Compute the simplicial homology groups of $S^1 \vee (S^1 \times S^1)$ in all dimensions.
I'm trying to practice simplicial homology, and want to make sure I understand at a technical level what's going on before I move onto singular homology or using various convenient theorems, Mayer-Vietoris Sequences, etc.
Let $X$ be the space in question, and equip it with this standard triangularization from Hatcher, with a $1$-simplex (loop) $d$ attached at the vertex $v$, to represent $S^1$ being wedged with the torus.
Under this triangularization, I believe the only change in computing $H_{n}^\Delta (X)$ from the homology groups of the torus would be in $H_1^\Delta (X)$ since we are not adding any $0$ or $2$-simplices but are attaching a $1$-simplex, $d$.
Hence $H_0^\Delta (X) = \mathbb{Z}$ and $H_2^\Delta (X) = \mathbb Z$. Of course we also know $H_i^\Delta (X) =0, \forall i\geq 3$ since $X$ has no simplices of dimension $3$ or higher.
That leaves $H_1^\Delta (X)$, which can be found by taking the kernel of the boundary map $\partial_1: C_1 (X) \to C_0 (X)$ modulo the image of the map $\partial_2: C_2 (X) \to C_1 (X)$: $$H_1^\Delta (X) = \frac{\ker{\partial_1}}{\text{Im } \partial_2}$$
I know from computing homology of the torus that $\text{Im } \partial_1 =0$, so therefore $\ker \partial_1 = C_1 (X) = \mathbb{Z}^4$ (free abelian group generated by the $1$-simplices $a,b,c,d$).
Now for finding $\text{Im } \partial_2$. Well, by definition $C_2 (X)$ is the linear combination of the $2$-simplies $U$ and $L$, so for $p,q \in \mathbb Z$ we have \begin{align*} \partial_2 (C_2 (X)) &= \partial_2 (p \partial_2(U) + q \partial_2 (L))\\ &= p(a+b+c+d)+q(a+b+c+d)\\ &= (p+q)(a+b+c+d)\\ \end{align*} And hence $\text{Im }\partial_2$ is generated by $(a+b+c+d)$ so equals $\mathbb{Z}$.
Now to find the quotient, I saw from one calculation of the homology of the torus that with a change of base, the generators of $C_1 (X)$ can be made to be, $a,b,(a+b+c),d$ and so when we modulo $\text{Im }\partial_2$ the generator $(a+b+c)$ cancels so we get $$H_1^\Delta (X) = \frac{\ker{\partial_1}}{\text{Im } \partial_2} = \mathbb{Z}^4 / \mathbb{Z} = \mathbb{Z}^3.$$
This seems to match my knowledge that $H_1(X)\cong \pi_1 (X) = \pi_1(S^1) \oplus \pi_1 (T)=\mathbb{Z} \oplus \mathbb{Z}^2 = \mathbb{Z}^3.$
Are the details of my calculation correct?