A Quick ACT: We know that a ring morphism $R \to S$ induces a map in the opposite direction $Spec S \to Spec R$. Well, the scheme morphism $X \to Y$ induces a map of Z-valued points $X(Z) \to Y(Z)$. These Z-valued points of an affine scheme are helpful, for instance, the rational solutions to $x^2 + y^2 = 9$ are the $\mathbb{Q}$-valued points of scheme $Spec \mathbb{Z} [x,y]/(x^2+y^2=0)$ and similarly, the integral solutions are the $\mathbb{Z}$-valued points.
Coming to the main ACT. Z-valued points of a projective scheme are a little bit harder. In fact, dealing with projective schemes is subtler and more troublesome when trying first time. Grothendieck only came up with the right interpretation of how the coordinates should look in the projective space, as they should be in a ring.
Now, as the map of rings induces a map of affine schemes in opposite directions, we can show that the map of graded rings induces a map of projective schemes in opposite directions. But again, it is a little subtle. So for a graded ring map
$$\varphi: R_\circ \to S_\circ$$
we have a map of the projective scheme map
$$(Proj S_\circ)/V(\varphi(S_+)) \to Proj R_\circ$$
and when $V(\varphi(S_+)) = \emptyset$ we have a morphism of projective schemes $Proj S_\circ \to Proj R_\circ$. However, it is very important to know that not every map of projective schemes comes from a map of graded rings in opposite directions. This is only a motivating case.
Moreover, the map $\varphi$ is a map of A-algebra and the induced morphism of projective schemes is a map of A-schemes.
