I want to do the following question (6.6k) in Vakli's Foundations of algebraic geometry

Suppose we have a contravariant functor $F$ from $Sch$ to $Groups$. Suppose further that $F$ composed with the forgetful functor is represented by the object Y. Show that the group operations on $F(X)$ uniquely determine $m:Y×Y \rightarrow Y$, $i:Y \rightarrow Y$, $e:Spec\mathbb{Z} \rightarrow Y$. And the group operation on $Mor(X,Y)$ is the same as on $F(X)$

Suppose $F$ composites with the forgetful functor is equal to $h_Y$ I know that $m:Y×Y \rightarrow Y$ must be the product of the projection morphism $π:Y×Y \rightarrow Y$ with itself and $e$ must be identity element of the group $Mor(spec\mathbb{Z} ,Y)$ For the morphism $i$, it composites with $f \in Mor(Y,Y) $ to give $f^{-1}$. But I have no idea why this morphism is unique. Also I don't know why $F(X)$ should have group structure induced by $Y$. I'm sorry if this is a stupid question. Any help or hints are appreciated.

  • $\begingroup$ What does "product of the projection morphism with itself" mean? $\endgroup$
    – Hoot
    Aug 5, 2016 at 4:01
  • $\begingroup$ I mean the group product in $Mor(Y×Y,Y)$ because $(\pi,\pi): Y×Y \rightarrow Y$ is the identity morphism and it composites with $m$ to give $m$ $\endgroup$ Aug 5, 2016 at 5:00

1 Answer 1


Consider the diagram given to you by the problem $$ \begin{matrix} & & \textbf{Groups} \\ F\nearrow & & \downarrow \\ \textbf{Schemes}^{op} & \xrightarrow{h_X} & \textbf{Sets} \end{matrix} $$ The question can be answered purely from a functor of points perspective since morphisms between representable functors can be turned into morphisms of schemes; so, what does it mean for $h_X$ to have the structure of a group schemes? You should have natural transformations $$ \begin{align*} &m:h_X\times h_X \to h_X \\ &\iota:h_X\to h_X \\ &e:h_\mathbb{Z} \to h_X \end{align*} $$ but what is a natural transformation? It's a map described on the functor evaluated on objects in your source category. So, because of the factorization of $h_X$ as $\text{Forget}\circ F$ we have for any morphism of schemes $S \to T$ a commutative diagram $$ \begin{matrix} h_X(T)\times h_X(T) & \xrightarrow{m(T)} & h_X(T) \\ \downarrow & & \downarrow \\ h_X(S)\times h_X(S) & \xrightarrow{m(S)} & h_X(S) \end{matrix} $$ where each row satisfies the diagrams for a group object in a category. You can do the same argument for the other group operations and call it a day.

  • $\begingroup$ You're welcome! This is an excellent theoretical example of the advantage of using the functor of points. $\endgroup$
    – 54321user
    Aug 6, 2016 at 18:02

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