Intuition behind the fundamental group $\pi_1(S^1)$

Question

I have gone over the fact that $\pi_1(S^1) \cong \mathbb{Z}$ and its proof, though I admittedly do not understand it in full. What I am struggling with is the intuition behind this fact, since as I understand it, every integer $n$ can be thought of as how many revolutions we make around the circle. However in view of the intuition behind homotopy and deformations, regardless of how many times we loop through the circle, is the deformation not going to be the same, that is, result in $S^1$? How then do different $n$ represent distinct homotopy classes?

Answer 1

I think the key to your misunderstanding is that a "loop" in $S^1$ in the definition of the fundamental group is a function $l:[0,1]\to S^1$, not a subset of $S^1$. This is quite a subtle point, but absolutely crucial. If a loop was defined as a sunset (in particular as the image of the function $l$, what is normally referred to as the trace of $l$) then you would be right, a loop which goes around $S^1$ once is exactly the same as a loop which goes around it a million times. However, a function which wraps $[0,1]$ around $S^1$ is completely different to a function which wraps $[0,1]$ around $S^1$ a million times.

Let's be concrete to illustrate this. Identify $S^1$ with the unit circle in $\mathbb{C}$ and define $l_1:[0,1]\to S^1:t\mapsto e^{2\pi i t}$, and define $l_2:[0,1]\to S^1:t\mapsto e^{2\pi i (2t)}$. These both trace out $S^1$, so their images are equal as sets. However, at $t=1/2$ for example, $l_1(1/2)=-1$, while $l_2(1/2)=1$, so these are different as functions. In particular we can think of $l_1$ as representing $1$ in $\pi_1(S^1)$, and $l_2$ as representing $2$.

Answer 2

This, I think, is quite a non-standard way of looking at this but I think it might help visualise why these loops are fundamentally different.

Split $S^1$ in half at the base point to form the unit line $[0,1]$, with the basepoint corresponding to $0$ and $1$. For ease lets assume that any loop in $S^1$ corresponds to starting at $0$ in $[0,1]$. Then a loop in $S^1$ that goes around just once is a path in $[0,1]$ that starts at $0$ and ends at $1$. A loop that goes around $S^1$ twice starts at $0$, goes to $1$ and then reappears at $0$ and continues to $1$. There is clearly no way to construct a homotopy that makes a "reappearing" loop into one that doesn't at all. For the "negative" loops in $S^1$ you can use the same argument but with the path starting at $1$ instead.

Like I said this reasoning is not standard and not at all precise, but I think it illustrates in some sense the differences between the homotopy classes in $S^1$.