I'm trying understand the construction of the lift of an analytic map $$f:\mathbb{C}/\Lambda_1\to \mathbb{C}/\Lambda_2$$ (with $\Lambda_i$ lattices in the complex plane) to a map $$F:\mathbb{C}\to \mathbb{C}$$ with $\pi_2\circ F=f\circ \pi_1$ with $\pi_i$ the projections $\mathbb{C}\to \mathbb{C}/\Lambda_i$.
The construction given does this by choosing $\mu\in\mathbb{C}$ s.t $\pi_2(\mu)=f(\pi_1(0))$ and then for any $z\in\mathbb{C}$ choosing path in $\mathbb{C}$ from $0$ to $z$. From here we uniquely lift the path $f\pi_1\gamma$ in $\mathbb{C}/\Lambda_2$ to $\mathbb{C}$ s.t $\tilde\gamma(0)=\mu$ and then define $F(z)=\tilde\gamma(1)$.
I'm struggling to see why $F(z)$ doesn't depend on the choice of $\gamma$, I thought about trying to apply the Monodromy theorem but I don't see how to do that. Any help would be great!
You can use compactness.
Look at an arbitrary $z\in\mathbb{C}$. Now do the following. Translate $\Lambda_1$ so that $z$ is in the interior of one of the $\Lambda_1$ parallelograms, then translate $\Lambda_2$ so that the points on $\mathbb{C}$ corresponding to $f(z+\Lambda_1)$ are also the interior of the $\Lambda_2$ parallelograms.
There will then be an open neighbourhood $B$ of $z$ such that $B$ is likewise in the interior of the $\Lambda_1$ parallelograms and $f(B+\Lambda_1)$ is in the interior of the $\Lambda_2$ parallelograms.
How many such open sets $B$ do we need to cover a closed $\Lambda_1$ parallelogram in $\mathbb{C}$? Of course finitely many will do, because the parallelogram is compact.
We now have a finite cover $B_1,\ldots,B_n$. On each $B_i$, if we fix the value of $F(z_i)$ for some $z_i\in B_i$, then there is a unique way of extending $F$ to the whole of $B_i$ so as to agree with $f$. That is because for this $z_i$, we fix the domain $B_i$ and range $F(B_i)$ as being contained in their respective parallelograms.
Now, fix some $z_1$, and use the overlaps between the $B_i$ to fix $F$ on the whole $\Lambda_1$ parallelogram in $\mathbb{C}$. We can do this because the parallelogram is connected. We now have enough information to fix $F$ on neighbouring $\Lambda_1$ parallelograms, and eventually on all $\Lambda_1$ parallelograms, meaning on the whole domain $\mathbb{C}$.