Suppose $f : \mathbb R \to \mathbb R$ is such that pre-image of every point under $f$ is dense in $\mathbb R$. This, of course, implies that $f$ is surjective, and hence has a right inverse $\mathbb R \to \mathbb R$.

My question is: does $f$ necessarily have a continuous right inverse?

This is a question I made up while thinking about another post. The connection is admittedly tenuous, but I find this question independently interesting anyway.

One can rephrase the above (almost equivalently) as follows:

Given a family $\{ X_{\alpha} \}_{\alpha \in \mathbb R}$ of dense subsets of $\mathbb R$, can we always find a continuous “representative map”, i.e., a function $f : \mathbb R \to \mathbb R$ such that $f(\alpha) \in X_\alpha$ for every $\alpha \in \mathbb R$?

I find the latter question even more natural. Stated this way, it is clear that while we have an strong restriction on any individual $X_\alpha$, there seems to be little to no relation between the different $\alpha$'s. In the absence of any such topological restrictions, I regard a positive answer to the above question extremely unlikely. However I am unable to construct a counter-example either.



1 Answer 1


No, in fact $f$ can't have a continuous right inverse . Suppose $f(g(x)) = x$. Then $g$ is one-to-one. Given $a < b$, if $g$ is continuous $g([a,b])$ must be an interval of positive length, and this must intersect each $f^{-1}(y)$, so $f(g([a,b])) = \mathbb R$, not $[a,b]$.


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