Simple question about the Brauer group of a variety

Question

If $X$ is a variety over a field $k$ and $\overline{k}$ is an algebraic closure of $k$, then there is a homomorphism $\text{Br}(X) \rightarrow \text{Br}(X \times_k \overline{k})$. Since $\text{Br}$ is a contravariant functor, is the homomorphism from $X_{\overline{k}}$ to $X$ the projection map? I would have thought that the inclusion morphism $X \rightarrow X_{\overline{k}}$ would give a natural homomorphism $\text{Br}(X_{\overline{k}}) \rightarrow \text{Br}(X)$.

Answer 1

There is no "inclusion map" $X\to X_{\overline{k}}$. The map $X_{\overline{k}}\to X$ you refer to is indeed the projection map: it's the base change of $\operatorname{Spec} \overline{k}\to \operatorname{Spec} k$ along $X\to \operatorname{Spec} k$, the structure map of $X$.