Let $G$ be a group. Suppose we are given the category $C$ where the objects are $G$-sets and the morphisms $f: X \rightarrow Y$ are $G$-equivariant maps, i.e. satisfy $f(g \cdot x) = g \cdot f(x)$ for all $g \in G$ and $x \in X$.
I have shown that given a family $\left\{A_i \mid i \in I \right\}$ of objects in $C$, their co-product is the disjoint union.
Also, I have shown that the initial object is the empty $G$-set.
But what would be the terminal object and what is the product in this category?
Best Answer
Let $(A_i)_{i \in I}$ be a family of objects in your category. Now, the cartesian product $\prod_i A_i$ has a $G$-set structure via $g(x_i)_{i \in I} = (gx_i)_{i \in I}$ which makes the projections $\pi_j : \prod_i A_i \to A_j$ to be $G$-equivariant. Now, let's see that $\prod_i A_i$ with these arrows is the product of $(A_i)_{i \in I}$ by verifying the universal property of the product. If you have arrows $h_j : X \to A_j$, then
$$ h(x) := (h_i(x))_{i \in I} $$
is an arrow $h: X \to \prod_j A_j$ since for each $g \in G$,
$$ gh(x) = g(h_i(x))_{i \in I} = (gh_i(x))_{i \in I} = (h_i(gx))_{i \in I} = h(gx). $$
It is also clear from the definition of $h$ that $\pi_j h = h_j$ for all $j$, and this is the only arrow that satisfies this propery: if $h' : X \to \prod_iA_i$ is such that $\pi_jh' = h_j$, then
$$ h'(x)_j = \pi_jh'(x) = h_j(x) = h(x)_j \quad (\forall j \in I) $$
and so $h'(x) = h(x)$ for all $x \in X$. Thus our candidate verifies the universal property of the product, which makes it the product of the $A_i$ (up to isomorphism, of course). Here we take advantage of the fact that our objects and arrows have an underlying structure as sets and functions.
As for the final object: again, our objects have an underlying set structure, so a natural candidate is a one element set $\{*\}$. Note that any other one element $G$-set is isomorphic to $\{*\}$ but this is not in contradiction with $\{*\}$ being final. In effect, if $X$ is any $G$-set, the constant map $c:X \to *$ is $G$-equivariant, and it is the only possible arrow $X \to *$ since it is the only existing function from $X$ to $\{*\}$ to begin with.