# The universal cover covers any connected cover

The universal cover (of the space $$X$$) covers any connected cover (of the space $$X$$).

I'm not able to find a proof of this result. Any idea how to find it or how to prove it?

Here the definition of universal cover is just that of a covering being simply connected. Moreover, a part from connectedness of the other cover there is no other hypothesis.

Sofie Verbeek's answer is perfect, but let me go back to the definition of a universal covering space. There are two different approaches: Some (perhaps most) authors define a universal covering $$p : E \to X$$ of a connected space $$X$$ to be one such that $$E$$ is simply connected, other authors define it by the property that it covers any covering $$p' : E' \to X$$ with a connected $$E'$$.

In my opinion the second definition is the better one because it is explains the name. If you accept this point of view, then it is a theorem that a simply connected covering of a connected and locally path connected $$X$$ is a universal covering.

A nice reference is Chapter 2 Section 5 of

Spanier, Edwin H. Algebraic topology. Vol. 55. No. 1. Springer Science & Business Media, 1989.

It contains a number of interesting results and examples (e.g. of a non-simply connected universal covering space of a of a connected and locally path connected space).

Hint. Covering spaces satisfy the path lifting property. That is, given a path in your base space $$X$$ from a point $$x$$ to a point $$y$$, together with a given lift of $$x$$ in your covering space $$E$$, there is a unique lift of the path to the covering space, starting at the lift of $$x$$. Prove that if your covering space $$E$$ is simply connected, the end-point of the lifted path does not depend on your initial choice of path from $$x$$ to $$y$$. It is at this point that the property of being simply connected is crucial.

Next, use this special property to manually define a map from the simply connected covering $$E$$ to any other covering space $$E'$$, and prove that the map $$E \to E'$$ that you constructed is a covering of $$E'$$ by $$E$$.