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One geometric way to see a continuous map (or any set function really) is as a "fiber bundle" with the usual picture of a comb - the base space indexes the fibers of the map and there's a nice picture in a relative setting.

I was wondering whether there is a nice way to visualize covering maps in this context, via the fibers. Maybe this is the standard way, but I have really poor geometric intuition so I wasn't able to figure that out.

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  • $\begingroup$ I don't understand the question. A covering map is a map, and as you say in your first paragraph you can visualize maps in the way you describe. $\endgroup$ – Mariano Suárez-Álvarez Apr 4 '16 at 13:42
  • $\begingroup$ @MarianoSuárez-Alvarez I'm trying to visualize the definition of a covering maps in terms of how the "fiber bundle" may or may not look like. $\endgroup$ – Arrow Apr 4 '16 at 13:45
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Yes, a covering map $f : X \to Y$ can be thought of as a fiber bundle, such that the fibers are discrete topological spaces. The path lifting lemma, which asserts existence and uniqueness of lifts of a path with a given lift of the initial endpoint, is the tool needed to verify the axioms of a fiber bundle.

One needs some assumptions on $Y$ to make sure this works well. Perhaps some weaker assumptions will work, but let me simplify matters and make a strong assumption that $Y$ can be covered by a collection $\mathcal{U}$ of open sets such that each $U \in \mathcal{U}$ is path connected, each pairwise intersection $U \cap U'$ with $U,U' \in \mathcal{U}$ which is nonempty is also path connected, and each $U \in \mathcal{U}$ satisfies the "evenly covered" property in the definition of a covering map. This assumption holds, for example, if $Y$ is a manifold or a CW complex.

The fiber bundle structure over each $U \in \mathcal{U}$ is not hard to describe. There exists a decomposition $f^{-1}(U) = \bigcup_i V_i$ such that each $V_i$ is open in $X$ and $f : V_i \to U$ is a homeomorphism. Pick $y \in U$. Equip the fiber $F = f^{-1}(y)$ with the discrete topology. Enumerate $F=\{x_i\}$ where $\{x_i\} = V_i \cap F$. One immediately obtains a homeomorphism $h_U : f^{-1}(U) \mapsto U \times F$ taking each point $x \in f^{-1}(U)$ to $(f(x),x_i)$ so that $x \in V_i$.

There is also a fiber bundle axiom one must check for each $U,U' \in \mathcal{U}$ such that $U \cap U'$ is nonempty. Pick points $y \in U$ and $y' \in U'$, and pick a path $\gamma : [0,1] \to U \cup U'$ from $y$ to $y'$. The path lifting lemma determines a bijection $g : F \to F'$ from the fiber $F = f^{-1}(y)$ to the fiber $F'=f^{-1}(y')$, so that for each $x \in F$ the point $g(x) \in F'$ is the opposite endpoint of the unique lift of $\gamma$ with intial point $y$. Then one obtains an "overlap homeomorphism" $$O^{U}_{U'} : h_U^{-1}(U \cap U') = (U \cap U') \times F \mapsto (U \cap U') \times F' = h_{U'}^{-1}(U \cap U') $$ taking $(x,y) \mapsto (x,g(y))$ for each $x \in U \cap U'$ and $y \in F$. Path connectivity of $U \cap U'$ is then used, together with uniqueness of path lifting, to check that for each $y \in F$ the image under $O^{U}_{U'}$ of $(U \cap U') \times \{y\}$ equals $(U \cap U') \times \{g(y)\}$.

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