Covering map associated to a fibration

139 Views Asked by At

I am reading about the generalization of the monodromy action to fibrations.

In this settings, the lift of a path is not unique, but it is unique up to free homotopy. In particular there is a well defined right action of $\pi_1(X,x)$ on $\pi_0(F_x)$ ($F_x$ is the fiber over $x$) which sends a path component of the fiber and a homotopy class of loops, to the path component of the ending point of a lift of a path representing the class in $\pi_1(X,x)$.

Of course if $F_x$ is discrete, then $\pi_0(F_x) \cong F_x$ and we recover the usual monodromy for coverings.

This got me thinking about the following construction.

Let $p: E \to X$ be a fibration (not sure which hypothesis you need). Consider the set: $$ \pi_0^v(E) := \bigsqcup_{x \in X} \pi_0(p^{-1}(x))$$ with the function $\pi_0^v(p) : \pi_0^v(E) \to X$ defined by $$\pi_0^v(p)|_{\pi_0(p^{-1}(x))} : [e] \mapsto x$$ I used a superscript $v$ for "vertical".

By putting some topology on $\pi_0^v(E)$ (quotient topology?) we could end up with a covering map. At this point, for example, the two notion of monodromy action will coincide.

This construction might be a functor from some category to the category of coverings.

Questions: Is this construction known/used? (First: Does it actually work?)

I am not asking to provide full details (but of course they are gladly accepted). I am mainly interested in knowing if this is known to work and used somewhere in mathematics.

1

There are 1 best solutions below

1
On BEST ANSWER

In general your construction does not work.

Consider fibrations with unique path lifting. These are more general than covering maps. A nice treatment an be found in

Spanier, Edwin H. Algebraic topology. Springer Science & Business Media, 1989.

It is easy to see that a fibration has unique path lifting if and only each fiber has only constant paths. In that case $\pi_0(p^{-1}(x))$ can be naturally identified with $p^{-1}(x)$ so that $\pi_0^v(E)$ is the same set as $E$. Thus the quotient map $E \to \pi_0^v(E)$ is a bijecton and (if you are not willing to give $\pi_0^v(E)$ an arbitrary topology) you will agree that the only reasonable topology on $\pi_0^v(E)$ is that making $E \to \pi_0^v(E)$ a homeomorphism.

Now there are fibrations with unique path lifting which are no covering maps, for example the infinite product $\Pi_{n=1}^\infty p_n : \Pi_{n=1}^\infty\mathbb R \to \Pi_{n=1}^\infty S^1$, where each $p_n(t) = e^{2\pi it}$ (see Example 9 in Spanier, Ch. 2 Sec. 2). For such maps your construction does not produce a covering map. Note that both $\Pi_{n=1}^\infty\mathbb R$ and $\Pi_{n=1}^\infty S^1$ are nice in the sense that they are path connected and locally path connected.

In this context also have a look at Theorem 10 in in Spanier, Ch. 2 Sec. 4 which says that if $p : E \to B$ is a fibration with unique path lifting such that $B$ is locally path connected and semilocally $1$-connected and $E$ is locally path connected, then $p$ is a covering map.