# Alternative definition of covering spaces.

in a lecture I have seen a definition of a covering space, different from what I would call the usual one (e.g. the one in Munkres): A surjective continuous map $p:E\rightarrow B$ of spaces $E$ and $B$ is a covering map (and $E$ a covering space of $B$) if for every $b\in B$ there is an open neighborhood $U$ of $b$, a discrete space $S$ and a homeomorphism $\varphi:p^{-1}(U)\rightarrow U\times S$ such that $p|_{p^{-1}(U)}=\varphi\circ\pi_U$ ($\pi_U$ the projection map)

The one in Munkres: a continuous surjective map $p:E\rightarrow B$ such that every $b\in B$ has a neighborhood $U$ evenly covered by $p$.

Now my question is: what's the connection between them, are these definitions equivalent? Has this something to with the fact that the fiber $p^{-1}(b)$ has the discrete topology as its subspace topology? Also, is there a book using the first definition?

• What is the definition of "evenly covered"? (Hint: it should be equivalent to the existence of a discrete $S$ such that ...) – Daniel Fischer Aug 1 '13 at 18:48
• Isn't the definition of $U$ being evenly covered that the preimage $p^{-1}(U)$ can be written as the disjoint union of open sets whose restriction to each open set is a homeomorphism? In that case, these are precisely the same definitions except that evenly covered is spelt out precisely in the former. Edit: Posted 11 seconds later and it looks like I'm answering that question :) – Tyler Holden Aug 1 '13 at 18:48
• The definition involving $U\times S$ comes from the more general definition of fiber bundle. The equivalence between the definitions tells you that a covering space is just a fiber bundle with discrete fibers. – Stefan Hamcke Aug 1 '13 at 18:58
• thanks guys, very helpful; in particular to Stefan to pointing me to this more general concept (in wikipedia they say that a covering space is a fiber bundle such that the projection is a local homeom., is this equivalent to the discreteness of the fiber?). So the point is that $p^{-1}=\cup_{\alpha\in S}V_\alpha\approx\bigsqcup_{\alpha\in S} U\approx U\times S$ (if $S$ is given the discrete topology)? Is this correct, in particular that $\bigsqcup_{\alpha\in S} U$ (disjoint union of sets, I mean formally disjoint union)? S ist just the fiber of $b$? – user88576 Aug 1 '13 at 23:04

You want to compare two definitions:

1. There is a continuous surjection $p:E\to B$ and each $b\in B$ has an open neighborhood $U$ evenly covered by $p^{-1}(U)$.

2. There is a continuous surjection $p:E\to B$ and for each $b\in B$ there is an open neighborhood $U$, a discrete space $S$ and a homeomorphism $\phi:p^{-1}(U)\to U\times S$ such that $\pi_U\circ\phi=p$.

So, let $p:E\to B$ be a continuous surjection and let $b\in B$.

Assuming definition $1$, there is a $U$ evenly covered by $(U_i)_{i\in S}$ and we can define $\sigma:\bigcup_{i\in S}U_i\to S$ by $\sigma(x)=i$ if $x\in U_i$. This map is continuous since $\sigma^{-1}(i)=U_i$ is open. Thus also the map $\phi:\bigcup_{i\in S}U_i\to U\times S, \phi=(p,\sigma)$ is continuous and has inverse $\phi^{-1}$ which maps $(y,i)$ to the unique point $x\in U_i$ which has $p(x)=y$. This $\phi^{-1}$ is continuous since $\phi(V)$ is a disjoint union of $p(V\cap U_i)\times\{i\}$. Hence $\phi$ is a homeomorphism.

Conversely, assume definition $2$. The preimage $p^{-1}(U)$ is the same as $\phi^{-1}(U\times S)$, which is a disjoint union of open $U_i=\phi^{-1}(U\times\{i\})$, each of which is mapped homeomorphically onto $U$.

If the fiber $S$ is constant for all $b\in B$, then definition $2$ is a special case of a fiber bundle. Then the equivalence says that a constantly sheeted covering space (for example if $B$ is connected) is just a fiber bundle with discrete fiber $S$.

Calling $S$ the fiber comes from the fact that $S\approx p^{-1}(b)$ for each $b\in B$, and these preimages of points are commonly known as "fibers". These fibers $p^{-1}(b)$ are discrete if $p$ is a local homeomorphism. On the other hand, each covering map is a local homeomorphism. This should answer your question in the comment above.

• Thank you, too kind! Everything is clear to me now. – user88576 Aug 3 '13 at 14:27