# Which properties determine the uniqueness of the global Artin map?

Let $L/K$ be a finite abelian extension of number fields. There is a well defined homomorphism $\Phi: J_K \rightarrow G(L/K)$, called the Artin map, with the following properties (among many):

(i) $\Phi$ is trivial on $K^{\ast}$.

(ii) $\Phi$ is continuous.

(iii) $\Phi$ is surjective.

(iv) If $v$ is an unramified finite place, and $\pi_v$ is a uniformizer or $K_v$ (interpreted as an idele $(...1, \pi_v, 1, ...)$), then $\Phi(\pi_v)$ is the Frobenius element at $v$.

(v) $\Phi$ is trivial on $N_{L/K}(\mathbb{I}_L)$.

These are all 'basic' properties, I say basic because even showing the existence of a well defined map $\Phi$ on the entire idele group is highly nontrivial. Usually one would begin with defining $\Phi$ on a certain subgroup $J_{\mathfrak c}$ (involving a so-called admissible cycle $\mathfrak c$) and then go about showing that $\Phi$ is trivial on $K^{\ast} \cap J_{\mathfrak c}$.

My question is what is the smallest combination of the above properties (if any) determine the uniqueness of $\Phi$? I have (I think) in my course notes a claim that if $\Phi$ is a homomorphism satisfying (i), (ii), and (iv), then $\Phi$ is unique. From just those three properties I can already get the other ones I mentioned, for example surjectivity is a consequence of (iv) and the cyclic global norm index equality.

Can anyone confirm or deny that (i), (ii), and (iv) uniquely determine the global Artin map?

A possible idea: let $A: \mathbb{I}_K \rightarrow G(L/K)$ be a homomorphism satisfying (i), (ii), and (iv). Let $\mathfrak c$ be an admissible cycle for $L/K$, and let $$H_{\mathfrak c} = \prod\limits_{v \mid \mathfrak c} W_v(\mathfrak c) \prod\limits_{v \nmid \mathfrak f}' K_v^{\ast}$$ where $W_v(\mathfrak c) = 1 + \mathfrak p_v^{\mathfrak f(c)}$ or $(0, \infty)$ depending on whether $v$ is finite or infinite. Since $K^{\ast} H_{\mathfrak c} = \mathbb{I}_K$ (a simple approximation theorem argument), given $\alpha \in \mathbb{I}_K$, we can find an $x \in K^{\ast}$ such that $\alpha x \in H_{\mathfrak c}$, and $A(\alpha)= A(\alpha x)$ by (i). So, we just have to show that $\Phi$ agrees with $A$ on some $H_{\mathfrak c}$.

What I think works: let $A: \mathbb{I}_K \rightarrow Gal(L/K)$ be a homomorphism satisfying (i), (ii), and (iv). Each $K_v^{\ast}$ inherits its topology as a subgroup of $\mathbb{I}_K$, so we can restrict $A$ to a map $A_v: K_v^{\ast} \rightarrow G(L/K)$. Then $A$ is just the product $\prod\limits_v A_v$. When $v$ is unramified and finite, $A_v: K_v^{\ast} \rightarrow Gal(L/K)$ does what we want by (iv).
When $v$ is ramified and finite, restrict $A_v$ to a continuous map $\mathcal O_v^{\ast} \rightarrow Gal(L/K)$. The preimage of $\{1\}$ is an open and closed subgroup of $\mathcal O_v^{\ast}$, necessarily containing $1 + \mathfrak p_v^n$ for some $n \geq 1$. We can enlarge $n$ to a number $n_v$ for which $1 + \mathfrak p_v^{n_v}$ is also contained in the group of local norms.
When $v$ is infinite, the preimage of $1$ under the map $K_v^{\ast} \rightarrow G(L/K)$ is an open and closed subgroup of $K_v^{\ast}$. If $v$ is real, this can either be all of $K_v^{\ast}$ or $(0, \infty)$. If $v$ is complex, this has to be all of $K_v^{\ast}$.
In any case, we can restrict $A$ to a homomorphism on $$H_{\mathfrak c} = \prod\limits_{v \mid \mathfrak c} W_v(\mathfrak c) \prod\limits_{v \nmid \mathfrak c}' K_v^{\ast}$$ for a suitable admissible cycle $\mathfrak c$, and here $A$ agrees with the global Artin map. Since $H_{\mathfrak c} K^{\ast} = \mathbb{I}_K$, $A$ agrees with the global Artin map everywhere by (i).