Let $X,Y$ be metrizable compact spaces, with $X$ connected, and $f\colon X \to Y$ a continuous function onto $Y$. Suppose $f$ is almost one-to-one in the following sense: there exists $Y' \subseteq Y$, a dense $G_\delta$, such that $\# f^{-1}(y) = 1$ for all $y \in Y'$. Then, can happen that $Y\setminus Y'$ is dense in $Y$?

I have this situation in another problem and it is hard for me to image a that $Y$ has both a dense set of point where $f$ has a unique preimage and a dense set of points where $f$ has two or more preimages.

  • $\begingroup$ "$Y\setminus Y'$ is either dense in $Y$ or empty". If it is empty, then $f$ is trivíally injective. You should omit this case because it only leads to irritation. What abou the case that $Y\setminus Y'$ is non-empty but not dense? $\endgroup$
    – Paul Frost
    Jul 1, 2020 at 8:19
  • $\begingroup$ @PaulFrost The case in which $Y\setminus Y'$ is non-dense and non-empty don't occur in my problem. $\endgroup$ Jul 1, 2020 at 13:54

1 Answer 1


Yes, it can. For instance, let $X=\{(x,y)\setminus [0,1]^2:y\le D_M(x)\}$, where $D_M(x)$ is a modified Dirichet function and $f$ be the projection of $X$, $(x,y)\mapsto x$.


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