Let $L/K$ be a finite Galois extension of nonarchimedean local fields, and let $A$ and $A^t$ be dual abelian varieties over $K$. Tate local duality tells us that $A^t(K)$ and $H^1(K, A)$ are Pontryagin dual locally compact Hausdorff abelian topological groups, and likewise over $L$. Under this duality, why does the dual of the restriction map $H^1(K, A) \rightarrow H^1(L, A)$ identify with the norm map $A^t(L) \rightarrow A^t(K)$? 


I am aware that this is "well-known" and that there are references that treat special cases, but I am interested in a proof or a reference in the stated generality.


Edit: Since $K$ is *any* nonarchimedean local field, the Tate local duality pairing $A^t(K) \times H^1(K, A) \rightarrow H^2(K, \mathbb{G}\_m)$ that I am using is defined using biextensions (as in III.7 of Milne's Arithmetic Duality Theorems).



Why is the norm map dual to restriction under Tate local duality?
Note first of all that the "norm" map you speak of does not make sense unless the field extension is separable. That is, for a separable field extension $k'/k$ of finite degree and a commutative $k$-group scheme $C$ there is a canonical "norm" homomorphism $C(k') \rightarrow C(k)$ induced by the passage to Galois-invariants on the homomorphism $C(k' \otimes\_k k\_s) \rightarrow C(k\_s)$ arising from the canonical $k\_s$-algebra decomposition of $k' \otimes\_k k\_s$ as a product of copies of $k\_s$ indexed by ${\rm{Hom}}\_k(k',k\_s)$. The same method defines a "norm" homomorphism $C(k' \otimes\_k S) \rightarrow C(S)$ for any $k$-algebra $S$, so it corresponds to a natural $k$-homomorphism ${\rm{R}}\_{k'/k}(C\_{k'}) \rightarrow C$ in terms of the Weil restriction when $C$ is finite type. This generalizes to when $k'/k$ is replaced with a finite etale cover. But beyond the separable setting there is no "reasonable" norm through finite flat maps whose formation commutes with base change.) So for this reason, we are going to assume that $L/K$ is separable. 


First, by the relationship between cup products and connecting homomorphisms (and the identification of $A^t(K)$ with ${\rm{Ext}}^1\_K(A, \mathbf{G}\_m)$ *functorially* in $A$) we see that the pairing $$A^t(K) \times {\rm{H}}^1(K,A) \rightarrow {\rm{H}}^2(K, \mathbf{G}\_m) = \mathbf{Q}/\mathbf{Z}$$ 
identifies covariant functoriality in degree-1 cohomology as adjoint to dual-functoriality of abelian varieties. 


Let $B$ denote the Weil restriction of scalars ${\rm{R}}\_{L/K}(A\_L)$; this is an abelian variety precisely because $L/K$ is separable. (If $L/K$ is not separable then $B$ is a smooth connected commutative $K$-group of dimension $[L:K]\dim(A)$ but is always non-proper if $A \ne 0$.) Let $j:A \rightarrow B$ be the natural inclusion. Ultimately we are going to transform your question into the above functoriality of the Tate pairing over $K$ applied to the $K$-homomorphism $j$.


By Shapiro's Lemma considerations, we naturally identify ${\rm{H}}^i(L,A)$ with ${\rm{H}}^i(K,B)$, and (check!) this identifies the restriction map on ${\rm{H}}^1$'s with ${\rm{H}}^1(j)$. Likewise, by the compatibility of Weil restriction with the formation of dual abelian variety (using the "norm" of the Weil restriction of the Poincare bundle), the norm map $A^t(L)\rightarrow A^t(K)$ is identified with the map on $K$-points induced by the dual homomorphism $j^t:B^t \rightarrow A^t$. Also, and most crucially, by a bit of diagram chasing (using the role of "norm of Poincare bundle" above) we see that the Tate pairing for $B$ over $K$ is identified with the composition of the Tate pairing for $A\_L$ over $L$ and the "norm" map on Brauer groups $${\rm{Br}}(L) = {\rm{H}}^2(L, \mathbf{G}\_m) =
{\rm{H}}^2(K, {\rm{R}}\_{L/K}(\mathbf{G}\_m))\rightarrow {\rm{H}}^2(K,\mathbf{G}\_m)={\rm{Br}}(K).$$
But when these flanking Brauer groups are identified with $\mathbf{Q}/\mathbf{Z}$, this composite map is the *identity*, as we see by analyzing pre-composition with the surjective restriction ${\rm{Br}}(K) \rightarrow {\rm{Br}}(L)$ (that intertwines with $[L:K]$ on $\mathbf{Q}/\mathbf{Z}$ via local class field theory, and the composition of $\mathbf{G}\_m \rightarrow {\rm{R}}\_{L/K}(\mathbf{G}\_m)$ with the "norm" map ${\rm{R}}\_{L/K}(\mathbf{G}\_m) \rightarrow \mathbf{G}\_m$ is $t \mapsto t^{[L:K]}$).


So putting it all together, the diagram you want to commute for $L/K$ separable really does translate into the elementary functoriality of the Tate pairing over $K$, applied to the map $j$ between abelian varieties over $K$.




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Remark: In case $L/K$ is inseparable there is still something one might wish to prove. Namely, focusing on the case when $L/K$ is purely inseparable of degree $p^n$ with $p = {\rm{char}}(K)>0$, so $L = K^{1/p^n}$ over $K$, we use the $p^n$-power isomorphism of local fields $L \simeq K$ to identify $A^t(L)$ with $(A^t)^{(p^n)}(K) = (A^{(p^n)})^t(K)$, so we could *define* the norm $A^t(L) \rightarrow A^t(K)$ to be the map on $K$-points induced by the $n$-fold relative Verschiebung homomorphism $V\_{A^t/K,n}:(A^t)^{(p^n)} \rightarrow A^t$. (The Verscheibung homomorphism is defined rather generally for commutative flat group schemes in section 4.2 of Exp. VII$\_{\rm{A}}$ of SGA3, and it is dual to the $n$-fold relative Frobenius isogeny by applying 4.3.3 of loc. cit. on $p^m$-torsion for all $m \ge 0$, or cheat and *define* it to be the dual of that Frobenius isogeny.) So you could ask if this "norm" is Tate-dual to the $L/K$-restriction map in degree-1 Galois cohomology.


The restriction ${\rm{H}}^1(K,A) \rightarrow {\rm{H}}^1(L,A) \simeq {\rm{H}}^1(K,A^{(p^n)})$ is likewise identified with ${\rm{H}}^1(F\_{A/K,n})$. Thus, since $V\_{A/K,n}$ is dual to $F\_{A/K,n}$, the elementary functoriality of the Tate pairing applied to the $K$-homomorphism $F\_{A/K,n}$ reduces the observation that that the Tate pairing for $A\_{K^{1/p}}$ over $K^{1/p}$ is equal to that of $A^{(p)}$ over $K$ via the isomorphism $K^{1/p} \simeq K$.


