In J. Roe's Winding Around and V. Arnold's Ordinary Differential Equations there is a beautiful story attributed to N. Konstantinov:
A pair of lovers travels from city $A$ to city $B$ via two different roads. If the distance between them is at least 10 kilometers, then they die. There is also a pair of haters – one travelling from $A$ to $B$ and the other travelling from $B$ to $A$. If the distance between them is at most 10 kilometers, then they die. Prove that the tragedy is inevitable.
Putting it rigorously, one measures the motion of each pair by continuous paths into the square and proves the following theorem:
Theorem Let $f, g\colon [0, 1]\to [0, 1]\times[0, 1]$ be continuous maps, such that $f(0) = (0, 0),\,f(1)= (1,1)$ and $g(0) = (0, 1),\,g(1) = (1, 0)$. Then $f(x)=g(y)$ for some $x, y\in [0, 1]$.
This is a special case of Exercise 0.5 of J. Rotman's Algebraic topology, with Bombyx mori's solution via Brouwer fixed point theorem. Another beautiful solution is possible – in Exercise 3.6.5 of Winding around, the author hints to close the image of $g$ to a loop:
and then says to use the properties of winding number.
Although both solutions – by Brouwer fixed point theorem or by winding number – work beautifully, I am looking for a third one:
Is there a solution that uses the Jordan curve theorem?
Surely, if $g$ is injective, the closed loop is homeomorphic to the circle and we immediately get the claim. I however don't know whether this can be generalized to an arbitrary $g$.
