The present question is inspired by an answer to this question of mine on applications of Category Theory in Abstract Algebra.
One of the answers stated that the universal property of free groups tells us that $Hom(F_n, G) \cong G^n$ where $F_n$ is a free group generated by n elements and G is any group.
I found that the universal property of free groups is usually formulated differently in terms of being able to uniquely extend a given set map $f:S \rightarrow G$ to a group homomorphism $\varphi :F_S \rightarrow G$ given a group $G$ and a set $S$, where $F_S$ denotes the free group on S.
My question is, are the two notions equivalent when $S$ is finite set of cardinality $n$? I think they are and I have a rough idea on how to prove it but I am not too sure if I am on the right track. Here's what I think:
Since $|S|=n$, the image set of each $f:S\rightarrow G$ determines an $n$-tuple in $G^n$, and we can straightway conclude that $|G^n|≥|Hom(F_n,G)|$ as the elements of $Hom(F_n,G)$ are nothing but extensions of all such $f:S \rightarrow G$. Conversely, each $n$-tuple $g \in G^n$ also uniquely determines a set-map $f_g:S \rightarrow G$ such that $f_g(i)=g_i$ and we can then always find a function $\varphi \in Hom(F_n,G)$ which uniquely extends $f_g$. This gives us the result that $|Hom(F_n,G)|≥|G^n|$. Thus, we can clearly construct a bijection between the two. Is this argument correct and is it sufficient to declare that the 2 are isomorphic in this case?
There is no need to argue in terms of cardinalities; you seem to be trying to use Cantor-Bernstein-Schroeder but we can just write down the bijection directly (and the arbitrary bijection CBS gives us wouldn't work for our purposes anyway since the bijection needs to be natural).
If you want to think in terms of a free group on a set $S$ then you can rewrite the universal property as
$$\text{Hom}(F_S, G) \cong G^S.$$
The point here is that $G^S$ is already exactly the set of functions $S \to G$. Now we can uniquely extend such a function to a homomorphism $F_S \to G$. That's the right-to-left part of the bijection. The left-to-right part is restricting a homomorphism $F_S \to G$ to the generators, getting a function $S \to G$. And the universal property says this is a bijection.