On the fundamental group of $S^{1}$: why is there a strict correspondence between maps from the helix and maps on the circle?

Question

I'm currently reading Nakahara's Geometry, Topology, and Physics, and he makes a claim that I find confusing in his derivation of the fundamental group for $S^{1}$. Instead of doing a standard proof via covering spaces, he argues that $S^{1}\cong \mathbb{R}/2\pi\mathbb{Z}$ by considering the map

$$p:\mathbb{R}\rightarrow S^{1}, p: x\mapsto e^{ix}$$

OK, that's fine. A natural equivalence class to consider is $$[x] = \{y | x-y=2\pi m, m\in\mathbb{Z}\}$$

and we say $x\sim y$. Now let $$\tilde f: \mathbb{R}\rightarrow\mathbb{R}$$

be a map which satisfies $\tilde f(0) = 0$ and $\tilde f(x+2\pi) = \tilde f(x)+2n\pi$, $n\in\mathbb{Z}, x\in\mathbb{R}$. It is easy to check that $x\sim y \Rightarrow \tilde f(x)\sim\tilde f(y)$. Hence $\tilde f$ and $p$ induce natural maps on $S^{1}$: $$f: S^{1}\rightarrow S^{1}, f([x]) = p(\tilde f(x))$$

Nakahara calls the point $1\in S^{1}$ the base point of $S^{1}$, as it is obtained by mapping $0\in\mathbb{R}$. Here comes the claim I find dubious: Nakahara says, and I quote:

"In summary, there is a one-to-one correspondence between the set of maps from $S^{1}$ to $S^{1}$ with $f(1)=1$ and the set of maps from $\mathbb{R}$ to $\mathbb{R}$ such that $\tilde f(0) = 0$ and $\tilde f(x+2\pi) = \tilde f(x)+2\pi n$."

I don't understand this at all. For example, what happens if I cook up a function which preserves the base point on the circle but winds around the image circle an irrational number of times while the pre-image circle is only wound once? In particular, the function

$$f(e^{i\theta}) = e^{i\pi\theta}$$

preserves the base point, but doesn't seem to be associated with an $\tilde f$ that satisfies the condition $\tilde f(x+2\pi) = f(x)+2\pi n$. Am I missing something, or is this just a glaring error? Many thanks for any assistance you can offer, and I apologize if this is somewhat inelegant - I'm really a physicist.

Answer 1

If your function winds around the circle an irrational (or merely non-integer) number of times as its input winds around the circle once, then it is not a well-defined continuous function on $S^1$. For instance, take your proposed function $$f(e^{i\theta})=e^{i\pi\theta}.$$ Note that $e^{i\cdot 0}=e^{i\cdot 2\pi}$, so for $f$ to be well-defined, it must give the same values for $\theta=0$ and $\theta=2\pi$. But it does not: $$e^{i\pi\cdot 0}=1\neq e^{2\pi^2i}$$ since $2\pi^2$ is not an integer multiple of $2\pi$.

That is, since $f$ is supposed to be a function on $S^1$, when you wind the input around the circle once, the output must be the same (since you have the exact same point in $S^1$). So the output must have wound around the circle an integer number of times, to end up back at the same output. (Of course, this is not a completely rigorous argument, but since you're really a physicist I'll leave the technical details for you to read elsewhere if you want them.)

Answer 2

You could try to cook up a map $S^1 \mapsto S^1$ that preserves the base point but winds around an irrational number of times.... but.... you would fail.

For example, your map $f(e^{i\theta})=e^{i\pi\theta}$ is not well-defined, because $$e^{i\theta}=e^{i(\theta+2\pi)} $$ but $$e^{i\pi\theta} \ne e^{i\pi(\theta+2\pi)} = e^{i(\pi\theta + 2 \pi^2)} $$

Maps $S^1 \mapsto S^1$ taking base point to base point can only wind around an integer number of times; that's what's going in this proof.