Let $U:\textbf{Gr}\rightarrow\textbf{Set}$ be the forgetful functor from the category of groups to the category of sets.
Let $X\in\textbf{Set}$. I want to construct the solution set $S_X\subset\text{Ob}(\textbf{Gr})$ associated to $X$. I propose the following correspondence:
$$S_X:=\left\{G\in\textbf{Gr}\ |\ (\exists i)(X\xrightarrow{i} U(G)\land\text{"$i(X)$ generates $G$"})\right\}.$$
The trouble is that I am not sure how to prove that the class $S_X$ is a set. In his answer to the question Adjoint Functor Theorem, Martin says,
The class of these groups is essentially small, since $U(G)$ admits a
surjection from $\coprod_n(X\times\mathbf{N})^n$.
He uses the same argument in his answer to the question Existence proof of the tensor product using the Adjoint functor theorem. to conclude once more that the solution class is a set; here, he says
Then $\#|A'|\leq\aleph_0\cdot\#|M|\cdot\#|N|$. Hence, up to
isomorphism, there is only a set of such $A's.$
What sort of an argument is this? I am not clear about why showing that $|U(G)|\leq|\coprod_n(X\times\mathbf{N})^n|$, for each $G\in S_X$, allows us to conclude that $S_X$ is a set. Can someone please flesh this out for me?
Best Answer
If you are just interested in finding some solution set, then there is an easier answer. Arguably this is cheating though, because it uses the description of the left adjoint $F$ of the forgetful functor $U: \mathbf{Gr} \to \mathbf{Set}$.
That being said, let's fix some set $X$. Then I claim that $\{F(X)\}$, where $F$ is the free group on $X$, is a solution set. Indeed, let $f: X \to U(G)$ be any function of sets. Then because $F(X)$ is the free group on $X$ we can extend $f$ to a group homomorphism $\bar{f}: F(X) \to G$ such that $\bar{f} i = f$, where $i: X \to F(X)$ is the obvious injection of the generators.
As I said, this is cheating because it uses a description of the left adjoint of $U$. Indeed, the free group functor $F: \mathbf{Set} \to \mathbf{Gr}$ is the left adjoint of $U$. So the above is just a specific instance of the following fact.
Proving this would be a nice exercise. Hint: use the unit $\eta_C: C \to RL(C)$.
The class $S_X$ you proposed is of course going to capture the free group as well, so it will definitely contain enough. As already mentioned in the comments, $S_X$ is technically a proper class. However, this is no issue, because it will be essentially small. That means that there is a set $S_X'$ such that for every group $G \in S_X$ there is $G' \in S_X'$ with $G' \cong G$. So technically $S_X'$ would be our solution set, but we might as well work with $S_X$ and suppress the isomorphisms in our notation (so this is purely for convenience).
To prove that this class is essentially small, you quote another answer where they give a cardinality argument. Basically what they say is that there is some cardinal $\kappa$ such that for all $G \in S_X$ we have $|U(G)| \leq \kappa$. If we fix a set $Y$ then there is only a set of group structures on $Y$. So for every cardinal $\lambda \leq \kappa$ there is, up to isomorphism, only a set of group structures where the underlying set has cardinality $\lambda$. Hence there is, up to isomorphism, only a set of groups where the underlying set has cardinality $\leq \kappa$.
In both answers you quoted it is explained how this cardinality bound is found (although I think in the first one it should be $\mathbb{Z}$ in place of $\mathbb{N}$, but this is of no real consequence).