Here is my main goal: I am trying to show that $\partial M$ is a disjoint union of $2$-spheres where $M$ a connected compact 3-dimensional orientable manifold with boundary satisfying $H_1(M;\mathbb{Z})=0$.
In the following let it be understood that coefficients are from $\mathbb{Z}$. Also let $i:\partial M\to M$ denote the inclusion.
Here is my progress so far, which to me seems like a promising direction: If the boundary is empty, let's say that the boundary is an empty union of 2-spheres, so assume that $\partial M\neq\varnothing$. Under these assumptions, we know that $\partial M$ is a closed orientable manifold of dimension $3-1=2$ (one dimension lower than $M$). Hence, by the classification of closed orientable surfaces, we know that $\partial M\cong\bigsqcup M_{g_\alpha}$ where $M_{g_\alpha}$ denotes the closed orientable surface of genus $g_\alpha\in\mathbb{N}$. It remains to show that $g_\alpha=0$ for all $\alpha$. My strategy was to show that $$H_1(\partial M)=0\text{ }(*)$$ because I have previously calculated the homology of the surface $M_g$ so by additivity of disjoint unions we have $H_1(\partial M)\cong\mathbb{Z}^K$ where $K:=2\sum_\alpha g_\alpha$. The problem have thus far been translated into the problem of showing $\mathbb{Z}^K=0$, from which it follows $g_\alpha=0$ for all $\alpha$. It seems like something which should not be too difficult, yet I have not successfully been able to show it. The strategy that I have tried is to show that $$\mathbb{Z}^K=H_1(\partial M)\stackrel{i_*}{\longrightarrow}H_1( M)=0$$ (which appears in the long exact sequence of the pair $(M,\partial M)$) is injective, which can only be the case if $K=0$.
Consider the following part of the long exact sequence in cohomology of the pair $(M,\partial M)$ $$ H^1(M)\stackrel{i^*}{\longrightarrow}H^1(\partial M)\stackrel{\delta}{\longrightarrow}H^2(M,\partial M) $$ where $H^2(M,\partial M)\cong H_1(M)=0$ as a consequense of Lefschetz duality, so $i^*$ is in fact surjective. Here is a specific question of (homological) algebraic nature, which is something I do not have a strong background in
Since the first homology groups of both $M$ and $\partial M$ are free, can we from the naturality of the universal coefficient theorem deduce that $i_*$ is injective because the vanishing of the Ext-term yields a natural isomorphism of the cohomology group and its dual?
I hope the question makes sense. I would also love to know if that approach is completely hopeless or not, and in case it is, I would appreciate it if someone could tell if there is a better way to go.
Continue from the long exact sequence of cohomology, $$\ldots\to H^1(M)\overset{i^*}{\to} H^1(\partial M)\to 0$$ By UCT, we also have $$H^1(M;\Bbb Z)\cong \operatorname{Hom}(H_1(M;\Bbb Z),\Bbb Z)\oplus \operatorname{Ext}(H_0(M;\Bbb Z);\Bbb Z)\cong\operatorname{Hom}(H_1(M;\Bbb Z);\Bbb Z)\cong 0$$ The second isomorphism is due to the fact that $H_0$ is free, and the third isomorphism is because $H_1(M)$ vanishes (assumption). So, the exact sequence above is actually $$0\overset{i^*}{\to}H^1(\partial M)\to0$$ which implies that $H^1(\partial M)\cong 0$. Apply UCT again, $\operatorname{Hom}(H_1(\partial M;\Bbb Z),\Bbb Z)\cong\Bbb Z^{b_1}\cong 0$, where $b_1$ is the betti number in dimension $1$, and it is immediate that $b_1=0$, which should proves your claim. (I'm not sure why you're thinking about showing the injectivity of $i_*$.)