Let $L/K$ be a Galois extension of degree $n$. On page 195, in the book Algebraic Number Theory, in the Chapter on Global class field theory by Tate, he writes that the inv map
$$H^2(Gal(L/K),J_L)\stackrel{inv}{\longrightarrow} \Big(\frac{1}{n}\mathbb{Z}\Big)/\mathbb{Z}$$
need not be surjective in general. This seems implicit in the sentence above diagram (9), where he says that, "... the point is that $H^2(L/K,J_L)_{reg}\supset Im(\epsilon_1)$, but they will not in general be equal".
My question: Is the inv map above surjective when $L/K$ is a finite abelian extension?
No. Let us first recap the argument behind the proof of p.196 : to show the final RESULT that $H^2(L/K, C_L)_{reg} = H^2 (L/K, C_L)\cong \frac 1 n \mathbf Z /\mathbf Z$, one first shows that for a certain type of finite "cyclic cyclotomic" extensions $L/K$, the exact sequence (11) is exact; then, by an inductive limit argument, that analogous sequences are exact at the level of ${\bar K}^*$; the bijectivity of $\beta_1$ follows, which allows to conclude. Actually this implies that (11) is nothing but the upper exact sequence of $H^2$ in diagram (2) coming from the exact sequence of $G(L/K)$-modules $1 \to L^* \to J_L \to C_L \to 1$ (using Hilbert's 90 which says that $H^1(L/K, L^*)=0$). One can naturally extend (11) further to the right: ...$\to H^2 (L/K, C_L) \to H^3(L/K, L^*) \to ...$ If $L/K$ is cyclic, $H^3(L/K,L^*)\cong H^1(L/K,L^*)=0$, but in general ($L/K$ abelian or not), $H^3(L/K,L^*)$ is cyclic of order $n/n_0$ = the global degree divided by the l.c.m. of the local degrees, and the cohomological exact sequence stops there because $H^3(L/K,J_L) = 0$ (look at the local components).