Maps from $D^n$ to $D^n$ with a single inverse set are open.

Solution 1:

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.