# Closure by Projective Limits of the category of Coverings of a Topological Space

Let $X$ be a connected topological space, and $C_{finite}$ the category of its finite coverings. Then I claim that the category $C$ of coverings of $X$ can be obtained by $C_{finite}$ taking projective limits and disjoint unions.

Since projective limits and disjoint unions commute the statement is reduced to say that $A:= \varprojlim A_i$, where each $A_i$ is a finite covering, is itself a covering space of $X$. For the proof this is my problem: Fixed a point $x\in X$, and a point $\{y_i \in A_i\} \in A$ of the fiber over $x$, then I must find an open neighborhood $x\in U \subset X$ such that it is homeomorphic to a convenient neighborhood $V$ of $\{y_i\}\in A$.

The passage is problematic because, if the limits is taken over an infinite index set $I$, then I'm not allowed anymore to restrict this neighborhood $U$ in a convenient way for each $A_i$ (Infinite intersection of opens can be not open).

There is a way to bypass this problem? Or my claim is simply wrong?

Probably in the solution it will be helpful (or necessary) to restrict to a base space $X$ which admits a universal covering, i.e. it is locally path connected and semi-locally simply connected.

Edit: I forgot to mention the following fact, which justify my claim. If $X$ admits a universal covering, then it can be obtained as projective limit of finite (normal) coverings of $X$.

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I don't see how your claim works, even in the simple case where $X$ is a circle. The universal covering space is the real line, winding infinitely often around the circle. The fiber over any point of the circle is a countably infinite discrete space. But a projective limit of finite coverings would, as far as I can see, have fibers that are projective limits of finite sets, so the fibers would be compact.