fundamental group of a sphere with 2 handles

Question

Prove that the fundamental group of a sphere with 2 handles contains a free group of 2 generators.

Intuitively, it is clear that we can cut a figure homotopically equivalent to a bouquet of two spheres with a plane, but I do not know how to formalize this and whether it follows that the free group of 2 generators is contained in the fundamental group of the pretzel.

Answer 1

I assume you meant "bouquet of two circles". Even if, the space is not homotopy equivalent to it.

The way we are going to do this is by the following:

Lemma. Let $X$ be a space and $A\subseteq X$ its retract. Then $\pi_1(A)$ embeds into $\pi_1(X)$.

Proof. Let $r:X\to A$ be retraction and $i:A\to X$ inclusion. Then by functoriality of $\pi_1$ we have $$id=\pi_1(id)=\pi_1(r\circ i)=\pi_1(r)\circ\pi_1(i)$$

and therefore $\pi_1(r)$ is surjective while $\pi_1(i)$ is injective. $\Box$

So it is enough to find a retract of our space with $\mathbb{F}_2$ as fundamental group. The obvious choice is bouquet of two circles, a.k.a. the eight shape. But we have to be careful. Not every eight shape on this surface will be its retract.

Have a look at this (horrible yet useful) drawing:

which is a 2d section of our space. The first step of constructing our retraction is by projecting entire upper half onto the lower. The blue arrows indicate that.

So we endup with half sphere with two half handles. The second step is to get rid of handles like this:

So we shrink/deform handles into the surface, and we end up with half sphere with two holes in it. If we flatten it then we get this:

Finally we can draw our eight shape around those holes:

with blue arrows indicating retraction. The rest follows from our lemma.

This of course is not truely formal solution, but putting concrete formulas is a tedious job, which I will leave to you. :)