This is an old qual question.
Let $K$ be a simplicial complex homeomorphic to the $3$-sphere. Let $L$ be a subcomplex homeomorphic to a manifold with nonempty boundary. Show that $H_1(L,\mathbb Z)$ has no torsion.
I am totally lost as to how to go about this question. Here is what I have tried/know:
First, the torsion in $H_1(L,\mathbb Z)$ is the same as the torsion in $H^2(L,\mathbb Z)$.
Second, since $L$ is a manifold, we have poincare duality $H_1(L,\mathbb Z) \cong H^2(L,\partial L)$ and $H^2(L,\mathbb Z) \cong H_1(L,\partial L)$.
Third, we have a long exact sequence for the pair $(L,\partial L)$.
However, I am not sure how to put this information together or how to use that it has a CW structure/substructre...
I would want to run Mayer-Vietoris on $L$ and $L' = \overline{S^3 \setminus L}$, which have $L \cap L' = \partial L$, some oriented surface. I prefer taking reduced homology.
Then we see $0 \to H_1(\partial L) \to H_1(L) \oplus H_1(L') \to H_1(S^3) \to H_2(\partial L)$ is exact. In particular, since $H_1(S^3) = 0$, $H_1(\partial L) \to H_1(L) \oplus H_1(L')$ is an isomorphism, and since $H_1(L)$ is a summand of a free abelian group, it must itself be free abelian.
To incorporate Steve D's comment, $L$ is a nice (compact, locally contractible) subset of $S^3$, so Alexander duality dictates that $H_k(L) \cong H^{3 - k - 1}(S^3 \setminus L)$. Because $S^3 \setminus L$ is homotopy equivalent to $L'$, this says $H_1(L) \cong H^1(L')$. (Note that this says more than the Mayer-Vietoris argument above!) Now the universal coefficient sequence dictates that $H^1(L') \cong \text{Hom}(H_1 L', \Bbb Z)$ - there is no Ext term because $H_0$ is always free. In particular, this term $\text{Hom}(H_1 L', \Bbb Z)$ is torsion free; we know it is a finitely generated free abelian group because $L'$ is compact.