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).

  • $\begingroup$ Is there only one non-singleton inverse set disjoint from the boundary (so there may be other non-singleton inverse sets, but they intersect the boundary), or is there only one non-singleton inverse set in total, and this set is disjoint from the boundary? $\endgroup$ – Stefan Hamcke Mar 23 '15 at 12:13
  • $\begingroup$ If I understand correctly, for $x \neq y$, $f^{-1}(x)$ is a singleton. So $f$ is surjective? $\endgroup$ – Najib Idrissi Mar 23 '15 at 16:34
  • $\begingroup$ The map need not be surjective, but it fails to be injective at exactly one point in the target and the preimage of this point is disjoint from the boundary. $\endgroup$ – PVAL-inactive Mar 23 '15 at 18:03
  • $\begingroup$ Is it easy to write down an example of such an $f$? I'm having trouble getting intuition as to why this should be true $\endgroup$ – Jason DeVito Apr 15 '15 at 20:01
  • $\begingroup$ Well a simple example would just be the map that sends a half open annulus homeomorphically to a punctured ball and maps the ball on the open side to the puncture point, though there are much more complicated examples. $\endgroup$ – PVAL-inactive Apr 15 '15 at 20:06

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.

  • $\begingroup$ +1 but I think you should explain better why $f^{-1}(\gamma)$ intersects $\partial D^n$ (because here you use the openness of $f(U)$) $\endgroup$ – Mizar Mar 27 '15 at 22:29
  • $\begingroup$ @Mizar yes, you're right, I left this point not completely clear. To prove this one has to parametrize $\gamma$ with $[0,1]$ and consider the connected component of $\gamma^{-1}(f(U))$ wich contains $0$. This is an open interval $[0,\tau)$ because $f(U)$ is open, and by continuity $\gamma(\tau)\in f(\partial D)$. $\endgroup$ – user126154 Mar 29 '15 at 11:40

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