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The fundamental group $\Pi_1(G)$ of a connected graph $G$ is defined to consist of all loops (i.e., closed paths) based at a given fixed basepoint/vertex $g \in G$ as elements, and concatenation as the group operation. (for $G$ connected, we can show that the group is independent of the choice of basepoint; given g,g' we construct an isomorphism between the two groups by joining g,g' with a path.) Also given a connected graph $G$, there is is a result that $G$ has a spanning tree $T$ -- a tree $T\lt G$ that uses all vertices in $G$ ).

Now, the main result (all else was a set-up until now) is that $\Pi_1(G) \sim \mathrm{Gp}({E(T-G)})$, i.e., $\Pi_1(G)$ is the free group generated by all edges in $G$ that are not in $T$.

My doubt is about this isomorphism; I do see how we need edges in $T-G$ in order to generate loops based at a point, since $T$ is acyclic (so that if $G=T$, the group is trivial). But I do not see how the group is isomorphic to the free group generated by edges in $T-G$. This is what I have so far: let $f_i$ denote a free edge, i.e., an edge in $T-G$. We can assume there is just one path joining any two edges in $T$ (otherwise, $T$ would contain a cycle) Given $g$ fixed in $G$ (assume WOLG $g$ in $T$), I think every closed loop based at $g$ can be realized as the concatenation of a path within $T$, and an edge $f_i$ (aka, a path in $G-T$), and then a(the) path from an endpoint of $f_i$ to $g$ again. Then, for any edge $f_i$, there is a (class) of loops based at $g$. Question: is the above the sketch of a proof for the fact that $\Pi_1(G)$ is the free group generated by all the edges in $G-T$?

Sorry, I did not know how to make the question any more concise; I hope I don't get any Courics for writing too-long of a question.

Thanks.

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    Note that your tree $T$ is contractible, which --- when contracted --- produces a bouquet (to see why, study how many vertices there will be)...and that's how this "spanning tree" approach to $\pi_{1}(G)$ relates to the "good, old-fashioned homotopy classes of loops" approach Qiaochu notes below.2012-10-06

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Your description of the fundamental group is not correct, and in fact does not describe a group (as it stands, concatenation is not associative and no non-identity element has an inverse). You need to consider homotopy classes of loops.

The proof I know of the result you want proceeds by showing that you can contract the graph along each edge of the spanning tree to get a wedge of circles, one for each edge not in the spanning tree. Once you know that the fundamental group is a homotopy invariant, and once you know how to describe the fundamental group of a wedge of circles (e.g. using Seifert-van Kampen), the conclusion follows.

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The argument given in Example 1.22 in Allen Hatcher's "Algebraic Topology", which computes the fundamental group of a particular planar graph, can be generalized to any connected graphs and used to answer your question.....