Here I want to share an interesting proof of the Lover and Hater Problem with you without using the theory of winding number.
In our course, we met with Lover and hater problem first, and after a long time we study the Brouwer Fixed Theorem. In fact, we can prove the Brouwer fixed theorem with a purely analytical method, if you want to see that, download this PDF file.
Click to access brouwer.pdf
Suppose we already know the Brouwer fixed theorem, as a special case, we have
Theorem. (Brouwer fixed point theorem) Let \(f: [-1,1]^2 \to [-1,1]^2\) be continuous, then there exists \(x \in [-1,1]^2\) such that \(f(x) = x\).
Using this theorem, we can prove the Lover and hater problem in a simple and analytical way.
Problem. Let \( c: [-1,1] \to [-1,1]^2\) and \(d: [-1,1] \to [-1,1]^2\) be continuous functions such that \(c(-1) \in \{-1\} \times \mathbf{R}, c(1) \in \{1\} \times \mathbf{R}, d(-1) \in \mathbf{R} \times \{1\}, d(1) \in \mathbf{R} \times \{-1\} \), then there exists \( s,t \in [-1,1]\) such that \( c(s) =d(t) \).
This is just another statement of the Lover and Hater problem.
Proof. (Sketch) Let
\(c(t)=(x_c(t),y_c(t)), d(t)=(x_d(t), y_d(t)).\)
Define a map
\( r: \mathbf{R}^2 – \{(0,0)\} \to [-1,1]^2, r(x,y) = \frac{(x,y)}{\textrm{max}\{|x|,|y|\}}.\)
Then it’s easy to check that \(r\) is continuous and the image of \(r\) is the boundary of the square \([-1,1]^2\).
Consider the map
\(g: [-1,1]^2 \to [-1,1]^2, g(s,t)=r(d(t)-c(s)).\)
Then \(g\) is continuous, now by the Brouwer Fixed point theorem, we know there is a point \((s,t) \in [-1,1]^2\) such that \( g(s,t)=(s,t)\).
But the image of \(r\) is the boundary of the square \([-1,1]^2\), so we have either \(|s|=1\) or \(|t|=1\). Without loss of generality, suppose \(|s|=1\), if \(s=1\), then we have
\(|x_d(t)-x_c(1)|= x_d(t)-x_c(1)=x_d(t)-1 \le 0\),
which is a contradiction!
Similarly if \(s=-1\), then we have
\(|x_d(t)-x_c(-1)|= -(x_d(t)-x_c(-1))=-1-x_d(t) \le 0\),
which is also a contradiction!
Hence we complete our proof.