The question is about path-lifting property for projections that possibly are not coverings.
Let $E, F$ be finite dimensional normed spaces, $A\subset E$ be a connected open set, $f:A\times F\to F$ be a continuously differentiable mapping. It is assumed that the set $\Gamma=f^{-1}(0)$ is non-empty and for each $(x,y)\in\Gamma$ the derivative $\partial_y f(x,y)$ is invertible; finally, $G$ is a connected component of $\Gamma$ such that $pr_1(G)=A$ ($pr_1$ being a local homeomorphism of $G$ onto $A$).
The problem is to show that any continuous mapping $\gamma:[0,1]\to A$ can be lifted to a continuous mapping $\tilde{\gamma}:[0,1]\to G$ (i.e. $pr_1\circ\tilde{\gamma}=\gamma$).
I know how to prove it under some additional assumptions. First is that $pr_2(G)$ is bounded, second (more general) is that $pr_1:G\to A$ is a covering. Without these I am not sure that the result is true, but on the other hand I can't find a counterexample with $A$ being an open set of $E$ (I am aware about examples of mappings onto $S^1$ without path-lifting property that were discussed here, but is it possible to adapt them to my case?)
Thanks for the help!
It turned out that the assertion in the question is false in general. An example follows.
Let $E=F=\mathbb{C},$ $A=\mathbb{C}\setminus\{0\},$ and $f(x,y)=x-(e^y-1)^2.$ Then the set of zeroes of $f$ is $$ \Gamma=\{((e^y-1)^2,y):y\in\mathbb{C}\setminus 2\pi i \mathbb{Z}\} $$ (values $y\in 2\pi i \mathbb{Z}$ are excluded because we must have $x\ne 0$). This set is connected as a continuous image of the connected set $\mathbb{C}\setminus 2\pi i \mathbb{Z}.$ Next, for every $(x,y)\in\Gamma$ the derivative $$ D_yf(x,y)=-2e^y(e^y-1)\ne 0 $$ (again becase $x\ne 0$).
Finally, $pr_1(\Gamma)=A.$ Indeed, every nonzero $x$ can be written in the form $x=e^{\log|x|+i\xi}.$ If $|x|\ne 1$ then $1+e^{\frac{1}{2}\log |x|+i\frac{\xi}{2}}\ne 0$ and the equation $e^y=1+e^{\frac{1}{2}\log |x|+i\frac{\xi}{2}}$ has solutions in $y.$ If $|x|=1$ then $x=e^{i\xi}$ for some $\xi\not\in 2\pi(2\mathbb{Z}+1)$ and again $$ 1+e^{\frac{1}{2}\log |x|+i\frac{\xi}{2}}=1+e^{i\frac{\xi}{2}}\ne 0 $$ So, $A,f$ and $\Gamma$ verify all assumptions (in particular, $pr_1$ is a local homeomorphism of $\Gamma$ onto $A$). However, the path $\gamma(t)=e^{it},$ $0\leq t\leq 2\pi,$ in $A$ can't be lifted to $\Gamma.$ Indeed, assume that there exists a path $\psi:[0,2\pi]\to\mathbb{C}$ such that $$ (e^{\psi(t)}-1)^2=e^{it}, \ 0\leq t\leq 2\pi. $$ Then for each $t$ $$ e^{\psi(t)}=1\pm e^{i\frac{t}{2}} $$ By continuity, there must be either $+1$ for all $t,$ or $-1$ for all $t.$ In the first case the equality $$ e^{\psi(t)}=1+ e^{i\frac{t}{2}} $$ fails for $t=2\pi$. In the second case the equality $$ e^{\psi(t)}=1- e^{i\frac{t}{2}} $$ fails for $t=0.$