I'm reading the wiki page of local Tate duality.
https://en.wikipedia.org/wiki/Local_Tate_duality#cite_note-2
Let $K$ be a non-archimedean local field, let $K^s$ denote a separable closure of $K$, and let $G_K = Gal(K^s/K)$ be the absolute Galois group of K.
Denote by $μ$ the Galois module of all roots of unity in $K^s$. Given a finite $G_K$-module $A$ of order prime to the characteristic of K, the Tate dual of A is defined as $A'=\mathrm{Hom}(A,μ)$.
The theorem states that the pairing
$$H^i(K,A) \times H^{2-i}(K,A') \to H^2(K,μ) =\mathbb{Q/Z}$$
given by the cup product sets up a duality between the $H^i(K,A)$ and $ H^{2-i}(K,A)$ for $i=0,1,2.$
My question is very naive, in my knowledge cup product is defined such that we have a bilinear map
$$H^i(K,A) \times H^{2-i}(K,A') \to H^2(K,A\otimes A')$$
But in the statement of the theorem it's not $A\otimes A'$ but $\mu$.
Since we have a pairing $A \times A' \to \mu$, so does it mean that the "cup product" in the statement is actually a composition of real cup product with a morphism of cohomology group $H^2(K,A\otimes A') \to H^2(K,μ) $ induced by $A\otimes A' \to \mu$?
Yes that’s exactly correct (although it’s a common thing that happens in duality theorems in other settings as well)