Maps from $D^n$ to $D^n$ with a single inverse set are open. Let $D^n$ denote the closed unit ball in $\Bbb R^n$. In multiple sources proving Brown's generalized Schoenflies theorem (including a version in the original paper), the following consequence of Brouwer's invariance of dimension is stated without proof.

If $f: D^n \rightarrow D^n$ with only one non-singleton inverse set $f^{-1}(y)$ disjoint from the boundary, then $y$ is in the interior of the image of $f$.

I am at a loss as to how to go from invariance of dimension to this. 
EDIT: I just went through the proof of generalized Schoenflies in Bing's book, and he doesn't makes use of this fact. I'm still interested how one proves this from invariance of dimension (or using similar homological techniques as such). 
 A: The invariance of domain theorem states that if $U$ is an open domain of $\mathbb R^n$ and $f:U\to\mathbb R^n$ is continuous and injective, then $f$ is open and in fact $f$ is an homeomorphism.  See here. 
Now we come to the question.
By continuity, $f^{-1}(y)$ is closed. Let $U=int(D^n)\setminus f^{-1}(y)$. $U$ is not empty otherwise $f(D^n)=y$ and therefore $f^{-1}(y)$ contains the boundary of $D^n$.
So $f(U)$ is open and homeomorphic to $U$. By definition, there is a sequence $u_n\in U$ so that $u_n\to f^{-1}(y)$. This provides a sequence $z_n=f(u_n)$ in the interior of the image of $f$ so that $z_n\to y$.
If $y$ is not an interior point, there is a second sequence of points
$w_n\to y$ not in the image of $f$. Let $\gamma_n$ be an arc between $z_n$ and $w_n$ wich do not contains $y$. $f^{-1}(\gamma)$ must contain a point of $\partial D^n$. So we have that there is $x_n\in\partial D^n$ so that $f(x_n)\to y$. $x_n$ has an accumulation point $x\in\partial D^n$ and by continuity $f(x)=y$. Thus, if $y$ is not an interior point, then $f^{-1}(y)$ intersects the boundary.
