algebraic-topologyfundamental-groupssurfaces
Let $n$ be a natural number. We denote by $F_n$ an orientable, compact & connected surface of genus $n$.
I now have to show that $\pi_1(F_4)$ contains $\pi_1(F_{10})$ as a subgroup of index $3$. However, I am not supposed to calculate the fundamental groups explicitly, but I should construct a properly discontinuous group action on $F_{10}$. But I have trouble finding such a suitable action.
Can someone give the group action or refer me to a work where this is shown? Thank you!
If $M$ is a connected closed (i.e. compact without boundary) 3-dimensional manifold, then $\pi_1(M)$ cannot be isomorphic to $\pi_1(S)$, where $S$ is an orientable surface of genus $\ge 2$. You can see this by first noting that $\pi_2(M)=0$ (otherwise, by the sphere theorem, $M$ is a nontrivial connected sum which will imply that $\pi_1(M)$ is a nontrivial free product, which is not the case). Then you observe that $\pi_k(M)=0, k\ge 3$ since the universal cover $\tilde M$ of $M$ is noncompact (here you use that $S$ has positive genus, which implies that the fundamental group is infinite): Thus, $H_k(M)=0$ for all $k\ge 2$. Now, use Hurewicz theorem to conclude that $\pi_k(M)=0$ for all $k\ge 3$.
Thus, $M$ would be homotopy equivalent to $S$ (they have isomorphic fundamental groups and contractible universal covers; now, use Whitehead's theorem). However, $H_3(M)\ne 0$ while $H_3(S)=0$. Contradiction.
What one may ask is about topological classification of compact orientable 3-dimensional manifolds such that $\pi_1(M)\cong \pi_1(S)$. You can still have manifolds of the form: $S\times I$ minus some disjoint open balls. However, such manifolds which are reducible, meaning that there exists a tame 2-sphere in $M$ which does not bound a balls.
It is a theorem (I think, due to Stallings), that if, $M$ is assumed to be oriented, $S$ is oriented, and $M$ is irreducible then, it indeed is homeomorphic to $S\times I$. You can find a proof of this in Hempel's book "3-manifolds". The key is that in this situation there exists a proper homotopy-equivalence $h: M\to S\times I$, i.e., a homotopy equivalence which is a homeomorphism on the boundary. One then proves that such $h$ is homotopic (rel. boundary) to a homeomorphism.
If one allows for connected boundary of $M$, or for nonorientable $M$ or $S$, then the conclusion is that $M$ is homeomorphic to an $I$-bundle over $S$.
Edit. Here is a construction of an open 3-manifold which is homotopy-equivalent to a closed surface $S$ but is not homeomorphic to an interval bundle over $S$. Start with an open contractible 3-dimensional manifold $W$ which is not homeomorphic to $R^3$, say, the Whitehead manifold. Embed (properly and smoothly) a ray $\rho$ into $W$ and then remove a small open tubular neighborhood of $\rho$ from $W$. The result is a contractible manifold with boundary $X$; $\partial X$ is homeomorphic to the open disk $D^2$. Now, take $M=S\times I$, pick an disk $D\subset \partial M$ and glue $M$ to $X$ identifying $\partial X$ and $D$ homeomorphically. Lastly, remove the remaining boundary from the result of gluing. You obtain an open 3-dimensional manifold $N$ which is homotopy-equivalent to $S$ but not homeomorphic to an open interval bundle over $S$. The reason is that $N$ is not "tame". Proving non-tameness of $N$ requires some work to prove (by appealing to its "fundamental group at infinity"); the proof is basically the same as the one showing that $W$ is not tame.
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:
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 $z_1\cdots z_b$.
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).
Best Answer
If we use the following to represent $F_{10}$, then it is clear that it admits a $3$-fold rotational symmetry, so $\Bbb Z_3$ acts on $F_{10}$ properly discontinuously, where the action is generated by rotation of angle $2\pi/3$. The quotient space $F_{10}/\Bbb Z_3$ is homeomorphic to $F_4$, and $q:F_{10}\to F_{10}/\Bbb Z_3$ is a $3$-fold covering. It follows from covering space theory that $\pi_1(F_{10})$ is a subgroup of $\pi_1(F_4)$ of index $3$.
In fact the following statement holds:
$\Rightarrow:$ If $p:F_k\to F_l$ is an $n$-sheeted covering, then $2-2k=\chi(F_k)=n\chi(F_l)=n(2-2l)$. Rearranging, we get $k=n(l-1)+1$.
$\Leftarrow:$ We can again use the "windmill" and exploit the $n$-fold rotational symmetry of $F_k$.