I know that the integral homology group of the manifold $M$ is given by $$ H_j(M,\mathbb{Z}) $$
I also have tried that $H_j(T^3,\mathbb{Z})$ is given by $$ H_0(T^3,\mathbb{Z})=\mathbb{Z}, $$ $$ H_1(T^3,\mathbb{Z})=\mathbb{Z}^3, $$ $$ H_2(T^3,\mathbb{Z})=\mathbb{Z}^3, $$ $$ H_3(T^3,\mathbb{Z})=\mathbb{Z}, $$
Could we or could you suggest how to derive $H_j(T^3 - D^2 \times S^1,\mathbb{Z})=?$ In particular $j=0,1,2,3$. Mostly: $$ H_1(T^3- D^2 \times S^1,\mathbb{Z})=?, $$ $$ H_2(T^3- D^2 \times S^1,\mathbb{Z})=?, $$ where we choose the $T^3=S^1_x \times S^1_y \times S^1_z$ of three circles. And we choose that $ D^2 \times S^1= D^2 \times S^1_z$ where its circle is along the same circle as $S^1_z$ of $T^3$, while the $D^2$ is a tiny 2-disk.
So by $T^3- D^2 \times S^1$, we make a small tubular neighborhood cut along a chosen $S^1_z$ circle, cut out from the $T^3$.
Thank you. <3