TITLE: Langlands paper on representations of abelian algebraic groups
QUESTION [5 upvotes]: I have been working through Langlands paper Representations of Abelian Algebraic Groups, and I can't understand why one of his maps is obvious and how it helps. First I'll give the notation
Take a algebraic torus over $F$ that splits over a Galois extension $K$ of $F$, then the torus corresponds to a lattice $L$ on which $G=Gal(K/F)$ acts. Then let $\widehat{L}=Hom(L,\mathbb{Z})$. Then we define $\widehat{T}=Hom(\widehat{L},\mathbb{C}^*)$, and we define the Weil group as the extension $$0 \longrightarrow C_K \longrightarrow W_{K/F} \longrightarrow Gal(K/F) \longrightarrow 0.$$ ($C_{K}$ is the idele class group) and finally let $N = \sum_{g \in G} g$
On page 13 of his paper hes shown that there is an isomorphism $$H^{1}(W_{K/F},\widehat{T}) \longrightarrow Hom(H_{1}(W_{K/F},\widehat{L}),\mathbb{C}^{*}).$$ And  now he wants to show that that image of a continuous cocycle will be continuous. To do this he proceeds as follows. He defines $U_{K}$ as the elements on norm 1 in $C_K$ and constructs the exact sequence $$1 \longrightarrow U_K \longrightarrow C_K \longrightarrow M_K \longrightarrow 1,$$ $M_K$ being $\mathbb{Z}$ or $\mathbb{R}$. He then uses $L$ to make the sequence $$0 \longrightarrow Hom(L,U_K) \overset{\lambda}\longrightarrow Hom(L,C_K) \overset{\mu}\longrightarrow Hom(L,M_K) \longrightarrow 0$$. And now this is where I get lost, he claims there is an obvious map from $$N(Hom(L,C_K)) \cap Hom(L,U_K)/N(Hom(L,U_K))$$ to $$\hat{H}^{-1}(G,Hom(L,M_K))/ \mu \hat{H}^{-1}(G,Hom(L,C_K)).$$ Here the $\hat{H}$ means Tate groups.
Is this map obvious? and why is the second group finite? which he claims. 
Thank you

REPLY [3 votes]: The map is induced by a map $N(Hom(L, C_{K})) \cap Hom(L, U_{K})$ to $\hat{H}^{-1}(G, Hom(L, M_{K}))$, which is obtained as follows. If $z$ is in $Hom(L, U_{K})$ and $z = Nx$, where $x \in Hom(L, C_{K})$, where should we send $z$?  Well, $N(\mu(x)) = 0$.  Therefore $\mu(x)$ maps to something in $\hat{H}^{-1}(G, Hom(L, M_{K}))$ by definition of the latter group.  And that element is where we want to send $z$. 
For the finiteness question, try writing down the long exact sequence in Tate cohomology for the short exact sequence $0 \rightarrow U_{K} \rightarrow C_{K} \rightarrow M_{K} \rightarrow 0$.