 Solution

A covering map is locally a disjoint union of homeomorphisms onto the base. A space is simply connected when it is path-connected and every loop is null-homotopic. By the uniformization theorem, the simply connected Riemann surfaces areup to analytic isomorphism.

A lattice is a discrete subgroupwith ${\omega }_1/{\omega }_2\notin \mathbb{R}$. The Weierstrass elliptic functionconverges normally on compact subsets of $\mathbb{C}\setminus L$, defines a meromorphic map $\mathbb{C}\to {\mathbb{C}}_{\infty }$, and is nonconstant because it has a double pole at every lattice point. Reindexing the normally convergent derivative series shows that ${\wp }_L^{\prime }$ is $L$-periodic; evenness fixes the integration constants, so ${\wp }_L(z+\omega )={\wp }_L(z)$.

The invariance makeswell-defined and unique on $\mathbb{C}/L$. It is analytic because the quotient projection is locally biholomorphic.

Neither map is a covering. For any nonzero $\omega \in L$, periodicity and oddness of ${\wp }_L^{\prime }$ giveso ${\wp }_L^{\prime }(\omega /2)=0$. The map is not a local homeomorphism there; since the quotient projection is locally biholomorphic, the descended map fails for the same reason.

There is no covering $\mathbb{C}/L\to {\mathbb{C}}_{\infty }$. The torus is connected, while every connected covering of the simply connected sphere is a homeomorphism. A torus is not homeomorphic to a sphere, for example because their fundamental groups are ${\mathbb{Z}}^2$ and $0$.