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

  • $\begingroup$ 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 $\delta$? $\endgroup$ – Nikodem Dyzma May 16 at 13:31
  • $\begingroup$ yes, and for arbitrary $t_3$. I think I know the answer... $\endgroup$ – annie heart May 16 at 15:49
  • $\begingroup$ Yeah, it deformation retracts to $S^1$. $\endgroup$ – Nikodem Dyzma May 16 at 17:36
  • $\begingroup$ what is your answer? $\mathbb{Z}^3$ for H_1 and $\mathbb{Z}^2$ for H_2? $\endgroup$ – annie heart May 16 at 17:45
  • $\begingroup$ OK, let me write a proper answer. $\endgroup$ – Nikodem Dyzma May 16 at 18:28

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