If the interior of a topological (n-1)-sphere in n-space is contractible, is it an n-ball? Consider a submanifold of $\mathbb{R}^n$ that is homeomorphic to the $(n-1)$-sphere.  By the Jordan-Brouwer separation theorem, removing that submanifold from $\mathbb{R}^n$ partitions what remains into two connected components, one of which is bounded.  Let $B$ be the bounded component, and let $\overline{B}$ be its closure.  Once $n>2$, additional assumptions are needed to ensure that $\overline{B}$ is homeomorphic to the closed $n$-ball, as discussed in the question DMiz.  
I originally asked --- or, at least, I intended to ask --- whether requiring $\overline{B}$ to be contractible would imply that it was a closed $n$-ball.  In fact, Bing showed that $\overline{B}$ is always contractible; so adding that requirement wouldn't change anything.
Lots of study has gone into this area recently, though mathematicians typically consider topological embeddings of the $(n-1)$-sphere in the $n$-sphere, rather than in $\mathbb{R}^n$, so that the closures of the two components can be studied on an equal footing.  Those closures are called "crumpled $n$-cubes", and they can be either closed $n$-balls or "wild".  Both of them can be wild.  If a crumpled $n$-cube is a manifold-with-boundary, then it is an $n$-ball.
The notes of Andrew Putman that Lee Mosher suggested are helpful, and this posting on MathOverflow has clear explanations and further pointers.
 A: I assume that your question is the one in the header:
Suppose that $f: S^{n-1}\to E^n$ is a homeomorphism to its image $\Sigma$ and the bounded component $B$ of $E^n - \Sigma$ is contractible. Is it true that $B$ is homeomorphic to $E^n$? 
The answer to this question is already negative for $n=3$. Namely, take a wild arc $a$ in $E^3$ connecting points $p, q$, which is only wild at $q$. See the figure below:
 
Then $S^3 -a$ is simply connected and, hence, contractible (for $i\ge 2$, $\pi_i(E^3-a)=0$ by Alexander Duality plus Hurewicz theorem). However, it is not "simply connected at infinity" (since $a$ is wild at $q$) and, hence, not homeomorphic to $E^3$. Now, "thicken" $a$ by replacing it with a 3-dimensional wild  ball $\bar{A}$ with a single wild point, again at $q$. The complement 
$B=S^3 - \bar{A}$ is homeomorphic to $S^3- a$ and, hence, contractible but not homeomorphic to $E^3$. The boundary of $\bar{A}$ is a topological sphere $\Sigma$. If you place a point, called $\infty$, in the interior  $A$ of $\bar{A}$ then $B$ becomes the bounded component of $E^3- \Sigma$, where $E^3= S^3- \{\infty\}$. 
