# Homotopy classes of $X$ and that of its universal covering space.

I am reading 1.3 Covering space of Hatcher's 'Algebraic Topology' and I cannot find out what I am missing.

In the book, he says

"The advantage of this is that, by the homotopy lifting property, homotopy classes of paths in $$\widetilde{X}$$ starting at $$\widetilde{x_0}$$ are the same at homotopy classes of paths in $$X$$ starting at $$x_0$$."

where $$\widetilde{X}$$ is a universal covering of $$X$$ ($$p:\widetilde{X}\rightarrow X$$).

I don't really get it. For example, I think, we can build two different lifts, $$f_1$$ and $$f_2$$ of a path $$f$$ in $$X$$ such that $$f_1\not\simeq f_2$$.

Let $$p : \tilde{X} \to X$$ be any covering (it may be the universal covering, but we do not require it). Hatcher proves that for any homotopy $$h : Y \times I \to X$$ and any lift $$\tilde{h}_0 : Y \to \tilde{X}$$ of $$h_0$$ (where $$h_t : Y \to X, h_t(y) = h(y,t)$$) there exists a unique lift $$H : Y \times I\to \tilde{X}$$ of $$h$$ such that $$H_0 = \tilde{h}_0$$.
This implies the unique path lifting property: Given a path $$u : I \to X$$ starting at $$x_0 \in X$$, then for any $$\tilde{x}_0 \in p^{-1}(x_0)$$ there exists a unique lift $$U : I \to \tilde{X}$$ such that $$U(0) = \tilde{x}_0$$.
You are right, a path $$u$$ in $$X$$ can have lifts which are not homotopic as paths ( a homotopy of paths is one keeping the end-points fixed). However, if both lifts start at the same point, then they are identical.