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Let $\mathcal{G}\rightrightarrows M$ be a groupoid and $G$ be a group acting on $M$. Then we might associate the action groupoid $G\ltimes M\rightrightarrows M$.

What are the groupoid morphisms from $\mathcal{G}\rightrightarrows M$ to the action groupoid?

I got to the conclusion that if $(F, f)$ is such a morphism then $F=(F^\prime, F^{\prime\prime})$ where $F^\prime:\mathcal{G}\longrightarrow G$ satisfies $$F^{\prime}(gh)=F^{\prime}(g)F^\prime(h)\quad \textrm{and}\quad F^\prime(1_x)=e_G,$$ and $F^{\prime\prime}:\mathcal{G}\longrightarrow M$ satisfies: $$F^{\prime\prime}(gh)=F^{\prime\prime}(g)\quad\textrm{and}\quad F^{\prime\prime}(1_x)=f(x).$$ Therefore $(F^\prime, f^\prime)$ is a morphism from $\mathcal{G}\rightrightarrows M$ to $G\rightrightarrows \{*\}$ where $f^\prime:M\longrightarrow\{*\}$ is the constant map $x\longmapsto *$.

What about $F^{\prime\prime}$, is it also a morphism between suitable groupoids?

PtF
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  • I've never seen the notation $\mathcal{G}\rightrightarrows M$ for a groupoid. Can you explain what it means? – anon Jul 10 '16 at 19:33
  • I can't tell you precisely but I interpret $\mathcal{G}\rightrightarrows M$ as follows: we have a groupoid whose class (in this case set) of morphisms is $\mathcal{G}$ and whose class (in this case set) of objects is $M$. Do you know category theory? Maybe the double arrows stand for the source and the target for sometimes we write the source above the top arrow and the target below the bottom arrow. But the notation is a nice shorthand for avoiding writing "let this be a groupoid which has $\mathcal{G}$ as class of morphisms and $M$ as a class of objects" every time we want to mention it. – PtF Jul 10 '16 at 22:17
  • That makes sense, thanks. – anon Jul 10 '16 at 22:34

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