Fundamental group of two circles joined If we consider the union of two $S^1$ joined at point, the fundamental group is a non abelian free group with two generators. Intuitively is clear but, how can I prove it formally?
Thanks in advance.
 A: As you assumably had to learn covering space theory to compute $\pi_1(S^1)$, this may be the best approach. If you can guess that the fundamental group of the wedge is $G=\mathbb{Z}*\mathbb{Z}$, then you can just observe that $G$ acts freely on the infinite tree with degree four at each node: the generator $a$ corresponds to moving right by one interval, and $b$ to moving up, if we visualize this graph embedded in the plane. The quotient by this action is $S^1\vee S^1$, and the graph is contractible, so it must be the universal cover of $S^1\vee S^1$ with deck transformations isomorphic to $\pi_1(S^1\vee S^1)$.
A: Since you want a proof that doesn't rely on the Seifert van-Kampen theorem, here is one that uses maximal spanning trees instead.
Let $\Gamma$ be the graph with a single vertex $b$ and two edges: $e_1$ and $e_2$.
That is, $\Gamma\cong S^1\vee S^1$.
Then a maximal spanning tree $T$ in $\Gamma$ consists just of the vertex $b$.
So the two edges are not in $T$, and hence $\pi_1(\Gamma,b)$ is a free group on two generators, where the $i^{\textrm{th}}$ generator is a loop going once around $e_i$.
Note that the above proof relies on knowing that the fundamental group of a connected graph is a free group, and also uses some ideas usually found in the proof of the statement.
If you haven't come across that either then let me know and I'll try to add a further explanation.
Pages 38 to 41 of these online notes might prove useful for you.
