Calculating $H_i(T^3,G)$ where $G= S^1 \times \{0,\frac{1}{2}\} \times \{0,\frac{1}{2}\}$.
Let $C_1 = S^1 \times \{0,\} \times \{0\}$
$C_2 = S^1 \times \{\frac{1}{2}\} \times \{\frac{1}{2}\}$
Then $G = C_1 \cup C_2 = S^1 \times \{0,\frac{1}{2}\} \times \{0,\frac{1}{2}\}$.
For the pair $(T,G)$, we have the following L.E.S.:
$0 \rightarrow H_3(T^3) \rightarrow^a H_3(T,G) \rightarrow^b H_2(G) \rightarrow^c H_2(T^3) \rightarrow^d H_2(T^3,G) \rightarrow^e H_1(G) \rightarrow^f H_1(T^3) \rightarrow^g H_1(T^3,G) \rightarrow^h H_0(G) \rightarrow^i H_0(T^3) \rightarrow^j H_0(T^3,G) \rightarrow 0$
According to Hatcher example 2.39 we have the following:
$H_3(T^3) = \mathbb{Z}$
$H_2(T^3) = \mathbb{Z}^3$
$H_1(T^3) = \mathbb{Z}^3$
$H_0(T^3) = \mathbb{Z}$
Also, all the zeroth homolgies are $\mathbb{Z}$ because each space is path connected. I also believe that $H_1(G) = \mathbb{Z}^4$ because it has the homology of four disjoint circles I believe.
Also, $H_2(G) = H_3(G) = 0$. Making these substitutions into our L.E.S. we get the following:
$0 \rightarrow \mathbb{Z} \rightarrow^a H_3(T,G) \rightarrow^b 0 \rightarrow^c \mathbb{Z}^3 \rightarrow^d H_2(T^3,G) \rightarrow^e \mathbb{Z}^4 \rightarrow^f \mathbb{Z}^3 \rightarrow^g H_1(T^3,G) \rightarrow^h \mathbb{Z} \rightarrow^i \mathbb{Z} \rightarrow^j \mathbb{Z} \rightarrow 0$
It is thus immediate that $H_3(T^3,G) \cong \mathbb{Z}$
Also, the image of all four generators in $H_1(G)$ are sent to the same generator of $H_1(T^3)$ by $f$, so $im(f)=\mathbb{Z}$ and $ker(f)=\mathbb{Z}^3=im(e)$
Since $d$ is injective, we have $ker(e) = im(d) = \mathbb{Z}^3$.
Since we are working with free ablian groups, the sequence splits and we have $H_2(T^3,G) = ker(e) \oplus im(e) = \mathbb{Z}^6$
I'm not sure if this is correct, and also I need assitance calculating $H_1(T^3,G)$ Thanks in advance!!