$3$-manifold whose boundary is composed of spheres and has nontrivial homology

Question

I would like to construct an example of a compact and orientable $3$-manifold $M^3$ with boundary such that its boundary is a union of $2$-spheres and $H_2(M, \partial M) \neq 0$. My guesses would be to consider $S^2 \times [0,1]$ and $\mathbb{R}P^3 \setminus B$, where $B$ is a small open ball, but I’m having trouble computing $H_2(M, \partial M)$ for these spaces.

Answer 1

Since $M$ is compact and orientable, Lefschetz Duality says that $H_2(M,\partial M) \cong H^1(M)$, so it's equivalent to find an example where $H^1(M) \neq 0$. Consider $M = T^3 \setminus B^3$: then by the Künneth theorem $H^1(T^3) \cong \mathbb{Z}^3$, and $M$ and $T^3$ have the same $2$-skeleton so it follows that $H^1(M)\cong \mathbb{Z}^3$ as well.

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Incidentally you can compute that $H_2(M, \partial M) = 0$ for your examples using the long exact sequence of a pair (and in particular without using cohomology). Suppose $M$ is a compact orientable $3$-manifold such that $\partial M \cong \sqcup^k S^2$. Then the long exact sequence of a pair gives us $$\dots \to H_2(\partial M) \to H_2(M) \to H_2(M,\partial M) \to H_1(\partial M)\cong 0$$

By exactness it follows that $H_2(M,\partial M)\cong 0$ if and only if $H_2(\partial M) \to H_2(M)$ is surjective.

For $M = \mathbb{R}P^3 \setminus B \simeq \mathbb{R}P^2$ we have $H_2(M) \cong 0$ so this space will not work. For $M = S^2 \times I$ the inclusion of the boundary is certainly surjective on $H_2$ so this space will also not work.