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I need to prove by induction $\pi_1(\Sigma_g)= \left\langle a_1,b_1,\dots ,a_g,b_g\mid \prod_i [a_i,b_i] \right\rangle$. For genus 1 this holds since $\pi_1(T^2)\cong \mathbb Z\times \mathbb Z$. For the general case I thought of using the connected sum formula $$\pi_1(\Sigma_g)\cong \pi_1(\Sigma_{g-1} \# T^2)\cong \pi_1 (\Sigma_{g-1}\setminus D)\amalg_\mathbb{Z}\pi_1(T^2\setminus D)$$ but I'm kind of lost now. I thin $\pi_1(T^2\setminus D)\cong \mathbb Z\ast \mathbb Z$ since its a bouquet of 2 spheres, but I don't know how to proceed...

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    $\begingroup$ A way to proceed is the following: As a first step, proof by induction that for a surface $\Sigma_{g}^{1}$ of genus $g$ with one boundary circle, and a point $x$ on the boundary, the fundamental group is a free group $\pi_1(\Sigma_{g}^{1},x) = \langle a_1, b_1, \ldots, a_{g}, b_{g} \rangle$, where the boundary circle maps to the element $\prod_{i} [a_i,b_i]$. In this case, the induction step is a direct application of the Seifert-van-Kampen theorem. $\endgroup$ – Oles Wohnzimmer May 27 '16 at 15:04
  • $\begingroup$ Do you know the polygonal representation of $\sum_g$? $\endgroup$ – Anubhav Mukherjee May 27 '16 at 15:06
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    $\begingroup$ @Anubhav.K doesn't the polygonal representation circumvents induction? $\endgroup$ – popo May 27 '16 at 15:09
  • $\begingroup$ @OlesWohnzimmer can you post this as an answer, with some details? $\endgroup$ – popo May 27 '16 at 15:13
  • $\begingroup$ @OlesWohnzimmer how do you see the boundary circle maps to the product of the commutators? $\endgroup$ – user153312 May 28 '16 at 9:11
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Let $\Sigma_{g}^{1}$ be denote a surface of genus $g$ with one boundary circle. Let $\iota_g\colon \textrm{S}^{1} \to \partial\Sigma_{g}^{1}$ be a homeomorphism on the boundary. For convenience later on, we write $\textrm{S}^{1} = [0,1]/\{0 \sim 1\}$, and we fix base points $x_g = \iota_g(0) \in \Sigma_{g}^{1}$.

I assume that the following statement is known:

$\pi_1(\Sigma_{1}^{1},x_1)$ is a free group, and if $a_1$ and $b_1$ denote the homotopy classes of the longitude and the latitude circle starting at $x_1$, one has $\pi_1(\Sigma_{1}^{1},x_1) = \langle a_1,b_1 \rangle$. In this notation, the boundary $\iota_1 \colon \textrm{S}^{1} \to \partial \Sigma_{1}^{1}$ represents the class $[a_1,b_1] = a_1 b_1 a_1^{-1} b_1^{-1}$.

One can prove this statement by constructing $\Sigma_{1}^{1}$ as a quotient of a square with a disc removed (e.g. $[-2,2]^{2} \setminus \{x \in \mathbb{R}^{2} \mid ||x||< 1\}$) with opposite edges identified.

Now one can inductively construct $\Sigma_{g}^{1}$:

Consider the intervals $\iota_1([0,\frac{1}{2}])$ , $\iota_{g}([0,\frac{1}{2}])$ in the boundaries of $\Sigma_{1}^{1}$ and $\Sigma_{g}^{1}$. Identifying these intervals yields a homeomorphism: $$ \phi_{g} \colon \Sigma_{1}^{1} \cup_{[0,\frac{1}{2}]} \Sigma_{g}^{1} \,\tilde{\to}\, \Sigma_{g+1}^{1}, \quad \phi_g(x_1) = \phi_g(x_g) = x_{g+1}$$

Now the intersection of the subspaces $\phi_{g}(\Sigma_{1}^{1})$ and $\phi_{g}(\Sigma_{g}^{1})$ in $\Sigma_{g+1}^{1}$ is an interval, hence contractible. The Seifert-van-Kampen theorem then computes $$ \pi_1(\Sigma_{g+1}^{1},x_{g+1}) \,\tilde{=}\ \pi_1(\Sigma_{1}^{1},x_1) \,\ast\, \pi_1(\Sigma_{g}^{1},x_g) \,\tilde{=} \, \langle a_1,b_1,\ldots,a_{g+1}, b_{g+1} \rangle,$$ where the indices of the generators of $\pi_1(\Sigma_{g}^{1},x_g)$ were shifted by $1$ to avoid conflicting notation. The new boundary circle is homotopic to the concatenation of the two old boundary circle, hence: $$[ \iota_{g+1} ] = [\iota_1] \cdot [\iota_{g}] = [a_1,b_1] \cdot \prod_{i=1}^{g} [ a_{i+1}, b_{i+1}] = \prod_{i=1}^{g+1} [ a_{i}, b_{i}] \in \pi_1(\Sigma_{g+1}^{1},x_{g+1})$$.

As a last step, one constructs $\Sigma_{g}$ by gluing a disk to $\Sigma_{g}^{1}$ at the boundary circle. Another application of the Seifert-van-Kampen theorem yields the desired result: $$ \pi_1(\Sigma_{g},x_g) = \langle a_1,b_1,\ldots, a_g,b_g\mid \prod_{i = }^{g} [a_i,b_i] \rangle$$

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