# Category of G-equivariant sets

I'm getting my very first introduction to categories. Recall that for $$G$$ a group and $$(X, \phi), (X', \phi')$$ $$G-$$sets, a $$G$$-equivariant map is a map $$f : X \longrightarrow Y$$ such that for all $$g \in G, x \in X: f(\phi(g,x)) = \phi'(g, f(x))$$.

I am now to show that there exists a category $$_G$$Set in which the objects are $$G$$-sets and the morphisms are equivariant maps.

Since I can think of examples of these, I certainly have some "non-empty collection" of $$G$$-sets with some corresponding $$G$$-equivariant maps. Why would this not be a category? i.e. what needs to be shown to promote this to a category?

• You need the existence of identity morphisms, and that composition of $G$-equivariant maps is $G$-equivariant. Also, but this is sort of trivial, you need to show that composition is associative. – jgon Mar 2 at 23:20
• I see. For $\mathcal{C}$ our desired category, can we then simply claim the collection of all $G$-equivariant sets to be ob($\mathcal{C}$)? – Jos van Nieuwman Mar 2 at 23:32
• Yes, and the morphisms are the $G$-equivariant maps between them. Then it just remains to verify @jgons claims. – Servaes Mar 2 at 23:33
• @JosvanNieuwman Look back at your definition of category. Is that a requirement for a category? (Is there a function between every pair of sets?) – Kevin Carlson Mar 3 at 0:23
• @JosvanNieuwman Most categories that are worked with are locally small and usually the "typical" definition of a category is specifically a locally small category. A locally small category is one where every hom-set is a set, not a (proper) class. (We talk about large categories or, rarely, metacategories when we want to allow proper classes of morphisms between two objects.) And yes, the empty set of morphisms is still a set of morphisms. – Derek Elkins Mar 3 at 1:41

## 1 Answer

In general it is not so hard to show that things are categories. Most of the time, the main axion, associativity of composition, comes down to associativity of function composition (i.e. when you are working over some category of sets with certain functions on them with maps that preserve their structure). If anything, the work to show a set based category is a category goes into showing the function composition is well defined.

As a reminder, to be a category, something must have a class of objects $$\text{Obj}(C)$$ and a class of morphisms $$\text{Mor}(C)$$, with two functions $$S$$ and $$T$$ (function of classes), source and target, from $$\text{Mor}(C)$$ to $$\text{Obj}(C)$$, one function $$\text{Id}$$ from $$\text{Obj}(C)$$ to $$\text{Mor}(C)$$, and one function $$M : \text{Mor}(C) \times_{\text{Obj}(C)} \text{Mor}(C) = \{ (g, f) \in \text{Mor}(C) \times \text{Mor}(C) : S(g) = T(f) \} \rightarrow \text{Mor}(C)$$ such that

1. $$S \circ \text{Id} (X) = X$$ and $$T \circ \text{Id}(X) = X$$ (henceforth, write $$\text{Id}(X) = \text{Id}_X$$, and think of it as the identity function). This is clear.

2. Associativity of the composition rule $$M$$. For our current example, this follows from associativity of functions on set.

3. $$\text{Id}_X \circ f = f = f \circ \text{Id}_X$$ for each $$f \in \text{Mor}(C)$$ such that $$S(f) = X$$ and $$T(f) = Y$$.

You may have seen a different definition, such as the one on wikipedia, which is slightly different. If so, it might also be a good exercise to show that these are the same.

For $$G$$-sets, choose $$\text{Obj}(G \text{-sets})$$ to be group actions of $$G$$ and $$\text{Mor}(G \text{-sets})$$ to be $$G$$-equivariant maps (it seems you already defined them). $$S$$ and $$T$$ of a map will simply be the source and target as usual. $$\text{Id}$$ will of course be the identity function (you could check that it's $$G$$-equivariant, if you want). The map $$M$$ here is simply composition (here we must show that it is well defined, or in other words that the composition of $$G$$-equivariant maps is $$G$$-equivariant). Note that it's defined only for maps whose domain and codomain match. It remains to check the three axioms.

1. The source and target of the identity map $$\text{Id}_X$$ are $$X$$ by construction.

2. This follows since function composition is associative. (it is true for all functions, so these in particular)

3. This follows since it is true for functions. (It is true for all functions, so these in particular).

• By Obj($G$-sets), do mean the underlying sets $X$ on which $G$ acts, or the actual actions $\phi : G \times X \rightarrow X$? – Jos van Nieuwman Mar 3 at 0:38
• The objects of $G$-sets are sets with the structure of the $G$-set. In other words, $G$-sets may be distinct even when they have the same underlying set. – Dean Young Mar 3 at 0:43
• Great, that makes a lot more sense. (Calling those the group actions confused me a bit though.) At first I wasn't feeling the maps $S$ and $T$. Now I think it simply states: "For every morphism there is an object which is the domain of the morphism, and there is an object which is the codomain of the morphism." The function Id then seems to state: "For every object $X$ there exists one (and only one) morphism of which $X$ is the domain and $X$ is the codomain". Am I making some sense here? I'll now turn to that last function $M$. – Jos van Nieuwman Mar 3 at 0:57
• "there is one (and only one morphism of which $X$ is the domain and $X$ is the codomain". There is one, but not necessarily exactly one. This special morphism $\text{Id}_X$ for an object $X$ is only unique after we require property 3 above. Think of it this way: will there be $G$-equivariant maps from $X$ to $X$ other than the identity map? Usually, there will be. If there are, then the identity is not the unique map with source $X$ and target $X$. – Dean Young Mar 3 at 1:01
• That's a good question (the class set distinction). Some people define a category to have these things $c(X, Y)$ to be sets. Some people have them to be classes and call the ones whose hom classes are sets "locally small". Almost always the category will be locally small in practice. This one is, since the hom-classes are subsets of $\text{Fun}(X, Y)$, the set of functions from $X$ to $Y$. A class contained in a set is a set by definition. – Dean Young Mar 3 at 1:23