Why is the transfer map Tate-dual to restriction ?  In  one of their papers (before Theorem 7.2), Benson and Carlson state that the transfer map is Tate-dual to the restriction homomorphisms (also see Remark 1.3 of this recent paper).  
More precisely: If $H \le G$ are finite groups and $k$ a field those characteristic divides the order of $H$, then the there should be a commutative diagramm
$$\begin{array}{ccc}
\hat{H}^{-s-1}(G,k) & \cong & \text{Hom}_k(\hat{H}^sS(G,k),k) \newline
res^G_H \downarrow &  & \downarrow (tr^G_H)^\ast \newline
\hat{H}^{-s-1}(H,k) & \cong & \text{Hom}_k(\hat{H}^sS(H,k),k)
\end{array}$$
where the horizontal isomorphisms are Tate duality.
Does anyone know a reference with a proof or can provide a proof of this statement ? Thanks in advance. 
 A: I don't know of a reference, but the duality in question can be proved by results from Brown's book on group cohomology. I'll show the case $s\ge 0$. First note that for each integer $j$ there is an isomorphism 
$$\psi: \hat{H}^j(G,k) \xrightarrow{\sim} \text{Hom}_k(\hat{H}_j(G,k),k)$$
(Brown, VI.7.2) and for $s\ge 0$ there is an isomorphism $$\varphi: \hat{H}_{-s-1}(G,k) \xrightarrow{\sim} H^s(G,k)$$
(Brown, VI.4). Denote the $k$-dual of a vector space or of a homomorphism by $(-)^\ast$. Tate duality is then the composition 
$$t = (\varphi^{-1})^\ast\circ \psi: \hat{H}^{-s-1}(G,k) \xrightarrow{\sim} \hat{H}_{-s-1}(G,k)^\ast  \xrightarrow{\sim} H^s(G,k)^\ast.$$
Hence we have to show the commutativity of the diagramm
$$\begin{array}{ccccc}
\hat{H}^{-s-1}(G,k) & \xrightarrow{\psi} & \hat{H}_{-s-1}(G,k)^\ast  & \xleftarrow{\varphi^\ast} & H^s(G,k)^\ast \newline
{\scriptstyle \widehat{res}} \downarrow & & \downarrow \scriptstyle res^\ast & & \downarrow \scriptstyle tr^\ast\newline  
   \hat{H}^{-s-1}(H,k) & \xrightarrow{\psi} & \hat{H}_t(H,k)^\ast & \xleftarrow{\varphi^\ast} & H^s(H,k)^\ast
\end{array}$$
($t$ stands for $-s-1$ which the editor doesn't accept!?) The commutativity of the left hand square follows right from the definition of the maps and the right hand square commutes if we can show the commutativity of the following square: 
$$\begin{array}{ccc}
\hat{H}_{-s-1}(G,k) & \xrightarrow{\varphi} &  H^s(G,k)\newline 
{\scriptstyle \widehat{res}} \uparrow  & & \uparrow \scriptstyle tr^G_H \newline 
\hat{H}_{-s-1}(H,k)  & \xrightarrow{\varphi} & H^s(H,k) 
\end{array}\tag{1}$$
In order to describe $\varphi$ on chain level, let $P \to k$ be a projective resolution over $kG$ and let $F$ be a complete resolution such that $F_i=P_i$ and $F_{-i-1} = P_i^\ast$ for $i \ge 0$. Then $\varphi$ is induced by the composition 
$$\varphi: F_{-i-1}\otimes_{kG}k=P_i^\ast \otimes_{kG}k \xrightarrow{\alpha\otimes id} \text{Hom}_{kG}(P_i,kG) \otimes_{kG} k\xrightarrow{\beta}\text{Hom}_{kG}(P_i,k)$$
where $\alpha(f)(x)=\sum_{g \in G}f(g^{-1}x)g$ (Brown, VI.3.4) and $\beta(f \otimes a)(x)=f(x)a$ (Brown, I.8.3). Hence 
$$\varphi(f \otimes a)(x)=\sum_{g \in G}f(g^{-1}x)(ga)=\sum_{g \in G}f(g^{-1}x)a=tr^G_E(f)(x)a\tag{2}$$
where $f \in P_i^\ast, a \in k, x \in P_i$ and $E=\{1\}$. 
On chain level $(1)$ is given by the diagramm 
$$\begin{array}{ccc}
P_i^\ast \otimes_{kG} k & \xrightarrow{\varphi_G} & \text{Hom}_{kG}(P_i,k) \newline 
{\scriptstyle \kappa} \uparrow  & & \uparrow \scriptstyle tr^G_H \newline 
P_i^\ast \otimes_{kH} k & \xrightarrow[\varphi_H]{} & \text{Hom}_{kH}(P_i,k) \newline 
\end{array}\tag{3}$$
where $\kappa(f \otimes_H a)=f \otimes_G a$. With $f,a,x$ as above, we obtain 
$$(tr^G_H \circ \varphi_H)(f \otimes_H a)(x)=\sum_{g \in G/H}\varphi_H(f\otimes_H a)(g^{-1}x)
\overset{(2)}{=}\sum_{g \in G/H}tr^H_E(f)(g^{-1}x)a$$
$$\qquad=tr^G_H(tr^H_E(f))(x)a=tr^G_E(f)(x)a$$
$$\qquad\qquad=\varphi_G(f \otimes_G a)(x)=(\varphi_G \circ res)(f \otimes_H a)(x)$$
Thus the commutativity of $(3)$ is shown. QED
