I have a dumb question about homologous cycles in a directed graph, viewed as a $1$-dimensional CW complex.
Let $G$ be the graph with vertex set $V=\{a,b,c,d\}$ and edges $E=\{(a,a), (a,b), (b,c), (b,d), (d,d), (d,c), (c,b), (c,a) \}$, which are oriented, i.e. the edge $(b,d)$ starts at $b$ and ends at $d$. We denote the paths in $G$ as sequences of edges.
To construct the homology group, we have to look at the boundary maps $\partial_i$ from the $i$-simplex to the $i-1$-simplex. Since there are no $2$-simplices, the first homology group is $H_1(G,\mathbb{Z})=ker \partial_1$, which should be $\mathbb{Z}^5$ with generators being the $1$-cycles $(a,a)$, $(d,d)$, $(c,a)(a,b)(b,c)$, $(c,b)(b,c)$, $(d,c)(c,b)(b,d)$.
Let $\gamma_1=(d,c)(c,b)(b,c)(c,a)(a,a)(a,b)(b,d)$ and $\gamma_2=(d,c)(c,a)(a,a)(a,b)(b,c)(c,b)(b,d)$, that is, $\gamma_1$ and $\gamma_2$ are closed paths passing through the same edges but in different orders. Both of them are elements of $H_1(G,\mathbb{Z})$ as they belong to $ker \partial_1$.
My question is: do $\gamma_1$ and $\gamma_2$ belong to the same homology class?
My answer would be yes. Intuitively because "they surround the same holes". Moreover, it looks to me that they can be written as the same linear combination of the generators. Is this correct?
Also, if my intuition is correct, can this be generalized? That is, given two oriented closed paths passing through the same edges but in different order, do they always belong to the same homology class?
Remember that the chain group $C_1(G)$ is abelian, in particular the cellular chain group is identified with the free abelian group $\mathbb{Z} \{ edges\}$. The chains represented by your paths $\gamma_1$ and $\gamma_2$ are actually equal in $C_1(G)$ since they are comprised of the same $1$-chains with the same coefficients.