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I'm trying to find the $\mathbb{Z}$-homology of a solid torus $S^1 \times D^2$ with $k$ points deleted from its interior. I think the $k = 1$ case would generalize without much difficulty, but I can't really visualize what should happen even in this case. Based on the analogous question for the hollow torus $S^1 \times S^1$, I should try deformation retracting to some more familiar space which is probably a wedge sum of circles and spheres. With one point $p$ deleted we can take an open ball $D^3$ centered at $p$ and retract it to its boundary $S^2$, creating a "hole" in the solid torus? I'm not sure how to visualize any further.

(If there's a way of doing this problem cleanly with Mayer-Vietoris/induction I'd be interested in seeing that too!)

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Use Mayer-Vietoris. Let $X=S_1\times D_2$ and $U$ by $X$ with $k$ interior points removed. Then define $V$ to be the union of $k$ small balls centred at the removed points. You can apply MV to $U$, $V$ and $U\cup V=X$. The homology of $V$ vanishes except in dimension zero where it is $\Bbb Z$. As $V$ is homotopy equivalent to the circle, then its $H_0$ and $H_1$ are $\Bbb Z$ and higher homology is zero. Finally $U\cap V$ is homotopy equivalent to $k$ disjoint $S^2$s, so $H_0$ and $H_2$ of $U\cap V$ are both $\Bbb Z^k$ and the other homology groups vanish.

The upshot is that $U$ has $H_0$ and $H_1$ being $\Bbb Z$ and $H_2$ being $\Bbb Z^k$ etc.

Another way to see this is to note that $U$ is homotopy equivalent to a "pearl necklace" of $k$ $S^2$s, that is a cycle of $S^2$s with the north pole of each attached to the south pole of the next.

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  • $\begingroup$ "As $V$ is homotopy equivalent...." I think you meant $V$ with the $k$ points removed? Because in the previous sentence you literally said $H_{n}(V)=0$ for any $n\neq0$ $\endgroup$ Commented Mar 11, 2021 at 9:21

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