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Let $\square=[0,1]\times[0,1]$ be the unit square and $f\colon\square\to \square$ is a continuous map that fixes the points on the boundary.

Assume $f$ is a limit of homeomorphisms $\square\to \square$. (By Moore's theorem it is equivalent to the condition that for any point $p\in \square$ the inverse image $f^{-1}\{p\}$ is connected and its complement is connected.)

Is it true that for most segments in $\square$, their inverse images are Jordan arcs?

Say, given $s\in [0,1]$ consider the vertical unit segment $I_s$ defined by $x=s$. Is it true that for a dense G-delta set of values $s\in [0,1]$ the inverse image $$J_s=f^{-1}I_s$$ is a Jordan arc?

Comments:

  • If $f$ is injective on $J_s$ then $J_s$ is a Jordan arc (evident). One may expect that $f$ is injective for most of values $s$, but this is not the case --- even if all $J_s$ are Jordan arcs, the map $f$ might map an arc in $J_s$ to one point for all $0<s<1$.

  • For sure $J_s$ has vanishing measure for all but countable set of values $s$.

  • The problem can be reformulated in terms of decomposition of $\square$ into the inverse images $f^{-1}\{x\}$. The main trouble comes from the sets $f^{-1}\{x\}$ that are not Jordan arcs nor single points. I do not know examples with uncountably many such sets. (I alos do not know if one can find uncountably may connected but not path connected compact disjoint sets in the plane.) These problems are related to the problem that there at most countable set of Y shapes in the plane, see "Ys in the plane" in "Mathematical Puzzles" of Peter Winker.

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  • $\begingroup$ "limit" is understood as pointwise limit? $\endgroup$
    – YCor
    Commented Sep 2, 2018 at 21:01
  • $\begingroup$ @YCor yes, sure. $\endgroup$
    – Anton Petrunin
    Commented Sep 2, 2018 at 21:04
  • $\begingroup$ @Anton If f is continuous on the unit square and fixes the boundary must then f be bijective on at least the boundary? $\endgroup$
    – Right
    Commented Sep 4, 2018 at 19:15
  • $\begingroup$ Sorry, I interpreted your question in the sense that the boundary as a whole is mapped into the boundary, it is obvious that you mean that surely f is bijective on the boundary, because it fixes each point on the boundary.. $\endgroup$
    – Right
    Commented Sep 4, 2018 at 19:19
  • $\begingroup$ @Right In general inverse image of a point $x$ on the boundary must contain $x$ and might be bigger. But you may assume that $f$ is injective on the boundary (it does not change the problem). $\endgroup$
    – Anton Petrunin
    Commented Sep 4, 2018 at 19:27

1 Answer 1

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Steve Ferry gave me an answer --- the answer is "no".

In fact according to "A continuous decomposition of the plane into pseudo-arcs" by Wayne Lewis and John Walsh, there is a continuous subdivision of the plane into pseudo-arcs. My Moore's theorem, the quotient space is also a plane. Therefore for the quotient map $f$ we have that $J=f^{-1}I$ is not a Jordan arc for any Jordan arc $I$ in the plane.

It is easy to modify the construction to get a counterexample to my question.

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