Can the fundamental group of an intersection of a homeomorphic image of a ball with a complement of a ball in $R^3$ be perfect? I have the following problem: Let $A, B\subset R^3$, $A$ is homeomorphic to a ball, while $B$ is a standard Euclidean ball. Can it happen that the fundamental group of $A\setminus B$ is a perfect group? I am interested in answers for $A$ and $B$ both closed and open, so in fact this is 4 questions. 
I am aware of disturbing examples like the infinite grope or the complement of the Alexander's horned sphere, but I still strongly believe that the answer should be no.
 A: Consider a smooth properly embedded  surface $P\subset \mathbb{R}^3$. Then $\mathbb{R}^3= X\cup Y$, where $X\cap Y=P$ and $X, Y$ are properly embedded submanifolds with $\partial X=\partial Y=P$. By Mayer-Vietoris, we have an exact sequence $0=H_2(\mathbb{R}^3)\to H_1(P)\to H_1(X)\oplus H_1(Y)\to H_1(\mathbb{R}^3)=0$, so we see that $H_1(X)\oplus H_1(Y)\cong H_1(P) \neq 0$ unless $P$ is a union of smoothly properly embedded planes and spheres. Therefore at least one component of $\mathbb{R}^3\backslash P$ does not have perfect fundamental group, or else $P$ is a union of planes and spheres (since $P$ is smoothly properly embedded, there's a nice collar neighborhood, so $H_i(X)\cong H_i(int(X))$, and same for $Y$). In the case that $P$ is a union of properly embedded planes and 2-spheres, by Seifert-Van Kampen's theorem, each $\pi_1(X,x)$ and $\pi_1(Y,y)$, for $x\in X, y\in Y$ injects into $\pi_1(\mathbb{R}^3)$, so is trivial. 
Let's apply this to your situation. I'll consider the case of $int(A)\backslash B$, since there's not issue of local connectivity for an open set. Then $P=\partial B \cap int(A)\subset int(A)\cong \mathbb{R}^3$ is a properly embedded smooth surface, so we have either there is a component of $int(A)\backslash B$ which does not have perfect fundamental group, or $P$ is a union of properly embedded planes in $int(A)$ (or a sphere), in which case each complementary region has trivial fundamental group. I think this answers at least one interpretation of your question.  
A: Let me deal with the following very special case: the closure of $B$ is contained in the internal part of $A$, and $A$ is bounded. In this case, using that $\partial A$ is compact one can show that there exists a standard open ball $B'$ which has the same center as $B$ but a strictly larger radius, and is such that the closure of $B'$ is contained in the internal part of $A$. Then, one may compute the fundamental group of $A$ by applying Van Kampen Theorem to the open covering $B'$, $A\setminus B$.  Since $A$ and $(A\setminus B)\cap B'=B'\setminus B$ are simply connected, we easily get that $A\setminus B$ is simply connected too.
