surface $X$ is smooth, projective and geometrically integral. The set $X(k)$ has exactly one point, whereas the set $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ is infinite. Furthermore, an infinite subset of $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ is contained in the Zariski open set $X\setminus X(k)$.

The inverse image of the unique $k$-point of $E$ in $X$ is $C^f$, hence $X$ has exactly one $k$-point. Since the étale Brauer–Manin obstruction is a functor from the category of $k$-varieties to the category of sets, the inclusion $\iota:C^f \hrar X$ induces an inclusion $$\iota_*: C^f({\mathbb{A}}_k)=C^f({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}} \hrar X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}.$$ We conclude that $X({\mathbb{A}}_k)^{\operatorname{\acute{e}t},\operatorname{Br}}$ contains infinitely many points that belong to the Zariski open set $X\setminus X(k)$.

Following a suggestion of Ambrus Pál made in response to the first version of this paper we now sketch a more elementary construction of a counterexample to weak approximation not explained by the étale Brauer–Manin obstruction. Consider any irreducible binary quadratic form $f(x,y)$ over $k$. Then our method produces a smooth, projective and geometrically integral surface $X$ fibred into conics over $E$. The fibre over the unique $k$-point of $E$ is the irreducible singular conic $C^f$, hence the set $X(k)$ consists of the singular point of $C^f$. Let $D$ be the discriminant of $f(x,y)$. If $v$ is a 