Suppose that we have the $\Delta$-complex $X$ (shown below), with $0$-simplices $p$ and $q$, $1$-simplices $a$,$b$,$c$ and $d$, and $2$-simplices $U$,$V$ and $W$.
I'd like to compute $H_1^{\Delta}(X)$, the first simplicial homology group of $X$.
We have the chain complex $\cdots \rightarrow C_2(X) = \langle U,V,W \rangle \cong \mathbb{Z}^3 \xrightarrow{\partial_2} C_1(X) = \langle a,b,c,d \rangle \cong \mathbb{Z}^4 \xrightarrow{\partial_1} C_0(X) = \langle p,q \rangle \cong \mathbb{Z}^2 \xrightarrow{\partial_0} 0$, where $C_n(X)$ is trivial for $n \geq 3$. Then, $\partial_2(U) = a+b-d$, $\partial_2(V) = a+d-c$, $\partial_2(W)=a+c-b$, $\partial_1(a) = p-p=0$, and $\partial_1(b)=\partial_1(c)=\partial_1(d)=q-p$.
Therefore, we know that $\ker(\partial_1) = \langle a, b-c,b-d \rangle$ and $\text{Im}(\partial_2) = \langle a+b-d, a+d-c, a+c-b \rangle$. But, is there then a better description for $H_1^{\Delta}(X)$ other than $H_1^{\Delta}(X) = \langle a,b-c,b-d \rangle / \langle a+b-d, a+d-c, a+c-b \rangle$ ?
We have that $H_1^{\Delta}(X) = \langle a, b-c, b-d | a+b-d, a+d-c, a+c-b \rangle$. The first relation says $a = -(b-d)$, the third relation says that $a = b-c$, and the second relation says that $a = -(b-c) + b-d$. Therefore, $a = -(b-d) = b-c = -(b-c) + b-d$.
Is it true then that $H_1^{\Delta}(X) = \langle b-c | 3(b-c) \rangle \cong \mathbb{Z}_3$?
Thanks!