I have been working through Langlands paper [Representations of Abelian Algebraic Groups](http://www.sunsite.ubc.ca/DigitalMathArchive/Langlands/pdf/AbelianAlg-ps.pdf), 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



Langlands paper on representations of abelian algebraic groups
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 $0 \rightarrow U\_{K} \rightarrow C\_{K} \rightarrow M\_{K} \rightarrow 0$. 


