Let $K$ be a non-archimedean local field and $\pi$ a uniformizer of $O_K$. Suppose I have two continuous homomorphisms $\phi$ and $\phi'$ from $K^*$ to $Gal(K^{ab}/K)$. Suppose $\phi(\pi) = \phi'(\pi)$. Then how does it follow that $\phi = \phi'$? Thank you!
(I am trying to understand the proof of Theorem 1.13 on page 24 of Milne http://www.jmilne.org/math/CourseNotes/CFT.pdf)
ps Here $K^{ab}$ is the union of all finite abelian extensions of $K$ in some fixed separable closure of $K$.
I think what they have shown is that $\phi(\pi) = \phi'(\pi)$ for any two homomorphisms $\phi, \phi'$ as well as for any uniformizer $\pi$.
Then it follows, since any element in $K^\times$ can be expressed as a product of uniformizers.