If $T$ is a covering space of $X$ then every loop in $T$ must be a lift of loop in $X$?

Question

I am trying to understand the proof of the following theorem (from the book: Homotopical Topology by Fomenko & Fuchs):

Before this theorem the book only talks about paths. Couldn't it be a loop $\alpha:I \rightarrow X$ such that the lift of $\alpha$ is only a path (not a loop)? Is this why this map is not an isomorphism? Could we have a loop $\beta: I \rightarrow T$ that is not a lift of any loop in $X$? Really would appreciate help here.

Answer 1

You’re correct that the lift of a loop is in general a path, so that the given map is not necessarily an isomorphism.

The image of any path (i.e. an image of some map from $[0,1]$) under the covering map is again a path, and if the path is a loop (i.e. its endpoints are the same point) then the image of this loop under the covering map is again a loop.

Answer 2

Of course loops in $X$ may have lifts which are no loops. As an example consider $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$. The loop $\alpha : I \to S^1, \alpha(t) = p(t)$, lifts to $\tilde \alpha : I \to \mathbb R, \tilde \alpha(t) = 2\pi t + 2k \pi$ with $k \in \mathbb Z$. These maps are non-closed paths.

However, any loop $\beta : I \to T$ is a lift of $p \circ \beta : I \to X$ which is a loop.