TITLE: Fibers of continuous maps of $\mathbb{R}^n$ which are injective at dense points
QUESTION [7 upvotes]: Question. Suppose that $f\colon\mathbb{R}^n \to \mathbb{R}^n$ is a continuous map and there is a dense subset $D \subset \mathbb{R}^n$ such that $f^{-1}(f(x)) = \{x\}$ for all $x \in D$.
Is every fiber of $f$ connected?
When $n =1$, it is easy to see that $f$ must moreover be a strictly increasing or decreasing function. 
I think that it it true in general, but I am not sure. An idea to prove: Suppose, to the contrary, that $f(x)^{-1}$ has two connected components.
Then pick an arc, in the domain, with endpoints in different connected components fo $f^{-1}(x)$. Then $f$ identifies the endpoints of the arc, and it seems unlikely that the resulting loop in the target  is null homotopic, which would be a contradiction.

REPLY [6 votes]: There is a continuous map $f:\mathbb{R}^2\to\mathbb{R}^2$ such that the fiber of the origin  $0$ is  the disconnected set $f^{-1}(0)=\mathbb{R}\times\{-1,+1\}$, and for any $x\in D:=\mathbb{R}\times\big (\mathbb{R} \setminus \{-1,1\}\big)$, there holds $f^{-1}(f(x))=\{x\}$. 
I didn't bother to write an equation (I'll do if compelled), but here is a picture: the loops with self-intersection at the origin are lines $f(\{a\}\times\mathbb{R})$, the short segments are lines $f(\mathbb{R}\times\{b\})$, which degenerates to a single point, the origin, for $b=\pm1$.  
[rmk] For a concrete realization of this picture, we may use the rational parametrization of the Conchoid of Nicomedes. One can even fill the hole in the picture.