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!
 A: There are many answers to this question, all pertaining to the fact that lifts are extremely important in homotopy theory. For instance, automorphisms of covering spaces can be used to calculate fundamental groups, and the homotopy lifting property is used to define fibrations. I don't want to bombard you with examples, so I'll leave it at this; but if you read ahead in your textbook you'll probably find liftings all over the place.
A: 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 $(\Bbb Z,+)$.
