The footpoint map is not a covering map I want to show that the footpoint map is not a covering map. It is defined as $\pi\colon \mathcal{O} \rightarrow \mathbb{C}, f_a \mapsto a $ where $\mathcal{O}$ is the sheaf of germs of holomorphic functions and $f_a$ is the germ of $f$ at $a \in \mathbb{C}$.  Therefore I need to show that it has not the curve lifting property.
Let us take the curve $\gamma \colon [0,1] \rightarrow \mathbb{C}, t \mapsto 1-t$. So I should show that this curve cannot be lifted to a curve $\tilde{\gamma}\colon [0,1]  \rightarrow \mathcal{O}$ for which $\tilde{\gamma}(0)$ is the germ  $\varphi$ at $1$ of the function $f\colon z \mapsto \frac{1}{z}$.
There is a hint given: Use the fact that for analytically continuable germs $f_a,g_a$ along a curve $\gamma$ with $\gamma(0)=a$ (in our case $a=0$) and a two-variable polynomial $P$ such that $P(f_a,g_a)=0$ it holds for the analytic continuations $F(t),G(t)$ that $P(F(t),G(t))=0$ for all $t\in [0,1]$.
Some thoughts. If I look at the function $g\colon z \mapsto z$, its germ $\theta$ at $1$ can be analytically continued along $\gamma$. Since I am supposed to use the above fact I thought that I need to construe a polynomial such that $P(\theta,\varphi)=0$ then for the representatives $(U,f),(V,g)$ we could take $f\cdot g-1=\frac{z}{z}-1=1-1=0$ on $U\cap V$. If $\phi$ were analytically continuable along $\gamma$ we would have $P(\Theta(t),\Phi(t))=0$ for all $t\in [0,1]$. Now if this turns out to be false by somehow using that $z\mapsto \frac{1}{z}$ is not defined at $z=0$ have I already shown that $\gamma$ cannot be lifted?
 A: We have $\gamma(t) = 1 - t$ for all $t\in [0, 1].$ We denote the germ of $(\mathbb{C}, z)$ at the footpoint $a$ by $z_a$.
Note that the map $\tilde{\gamma}: t \mapsto z_{\gamma(t)}$ for all $t \in [0, 1]$ is a lift of $\gamma$, with respect to the footpoint map, that starts at the germ of $(\mathbb{C}, z)$ at the footpoint $\gamma(0)= 1.$
Suppose, towards a contradiction, there were a lift, $\tilde{\Gamma} : [0, 1]\mapsto \mathcal{O}$ of $\gamma$ such that $\tilde{\Gamma}(0)$ is the germ of $(\mathbb{C}^{\times}, \frac{1}{z})$ at the footpoint $a = \gamma(0) = 1$. (If $\pi$ were a covering map then there has to be such a lift because $\pi\left(\left(\frac{1}{z}\right)_{a = 1}\right) =  \gamma(0) = 1.$ We shall show that no such lift exists, which will enable us to conclude that $\pi$ is not a covering map.)
Consider the polynomial $P(X, Y) = XY - 1.$
Then note that $$P(\tilde{\Gamma}(0), \tilde{\gamma}(0)) = P\left(\left(\frac{1}{z}\right)_{a = 1}, (z)_{a = 1}\right) = 0.$$ This can be verified by recalling the definition of multiplication in the ring $\mathcal{O}_{a = 1}.$
By permanence of relations, we must have $$P(\tilde{\Gamma}(t), \tilde{\gamma}(t)) = 0 \text{ for all }t \in [0, 1].$$ This means that, in particular, we must have $$P(\tilde{\Gamma}(1), \tilde{\gamma}(1)) = \tilde{\Gamma}(1)\tilde{\gamma}(1) - 1 = 0.$$
Now, the germ $\tilde{\gamma}(1)$ is the germ of $z$ at the footpoint $a = 0.$ Let us denote $\tilde{\Gamma}(1) := h_{a = 0}$ for some holomorphic function $h$ defined in a neighborhood of $0$. Thus, we get $$h_{a = 0}z_{a = 0} = 1.$$ But this means that the function $z \mapsto z$ has an inverse in a neighborhood of $0$, which  is a contradiction.
So, there cannot be a lift of $\gamma(t) = 1 - t$, with respect to the map $\pi : \mathcal{O}\rightarrow \mathbb{C}$, that starts at the germ of $(\mathbb{C}^{\times}, \frac{1}{z})$, at the footpoint $a = 1.$
