The task:
Take integers $p,q\ge1$ and define $S^{p-1}\subset S^{p+q-1}\subset\Bbb R^{p+q}$ to be the subsphere consisting of points of $S^{p+q-1}$ whose last $q$ coordinates are zero, and define $S^{q-1}\subset S^{p+q-1}$ to be the subsphere consisting of points whose first $p$ coordinates are zero.
Show that $S^{p+q-1}\setminus S^{p-1}\cong S^{q-1}\times\Bbb R^p$ and that it strongly deformation retracts onto $S^{q-1}$. Conclude, using the previous exercise, that:
$S^{p-1}$ and $S^{q-1}$ are not the boundaries of any pair of disjointly embedded disks $D^p$ and $D^q$ in $D^{p+q}$
The preceding exercise:
Say $n\ge0$ and $(D,S)\subseteq(D^n,S^{n-1})$ has: (i) $D\cap S^{n-1}=S$ and (ii) $(D,S)\cong(D^k,S^{k-1})$ for some $k\ge0$.
Show that the inclusion $S^{n-1}\setminus S\hookrightarrow D^n\setminus D$ induces an isomorphism on homology.
This is a fun problem with very strong intuitive justification. I was able to solve it... under a suitable interpretation of "boundaries". Specifically, if $h:D^p\to D^{p+q}$ and $h':D^q\to D^{p+q}$ are the disjoint embeddings in question I assumed that $S^{p-1}=h(S^{p-1})=h(D^p)\cap S^{p+q-1}$ and I assumed that $S^{q-1}\subseteq h'(D^q)$. Note the asymmetry here; I could get away with making weaker assumptions on $h'$ but seemingly not with $h$.
However, it would be reasonable to interpret Hatcher as just saying, $h(S^{p-1})=S^{p-1}$ (two different meanings of $S^{p-1}$, but what can you do?) and similarly for $h'$. These are weaker assumptions and my proof strategy does not seem to work in this case. However, the preceding exercise does not seem to apply in this case either... so what's going on? I am not sure what the most general correct form of the statement is, and if one can weaken the hypotheses successfully I would appreciate a proof or proof sketch of how to achieve that.
I have faith the stronger form of the theorem holds because, playing with examples in $D^2$ and $D^3$, it's "obvious" - in the naive, visual sense - that the strong version is correct. But visual intuition only gets you so far.
Below I write my solution and discussion for my interpretation of the exercise.
The deformation retraction thing is easy. It follows that $S^{q-1}\hookrightarrow S^{p+q-1}\setminus S^{p-1}$ induces an isomorphism on (reduced) homology. It also follows from the preceding exercise that $S^{p+q-1}\setminus S^{p-1}\hookrightarrow D^{p+q}\setminus h(D^p)$ induces an isomorphism on (reduced) homology, so the overall composite $S^{q-1}\hookrightarrow D^{p+q}\setminus h(D^p)$ induces an isomorphism on (reduced) homology.
However, we can factor this inclusion as: $$S^{q-1}\hookrightarrow h'(D^q)\hookrightarrow D^{p+q}\setminus h(D^p)$$Using the fact that $h'$ and $h$ have disjoint image (observe we can get away with $h(D^p)$ being a subset of $h'(D^q)$ which avoids $S^{q-1}$). Since the middle term has trivial reduced homology in all degrees we conclude (from this being an isomorphism and $S^{q-1}$ not having trivial reduced homology in all degrees) that something is wrong with our premises and no such $h,h'$ can exist.
This crucially relies on $S^{p+q-1}\setminus S^{p-1}\hookrightarrow D^{p+q}\setminus h(D^p)$ inducing an isomorphism on homology, which is horribly false if we relax the assumption $S^{p-1}=h(D^p)\cap S^{p+q-1}=h(S^{p-1})$ to the assumption $h(S^{p-1})=S^{p-1}$. There is a partial fix to this, namely that I am confident we can amend the previous exercise to get that: $S^{p+q-1}\setminus(h(D^p)\cap S^{p+q-1})\hookrightarrow D^{p+q}\setminus h(D^p)$ is an isomorphism on homology (Hatcher doesn't state this, though I think my proof can be adapted). From this perspective, we can then work with the slightly weakened hypotheses:
$h,h'$ are embeddings and $S^{q-1}\subseteq h'(D^q)$ and $S^{p-1}=h(D^p)\cap S^{p+q-1}$ (i.e. we do not care what $h(\partial D^p)$ is)
But working with the reasonable alternative interpretation of what Hatcher meant, we'd have an issue in that $S^{p+q-1}\setminus(h(D^p)\cap S^{p+q-1})$ doesn't, in general, deformation retract onto $S^{q-1}$!