the natural map $\operatorname{Br}(k)\to \operatorname{Br}(D)$ is injective. By Corollary \[c4\] we need to prove that $\operatorname{Br}(k)$ is the kernel of the map $$\operatorname{Br}(L)\oplus \operatorname{Br}(F)\oplus \bigoplus_{i=1}^n\operatorname{Br}(K_i)
\lrar \bigoplus_{i=1}^n\operatorname{Br}(LK_i) \oplus \bigoplus_{i=1}^n\operatorname{Br}(FK_i).$$ Recall that the maps $\operatorname{Br}(L)\to \operatorname{Br}(LK_i)$ are the restriction maps, whereas $\operatorname{Br}(K_i)\to\operatorname{Br}(LK_i)$ are opposites of the restriction maps. The same convention applies with $F$ in place of $L$.

Suppose that we have $\alpha_i\in\operatorname{Br}(K_i)$, $i=1,\ldots,n$, $\beta\in\operatorname{Br}(L)$ and $\gamma\in\operatorname{Br}(F)$ such that $(\alpha_i,\beta,\gamma)$ goes to zero. Let $v$ be a place of $k$, and let $k_v$ be a completion of $k$ at $v$. Since $Z^f(k_v)\not=\emptyset$, there is an index $i$ such that $v$ splits in $K_i$. Let $w$ be a degree 1 place of $K_i$ over $v$, so that the natural map $k_v\to K_{i,w}$ is an isomorphism. Let $a_v\in \operatorname{Br}(k_v)=\operatorname{Br}(K_{i,w})$ be the image of $\alpha_i$ under the restriction map $\operatorname{Br}(K_i)\to \operatorname{Br}(K_{i,w})$. This defines $a_v\in\operatorname{Br}(k_v)$ for any place $v$ of $k$, moreover, we have $a_v=0$ for almost all $v$ since each $\alpha_i$ is unramified away from a finite set of places (and the Brauer group of the ring of integers of $k_\nu$ is trivial). We have a commutative diagram of restriction maps $$\begin{array}{ccccc}
\operatorname{Br}(L)&\lrar&\operatorname{Br}(L\otimes_k K_i)&\longleftarrow&\operatorname{Br}(K_i)\\
\downarrow&&\downarrow&&\downarrow\\
\operatorname{Br}(L\otimes_k k_v)&\tilde\lrar&\operatorname{Br}(L\otimes_k K_{i,w})
&\longleftarrow&\operatorname{Br}(K_{i,w})
\end{array}$$ Here for a family 