# Homology of solid torus with k points removed

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!)

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.
• "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 $n\neq0$ Commented Mar 11, 2021 at 9:21