Vtyjkyuk

I am trying to understand the paragraph from Wikipedia on group-scheme action (to make this paragraph complete or "self-contained", I put the definition and basic settings at the end of the post):

One can also consider (at least some special case of) an action of a group functor: viewing $G$ as a functor, an action is given as a natural transformation satisfying the conditions analogous to the above.

In details, given a group-scheme action $sigma$, for each morphism $T to S$, $sigma$ determines a group action $$G(T) times X(T) to X(T);$$ i.e., the group $G(T)$ acts on the set of $T$-points $X(T)$.

Conversely, if for each $T to S$, there is a group action $$sigma_T: G(T) times X(T) to X(T)$$ and if those actions are compatible; i.e., they form a natural transformation, then, by the Yoneda lemma, they determine a group-scheme action $$sigma: G times_S X to X.$$

I understand the direction from group-scheme action $sigma$ to natural transformation. But I am confused about how to determine $sigma$ by given natural transformation. Here is my unsuccessful attempt:

We know $sigma$ is the transformation between $G(-)times X(-)$ to $X(-)$, where $$G(-)=h_G=mathrmHom(-,G).$$

However, the Yoneda Lemma says
$$F(A)cong mathrmHom(h_A,F)$$
where $Fin mathrmFun(mathcalC^mathrmop,mathbfSet)$.

Take $F=h_Gtimes h_X$ and $A=X$, we have
$$mathrmHom(X,G)times mathrmHom(X,X)cong mathrmHom(h_X,h_Gtimes h_X).$$
If the "arrows" are reversing, that is, $$sigmain mathrmHom(h_X,h_Gtimes h_X)$$ (in fact, it is in $mathrmHom(h_Gtimes h_X,h_X)$), so we can find a unique corresponding element in $$hatsigmainmathrmHom(X,G)times mathrmHom(X,X).$$

By the universal property of fiber product, we have
$$mathrmHom(X,G)times mathrmHom(X,X)cong mathrmHom(X,Gtimes_S X)$$
If we treat the arrow as its reverse again, we have $sigma$ corresponds to
$$Gtimes_S Xto X$$
(in fact, it should be $ X to Gtimes_S X$)

Obviously, it is absurd to reverse the arrows. So probably I was missing something small but important. Or my try is totally wrong and meaningless.
Any hint and answer are welcome! Thanks!

The definition of group-scheme action is as follows:

given a group $S$-scheme $G$, a left action of $G$ on an $S$-scheme $X$ is an $S$-morphism
$$sigma: G times_S X to X$$
such that

• (associativity) $sigma circ (1_G times sigma) = sigma circ (m times 1_X)$, where $m: G times_S G to G$ is the group law,


• (unitality) $sigma circ (e times 1_X) = 1_X$, where $e: S to G$ is the identity section of $G$.

The Yoneda lemma, when applied to the case where $F$ is itself a representable functor $h_B$, gives you an isomorphism $mathrmHom(A,B)=h_B(A)cong mathrmHom(h_A,h_B)$ (note that the two $mathrmHom$ don't apply to the same category); this means that the Yoneda embedding
$$mathcalCto mathrmFun(mathcalC^mathrmop,mathbfSet):Ato h_A=mathrmHom(-,A)$$
is fully faithful, and sometimes this fact is also called the Yoneda lemma.

In your case, you have a natural transformation $sigma :h_Gtimes h_XRightarrow h_X$, and as you note, the universal property of the fiber product implies that $h_Gtimes h_Xcong h_Gtimes_S X$; so you can also see $sigma$ as a natural tansformation $h_Gtimes_S XRightarrow h_X$, and as explained above, this must be induced by a map $Gtimes_S Xto X$.

To see that this map satisfies the associativity and unitality axiom, you apply the Yoneda lemma again, more precisely, the fact that the map inducing a specific transformation must be unique : indeed, this implies that in order to check for example that $sigma circ (1_G times sigma) = sigma circ (m times 1_X)$, it is enough to check that they induce the same natural transformation $h_Gtimes h_Gtimes h_XRightarrow h_X$; and this will just follow from the fact that every
$$sigma_T:G(T)times X(T)to X(T)$$
is itself a group action.