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Let $M^3$ be a closed, connected and orientable smooth $3$-manifold and let $\mathring{M}$ denote the manifold $M$ with $n$ disjoint open balls $B_1, \dots, B_n$ removed. I am trying to compute $H_2(\mathring{M})$ in terms of $H_2(M)$, with the assumption that $H_2(M)$ is abelian free (everything with integer coefficients).

What I've done so far: by Mayer-Vietoris with $U = B_1' \cup \cdots \cup B_n'$ a collection of slightly larger balls ($B_i \subset B_i'$) and $V = M \setminus (B_1 \cup \cdots B_n)$, we have a long exact sequence in homology

$$0 \to H_3(M) \cong \mathbb{Z} \stackrel{j}{\to} H_2(U \cap V) \cong \mathbb{Z}^n \stackrel{f}{\to} H_2(\mathring{M}) \stackrel{g}{\to} H_2(M) \to 0$$

By the comment made in this previous post, it follows that $H_2(\mathring{M}) \cong H_2(M) \oplus \operatorname{coker}(j)$. Can we compute $\operatorname{coker}(j)$ in this case?

My guess is that $\operatorname{Im}(j)$ is generated by the element $(1, \dots, 1) \in \mathbb{Z}^n$, but I am not sure of this.

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  • $\begingroup$ Your guess follows from an explicit description of these groups and their boundary maps. $H_3(M)$ is generated by the fundamental class, a sum of oriented simplices. $H_2(U \cap V) \simeq \sqcup_n S^2$ is generated by the $n$ fundamental classes of these distinct spheres. And the boundary map is obtained by breaking up a chain on $M$ as the sum of a chain in $U$ and a chain in $V$, taking the boundary, and seeing what class in $U \cap V$ gives that boundary. . $\endgroup$
    – mme
    Commented Dec 8, 2020 at 19:08
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    $\begingroup$ Doing this explicitly (break it up into the fundamental chain for $M \setminus B$ and the fundamental chain for $B$) sends the fundamental class of $M$ to the sum of the fundamental classes of the spheres. $\endgroup$
    – mme
    Commented Dec 8, 2020 at 19:08
  • $\begingroup$ @MikeMiller Yes, this is exactly what I have to prove. Is it obvious? $\endgroup$
    – Eduardo Longa
    Commented Dec 8, 2020 at 19:11
  • $\begingroup$ I am not sure what confusion remains. Are you worried about orientations? Ignoring orientations it should be clear that $\partial(M \setminus B)$ is precisely the union of spheres, and the simplices appearing in the boundary operator are precisely the simplices in the triangulation of that sphere. $\endgroup$
    – mme
    Commented Dec 8, 2020 at 19:28
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    $\begingroup$ It is perhaps not helpful to delete all the balls at once. If you delete a single ball, you find that $H_2(M \setminus B) = H_2(M)$; the cost of deleting $B$ is that the third homology dies. After that, every new ball we delete adds to second homology, and in fact, if $X$ is a noncompact $n$-manifold and $D \subset X$ is an n-disc, then $X \setminus D^\circ \simeq X \vee S^{n-1}$. I do not really want to write down the proof, though. To see this, when you are in the process of contracting $X$ to a complex one dimension lower, you should delete the last n-cell you contract. $\endgroup$
    – mme
    Commented Dec 8, 2020 at 19:30

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