Let $M$ be the Möbius band and $\partial M$ its boundary. Consider an embedding $g:M\to S^3$. I have to determine the homomorphism $i_*:H_1(S^3-g(M))\to H_1(S^3-g(\partial M))$ induced by the inclusion.
Firs of all, I've computed both homology groups. For $H_1(S^3-g(M))$ I divided $M$ into rectangles $M_1$ and $M_2$ with intersection homeomorphic to $D^2\sqcup D^2$. Then, I can write $S^3-g(D^2\sqcup D^2)$ (of which I had already computed its homology using proposition 2B.1(a) from Hatcher and Mayer-Vietoris) as the union $(S^3-g(M_1))\cup (S^3-g(M_2))$. The intersection of these spaces is precisely $S^3-g(M)$, so I can compute its homology via Mayer-Vietoris, obtaining $H_1(S^3-g(M))=\mathbb{Z}$.
For $H_1(S^3-g(\partial M))$ I can use 2B.1(b), which tells me that $H_1(S^3-g(\partial M))=\mathbb{Z}$. Unfortunately, I cannot get explicit generators for any of this groups as they depend on $g$ in a not at all explicit way.
How can I determine $i_*:H_1(S^3-g(M))\to H_1(S^3-g(\partial M))$?
An idea that had come to my mind was using the LES for the pair $(S^3-g(\partial M),S^3-g(M))$, but I would have to compute somehow the relative homology groups of dimension $1$ and $2$, and I don't know how.
Edit I would like to solve it without using any kind of duality. Specifically, I can only use the tools from chapters 2 and 3 (before Poincaré Duality) of Hatcher.
You can work again with Alexander duality. It is known that the isomorphism $\alpha_A : \tilde{H}_i(S^n \setminus A) \to \tilde{H}^{n-i-1}(A)$ is natural with respect to inclusions, i.e. if $A \subset B \subset S^n$ and $i : A \to B$, $j : S^n \setminus B \to S^n \setminus A$ are the inclusions, then $i^* \alpha_B = \alpha_A j_*$.
Thus you have to understand what $i^*$ is for $i : \partial M \to M$. Write $M = [0,1] \times [-1,1]/\sim$, where $(0,t) \sim (1,-t)$ for $t \in [-1,1]$ and let $p : [0,1] \times [-1,1] \to M$ denote the quotient map. Then $S = p([0,1] \times \{ 0 \})$ is a copy of $S^1$ (embedded as the "middle circle" of $M$). It is easy to see that $S$ is a strong deformation retract of $M$. A strong deformation retraction $r : M \to S$ is given by $r([x,t]) = [x,0]$. Also $\partial M$ is a copy of $S^1$ and we easily see that $ri : \partial M \to S$ is a map of degree $2$. Hence $i^* : \tilde{H}^1(M) \to \tilde{H}^1(\partial M) $ is a map between infinite cyclic groups which corresponds to multiplication by $2$.