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.
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.