# Some questions about inclusions of Brauer groups/sets

I have a feeling this question will be left hanging on StackExchange, so I've decided to just post it here, let me know if it's not suitable.

First of all, I'm trying to show that given an immersion of varieties $$j:X \rightarrow Y$$ defined over a number field $$k$$, we have an inclusion $$j(X(\mathbb{A}_k)^{\mathrm{Br}_1(X)}) \subset Y(\mathbb{A}_k)^{\mathrm{Br}_1(Y)}$$. Here $$X(\mathbb{A}_k)$$ denotes the topological space of adelic points of $$X$$ and $$\mathrm{Br}_1(X)$$ is the algebraic Brauer group, i.e., the kernel of the map $$\mathrm{Br}(X) \rightarrow \mathrm{Br}(X \times _k \bar{k})$$. A point $$(P_v)$$ in $$X(\mathbb{A}_k)^{\mathrm{Br}_1(X)}$$ is defined to be orthogonal to any $$A \in \mathrm{Br}_1(X)$$ under the Brauer-Manin pairing.

Given a $$k$$-rational point $$p\in X(k)$$, we have an induced morphism $$\bar{p}:\mathrm{Spec}\,k \rightarrow X$$, and so the composition $$j \circ \bar{p}$$ corresponds to a $$k$$-rational point of $$Y$$, i.e., $$j$$ maps $$k$$-rational points to $$k$$-rational points. Thus for a place $$v$$, we see that the same is true for the $$k_v$$-rational points, and so we have an inclusion $$X(\mathbb{A}_k) \rightarrow Y(\mathbb{A}_k)$$.

For a point $$(P_v) \in X(\mathbb{A}_k)^{\mathrm{Br}_1(X)}$$, we have $$((P_v),A) = 0$$ for any $$A \in \mathrm{Br}_1(X)$$. Assuming we have an injective map $$\mathrm{Br}(Y) \rightarrow \mathrm{Br}(X)$$, it would follow that $$\mathrm{Br}_1(Y) \rightarrow \mathrm{Br}_1(X)$$ is also injective, and so we have $$(j(P_v),A')=0$$ for any $$A' \in \mathrm{Br}_1(Y)$$. In particular, this means that $$j(P_v) \in Y(\mathbb{A}_k)^{\mathrm{Br}_1(Y)}$$ and we are done.

Question 1. Is the above assumption true? The restriction map $$\mathrm{Br}(Y) \rightarrow \mathrm{Br}(X)$$ is known to be an isomorphism in some situations (for example, in a purity result following the works of Gabber and then Cesnavicius), but for a general immersion of varieties, does injectivity hold?

Question 2. Now assuming my assumption is indeed true and so we have an inclusion $$j(X(\mathbb{A}_k)^{\mathrm{Br}_1(X)}) \subset Y(\mathbb{A}_k)^{\mathrm{Br}_1(Y)}$$, under what conditions would equality follow?

Remark. When $$E$$ is an elliptic curve of positive rank and $$C$$ is one $$k$$-rational point removed from $$E$$ and $$j:C \rightarrow E$$ is the obvious embedding, we can indeed show that we have an equality $$j(C(\mathbb{A}_k)^{\mathrm{Br}_1(C)}) = E(\mathbb{A}_k)^{\mathrm{Br}_1(E)}$$. But this follows because we have the stronger condition $$\mathrm{Br}(E) \cong \mathrm{Br}(C)$$.

Any references will be appreciated.