Is this variant of the Jordan Curve Theorem true? 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.
 A: 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.
