Is the outer boundary of a connected compact subset of $\mathbb{R}^2$ an image of $S^{1}$? A connected compact subset $C$ of $\mathbb{R}^2$ is contained in some closed ball $B$. Denote by $E$ the unique connected component of $\mathbb{R}^2-C$ which contains $\mathbb{R}^2-B$. The outer boundary $\partial C$ of $C$ is defined to be the boundary of $E$.
Is $\partial C$ always a (continuous) image of $S^{1}$?
 A: Even in two dimensions this is false.
Let $C$ be the closed topologist's sine curve.  It is compact and connected.  It is also not hard to see that the outer boundary $\partial C$ is again $C$ itself.  But $C$ is not the continuous image of $S^1$, since it is not path connected.
For a path connected example,  consider the "extended closed topologist's sine curve" $C'$ which is the union of $C$ with the segment $[0,1] \times \{1\}$.  $C'$ is compact, path connected, and $\partial C' = C'$.  Suppose to the contrary that $f : S^1 \to C$ is continuous and surjective.  But $C'$ is not the image of $S^1$, as I now argue. 
Let $y_n = (1/(2 \pi n + \frac{3\pi}{2}), -1)$ be the sequence of points at the bottom of the sine curve, and $y = (0,-1)$ so that $y_n \to y$.  Also let $U = C' \cap (\mathbb{R} \times (-\infty,0))$ be the open lower half of the sine curve.  By surjectivity, for each $n$ there is an $x_n \in S^1$ with $f(x_n) = y_n$.  By compactness, the set $\{x_n\}$ has a limit point $x$, and by continuity of $f$, we must have $f(x) = y$.  $U$ is open and contains $y$, so $f^{-1}(U)$ is open and contains $x$.  Hence we can find an open arc $V$ with $x \in V \subset f^{-1}(U)$.  Being open, $V$ contains some $x_k$ (indeed, infinitely many), so $f(V)$ contains $y_k$.
To summarize, $V$ contains $x$ and $x_k$, is contained in $f^{-1}(U)$, and is connected.  Hence $f(V)$ contains $y$ and $y_k$, is contained in $U$, and is connected.  This is absurd since $y$ and $y_k$ are in different components of $U$.
Actually, the same argument shows that $C'$ is not the continuous image of any compact, locally connected topological space.  Moreover, I just saw this question which asserts that any Hausdorff space which is the continuous image of some compact, locally connected space must again be locally connected.
A: No; the outer boundary of any handle body is a $n$-holed torus for some $n$. But all sorts of other things can happen; if $C$ is the cantor set, then the boundary is the Cantor set. So it can be all sorts of things.
