A theorem of Auslander and Goldman states that for a regular integral scheme $X$ the inclusion of the generic point $\mathrm{Spec}\ K \to X$ induces an injective map $Br(X) \to Br(\mathrm{Spec}\ K)$.


Let $U \subset X$ be a dense open subset of a complex analytic space. In which situations (i.e. under which conditions on $U$ and $X$) is it known that the map $Br(X) \to Br(U)$ is injective?



when is a map of analytic Brauer groups induced by inclusion injective?
This got a bit long for a comment, so here's an answer. The answer is it is almost never injective. In general, for compact $X$, $Br(X)$ is infinite (at least if $H^2(X,O\_X)\neq 0$), while $Br(U)$ is finite if $U$ is Stein. I'm not sure how Jason's comment above gives a counterexample, as both $\mathbb{C}^2$ and $\mathbb{CP}^2$ have trivial analytic Brauer group. But, you could do something similar. Let $X$ be an K3 surface, and let $U\subset X$ be a Stein submanifold. Then, $U$ has the homotopy type of a $2$-dimensional CW-complex, so by the Oka principle, the analytic Brauer group of $U$ is zero. But, the analytic Brauer group of $X$ is pretty big (it's a copy of several $\mathbb{Q}/\mathbb{Z}$. For more details about this sort of thing, see Section 2 of Schroer's 2005 paper on topological methods for analytic Brauer groups.


