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?

  • 1
    $\begingroup$ It's the same as in sets. $\endgroup$ Nov 3, 2018 at 20:10
  • $\begingroup$ How would you define the unique morphism in the definition then? If the product is the Cartesian product? because this morphism must be G-equivariant. $\endgroup$
    – Kamil
    Nov 3, 2018 at 20:11
  • 1
    $\begingroup$ It is automatically $G$-equivariant. I should mention that the $G$-action on the product is the diagonal action: $g(x, y) = (gx, gy)$. This action is uniquely determined by the condition that the projection maps are $G$-equivariant. $\endgroup$ Nov 3, 2018 at 20:12
  • $\begingroup$ @Kamil The category of $G$-sets can be identified with the functor category $[G,\mathbf{Set}]$, treating $G$ as a one object groupoid, or we can make an algebraic theory whose models are $G$-sets. Either way, we get that limits are computed pointwise (and also that all colimits exist, and in the first case, immediately that they are computed pointwise, in the second case, we only immediately have that sifted colimits are computed pointwise). $\endgroup$ Nov 3, 2018 at 20:17
  • 1
    $\begingroup$ @Kamil I recommend proving to yourself that a $G$-set is just a functor from $G$ to $\mathbf{Set}$ where $G$ is viewed as a one-object category. Then show that natural transformations between such functors are exactly $G$-equivariant functions, i.e. the naturality condition is exactly equivariance. This just involves spelling out the definitions of functor and natural transformation and comparing them to the definitions of $G$-sets and $G$-equivariant functions. You can then show that the universal property of a product of such functors is witnessed by a functor that produces products of sets. $\endgroup$ Nov 4, 2018 at 1:06

1 Answer 1


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.

  • 1
    $\begingroup$ A final object is the same thing as a product of the empty family, so the description of the final object in this answer is already contained in the earlier part of the answer, as the special case $I=\varnothing$. (Of course the separate discussion of the final object is likely to still be useful for some readers, perhaps including the OP.) $\endgroup$ Nov 3, 2018 at 22:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .