Does $\partial A$ determine $A$? Given a bounded closed set $A$ in $\mathbb R^n$, can $A$ be uniquely determined by $\partial A$, except for the boundary itself?
Or, use it differently, given two bounded closed sets $A_1, A_2$ in $\mathbb R^n$ with $\partial A_1 = \partial A_2$, $A_1 \ne\partial A_1$, and $A_2 \ne\partial A_2$, is it true that $A_1 = A_2$?
Without the boundedness assumption the assertion is clearly false: the sphere $\{ x \in \mathbb R^n : |x|=1\}$ is the common boundary to $\{ x \in \mathbb R^n : |x| \ge 1\}$ and $\{ x \in \mathbb R^n : |x| \le 1\}$.
note) It was originally intended for the Euclidean compact(bounded and closed) set, but it was incorrectly modified as an open set. I am sorry.

Consider a torus $\mathbb{T}^2$ in $\mathbb{R}^3$ and a circle $S^1$ on it such that it is null homotopic in $\mathbb{T}^2$. Then you get a disc on one side of the circle and a surface with nontrivial fundamental group on the other which share the same border. So the answer to the question is negative.
