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?
Best 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
$S \circ \text{Id} (X) = X$ 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.
Associativity of the composition rule $M$. For our current example, this follows from associativity of functions on set.
$\text{Id}_X \circ f = f = f \circ \text{Id}_X$ 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.
The source and target of the identity map $\text{Id}_X$ are $X$ by construction.
This follows since function composition is associative. (it is true for all functions, so these in particular)
This follows since it is true for functions. (It is true for all functions, so these in particular).