# Covering map in the context of Riemann Surfaces and Algebraic Topology

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?

• Find a branched covering from a Riemann surface to the projective line $\Bbb P^1$. There is no actual nontrivial connected cover of $\Bbb P^1$. – user98602 Feb 12 '16 at 2:01
• In the study of Riemann surfaces it's typical to study branched covers, not covers. Branched covers fail to be local homeomorphisms at a finite set of points. It's unclear to me if this is what your professor is talking about, or if they're talking about the fact that in general being a local homeomorphism isn't enough to be a covering map. – Qiaochu Yuan Feb 12 '16 at 2:14
• Sorry, I don't think branched covering has been mentioned so I don't think it's the same notion. Please see the above edit above where I quoted my notes from my Riemann Surfaces course. – daruma Feb 12 '16 at 9:28

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").

• Yes, a topologist's covering is what in complex analysis is/used to be called an unbranched unlimited covering. – Daniel Fischer Feb 12 '16 at 14:12
• I am just checking I understood the examples correctly. With the first example, if you take a point on the boundary then any open set would intersect with the interior which in this case is the open ball, B(0,10). Then it is not a homeomorphism because of the point on the boundary. – daruma Feb 12 '16 at 17:12
• I don't believe I understood the second one correctly. Take a point $t\in X$ and then $p^{-1}(t)$ is a finite set of points. Each of the individual points in $p^{-1}(t)$ have an open neighbourhood where its restriction would make it a homeomorphism with the image and so if we take the interesection of all the images of the neighbourhoods, it will be evenly covered. (Finite intersection of open sets) – daruma Feb 12 '16 at 17:19
• Regarding your first comment, a point on the boundary of the open ball is not in the subset, so local homeomorphism is not violated at that point because that point is not in the domain of the function. – Lee Mosher Feb 12 '16 at 20:46
• Regarding your second comment, the requirement of being evenly covered is that each $t \in X$ has a neighborhood which is evenly covered. But, if $t$ is equal to the image of a point $\tilde t$ that is contained in the boundary of $B(0,10)$ then no neighborhood of $t$ is evenly covered. – Lee Mosher Feb 12 '16 at 20:47