Covering map in the context of Riemann Surfaces and Algebraic Topology

Question

I am taking a course in Riemann surfaces and our lecturer has warned us that the definition of covering maps in the context of Riemann surfaces is strictly weaker than the ones used in Algebraic Topology. In both contexts, the map is locally a homeomorphism.

In the context of Riemann surfaces, the spaces are both Hausforff and path-connected whilst the spaces don't have any explicit specified properties in the Algebraic Topology from what I gather.

The other difference is that in the Algebraic Topology course the space downstairs always has a neighbourhood for any point that has a preimage which covers the entire space upstairs.

i.e. $p : \tilde{X} → X$ such that for any $x\in X $ there is an open neighbourhood $U$ of $x$ such that $p^{−1}(U)$ is a disjoint union of open sets of $Y$ each of which is mapped homeomorphically onto $U$.

However in the Riemann surfaces course for every point upstairs, there is a neighbourhood that is locally a homeomorphism.

i.e. A covering map of a topological space is a continuous map $\pi:\tilde{X}\to X$ where $\tilde{X},X$ Hausdorff path-connectedd topological spaces and $\pi$ is a local homeomorphism. \i.e. For each $\tilde{x}\in\tilde{X}$ then there is an open neighbourhood $\tilde{N}$ of $\tilde{x}$ such that $\pi|_{\tilde{N}}:\tilde{N}\to N$ is a homeomorphism where $N$ is an open neighbourhood of $x=\pi(\tilde{x})$

$X$ is called the base space.

$\tilde{X}$ is called the covering space.

Could someone describe an example where the notion can be distinguished to show that one is weaker than another?

Answer 1

Here are a couple of examples which satisfy the weaker condition and not the stronger one:

• The inclusion map $B(0,10) \hookrightarrow \mathbb{C}$ , where $B(0,10) \subset \mathbb{C}$ is the open ball of radius $10$ centered on $0$. This example is not surjective.
• The map $B(0,10) \hookrightarrow \mathbb{C} / (\mathbb{Z} \oplus \mathbb{Z})$ which is the composition of the inclusion map above with the (topologists) covering map $\mathbb{C} \mapsto \mathbb{C} / (\mathbb{Z} \oplus \mathbb{Z})$, where $\mathbb{Z} \oplus \mathbb{Z}$ acts additively on $\mathbb{C}$ in the obvious fashion, namely, $(m,n) \in \mathbb{Z} \oplus \mathbb{Z}$ takes $z=x+iy \in \mathbb{C}$ to $(x+m) + i (y+n) \in \mathbb{C}$. This example is surjective.
• If you don't like the quotient construction of the previous example, simply take the map $\mathbb{C} \mapsto \mathbb{C}$ given by $z \mapsto e^z$. This example is not surjective. You can turn it into a surjective example by replacing the domain with the set $0 < \text{Im(z)} < 4 \pi$ and the range with $\mathbb{C} - 0$.

I'll say that I was doubtful of the statement from your lecturer that this weaker definition was the one used in the study of Riemann surfaces. But, then I looked up the definition in Section I.2.4 of the book by Farkas and Kra, and sure enough, it was just what you said (outside of the more general issues of branched covering maps). By the terminology of that book, there is stronger concept called "unlimited covering map" which looks to me like it might be equivalent to the ordinary topologist's "covering map".

Nonetheless, except for the context of reading and learning from a textbook where the terminology is used, I would hesitate to recommend adopting this weaker usage of the terminology of covering maps. Nowadays I think that Riemann surface people and topology people have merged their terminology more consistently. The weaker notion is well described by simplying using the topology terminology of a "local homeomorphism". In the context of Riemann surfaces one could say "holomorphic local homeomorphism" (or one could just say "conformal map").