Homology of a graph. Let $\Gamma$ be a graph with $V$ vertices and $E$ edges. If we orient the edges, we can form the incidence matrix of the graph.
This is a $V\times E$ matrix whose $(i j)$ entry is $+1$ if the edge $e_j$ starts at $v_i$, $-1$ if the edge ends at $v_i$, and $0$ otherwise.
Let $C_0$ be the free $R$-module on the vertices, $C_1$ the free $R$-module on the edges, $C_n=0$ if $n\ne 0,1$, and $d: C_1\rightarrow C_0$ be the incidence matrix.
If $\Gamma$ is connected, show that $H_0(C)$ and $H_1(C)$ are free $R$-modules of dimensions $1$ and $V-E-1$ respectively.
I am given as a hint to choose as a basis $\{ v_0 , v_1-v_0,\ldots , v_V-v_0 \}$ for $C_0$ and use a path from $v_0$ to $v_t$ to find an elemenet of $C_1$ mapping to $v_i-v_0$.
I don't see how to use the hint, can someone please enlighten me? thanks in advance.

Edit (not by OP): The exact source is Exercise 1.1.6. of Charles A. Weibel's An Introduction to Homological Algebra.
 A: Let $M$ be the matrix of the graph, $t:E\rightarrow V$ be the target map and $s:E\rightarrow V$ be the source map, so that if $e$ is an edge linking $v_i$ to $v_j$ then $t(e)=v_j$, $s(e)=v_i$ and $Me=t(e)-s(e)$.
If $e$ is any edge from $v_i$ to $v_j$, interpret $-e$ as the edge $e$ with the reverse orientation. That is the same edge but from $v_j$ to $v_i$. Hence $t(-e)=s(e)$ and $s(-e)=t(e)$ so that $M(-e)=-Me$ as expected.
Let $e_1,\dots,e_t$ be a path between two vertices $v_i$ and $v_j$. By this I mean, $e_k$ are either an edge $e\in E$ or its opposite $-e\in E$, and that 


*

*$s(e_1)=v_i$ (the path start at $v_i$)

*$t(e_k)=s(e_{k+1})$ (the edges $e_k$ and $e_{k+1}$ are consecutive)

*$t(e_t)=v_j$ (the path ends at $v_j$)
Now compute 
$$M(\sum_{k=1}^t e_k)=\sum_{k=1}^t (t(e_k)-s(e_k))=t(e_t)-s(e_0)=v_j-v_i$$
Hence, you see that elements of the form $v_j-v_i$ are in the image of $M$ if there is a path between the vertices $v_j$ and $v_i$. If $\Gamma$ is path-connected, every element of the form $v_i-v_0$ are in the image. Since $(v_0,v_1-v_0,\dots,v_{V-1}-v_0)$ is a basis, you see that $v_0$ generate the homology group $H_0=C_0/\operatorname{im} M$.
To see that $H_0$ is freely generated by $v_0$, note that the degree map $\deg:C_0\rightarrow \mathbb{Z}, \sum n_i v_i\mapsto \sum n_i$ satisfies $\deg\circ M=0$ hence define a morphisme $H_0\rightarrow \mathbb{Z}$. Since $v_0$ is of degree 1, it is an isomorphism. We have proven that $H_0$ is free of rank 1.
Finally, for $H_1=\ker M$, note that it is free as a subgroup of free abelian group. To compute its rank, just use the rank theorem : 
$$E=\operatorname{rk} C_1=\operatorname{rk}\ker M+\operatorname{rk}\operatorname{im}M=\operatorname{rk}H_1+\operatorname{rk}C_0-\operatorname{rk}H_0=\operatorname{rk}H_1+V-1$$
Hence $\operatorname{rk}H_1=E-V+1$.
