Computing homology of complement of an embedding Let $I=[0,1]$ and $S^3$ the $3$-sphere. Assume we have injective maps $f_1,f_2:I\to S^3$ such that $\mathrm{Im}f_1\cap\mathrm{Im}f_2=\emptyset$. I have the following problem:

Compute $H_*(S^3-\mathrm{Im}f_1\cup\mathrm{Im}f_2)$, giving exlicit generators.

I am only allowed to use chapter 2 and chapter 3 before Poincaré Duality from Hatcher's Algebraic Topology. 
I know by compactness and Hausdorffness that $f_1$ and $f_2$ are embeddings.
What I first did was considering tubular neighborhoods $N_1$ and $N_2$ of $\mathrm{Im}f_1$ and $\mathrm{Im}f_2$, respectively, which must be contractible, and then Mayer-Vietoris would show everything via decomposition $S^3=(S^3-\mathrm{Im}f_1\cup\mathrm{Im}f_2)\cup (N_1\cup N_2)$. However, it seems that stronger conditions are needed in order to consider tubular neighborhoods, so this method was wrong. 
From Alexander Duality I know what the homology groups are, but that theorem comes after Poincaré Duality, so I cannot use it. 
Question
Can I somehow justify the existence of contractible neighborhoods $N_1$ and $N_2$ even though they are not tubular? Is there a better way to approach this problem? If possible, I would like a method that also works for embeddings of other compact manifolds such as the Möbius band and $S^1$. 
Edit In section 2.B of Hatcher there is a proof for the case of an embedding $h:D^k\to S^n$. I don't know if it is possible to adapt this theorem to an arbitrary (at least finite) number of copies of $D^k$ being embedded by different embeddings, but that would definitely help. 
 A: You can indeed make use of the proof from Hatcher Section 2.B. Presumably what that result gives you is trivial reduced homology in all dimensions for $S^n - h(D^k)$.
The way you use it is not to decompose $S^3 - (\text{Im}(f_1) \cup \text{Im}(f_2))$ but instead to decompose $S^3$ itself.
Let $U_1 = S^3 - \text{Im}(f_1)$ and let $U_2 = S^3 - \text{Im}(f_2)$, and so $U_1 \cap U_2 = S^3 - (\text{Im}(f_1) \cup \text{Im}(f_2))$.
So you want to compute the homology of $U_1 \cap U_2$. You can do this using the Mayer-Vietoris sequence for the union $S^3 = U_1 \cup U_2$. 
Consider for example this portion of the sequence:
$$\underbrace{H_3(U_1)}_{\approx 0} \oplus \underbrace{H_3(U_3)}_{\approx  0} \to \underbrace{H_3(U_1 \cup U_2)}_{H_3(S^3)\approx\mathbb Z} \mapsto H_2(U_1 \cap U_2) \mapsto \underbrace{H_2(U_1)}_{\approx  0} \oplus \underbrace{H_2(U_2)}_{\approx  0}
$$
It follows that $H_2(U_1 \cap U_2) \approx \mathbb Z$. Furthermore, the proof of the Mayer Vietoris sequence is sufficiently explicit that you should be able to use it to produce an explicit 2-cycle representing the generator of $H_2(U_1 \cap U_2)$. The word "explicit" must be taken with a grain of salt, of course, because you are not being given the maps $f_1$ and $f_2$ explicitly. Really what you will do is to produce a formula for that 2-cycle that is expressed in terms of $f_1$ and $f_2$.
A similar method for $H_1(U_1 \cap U_2)$ works as well, with an even simpler outcome. It also works for $H_0(U_1 \cap U_2)$ but make sure to use reduced homology to save yourself some headaches.
