# Definition of double covers

It's unusual that I have to ask for a definition on this site, but it in the context of spin structures, it seems common to talk about double covers without giving the definition. I understand the definition of a covering space, but the definition of double covers (more generally, $$n$$-fold covers) on Wikipedia is based on an alternative definiton of covering spaces I don't understand:

Equivalently, a covering space of $$X$$ may be defined as a fiber bundle $$p\colon C\to X$$ with discrete fibers.

And further below:

A covering is a double cover if every fiber has two elements.

I don't know what they mean by "discrete fibers". The article links to the discrete topology of a set, but I don't see the correlation. In the case of a double cover, does it simply mean that each fiber $$p^{-1}(\{x\})$$ consists of two elements for all $$x\in X$$?

Your first bullet point: To repeat the definition you most likely already know, for any function $$p : A \to B$$, a "fiber" of $$p$$ is a certain kind of subset of $$A$$, namely one which can be written in the form $$p^{-1}(\{b\})$$ where $$b \in B$$.

Assuming now that $$A$$ and $$B$$ are topological spaces, each fiber, being a subset of $$A$$, inherits the subspace topology relative to $$A$$

So, to say that a continuous map of topological spaces $$p : A \to B$$ has "discrete fibers" means that each fiber is a discrete topological space with respect to the subspace topology relative to $$A$$.

Your second bullet point (regarding older version of post): I'm not sure why you are concerned with whether a two point set is a union open sets. Nothing in the definitions requires a fiber $$p^{-1}(x)$$ of a covering map $$p : C \to X$$ to be a union of open set; in fact that is practically never the case.

Perhaps you are misreading the definition? You have only partially typed out the relevant part of the definition, leaving (...) for the part which explains this issue in full. Suffice it to say that in that portion, the definition is not referring to $$p^{-1}(\{x\})$$ being a union of open sets for a point $$x \in X$$. Instead, it is referring to $$p^{-1}(U)$$ being a union of open subsets of $$C$$, where $$U$$ is itself an open subset of $$X$$.

Perhaps one thing that might help is to learn the following theorem/definition (which you ought to be able to find in most textbooks where covering maps are defined, such as Munkres "Topology"):

If $$p : C \to X$$ is a covering map and if $$X$$ is a path connected then for all $$x_1,x_2 \in X$$ the fibers $$p^{-1}(\{x_1\})$$ and $$p^{-1}(\{x_2\})$$ have the same cardinality. That cardinality is called the degree of $$f$$.

Once that theorem is proved, another definition of a double covering map is a covering map of degree $$2$$.

Perhaps another thing that might help is to see some familiar examples of double covering spaces:

• The circle $$S^1$$ is a double covering space over itself. Using a complex coordinate $$S^1 = \{z \in \mathbb C \mid |z|=1\}$$, the self-double covering map $$p : S^1 \to S^1$$ is $$p(z)=z^2$$.
• The 2-sphere $$S^2$$ is a double covering space over the real projective plane $$\mathbb R P^2$$. The description of the double covering map depends on how $$\mathbb R P^2$$ is defined, but it is often defined as the set of lines through the origin of $$\mathbb R^3$$ (with an appropriate description of the topology), where the map $$p : S^2 \mapsto \mathbb R P^2$$ is defined by taking $$p(x)$$ to be the line through $$x$$ and the origin (for each $$x \in S^2$$).
• Thank you very much for the detailed explanation! "Second bullet point": Indeed, I misread the definition. I realized that and decided to edit the question, however, while I was editing, you posted your answer. Is it okay for you if leave it that way? I think the second bullet was just a dumb mistake and would distract from the actually important question. I'm sorry that I didn't realize that earlier. In any case, your examples are great! – Filippo Apr 17 at 17:13
• What I did was to add a parenthetical mark to my comment on that point. – Lee Mosher Apr 17 at 20:56

In fact discrete fiber means a space with the discrete topology. Therefore a fiber bundle $$p : C\to X$$ with discrete fibers is characterized by the property that for each $$x \in X$$ there exist an open neigborhood $$U$$, a discrete space $$F$$ and a homeomorphism $$h : p^{-1}(U) \to U \times F$$ such that $$p\mid_{p^{-1}(U)} = p_U \circ h$$, where $$p_U : U \times F \to U$$ denotes the canonical projection map.

1. If $$p : C \to X$$ is a fiber bundle with discrete fibers and $$U, F, h$$ are as above, then $$p^{-1}(U) = \bigcup_{f \in F}h^{-1}(U \times \{f\})$$. The $$V_f = h^{-1}(U \times \{f\})$$ are pairwise disjoint open subsets of $$p^{-1}(U)$$ (and hence open subsets of $$C$$) which are mapped by $$p$$ homeomorphically onto $$U$$. This means that $$p$$ is a covering map.

2. If $$p : C\to X$$ is a covering map, then $$p^{-1}(U) = \bigcup_{f \in F}V_f$$ with pairwise disjoint open subsets of $$C$$ which are mapped by $$p$$ homeomorphically onto $$U$$. These are called sheets over $$U$$. Give $$F$$ the discrete topology. Then $$h : \bigcup_{f \in F}V_f \to U \times F, h(c) = (p(c),f)$$ for $$c \in V_f$$, is a homeomorphism. Clearly $$p\mid_{p^{-1}(U)} = p_U \circ h$$.

Note that we do not require that a fiber bundle with discrete fibers has the same fiber $$F$$ over all of $$X$$, it may vary with $$U$$. This reflects the fact that for a covering map $$p$$ the cardinality of the index set $$F$$ in $$p^{-1}(U) = \bigcup_{f \in F}V_f$$ may vary with $$U$$. In fact the cardinality of $$F$$ is the same as that of all fibers $$p^{-1}(x)$$ with $$x \in U$$. Just note that $$p^{-1}(x) \hookrightarrow p^{-1}(U) = \bigcup_{f \in F}V_f \stackrel{h}{\to} U \times F \stackrel{p_F}{\to} F$$ is a bijection. The cardinality of $$p^{-1}(x)$$ is therefore locally constant, but not necessarily constant on all of $$X$$. However, $$X$$ can be decomposed in the disjoint union of open $$X_\alpha \subset X$$ such that each $$p_\alpha : p^{-1}(X_\alpha) \to X_\alpha$$ is a covering map whose fibers $$p^{-1}(x)$$ all have the same cardinality. In that case the "fiber bundle interpretation" works with a fixed discrete fiber $$F$$ - as one usually expects for fiber bundles.

A double cover is one having exactly two sheets everywhere. This is equivalent to $$p^{-1}(x)$$ having two elements for all $$x \in X$$.