A question related to the index of a Jordan curve Problem. Let $f,g : [0,1]\to\mathbb R$ continuous, such that
$$
f(0)=f(1)\qquad\text{and}\qquad g(1)-g(0)=2.
$$
Show that there exist $x,y\in [0,1]$, such that $x<y$ while
$$
f(x)=f(y)\qquad\text{and}\qquad g(y)-g(x)=1.
$$
A proof of this can be derived from the fact that the index, with respect to a Jordan curve, of a point not belonging to the curve can only be $-1,0$ or $1$.
I was wondering whether this problem can be shown in and elementary fashion, and without the use of Algebraic Topology.
 A: Here’s a proof where analysis is used in place of algebraic topology (although it’s clear that it’s really algebraic topology in disguise).
By approximation, it is enough to show the result when $f,g$ are $C^2$. Define the paths $\alpha(t)=(f(t)-f(0),g(t)-g(0)-1)$ and $\beta(t)=(f(1)-f(1-t),g(1)-g(1-t)-1)$. Both paths are $C^2$, with values in $\mathbb{C}$, from $-i$ to $i$. Moreover, $\alpha(t)+\beta(1-t)=0$.
Let $\psi(x,y)=(f(y)-f(x))+i(g(y)-g(x)-1)$ for $0 \leq x,y \leq 1$, and $\gamma_s(t)=\psi(s(1-t),(1-s)t+s)$ for $0 \leq s,t \leq 1$. One can check that $\gamma_s$ is $C^2$, $\gamma_s(0)=\psi(s,s)=-i$, $\gamma_s(1)=\psi(0,1)=i$, $\gamma_0(t)=\psi(0,t)=\alpha(t)$, $\gamma_1(t)=\beta(t)$.
If some $\gamma_s(t)$ is zero, we are done.
Assume this doesn’t happen. Let $I(s)=\int_0^1{\frac{\gamma_s’(t)}{\gamma_s(t)}\,dt}$. Write $\alpha(t)=r_1(t)e^{i\phi(t)}$, then $\phi(0)=-\pi/2$, $\phi(1)=\pi/2+2ik\pi$, and then $I(0)=(2k+1)i\pi$. But since $\beta(t)=-\alpha(1-t)$, $I(1)=-I(0)$ so $I(1) \neq I(0)$.
Now we’ll show that $I$ is constant. Because $\gamma$ was chosen regular enough, $I$ is $C^1$ and $I’(s)=\int_0^1{\left(\frac{(\partial_{st}\gamma_s)(t)}{\gamma(t)}-\frac{\gamma_s’(t)(\partial_s\gamma_s)(t)}{\gamma_s^2(t)}\right)\,dt}$. But what is in the integral is actually $\partial_t\left(\frac{(\partial_s\gamma_s)(t)}{\gamma_s(t)}\right)$, and thus $I’(s)=\frac{(\partial_s\gamma_s)(1)}{\gamma_s(1)}-\frac{(\partial_s\gamma_s)(0)}{\gamma_s(0)}=0$ since $\gamma_s(1)=i,\gamma_s(0)=-i$. So $I’=0$ and thus $I$ is constant. Contradiction.
