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It is well known fact about fundamental group of orientable compact surface:

Letting $g$ be the genus and $b$ the number of boundary components of surface $M$. There is a generating set $S=\{\alpha_1, \beta_1, ... , \alpha_g, \beta_g, x_1, ..., z_b\}$ for $\pi_1(M)$ such that $$ \pi_1(M, *) = \langle a_1, b_1, ... , a_g, b_g, x_1, ... , x_b \mid [a_1, b_1]\cdots [a_g, b_g]= x_1\cdots x_b \rangle . $$

but I don't know where find a book with proof of it. I was looking in classical positions as: Hatcher, Massey, May, Greenberg without success. The best what I found is calculation of fundamental group of surface without boundary, but boundary is important for me. Could anybody help me?

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OK, my bad, Fulton's Algebraic topology: A First Course only deals with the closed case. I'll suppose that you know this case quite well.

Let's do the bounded case by hand.

First case: one boundary component

Keep in mind the classical decomposition of the closed surface $F_{g,0}$ of genus $g$ : you have 1 vertex, $2g$ edges, and that $2$-cell whose boundary gives the complicated $[a_1,b_1]\cdots[a_g,b_g] = 1$ relation.

Now, take a needle, and pierce a hole in the middle of the 2-cell. You get $F_{g,0} \setminus \textrm{a point}$. Deformation retract the pierced 2-cell on its boundary: that creates a movie whose opening scene is this pierced surface, and whose closing scene is the $1$-skeleton, which is a wedge of $2g$ circles (the $a_i$'s and the $b_i$'s). What happens in the middle of the movie? Well, you have a surface with a disc-shaped hole which expands with time. Topologically, it's exactly the surface $F_{g,1}$ of genus $g$ with 1 boundary component.

So we have learned two things:

  • Piercing a surface (i.e. taking a point out) or making a true hole in it (i.e. take an open disc out) gives the same result up to homotopy equivalence [that's quite irrelevant for our discussion, but it's good to know nevertheless. Of course it works for many other spaces: they only have to be locally not too complicated].
  • A pierced surface has the homotopy type of a graph. This is quite important for the study of surfaces. In particular, it gives the wanted presentation: $$ \pi_1(F_{g,1}) = \left\langle a_1, \ldots, a_g, b_1, \ldots, b_g\right\rangle.$$

Of course, because the boundary of the surface is associated to the word $[a_1, b_1]\ldots[a_g, b_g]$, you can choose to write this group $$ \pi_1(F_{g,1}) = \left\langle a_1, \ldots, a_g, b_1, \ldots, b_g,x \middle| x = [a_1,b_1]\cdots [a_g, b_g]\right\rangle$$ but this quite obfuscates the fact that this group is free.

Second case: the sphere with holes

Take now $F_{0,b+1}$, the sphere with $b+1 > 0$ boundary components. You can see it as the disc with $b$ boundary components. This amounts to choosing one of the boundary components and declaring it the "outer" one. It's quite easy to retract that on a wedge of $b$ circles, so that $$\pi_1(F_{0, b}) = \left\langle z_1, z_2, \ldots, z_{b}\right\rangle.$$ In this presentation, the (carefully oriented) bouter boundary component is simply the product $z_1\cdots z_b$.

enter image description here

The general case $F_{g,b}$

You can write the surface $F_{g,b}$ of genus $g$ as the union of $F_{g,1}$ and $F_{0,b+1}$, gluing the boundary of the former with the outer boundary of the latter. Since we have computed the fundamental groups of the two pieces and that we know the expression of the gluing curve in both of them ($[a_1, b_1]\cdots[a_g,b_g]$ and $z_1\cdots z_b$, respectively), the Van Kampen theorem gives us the answer $$\pi_1(F_{g,b}) = \frac{\left\langle a_1, \ldots, a_g, b_1, \ldots, b_g\right\rangle * \left\langle z_1, \ldots, z_b \right\rangle}{\langle\langle [a_1, b_1]\cdots[a_g,b_g] \cdot (z_1\cdots z_b)^{-1}\rangle\rangle} = \left\langle a_1, \ldots, a_g, b_1, \ldots, b_g, z_1, \ldots, z_b \middle| [a_1, b_1]\cdots[a_g,b_g] =z_1\cdots z_b \right\rangle.$$

It is probably worth noting that you can rewrite the relation so that it expresses $z_b$ (say) as a word in the other generators. You can then eliminate it and notice that this is also a free group (again, as long as $b > 0$, $F_{g,b}$ deformation retracts to a graph).

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  • $\begingroup$ ok, i didn't find it in this book. Of course I found proof/calculation of fundamental group for surface without boundary. Are you sure you found this? $\endgroup$ – Filip Parker Aug 27 '14 at 14:52
  • $\begingroup$ You are absolutely right. Hope the edit helps. $\endgroup$ – PseudoNeo Aug 28 '14 at 13:40

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