 Solution

Choose a spanning tree $T$ of the finite connected graph $X$. It contains every vertex of $X$ and has $v(X)-1$ edges by part (i). Collapsing $T$ to a point is a homotopy equivalence: $T$ is a contractible subcomplex, and collapsing a contractible cofibration in a CW complex preserves homotopy type.

Every edge outside $T$ then becomes a circle attached to the common point. Hence the quotient space $X/T$ is a bouquet of circles withpetals. The fundamental group of a bouquet of circles is the free group with one member of a free basis of a group for each petal. Therefore the rank of the fundamental group of a finite connected graph isMore concretely, choose a vertex as base point; follow the unique path in a graph in $T$ to one end of a non-tree edge, traverse it, and return through $T$. These based loops give the free basis of a group. Thus $r$ is the number of free group generators, equivalently the minimum size of a generating set of a group, rather than the size of every possible redundant generating set of a group. Minimality also follows by abelianization, since $F_r$ has abelianization ${\mathbb{Z}}^r$.