cdot\ldots\cdot c_{2g+1}^2=e\rangle$, where $P_g$ is $\mathbb{R}P^2$ with $g$ handles, using the van Kampen theorem. So far I have shown that for $g=1$ the fundamental group is $\langle a,b,c\ |\ a^2=bcb^{-1}c^{-1}\rangle$.
I divided $P_g$ into $A_1$ and $A_2$ such that $T^2$ belongs to $A_2$, $\mathbb{R}P^2$ belonds to $A_1$, and $A_1\cap A_2$ is a small neighbourhood around the intersection of $T^2$ and $\mathbb{R}P^2$ homeomorphic to an open cylinder.
Denote $i_{k}: \pi_1(A_1\cup A_2)\to \pi(A_k)$. Then $\pi_1(A_1)=\langle a \rangle$, $\pi_1(A_2)=\langle b,c \rangle$ ($b,c$ are the loops around the 2 generating circles), $\pi_1(A_1\cup A_2)=\langle z \rangle$, $i_1(z)=a^2$, $i_2(z)=bcb^{-1}c^{-1}$. Then we obtain the aforementioned result by van Kampen.
As long as the above is correct, it remains to show that by choosing a different basis we can prove that $\langle a,b,c\ |\ a^2=bcb^{-1}c^{-1}\rangle$ is isomorphic to $\langle c_1,c_2, c_3\ |\ c_1^2c_2^2c_{3}^2=e\rangle$, but I have no idea how to do that.
I would also like to find a solution using van Kampen only, without gluing arguments.