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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?

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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 $$s \cdot (gh) = F(gh)(s) = \big( F(h) \circ F(g) \big)(s) = F(h) \big(F(g)(s) \big) = F(h)(s \cdot g) = (s \cdot g) \cdot h.$$

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If you wite

$$ x \cdot g = F(g)(x) $$

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.

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