Free objects in $\mathrm{Set}(G).$

Question

What are the free objects in the category of $G$-sets for a group $G$?

After considerable deliberation (I'm not very bright), I'm pretty sure they are the $G$-sets $X$ on which $G$ acts freely, that is in such a way that only $e$ fixes any elements in $X$. I can prove it -- almost.

Suppose $X$ is a $G$-set on which $G$ acts freely. Let's take a transversal $B$ of the set of orbits $X/G$. I will show that $X$ is a free object in the category of $G$-sets, and $B$ is its basis. Let $Y$ be an arbitrary $G$-set, and let $f:B\to Y$ be any function. Suppose $\tilde f:X\to Y$ is a $G$-set morphism extending $f$. Let $x\in X$. Then there exist $b\in B$ and $g\in G$ such that $gb=x.$ Then $$\tilde f(x)=\tilde f(gb)=g\tilde f(b)=gf(b).$$ Hence the extension $\tilde f$ is unique.

For the existence of $\tilde f$, it's enough to check that the above is a good definition, or that it doesn't depend on the choice of $g$. Suppose $gb=x$ and $g'b=x$. Then $b=g^{-1}g'b$, and since $G$ acts freely on $X$, we must have $g=g'.$ Therefore the extension $\tilde f$ exists and is unique, and so $X$ is a free object in the category of $G$-sets.

Now suppose $X$ is a free object in the category of $G$-sets. Suppose for a moment that the following lemma is true.

Lemma. If $X$ is a free object in the category of $G$-sets, then its bases are exactly the transversals of $X/G.$

With this lemma I can show that $G$ must act freely on $X$. For let $x\in X.$ Then there exist a (unique, but I don't think I need this) $b\in B$ and $g\in G$ such that $gb=x.$ Suppose for an $h\in G$ we have $hx=x.$ This means that $$g^{-1}hgb=b.$$ Now let $f:B\to G$ be any function. Since $B$ is a basis, we can extend it uniquely to a $G$-set morphism $\tilde f$. ($G$ acts on itself by multiplication.) This implies that $$g^{-1}hgf(b)=f(b).$$ But $f(b)$ can be chosen arbitrarily, which means that $g^{-1}hg$ must fix every element of $G$, and so $h=e.$

So what I need to do is prove the lemma. Actually, I just need the "exactly" part. I just need to know that if $X$ is a free object in the category $G$-sets, then its basis must be a transversal of $X/G.$

Suppose $X$ is a free object. A transversal is a subset of $X$ such that there is exactly one element of each orbit in it. I think the uniqueness is analogous to linear independence in a vector space, and the existence is analogous to spanning the whole space. I can show the uniqueness (but as I said, I don't think I need it actually). Suppose $B$ is a basis of $X$. Suppose there are $b,b'\in B$ such that $b'=gb$ for some $g$. Again, by considering the functions $f:B\to G$, I can show that $g=e.$

But I don't see how to show that every orbit must have a representative in $B$. (Please keep in mind that at this point I still haven't proven that $G$ acts freely on $X$.) Since this is analogous to the fact that a vector space basis must generate the whole space, I've looked at the proof of that, the one in which we generate a subspace by the basis, quotient it out and show that the quotient is $0$. Can it be done here? I don't know how to define a quotient of a $G$-set. Or maybe I just need to prove it in some other way?

Answer 1

Sorry I've only read the first line of your question.

If $X$ is a set, the free $G$-set on $X$ is $G \times X$ with the $G$-action $g(h,x)=(gh,x)$. This is because for every $G$-set $Y$ every map $\alpha : X \to Y$ of sets extends uniquely to a $G$-map $G \times X \to Y$ via $(g,x) \mapsto g\, \alpha(x)$.

A $G$-set is free in the sense of category theory iff it is free in the usual sense: $\Rightarrow$ follows since $G$ acts freely on $G \times X$. For $\Leftarrow$ assume that $G$ acts freely on a set $X$. Let $B$ be a system of representatives of $G/X$. Then $B \hookrightarrow X$ extends to a $G$-map $G \times B \to X$, which is surjective by construction. It is also injective: If $gb=g'b'$, then $b=b'$ (by definition of $B$), and then $g=g'$ (since the action is free).

Good that you asked this, I didn't know this up to now!

(More generally, if $T : C \to C$ is a monad in a category, we can consider the forgetful functor $\mathsf{Mod}(T) \to C$, and it's left adjoint is given by $x \mapsto (T(x),\mu_x)$. For example, when $G$ is a monoid object in $C$ and $C$ has products, then $T(X) = G \times X$ is a monad and $\mathsf{Mod}(T)=\mathsf{Mod}(G)$ is the category of objects of $C$ together with a left $G$-action, for short $G$-objects. The free $G$-object on an object $X \in C$ is therefore $G \times X$ with the obvious $G$-action.)

Answer 2

I want to show that if a $G$-set $X$ is a free object in the category of $G$-sets, then its basis $B$ has to generate $X$. In other words,

$$\langle B\rangle=\bigcup_{b\in B}\mathrm{orb}(b)=X.$$

Let's say that an equivalence relation $\sim$ on $X$ is a congruence whenever for any $x,x'\in X$ and $g\in G$ we have $$a\sim x'\implies gx\sim gx'.$$

It's easy to show that the usual things hold: the kernel of a $G$-morphism is a congruence relation; quotient $G$-sets are well-defined; the first isomorphism theorem holds.

Now let $X$ be free with basis $B$. It's clear that the relation $\sim$ on $X$ defined by $$x\sim x'\iff(x=x'\vee x,x'\in\langle B\rangle)$$

is a congruence on $X$. It's a relation that contracts $\langle B\rangle$ to a single element and doesn't touch the rest of the elements. Let's denote that single element by $0$. Let $\tilde f:X\to X/\sim$ be the canonical $G$-epimorphism. Now clearly for any $b\in B$, we have $\tilde f(b)=0.$ But the function $\tilde f':X\to X/\sim$ assigning $0$ to every element of $X$ is also a $G$-morphism. $\tilde f$ and $\tilde f'$ are equal when restricted to $B$, and therefore must be equal, since $B$ is a basis. Hence for any $x,x'\in X$ we must have $x\sim x'$ and so every element of $X$ belongs to $\langle B\rangle.$