Let $X$ and $Y$ be schemes over a field $K$. We assume, moreover, $X$ and $Y$ to be of finite type, separated and geometrically integral. Let $f:X \rightarrow Y$ be a surjective étale morphism. Is it true that then the induced morphism
$$f:X(K_s) \rightarrow Y(K_s)$$
is surjective? (With $K_s$ we denote a separable closure of $K$).
Please provide a suitable reference with any answer.
Let $\mathrm{Spec}(K_s) \to Y$ be a morphism. Then $X \times_Y \mathrm{Spec}(K_s)$ is non-empty (since this schematic fiber product surjects onto the topological fiber product, which is clearly non-empty because $f$ is surjective). Since $X$ étale over $Y$, we have that $X \times_Y \mathrm{Spec}(K_s)$ is étale over $K_s$. Since it is of finite type, it has a closed point $z$. Then $\kappa(z)$ is a finite separable field extension of $K_s$, hence $\kappa(z)=K_s$. Thus we get a $K_s$-morphism $\mathrm{Spec}(K_s) \to X \times_Y \mathrm{Spec}(K_s)$, i.e. a morphism $\mathrm{Spec}(K_s) \to X$ which lifts $\mathrm{Spec}(K_s) \to Y$.