friends.
I'm working on a problem, the broad scope of which is to show that given a map $f:S^1\rightarrow \mathbb{R}^3$ be a smooth embedding, and a projection map $\pi_v:S^2\rightarrow P_v$, where $P_v$ is the plane orthogonal to the vector $v\in S^2$ the map $f_v=\pi_v\circ f:S^1\rightarrow P_v$ satisfyies $\vert f^-1(v) \vert =2$ for almost every value of v.
The problem defines the following function, $G:S^1\times S^1 - \Delta \rightarrow S^2$ given by $G(x,y)=\frac{f(x)-f(y)}{\vert f(x)-f(y)\vert}$ and another function, $\Phi(x,y,z,t):S^1\times S^1 \times S^1 \times \mathbb{R}\rightarrow \mathbb{R}^3$ given by $\Phi(x,y,z,t)=f(x)+(t-1)f(y)-tf(z)$.
I'm asked to show that if $v\in S^2$ is a regular value of $G(x,y)$ and $\Phi(x,y,z,t)=0$, $x\neq y\neq z$, then $(x,y,z,t)$ is a regular point of $\Phi(x,y,z,t)$
I figured this could be done by direct computation. We need to show that $D\Phi_{(x,y,z,t)}:T_{(x,y,z,t)}S^1\times S^1 \times S^1 \times \mathbb{R}\rightarrow T_{\Phi(x,y,z,t)}S^1\times S^1 \times S^1 \times\mathbb{R}$ is surjective if every point in the pre-image of $G^-1(v)$ has surjective derivative $DG_{(x,y)}:T_{(x,y)} S^1\times S^1 -\Delta \rightarrow T_{(x,y)}S^1\times S^1\ -\Delta$. The condition that $\Phi(x,y,z,t)=0$ gives us that $f(x)-f(y)=f(z)-f(y)$ an hence $G(x,y)=G(z,y)$ Computing the jacobian of $\Phi(x,y,z,t)$ hasn't gotten me any closer to a solution though, since I get two matrices of different dimension that I have no idea how to relate. Hence, I am here asking for assistance.
Will upvote as always, thanks in advance.