# Integral homology group of a 3-torus cut out a donut

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

• I don't understand what are you actually cutting out. Is it the set $\{(e^{it_1},e^{it_2},e^{it_3}) \in T^3\; |\; t_1^2 + t^2_2 < \delta \}$ for some small $\delta$? – Nikodem Dyzma May 16 at 13:31
• yes, and for arbitrary $t_3$. I think I know the answer... – annie heart May 16 at 15:49
• Yeah, it deformation retracts to $S^1$. – Nikodem Dyzma May 16 at 17:36
• what is your answer? $\mathbb{Z}^3$ for H_1 and $\mathbb{Z}^2$ for H_2? – annie heart May 16 at 17:45
• OK, let me write a proper answer. – Nikodem Dyzma May 16 at 18:28