Homology of $S^n-K$ is acyclic, where $K$ is union of 2 disks Let $K=A\cup B$ be a subset of $S^n$ and assume that $A,B$ and $A\cap B$ are all homeomorphic to disks.
Show that $S^n-K$ is acyclic.
I am not sure how to approach this problem.
I somehow want to use the fact that $\tilde H_*(S^n-f(D^r))=\mathbb Z[n-1]$ for any $r$ and $f$ such that $f:D^r\rightarrow S^n$ is an embedding.
I think with
$$S^n-K\hookrightarrow S^n-A\cap B\hookrightarrow S^n,$$
one might be able to derive an exact sequence in homology
$$0\rightarrow H_n(S^n)\rightarrow H_n(S^n-K)\rightarrow H_{n-1}(S^n-A\cap B)\rightarrow 0,$$
but I do not know how to continue here.
EDIT: I think I found a way to make Mayer-Vietoris work.
Let $U=S^n-A,\,V=S^n-B$.
Then $U\cap V=S^n-A\cup B=S^n-K$ and $U\cup V=S^n-A\cap B$.
By Mayer-Vietoris we get a long exact sequence
$$...\rightarrow H_{k+1}(S^n-A\cap B)\rightarrow H_k(S^n-K)
\rightarrow H_k(S^n-A)\oplus H_k(S^n-B)\rightarrow H_k(S^n-A\cap B)\rightarrow...$$
With exactness the computation of $H_k(S^n-K)$ is straight forward, since $A,B,A\cap B$ are homeomorphic to disks.
So I guess one can have arbitrary dimensional closed balls.
But won't I have an issue in degree $n-1$ because of the direct sum?
 A: Let $A\approx B\approx A\cap B\approx D^k\subseteq S^n, k\le n$ or even more general, $A\approx D^l,B\approx D^m,A\cap B\approx D^p$, where $m,n,p$ don't need to be equal.
Actually, the most simplest argument doesn't even need MV sequence. We know that $\{*\}\simeq\Bbb{R}^n\approx S^n-\{*\}$ by stereographic projection, and $S^n-\{*\}\simeq S^n-D^j$ by deformation retraction. Then, choose a point $x_0\in A\cup B$, then $\forall x\in A\approx D^j$, it can slide through a straight line to $x_0$. And $\forall y\in B$, we can do the same thing to $B$. Now, if $f$ is a deformation retraction defined as deforming all $x\in A$ to $x_0$ and $g$ is defined as deforming all $y\in B$ to $x_0$, then we combine these $f,g$ to get a new map $h$ s.t. $h$ defomormation retracts all points in $A\cup B$ continuously onto $x_0\in A\cup B$. Because of this homotopy and path-connectedness, we can also define a deformation retraction $k:A\cup B-\{x_0\}\times[0,1]\to\partial(A\cup B)$ that sends all $(A\cup B)-\{x_0\}$ to $\partial(A\cup B)$ through a path.
So, extends $k$ to $S^n$  so that it expands the hole on $S^n-\{*\}$ to $S^n-(A\cup B)$ homotopically. Define the map $\tilde{k}:(S^n-\{x_0\})\times[0,1]\to S^n-(A\cup B)$, $\forall z\in S^n-\{x_0\}$:
$$
\tilde{k}(z,t)=
\begin{cases}
z & \text{ if }z\in S^n-\text{Int}(A\cup B)\\
k(z,t)  & \text{ if }z\in A\cup B
\end{cases}
$$
This is continuous and well-defined. When $z_1\in\partial(A\cup B)$, it get mapped to itself by branch one and also branch two because $k$ is also a deformation retraction.
Hence by the invariance of homology groups $\tilde{H}_i(S^n-(A\cup B))\cong\tilde{H}_i(\Bbb{R})=0\implies$ it's acyclic.

Also, I looked at your MV sequence and found that there is a problem: $S^n-(A\cap B)\supset S^n-(A\cup B)$, so it's actually not a MV sequence. Besides that, If you want to use MV sequence, then you have to choose two set $K_1,K_2$ s.t. $K_1\cup K_2= S^n-(A\cup B)\subset S^n-A$, which isn't easy to work out.


Case 1: Let $A\approx B\approx A\cap B\approx D^k\subseteq S^n, k\le n$.

Hence $A\cup B$ must be a proper compact locally contractible subset of $S^n$. Then, I think Alexander duality tells us that for $k\ge0$,
$$\tilde{H}_{k}(S^n-(A\cup B))=\tilde{H}^{n-k-1}(A\cup B)=0$$
The RHS is trivial, which can be shown by a MV sequence argument on cohomology if you want.

Generalize the situation above to $A\approx D^l,B\approx D^m,A\cap B\approx D^p$, where $m,n,p$ don't need to be equal. Actually, the argument seems to be plausible because I couldn't find any counterexample, but I'm not really sure about it...

We can argue using the cohomology version of MV sequence, I think: for $i\ge 0$,
$$...\to H^{i-1}(A\cap B)\to H^{i}(A\cup B;\mathbb{Z})\to H^i(A)\oplus H^i(B)\to H^i(A\cap B)\to...$$
Because $A\simeq B\simeq A\cap B\simeq\{*\}$, the sequences can be reduced to
$$0\overset{\phi}{\to} H^i(A\cup B)\overset{\varphi}{\to}0$$
which implies $\text{ker}(\varphi)=H^i(A\cup B)=\text{im}(\phi)=0$ if $i>0$. When $i=0$, $H^0(A\cup B)\cong\mathbb{Z}$ because of its connectedness. Hence the reduced cohomology groups $\tilde{H}^i(A\cup B)=0,\forall i$.
Now, we can topologize $A\cup B$ as a subspace of $\Bbb{R}^{n+1}$ so that if $x\in \text{Int}(A)$, Any open neighborhood $V\approx\Bbb{R}^l$ of $x$ contains an even smaller ball $U=B(x,\epsilon)\simeq\{*\}$ (If $x\in\partial A$, then replace $\Bbb{R}^l$ by $\Bbb{E}_+^l$, a half space) A similar argument also works for $B$.
Finally, Apply the duality to obtain the result: $\tilde{H}_k(S^n-(A\cup B))=0\implies$ it's acyclic.
