# The fundamental group of closed orientable surface of genus 2 contains a free group on two generators

Let $$S$$ be the closed orientable surface of genus $$2$$. It is well known that its fundamental group is given by $$\pi_1(S)=\langle a,b,c,d:[a,b][c,d]=1\rangle.$$

How can we show that this group has a subgroup isomorphic to a free group on two generators?

Let $$F=\langle x,y\rangle$$ be a free group on two generators $$x,y$$. Intuitively, it seems that the map $$F\to \pi_1(S)$$ determined by $$x\mapsto a$$, $$y\mapsto b$$ is injective, but I can't see how to prove that this map has trivial kernel. Or can we use covering space theory? If there is a covering map $$S'\to S$$ such that $$\pi_1(S')\cong F$$ then we'll be done. Any hints?

Here's a proof that avoids covering space arguments, using a bit of differential topology, and using some surface topology arguments including the (smooth) Schönflies Theorem.

One can cut $$S$$ along a simple closed curve $$\gamma$$ (representing $$[a,b]=[c,d]^{-1}$$) into two compact subsurfaces $$S_0,S_1$$ with boundary, each a one-holed torus, each with fundamental group free of rank 2, so that the inclusion map $$S_0 \hookrightarrow S$$ takes a free basis of $$\pi_1(S_0)$$ to the two group elements $$a,b$$, so that the other inclusion map $$S_1 \hookrightarrow S$$ takes a free basis of $$\pi_1(S_1)$$ to $$c,d$$, and so that $$\gamma = \partial S_0 = \partial S_1$$ represents infinite order elements of $$\pi_1(S_0)$$ and of $$\pi_1(S_1)$$, namely $$[a,b]$$ and $$[c,d]^{-1}$$ respectively.

It suffices to prove that the inclusion map $$S_0 \hookrightarrow S$$ induces an injection on fundamental groups.

Since the inclusion of $$\text{interior}(S_0)$$ into $$S_0$$ is a homotopy equivalence, it suffices to prove that $$\text{interior}(S_0) \hookrightarrow S$$ induces an injection on fundamental groups. And for this, it suffices to prove that for every continuous closed curve $$f : S^1 \to \text{interior}(S_0)$$, if $$f$$ extends to a continuous function $$F : D^2 \to S$$ then $$f$$ also extends to a continuous function $$F_0 : D^2 \to \text{interior}(S)$$. By approximation methods in differential topology, we may assume that $$f$$ and $$F$$ are both smooth.

Now we apply transversality methods of differential topology. We know that $$F^{-1}(\gamma) \subset \text{interior}(D^2)$$, and so (applying perturbation methods in differential topology) we may assume that the map $$F$$ is tranverse to the curve $$\gamma$$. The subset $$F^{-1}(\gamma) \subset \text{interior}(D^2)$$ is therefore a pairwise disjoint collection of smooth, simple closed curves in the interior of $$D^2$$. Let $$n$$ be the number of those curves, denoted $$C_1,...,C_n$$.

By applying the Schönflies Theorem, each $$C_i$$ subdivides $$D^2$$ into two pieces: $$\text{inside}(C_i)$$ is diffeomorphic to $$D^2$$ with boundary $$C_i$$; and $$\text{outside}(C_i)$$ is diffeomorphic to an annulus with boundary $$C_i \cup S^1$$. Pick a collar neighborhood of $$C_i$$, parameterized as $$C_i \times [-1,+1] \to D^2$$ so that $$C_n \times \{0\} \to C_i$$ and $$C_i \times \{+1\}$$ is a simple closed curve $$C'_i \subset \text{outside}(C_i)$$.

The proof now proceeds by induction on $$n$$. The relation $$C_i \subset \text{inside}(C_j)$$ is a partial ordering on the finite set $$\{C_1,\ldots,C_n\}$$, and so there is a minimal element. Let's permute the notation so that $$C_n$$ is a minimal element, hence $$\text{inside}(C_n)$$ contains no other $$C_i$$. It follows that $$F(\text{inside}(C_n)) \subset S_j$$ for $$j=0$$ or $$1$$, and therefore $$F | C_n$$ is null homotopic in $$S_j$$. But $$F | C_n \subset \gamma$$ and $$\gamma$$ represents an infinite order element of $$\pi_1(S_j)$$, and so it follows that $$F | C_n$$ is null homotopic in the circle $$\gamma$$ itself. From this it follows that $$F \mid C'_n$$ is null homotopic in $$\text{interior}(S_k)$$ (namely in the complement of $$S_j$$, i.e. $$k=1-j$$). We can therefore redefine $$F \mid \text{inside}(C'_n)$$ to be a null homotopy of $$F \mid C'_n$$ in $$\text{interior}(S_k)$$. This redefinition operation removes $$C_n$$ from $$F^{-1}(\gamma)$$ with affecting the remainder of $$F^{-1}(\gamma)$$, thus reducing the number of components of $$F^{-1}(\gamma)$$, and completing the induction step.

It is a bit easier to instead show that $$a$$ and $$c$$ freely generate a subgroup. Consider the map $$F\to \pi_1(S)$$ sending the generators $$x$$ and $$y$$ to $$a$$ and $$c$$. To show this map is injective, you just have to construct a left inverse $$F\to\pi_1(S)$$ to it. You can do this by just mapping $$a$$ to $$x$$, $$c$$ to $$y$$, and $$b$$ and $$d$$ both to $$1$$.

You can also see this topologically: $$S^1\vee S^1$$ is a retract of $$S$$, where you take an appropriate copy of $$S^1\vee S^1$$ in $$S$$ representing the generators $$a$$ and $$c$$. For instance, embed $$S^1\vee S^1$$ in $$S$$ as a loop that traces the "figure 8" shape of the double torus. To construct a retraction of $$S$$ onto this figure 8, first collapse a simple closed curve $$\gamma$$ representing $$[a,b]=[c,d]^{-1}$$ to get a quotient of $$S$$ that is a wedge sum of two tori. We may furthermore choose this $$\gamma$$ so that it intersects our copy of $$S^1\vee S^1$$ at only the intersection point of the two circles. Then our copy of $$S^1\vee S^1$$ maps injectively to the quotient and its image is just a wedge of two circles, one in each torus. These circles are each retracts of the torus they are in (they are just one of the two "coordinate axis" circles in the torus), and so the wedge of two tori retracts onto the wedge of two circles. Composing this with the quotient map gives a retraction from $$S$$ to $$S^1\vee S^1$$.

Alternatively, you can also use similar methods for the subgroup generated by $$a$$ and $$b$$. This time, the left inverse $$F\to\pi_1(S)$$ can map $$a$$ to $$x$$, $$b$$ to $$y$$, $$c$$ to $$y$$, and $$d$$ to $$x$$. This one works topologically as well: $$S$$ retracts onto the punctured torus $$S_0$$ as in Lee Mosher's answer by just reflecting $$S_1$$ across $$\gamma$$ to map it to $$S_0$$.

• Sorry to revive this somewhat old post but what I am having trouble figuring out what definition of $[a,b]$ is being used here. The one I am most familiar with is $[a,b]=aba^{-1}b^{-1}$ in which case the left inverse in the "alternatively" case should be sending $c$ to $y$ and $d$ to $x$ as $[x,y]^{-1}=[y,x]$.
– Math
Commented Apr 22 at 13:29
• @Math: Oops, thanks, I've corrected that. Commented Apr 22 at 14:41