This is Example 2.13 (b) from Lee's Introduction to Smooth Manifolds.

If the circle $\mathbb{S}^1$ is given its standard smooth structure, the map $\epsilon: \mathbb{R} \to \mathbb{S}^1$ defined by $\epsilon(t)=e^{2\pi it}$ is s mooth, because with respect to any angle coordinate $\theta$ for $\mathbb{S}^1$ it has a coordinate representation of the form $\hat{\epsilon}(t)=2\pi t+c$ for some constant $c$, as you can check.

Here $\theta$ is a function from some open subset $U \subset \mathbb{S}^1$ to $\mathbb{R}$ such that $e^{i\theta (z)}=z$ for all $z \in U$. So we have $\hat{\epsilon} = \theta \circ \epsilon$. I can't see why we have $\hat{\epsilon}(t)=2\pi t+c$ for some constant $c$ here. I would appreciate any help.


Well, $e^{i\theta (z)}=z\implies \theta(z)=\ln z/i$.

Note, your $S^1$ sits in $\Bbb C$.

Now, we need a branch of log, because the complex logarithm is not a single valued function. The principal branch can be assumed, for instance, in which case, we have $\ln z=\ln r +i\theta$, where $z=re^{i\theta}$.

In this case your $c$ will be $0$. For a different branch of log, it will be $\ln z=\ln r+i\theta+2\pi ki$. See https://math.stackexchange.com/a/1475592/403337.

So let's compose and see what we get: $\hat\epsilon(t)=\theta\circ\epsilon(t)=\theta(e^{2\pi i t})=\ln(e^{2\pi i t})/i=2\pi t+2\pi k =2\pi t+ c$, where $c=2\pi k $.


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