In a comment to this [question](https://mathoverflow.net/questions/429009/non-existence-of-higher-artin-map), David Loeffler asked if one can show that the (local) Artin map
$$K^\times \to G\_K^{ab}$$
does not have a lift to $G\_K$. Probably this wouldn't be canonical, but I can't show that such a lift doesn't exist.


So does there exist such a lift?



Existence of lift of (local) Artin map
No, there does not, except possibly a few small corner cases. The problem is finding somewhere for the torsion in $K^\times$ to go.


If $K$ is a finite extension of $\mathbf{Q}\_p$, then there exists a number field $\mathscr{K}$ and a prime $v$ of $\mathscr{K}$ above $p$ such that $\mathscr{K}\_v = K$; and hence we obtain an embedding $G\_K \hookrightarrow G\_{\mathscr{K}} \hookrightarrow G\_{\mathbb{Q}}$. But $G\_{\mathbb{Q}}$ is known to have no nontrivial elements of finite order except the conjugacy class of complex conjugation of order 2 (see [this question](https://mathoverflow.net/questions/6802/element-in-the-absolute-galois-group-of-the-rationals)).


So $K^\times$ cannot embed into $G\_K$ unless $(K^\times)\_{\mathrm{tors}}$ has order exactly 2, which can only happen if $p = 2$ or $p = 3$ (and I am too tired this evening to think about those cases).


