Nonzero class in integral homology implies orientability

Question

Let $M^3$ be a compact, connected and oriented $3$-manifold with nonempty boundary and let $\Sigma^2$ be a compact and embedded surface such that $\Sigma \cap \partial M = \partial \Sigma$. If $\Sigma$ determines a nonzero class in $H_2(M, \partial M;\mathbb{Z})$, does it follow that $\Sigma$ is orientable?

Answer 1

Let me see if you agree with this answer. Let $\phi : (\Sigma, \partial \Sigma) \to (M, \partial M)$ be the embedding of $\Sigma$ into $M$. This induces a map $$\phi_* : H_2(\Sigma, \partial \Sigma; \mathbb{Z}) \to H_2(M, \partial M; \mathbb{Z})$$ in homology. Since $\Sigma$ determines a nonzero class in $H_2(M, \partial M; \mathbb{Z})$, this implies that there exists a nonzero class $\alpha \in H_2(\Sigma, \partial \Sigma; \mathbb{Z})$ such that $\phi_*(\alpha) \neq 0$. In particular, $H_2(\Sigma, \partial \Sigma; \mathbb{Z}) \neq 0$, which means precisely that $\Sigma$ is orientable.