What does the degree of a loop in $S^1$ means *intuitively? I'm trying to understand what the degree of a loop in $S^1$ means intuitively from the definition and how it relates to the winding number of a loop in $\mathbb{R^2}$.
My lecture notes define degree to be

Let $p$ be a loop in $S^1$, and let $\tilde p$ be a lift of $p$.
  Define the degree of $p$ to be $\tilde p(1) - \tilde p(0)$.

What does this definition actually mean intuitevly, and how does it relate to 'the number of times it winds round itself'?
Also, either I've missed something or my lecture notes seem to assume that the degree in in $\mathbb{Z}$, is this an obvious fact that I'm just not seeing? 
Thank!
 A: The universal covering space for $S^1$ is $\mathbb{R}$. You can think of this as coiling $\mathbb{R}$ up into a big helix, with any two points sitting directly above/below one another if they differ by an integer. This helical $\mathbb{R}$ sits above $S^1$, with $0 \in \mathbb{R}$ (and thus all integers) sitting above the point $(1,0) \in S^1$; and more generally, $\frac{\theta}{2\pi} \in \mathbb{R}$ sits above $(\cos \theta, \sin \theta) \in S^1$.
Here's a picture I drew in my lecture notes illustrating this: (Careful, we used '$p$' to mean different things: my $p$ is the covering map, not a loop.)

Now, the lift of a path in $S^1$ is, intuitively, the path traced by starting at some point in $\mathbb{R}$ and following it round on the helix continuously. Moving anticlockwise around the circle makes you move up the helix, and moving clockwise makes you move down. Thus, for instance, if you start at $(1,0) \in S^1$ and do one anticlockwise loop around the circle, then the path you trace on $\mathbb{R}$ (i.e. the lift of the loop) starts at $0$, winds up the helix and ends up at $1$. If you go round the circle $n$ times then you end up at $n$... and so on.
So if you have a loop in $S^1$ based at $(1,0)$, and you lift it to a path in $\mathbb{R}$ starting at $0$, then the place $n$ where it finishes in $\mathbb{R}$ is going to be the number of times you performed a whole loop around the circle clockwise, minus the number of times you performed a whole loop of the circle anticlockwise. This is $\tilde p(1) - \tilde p(0)$ in the case when your loop starts at $(1,0)$ and the basepoint lifts to $0$. But more generally, the same holds: to see why, just translate everything by rigid motion to the above scenario!
Sorry for my lack of decent illustrations $-$ if you need more explanation then please let me know.
