showing the fundamental group of a graph $G$ with spanning tree $T\subset{G}$ s.t $G - T = \{e\}$

Question

Let $G$ be a path-connected graph with $T\subset{G}$ a spanning tree of $G$ s.t. $G\setminus{T} = \{e\}$ where $e$ is some edge of $G$. I want to show that $\Pi_1(G)\cong{\mathbb{Z}}.$

my attempt.

I've shown that G must contain 1 closed loop, label it $L\subset{G}$. I've also shown that G can be retracted to $L$. So there is a deformation retract from a path in $G$ to a path in $L$. Using the fact that L is homeomorphic to $S^1$, we can say (using the fact that the fundamental group of $S^1$ is isomorphic to $\mathbb{Z}$).

$\Pi_1(G)\cong{}\Pi_1(S^1)\cong{\mathbb{Z}}.$

This seems incomplete or incorrect. Any guidance would be appreciated.

Answer 1

Your proof looks good.

You might want to explicitly mention the fact that if $A$ is a deformation retract of $B$, then $\pi_1(B) \cong \pi_1(A)$, but other than that, you're good.

As a fun aside, this result can be generalized: for any spanning tree $T$ of a graph $G$, $\pi_1(G)$ is the free group generated by $G\setminus T$.