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This feels as though it should be falsifiable, but it's not immediately obvious to me. The informal version of the statement is 'for every non-intersecting curve between two opposite corners of a square, there's a curve between the other two corners that only intersects it once.' Formally:

Let $f(): [0, 1]\mapsto [0,1]^2$ be a non-self-intersecting curve with $f(0) = \langle0,0\rangle$, $f(1) = \langle1,1\rangle$, and $f(t)\in (0,1)^2$ for $t\in(0,1)$. (Note that I'm requiring that all but the endpoints of the curve lie in the open square!) Then there exists a non-self-intersecting curve $g(): [0, 1]\mapsto [0,1]^2$ with $g(0) = \langle1,0\rangle$, $g(1) = \langle0,1\rangle$, and $g(t)\in (0,1)^2$ for $t\in(0,1)$ such that there are unique $t_0$ and $ t_1$ with $f(t_0) = g(t_1)$.

This feels like it ought to be a consequence of the JCT and/or Schoenflies' theorem, but the catch is that I don't see any clean ways of ensuring that a $g()$ constructed by those theorems actually maps back to the interior of the square as opposed to its boundary.

Steven Stadnicki
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1 Answers1

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This is either right or, you know, just wrong. But here goes:

Let $L,R,T,B$ be the left, right, top, and bottom edges of the square, and let $F$ be the image of $f$. $C_{1}=F\cup R\cup B$ is a simple closed curve, so by Jordan–Schönflies, $C_{1}$ together with the region $R_{1}$ that it encloses is homeomorphic to a closed disk $D_1$. The homeomorphism can be chosen so that $(1,0)$ is mapped to the circle's north pole while $F$ is mapped onto the lower semicircle.

Similarly, $C_{2}=F\cup T\cup L$ together with the region $R_{2}$ that it encloses is homeomorphic to a closed disk $D_2$, where $(0,1)$ is mapped to the north pole and $F$ to the lower semicircle. This homeomorphsim can be further chosen so that the point on $F$ mapped to the south pole of $D_2$ is the same as the point on $F$ that was mapped to the south pole of $D_1$.

For $i=1,2$, let $V_i$ be the vertical diameter of $D_i$, and let $G_i$ be the image of $D_i$ under the homeomorphism sending $D_i$ to $C_i\cup R_i$. Then the concatenation $G=G_1\cup G_2$ is the desired non-self-intersecting curve.

Greg Martin
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  • This looks pretty clean to me! The one thing that I'm not sure of: Can we in fact be sure that $F\bigcup R\bigcup B$ is actually a simple closed curve? I can easily believe this, but it's not clear to me that $F$ can't come arbitrarily close to the sides of the square - or is this the essence of the JCT in action, that we can essentially bound $F$ away from an edge everywhere but a neighborhood of a corner? – Steven Stadnicki Jun 05 '14 at 00:04
  • @StevenStadnicki: The fact that $F \cup R \cup B$ is a simple closed curve follows from your hypothesis that $f(t)\in (0,1)^2$ for $t\in(0,1)$. From this it follows that $F \cup R \cup B$ is the continuous image of a 1--1 map of the circle, and every continuous injection from a compact space to a Hausdorff space is a homeomorphism onto its image, therefore $F \cup R \cup B$ is homeomorphic to a circle. – Lee Mosher Jun 05 '14 at 14:56
  • And, by the way, this answer looks perfect to me. – Lee Mosher Jun 05 '14 at 14:57
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    @LeeMosher Thank you! That's the only gap I had. – Steven Stadnicki Jun 05 '14 at 15:24