Fundamental group of graphs If $G$ is a connected graph with a maximal tree $T \subset G$ such that:
$G-T$ consists of only a single edge $e$, then how would i find the fundamental group $\pi_1(G)$ and show that it $\pi_1(G)\cong \mathbb{Z}$ 
i know that the fundamental group is defined to consist all loops based at a given fixed vertex $v \in V$ but i am struggling to show that this is isomorphic to $\mathbb{Z}$
 A: Let $v\in V$. In order to get a loop you need to use $e$ that single edge, that gives the graph a loop.
You need to show that a loop based at $v$ is precisely characterized by how many times it crosses $e$ (this means $0,1,2$ times or also $-1,-2,\cdots$ if you cross it with reverse orientation). To show this you can construct a homomorphism $\pi_1(M) \to \mathbb Z$ and show that it is surjective (you can choose loops crossing $e$ as often as you wish) and injective (i.e. if you have a loop mapping to zero it has to be trivial, or equivalently, if two loops map to the same integer, they should be the same). The last part is harder. Note that the multiplication on $\pi_1$ is defined by concatenating and then reduction of backtracking (throw out everything which causes backtracking).
The injectivity part is harder. You have to make use of the fact that (say $e=\{x,y\}$) in $T$ there are unique geodesics from $v$ to $x$ and from $v$ to $y$ and vice versa. This will give you necessary backtracking if you construct a different loop than the standard element. (with standard element, I mean that you can explicitely write down all elements in $\pi_1$ by using the geodesics in $T$. Just because a geodesic from $x$ to $y$ is unique in $T$ and vice versa. Those will give you all elements)
If you read carefully my answer you should be able to figure out the answer, with the standard graph theoretic techniques. Good luck!
