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I'm struggling with the definition of an action in category theory, this is nLab's definition:

An action of a group $G$ on an object $x$ in a category $C$ is a representation of $G$ on $x$, that is a group homomorphism $\rho : G \to Aut(x)$, where $Aut(x)$ is the automorphism group of $x$.

As indicated above, a more sophisticated but equivalent definition treats the group $G$ as a category denoted $\mathbf{B} G$ with one object, say $*$. Then an action of $G$ in the category $C$ is just a functor $$\rho : \mathbf{B} G \to C.$$
Here the object $x$ of the previous definition is just $\rho(*)$.

The definition indicating that each object in $G$ specifies an automorphism on $x$ makes sense to me, but I don't understand how this is equivalent to a functor that just identifies $x$.

And from Category Theory in Context:

Let $G$ be a group, regarded as a one-object category $\mathbf{B} G$. A functor $X : \mathbf{B} G \to C$ specifies an object $X \in C$ (the unique object in its image) together with an endomorphism $g^∗ : X \to X$ for each $g \in G$. This assignment must satisfy two conditions:

(i) $h^∗g^∗$ = $(hg)^∗$ for all $g, h \in G$.

(ii) $e^∗ = 1_X$, where $e \in G$ is the identity element.

This sounds like the second definition from nLab, except that it adds the endomorphism $g^*$, which seems to make it more compatible with the $\rho : G \to Aut(x)$ definition. How are these two definitions equivalent?

Ross Griebenow
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    It seems to me that the first "more sophisticated" approach winds up mapping the arrows of category $\mathbf B G$ to arrows in category $C$ from $x$ to $x$, i.e. the "automorphisms" of $x$. Note that the single object $*$ of $\mathbf B G$ is assumed to be mapped by the functor to $x$. To be a functor, $\rho$ has to respect arrow composition, i.e. associativity and preservation of the identity element built into the construction of $\mathbf B G$. Is that much clear? – hardmath Aug 25 '19 at 17:37
  • Thank you, yes, that makes sense. But then why is the endormorphism $g^*$ necessary in the third definition? – Ross Griebenow Aug 25 '19 at 17:42
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    It's just spelling out what it means to be a functor. – Berci Aug 25 '19 at 17:45
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    What @Berci said! The endomorphism $g^*$ is what we called $\rho(g)$ for an arrow $g$ (group element of G) in category $\mathbf B G$. – hardmath Aug 25 '19 at 17:49
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    Aaaah. Neat! Thank you. – Ross Griebenow Aug 25 '19 at 17:50

1 Answers1

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As the discussion at the linked nLab page notes, the word action can have various meanings and related applications in category theory. The phrase action of a group is more precise, but still susceptible of two distinct but equivalent framings.

We are first given a group $G$ and a category $C$ that has an object $x$. What is defined is an action of group $G$ on such an object $x$. It is said to be "a representation of $G$ on $x$, that is a group homomorphism $\rho:G \to \operatorname{Aut}(x)$". Note that this allows the trivial homomorphism, one that sends all elements of $G$ to the identity arrow on $x$.

Also the category $C$ might as well be reduced to just the one object $x$ in this context, because $\operatorname{Aut}(x)$ depends only on certain arrows from $x$ to itself.

The "more sophisticated" approach restates this idea by "treat[ing] the group $G$ as a category denoted $\mathbf B G$ with one object, say $*$." Here the arrows of category $\mathbf B G$ are the elements of $G$, the composition of arrows is given by group multiplication, and the identity arrow on $\mathbf B G$ corresponds to the identity element of $G$. This construction allows us to concisely define an action of group $G$ as a functor $\rho:\mathbf B G \to C$.

Verifying the equivalence of these two approaches is straightforward once the claim is established that a group essentially amounts to a one-object category in which every arrow has an inverse, as previously shown on Math.SE.

We are able to uniquely identify the object $x$ in category $C$ as the image of object $*$ in category $\mathbf B G$, and likewise the identity arrow on $x$ must correspond to the identity arrow on $*$. Because the arrows of $\mathbf B G$ are the elements of group $G$, the functor $\rho$ sends arrows of $\mathbf B G$ to arrows from $x$ to $x$ in category $C$, and because the group elements are invertible, these arrow images have inverse arrows in $C$, also from $x$ to $x$, and thus these arrow images belong to $\operatorname{Aut}(x)$.

It's hard to be sure I'm not overlooking an important point that causes doubts here about equivalence of the two approaches. A rigorous step-by-step demonstration strikes me as overkill, but if more detail is needed I'd be glad to supply them.

hardmath
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