Let $a,b:[0,1]\to D^2$ be injective continuous curves such that $a(0)=(-1,0),a(1)=(1,0)$ and $b(0)=(0,-1),b(1)=(0,1)$. How to show that these two curves must intersect?
From intuition, this highly depends on the choices of initial points.
Thanks in advance for answering.
Consider the red points joined by the blue path If any of the green points is on the blue path, done. Otherwise, consider the purple circle with center the right green point $R$ through the red points, call it $\eta$. The indexes of the green points $L$, $R$ with respect to $\eta$, are clearly $0$, and $1$. We can deform $\eta$, into the closed path $\gamma$, by deforming the purple arc into the blue path.The indexes with respect to $\eta$ equal the indexes with respect to $\gamma$, since the deformation does not touch the green points ( the disk is convex, the deformation is done by moving along segments, in a standard way).
Therefore $I_{\gamma}(L)=0$, $I_{\gamma}(R)=1$. Since the index $P\mapsto I_{\gamma}(P)$ is locally constant in $P$ on $\mathbb{R}^2 \backslash \gamma$, any path from $L$ to $R$ will intersect $\gamma$. If this path is inside the disk, it cannot intersect the purple part of $\gamma$, hence it must intersect the blue path.
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Notes:
the paths in the statement of the problem do not need to be injective.
The statement can be given for an arbitrary Jordan curve $\omega$. Assume the points $U$, $D$, $L$, $R$ are on $\omega$, such that $U$, $D$ separate $L$, $R$. That means $\omega \backslash \{U,D\}$ has two connected components, and $L$, $R$ are in different components. Then any two path joining $U$, $D$, respectively $L$, $R$, and inside $\omega$ intersect.