Consider a continuous map $\alpha:A\to B$ with the homotopy lifting propery and the unique path lifting property. Consider the induced fiber functor $$F:\pi_1B\longrightarrow \mathsf{Set}$$ taking a point $b\in B$ to its fiber $\alpha^{-1}(b)$ and acting on homotopy classes by unique path lifting.

If $A,B$ are locally path connected and moreover $B$ is semilocally simply connected, HLP+UPL are equivalent to $\alpha$ being a covering map. Under these conditions:

Is $F$ being fully faithful by any chance equivalent, or at least related to $\alpha$ being a universal covering?

Fullness means every set function $\alpha^{-1}(b)\to \alpha^{-1}(b^\prime)$ is induced by some homotopy class of path while faithfulness means distinct homotopy classes induce distinct lifting functions.


This is never going to be full, if the fibers aren't singletons, because the action is by automorphisms. It won't even be full onto the core of automorphisms in essentially any case. In the universal cover, the fiber is identified with $\pi_1,$ and the fiber functor is identified with the left multiplication action of $\pi_1$ on itself. So this is only going to be full if every set automorphism of $\pi_1$ is induced by left multiplication by some element, which is almost always impossible, which you can see for finite groups just by cardinality counting. The functor is faithful for the universal cover, though, since no nonidentity group element fixes the group under left multiplication.

A more robust fiber functor lands in $\pi_1$-sets, and in that case your claim is true.

  • $\begingroup$ Thank you for the answer. Could you please write which "more robust fiber functor" would satisfy the equivalence "fully faithful iff universal cover"? $\endgroup$ – Arrow Dec 14 '17 at 20:07
  • $\begingroup$ Actually, I don't think the idea I had in mind works. The nearby idea that is true is that the category of coverings of $X$ is equivalent to the category of $\pi_1X$-sets. In particular, the universal coverings are exactly those whose fiber, as a $\pi_1X$-set, is $\pi_1X$. $\endgroup$ – Kevin Arlin Dec 14 '17 at 20:32
  • $\begingroup$ It is also useful to think of covering morphisms of groupoids. See for example mathoverflow.net/questions/159663/functors-and-coverings/… $\endgroup$ – Ronnie Brown Dec 23 '17 at 11:40

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