I have been asked to conjecture a relationship between the Euler characteristic and the fundamental group of finite connected graphs and, if possible, to prove the conjecture.
We know that if the finite connected graph $G$ is a tree, then its Euler characteristic is $\chi(G)=1$ and its fundamental group is trivial $\pi_1(G)=\{e\}$.
If $G$ is the triangulation of an $n$-bouquet, $\textit{i.e.}$ a union of $n$ circles all intersecting at one point, then we may observe that its Euler characteristic is given by $\chi(G)=1-n$, and its fundamental group is known to be $\pi_1(G) \cong \mathbb{Z}^n$. This gives the relationship $\pi_1(G) \cong \mathbb{Z}^{1-\chi(G)}$ in the case of an $n$-bouquet.
But, of course, not every finite connected graph falls neatly into one of the above two categories, and, in general, the fundamental group is not free. I suppose one might consider every finite connected graph as a union of graphs in one of the above two categories, and hence arrive at a general statement, but I cannot seem to see how one would go about this. I also feel that this must be a well-known result, but none of the resources I have consulted have offered any insight.
All help or input would be highly appreciated.
Edit in response to comments: I stated above that the fundamental group of the $n$-bouquet is isomorphic to $\mathbb{Z}^n$. This is incorrect. In fact the fundamental group of the $n$-bouquet is the free group on $n$ generators, here denoted by $F(n)$. I also stated that in general the fundamental group of a finite connected graph is not free (in the erroneous belief that loops might introduce relations). This is incorrect, as was pointed out in a comment, and indeed, as follows from the proof given in the answer below.
Any connected graph is homotopy equivalent to wedge of $S^1$'s Thus let $G \simeq \lor _{1\leq i\leq n} S^1$. Euler characteristic is defined in terms of homology groups which are homotopy invariant. Fundamental group is also homotopy invariant. Thus it suffices to solve the problem in this case. Now $H_m(\lor _{1\leq i\leq n} S^1)= 0 $ if $m>1$ Also $H_1(\lor _{1\leq i \leq n} S^1) = \mathbb Z^n $ and the zeroth homology group is $\mathbb Z$ .so $\chi(G)= 1-n $ . Thus u get $\pi_1(G)= ●_{i=1} ^{1-\chi(G)} \mathbb Z$ where $●$ denotes free product of groups.