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A covering map $p:E \rightarrow B$ is continuous map satisfying the following:

  • For every point $b\in B$, there exists a neighborhood $U$ of $b$, such that $p^{-1}(U)$ is a disjoint union of open sets, each of which is mapped homeomorphically onto $U$ by $p$. (see Covering Spaces)

A classic (non-trivial) example of such a map is $p:\mathbb{R} \rightarrow S^1$ defined by $x \longmapsto e^{2\pi ix}$.

These maps have some amazing properties, for example:

  • Path-lifting: for every path $f:[0,1] \rightarrow B$, and for every $e \in p^{-1}(f(0))$ there exists a (unique) continuous map $\tilde{f}:[0,1] \rightarrow E$ with $\tilde{f}(0) = e$

There is also the Homotopy lifting property, the Uniqueness of lifts, and the "General lifting property", see lifting properties.

My question is: Do the lifting properties, under suitable conditions on $E$, characterize covering maps?

Since this is my first question on Math Stack Exchange, my question may need further elaboration, which I would happy to do.

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Yes, lifting information classifies covering spaces (assuming the base space is nice, e.g., connected, locally path connected and semilocally simply connected - just start with a CW complex if you don't want to fuss about the niceness conditions)

If $p : (E, \tilde{x_0}) \to (B, x_0)$ is a (based) covering space of $B$, then given an open cover $\{U_i\}$ of $B$, $E$ can be built from gluing copies of $U_i \times p^{-1}(x_0)$ appropriately. That is to say, $E$ is a quotient space of $\bigsqcup U_i \times p^{-1}(x_0)$ such that the map given by projection $U_i \times p^{-1}(x_0) \to U_i$ for each copy in the disjoint union induces the map $p$ from the quotient $E$ - this follows immediately from the trivialization property of covering spaces.

So how does one glue, say, $U_i \times p^{-1}(x_0)$ and $U_j \times p^{-1}(x_0)$? Suppose $U_i \cap U_j$ is nonzero in $B$ - then and only then I'll glue $U_i \times p^{-1}(x_0)$ to $U_j \times p^{-1}(x_0)$, along $(U_i \cap U_j) \times p^{-1}(x_0)$ by a self-homeomorphism $(U_i \cap U_j) \times p^{-1}(x_0) \to (U_i \cap U_j) \times p^{-1}(x_0)$. $p^{-1}(x_0)$ is a discrete set, so this amounts to saying we're choosing an automorphism/self-bijection of $p^{-1}(x_0)$ for each nonempty overlap $U_i \cap U_j$, aka, an assingment $U_i \cap U_j \to \text{Aut}(p^{-1}(x_0))$ - such an assignment for each nonempty overlap determines the cover. These are known as the transition functions of the cover.

We'll see how the transition functions of the cover are determined by lifting property of the cover. Suppose $\gamma$ is a loop based at $x_0$ in $B$, and I lift it in $E$ to $\tilde{x_0}$. Then we know we get a path starting at $\tilde{x_0}$ and ending at something else in $p^{-1}(x_0)$, so going around a loop downstrairs you may not return upstairs - why does this really happen? Well, precisely because of the transition functions! $\gamma$ runs through various open sets in the cover $\{U_i\}$ in $B$ - passing each overlap changes it's endpoint in upstairs due to nontriviality of the transition functions, hence when you come back you might end up at something entirely different!

More formally, lifting loops based at $x_0$ in $B$ to paths starting at different points in $p^{-1}(x_0)$ gives me an action of $\pi_1(B, x_0)$ on $p^{-1}(x_0)$ given by $[\gamma] \cdot x \mapsto [\tilde{\gamma}_x(1)]$ where $\tilde{\gamma}_x$ is the lift of $\gamma$ to $E$, with starting point $x \in p^{-1}(x_0)$. This action in turn gives me a map $\pi_1(X, x_0) \to \text{Aut}(p^{-1}(x_0))$ - this is called the monodromy of the cover. By now you're probably convinced that this is the homomorphism which completely determines the cover, which it indeed does.

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    $\begingroup$ Many thanks! I found your answer very useful, although I will have to clarify all the details for myself. $\endgroup$ – Mike Jul 8 '16 at 13:56
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    $\begingroup$ @MichaelHoefnagel What I gave was more of an intuition than a proof of the fact. But you can find a rigorous, although maybe not quite illuminating, proof in Hatcher's algebraic topology book. $\endgroup$ – Balarka Sen Jul 8 '16 at 14:52

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