What is the point of a lift in topology?

I've just covered 'lifts' in topology and also homotopy lifting to a covering map but I'm struggling to understand the intuition behind lifts and essential the 'point' of them.

Could someone please explain this intuitively? Thanks!

• Liftings of paths are closely related to actions of the fundamental group. As for covering spaces, they are roughly analogous to field extensions in algebra, with a universal covering space being analogous to an algebraic closure of a field. – KCd Dec 20 '12 at 13:21

Probably one of the first examples of a covering map you'll see is $p:\Bbb R\to S^1$, where $$p(x)=e^{2\pi ix}$$ (that is, when $x$ goes from $n$ to $n+1$ in $\Bbb R$, $p(x)$ makes one round around $S^1$ counter clockwise). Now take a path $\alpha:I\to S^1$ such that $\alpha(0)=1$ (where $I=[0,1]$). We can prove that there exists a unique lift $\tilde\alpha:I\to\Bbb R$ such that $\tilde\alpha(0)=0$ and $p\circ\tilde\alpha=\alpha$. This lift basically "straightens out" the path $\alpha$.
Start at $t=0$, $\alpha(0)=1\in S^1$. There is now a covering neighborhood for $1$ which means that if you change $t$ little enough so that $\alpha(t)$ stays in this neighborhood, there is only one possible way to define $\tilde\alpha$ such that $p\circ\tilde\alpha(t)=\alpha(t)$ still holds, namely $\tilde\alpha(t)=p^{-1}\circ\alpha(t)$ (since $p^{-1}$ exists in this neighborhood). Notice that if $\alpha(t)$ starts going counter clockwise, $\tilde\alpha(t)$ must go to the right in $\Bbb R$, and vice versa.
You continue like this, always getting a little bit forward, until you get to $t=1$. Notice that if $\alpha$ makes a full trip around $S^1$, $\tilde\alpha$ will travel one unit (e.g. from $0$ to $1$). In the end, $\tilde\alpha(1)$ will tell you how many times, and in which direction, $\alpha$ has travelled around $S^1$ in total. With this you can then prove that the fundamental group of $S^1$ is isomorphic to $(\Bbb Z,+)$.