# Fundamental group of a connected graph

Is this a legit way to prove that a fundamental group of a connected graph $\Gamma$ is a free group? Without using quotient and homotopy extension property from Hatcher's "Algebraic Topology":

Take a maximal tree $T \subset \Gamma$, built on a vertex $v_0$. Take a set of edges $\{ e_\alpha | \alpha \in A\}$ s.t. $e_ \alpha \in\Gamma \setminus T$. First, suppose $A=\{1\}$. Then there is a unique reduced cycle in $\Gamma$ containing $e_\alpha$. Call it $L$. As a loop subgraph, $L \cong S^1$ (homotopy equivalent).

Now we prove that $\Gamma \cong L$: all components of $\overline{\Gamma \setminus L}$ are trees, and if any component intersects $L$ in more than one point, then it either forms a second reduced cycle in $\Gamma$ containing $e_\alpha$, or we get a reduced cycle in $T$. Both are a contradiction, so each component of $\overline{\Gamma \setminus L}$ has a unique intersection point with $L$, and since $T$ is a tree, we can deformation retract this component to its unique intersection point. So $\Gamma$ deformation retracts to $L$, and $\Gamma \cong L$.

Now for general $A$, let $m_ \alpha$ be some midpoint of $e_\alpha$. Let $\Gamma_\alpha= (\Gamma \setminus \cup_{\beta \in A} m_\beta) \cup m_\alpha$. From case 1 we have that $\Gamma_\alpha$ has a fundamental group $\mathbb Z$. And $\Gamma _ \alpha$ satisfy the conditions of Seifert-Van Kampen theorem: open, path-connected, intersection non-empty, double and triple intersections simply connected. Then $\pi _1 (\Gamma) \cong *_{\alpha \in A} \mathbb Z \cong F(e_\alpha, \alpha \in A)$ - free group on $e_\alpha$.

The relation with topology is also covered in Topology and Groupoids, as it was in the 1968 edition of this book, "Elements of Modern Topology". A graph can be regarded as a quotient of a disjoint union of unit intervals $[0,1]$ by identifying various vertices. This gives a $1$-dimensional CW-complex. Now the fundamental groupoid $\pi_1([0,1], \{0,1\})$ on the two base points $0,1$ is a groupoid which we sometimes write $\mathcal I$; a free groupoid is obtained from a disjoint union of copies of $\mathcal I$ by identifying some of the vertices. This is a special case of a groupoid construction in which given a groupoid $G$ with object set $X$ and a function $f:X \to Y$ to a set $Y$ one obtains a groupoid $U_f(G)$, which I also like to write $f_*(G)$, with object set $Y$ and a universal morphism'' $G\to U_f(G)$.
Any connected groupoid $H$ can be written as $H(a) \ast T$ where $H(a)$ is the vertex group of $H$ at the object $a$ and $T$ is a "tree groupoid'', i.e. each $T(x,y)$ has exactly one element. If $H$ is a free groupoid, then $H(a)$ is a free group.