# Realizing a right group action as a contravariant functor

Let $$G$$ be a group regarded as a one-object category $$\mathcal{G}$$. Then a functor $$F:\mathcal{G}\rightarrow \text{Set}$$ consists of a set $$S$$ (the value of the unique object $$G$$), together with for each $$g\in G$$ a function $$F(g) :S\rightarrow S$$ satisfying the functoriality axioms. So, writing $$F(g)(s)=g\cdot s$$ it follows that $$F$$ amounts to a set $$S$$ together with a function $$G\times S\rightarrow S$$, $$(g,s)\mapsto g\cdot s$$ satisfying $$(g'g)\cdot s = g'\cdot(g\cdot s)$$ and $$1\cdot s= s$$ for all $$g,g'\in G$$ and $$s\in S$$. So we see that $$F$$ describes a set equipped with a left group action.

Now, I'm having trouble seeing why a functor $$F:\mathcal{G}^{\rm op}\rightarrow \text{Set}$$ describes a right action of $$G$$ on a set $$S$$. If $$g,g'\in G$$, then $$F(gg')=F(g')g(g)$$ since $$F$$ is now a contravariant functor. However, I'm having trouble seeing why this satisfies the axioms of a right group action. What am I missing?

Defining $$S \times G \to S$$ by $$(s,g) \mapsto s \cdot g := F(g)(s)$$, the axiom $$(s \cdot g) \cdot h = s \cdot (gh)$$ follows from $$F(gh) = F(h) \circ F(g)$$, since $$s \cdot (gh) = F(gh)(s) = \big( F(h) \circ F(g) \big)(s) = F(h) \big(F(g)(s) \big) = F(h)(s \cdot g) = (s \cdot g) \cdot h.$$
$$x \cdot g = F(g)(x)$$
then $$x \cdot 1 = x$$ and
$$x \cdot gg' = F(gg')(x) = F(g')(F(g)(x)) = (x \cdot g) \cdot g'.$$
A more conceptual way to think about this would be to realize that $$B(G^{op}) \simeq (BG)^{op}$$ and a right $$G$$-action is a left $$G^{op}$$-action.