The map $f:S^1 \to S^1, f(z) = z^k$ is well-defined only if $k \in \mathbb{Z}$ Is the map $f:S^1 \to S^1, f(z) = z^k$ well-defined only if $k$ is an integer? 


I believe that $k$ needs to be at least a rational number, because $(-1)^k$ is not really defined when $k$ is irrational. However, I don't know how to continue from here.
You need $k$ to be an integer for $f(e^{0i})=f(e^{2\pi i})$. And this equality needs to hold for well-defined-ness.
You need $k$ to be an integer for $f(e^{0i})=f(e^{2\pi i})$. And this equality needs to hold for well-defined-ness.