Action groupoid as $G\rightrightarrows \textrm{Bij}(X)$?

Question

Let $G$ be a group and $X$ a set. A left action of $G$ on $X$ can be thought either as a map $G\times X\longrightarrow X$, $(g, x)\longmapsto g\cdot x$, satisfying:

$(i)$ $g\cdot (h\cdot x)=(gh)\cdot x$;

$(ii)$ $e_G\cdot x=x$,

or as a group homomorphism $\rho:G\longrightarrow \textrm{Bij}(X)$ where $\textrm{Bij}(X)$ is the group of bijections of $X$ (under composition).

Given an action $G\times X\longrightarrow X$ we can associate the action gropoid as follows:

$(a)$ The base is $X$;

$(b)$ The morphism set is $G$;

$(c)$ The source is $\textrm{pr}_2$;

$(d)$ The target is given the action map;

$(e)$ The identity-assigning map is the diagonal of $G$;

$(f)$ The inversion map is induced by the inversino of $G$.

Let us denote it by $G\times X\rightrightarrows X$. Is it possible to construct a groupoid $G\rightrightarrows \textrm{Bij}(X)$ using the map $\rho:G\longrightarrow \textrm{Bij}(X)$?

I think so and I assume the result will be isomorphic to $G\times X\rightrightarrows X$.

Thanks.

Answer 1

If I understand correctly, you want to take advantage of the curryfication of the maps $G\times X \rightrightarrows X$ to define a groupoid $G \rightrightarrows \operatorname{Bij(X)}$. This can not be done. I will try to explain why.

Recall that the curryfication of an arrow $f\colon A\times B \to C$ is given by $$ \bar f \colon A \to C^B \colon a \mapsto f(a,\cdot) $$ So, given your groupoid $G \times X \overset s {\underset t \rightrightarrows} X$, the curryfication of the target map $t$ is exactly $\rho$: sweet! But the curryfication of the source map is define for all $g\in G$ by $\bar s (g) = \mathrm{id}_X$. Hence, in your supposed groupoid $$G \overset{\bar s}{\underset{\bar t}\rightrightarrows}\operatorname{Bij(X)},$$ every arrows have the same source! That can't be good: for example, the arrows of a groupoid are required to be invertible. In fact this would imply that $\rho$ is the trivial morphism...

(And even in the particularly boring case of the trivial action, you wouldn't be able to define the identity map $\operatorname{Bij}(X) \to G$.)