How does a morphism of k-varieties induce a morphism of their k-rational points? Let $k$ be an infinite field, not necessarily algebraically closed. By a k-variety I mean a k-scheme which is separated and of finite type. By a $k$-rational point of a $k$ variety I mean a point $x$ in $X$ such that the field $\kappa(x)$ is isomorphic to $k$ where $\kappa(x)$ is the stalk space of $X$ at $x$ quotiented by its maximal ideal. I have read in a paper that a morphism $f\colon X\longrightarrow Y$ of $k$-varieties induces a map from the $k$-rational points of $X$ to the $k$-rational points of $Y$. I suppose that this is simply the map $f$ restricted to the $k$-rational points of $X$. If that's the case how do I show that if $x$ is $k$-rational, so is $f(x)$?
 A: One way to describe a $k$-point is as a morphism $x:\operatorname{Spec}k \to X$, so the composition $f \circ x:\operatorname{Spec}k \to Y$ is also a $k$-point. Algebraically, this looks like taking the homomorphism $\mathcal O_{X,x} \to \kappa(x)$ and precomposing with $\mathcal O_{Y,f(x)} \to \mathcal O_{X,x}$.
If you replace the closed point $x$ by the spectrum of some ring (i.e. replace $k$ by a ring $A$), one sees that the relative analogue of a rational point is a rational section of a family of schemes (a.k.a. a rational point of the generic fiber) over an affine base, at which point it should be clear that this definition extends without complication to families over an arbitrary base scheme.
So the upshot is that at the categorical level, all of these are just geometric ways of describing right inverses of structure maps.
A: Question: "How does a morphism of k-varieties induce a morphism of their k-rational points?"
Answer: As a particular case if $X,Y$ are affine schemes you may argue as follows:
Let $f:X:=Spec(B) \rightarrow S:=Spec(A)$ where $k \rightarrow A \rightarrow B$ are maps of $k$-algebras. and let $\mathfrak{m}\subseteq B$ be a maximal ideal with $B/\mathfrak{m}\cong k$. Note: Any $k$-rational point in $B$ must be a closed point. You get a sequence
$$ k \rightarrow A/A\cap \mathfrak{m} \rightarrow B/\mathfrak{m}=k$$
hence we must have $A/A\cap \mathfrak{m} \cong k$ since $k$ is a field. Hence you get an induced set theoretic map
$$f(k): X(k)\rightarrow S(k).$$
