Let $f:X\rightarrow Y$ be a morphism of schemes. I know that the surjectivity of $f$ may be tested via the functor of points of $X$ and $Y$ in the following way:
The morphism of schemes $f$ is surjective if and only if for any field $K$ and $y\in Y(K)$, there is a field extension $L/K$ and $x\in X(L)$ whose image by $X(L)\rightarrow Y(L)$ is the image of $y$ under $Y(K)\rightarrow Y(L)$.
In particular, if the induced map $X(K)\rightarrow Y(K)$ on points over a field $K$ is surjective for every field $K$, then $f$ is surjective.
Is there an equivalent statement for the bijectivity of $f$?
I thank you for your help.