Showing that $F(S)$ is unique up to unique isomorphism and the universal property. Some nomenclature and the universal property.

Definition: Let $S$ be a nonempty set, the free group on $S$, denoted by $F(S)$ is the collection of all reduced words on $S$. Equivalently if we define an equivalence relation $\sim$ by stating that words are equivalent if one can be reduced to the other by eliminating terms of the form $ss^{-1}$ or $s^{-1}s$ from their words then they are equivalent, we may write $F(S) = S/\sim$.


Theorem (Universal Property): Let $G$ be a group and $S$ a set. For any map $f:S \rightarrow G$ there is a unique homomorphism $\varphi:F(S) \rightarrow G$  such that $\varphi|_S = f$.

Proposition: Show that $F(S)$ is unique up to unique isomorphism.
I know that this is supposed to use the universal property but I'm not quite understanding the steps taken in this proof.

Proof: Let $F'$ be another group containing $S$ and satisfying the conclusions of the above theorem. Applying the theorem to $F(S)$ and $F'$ give a unique homomorphism $\varphi: F(S) \rightarrow F'$ and $\varphi': F' \rightarrow F(S)$ satisfying $\varphi(s) = s$ and $\varphi'(s) = s$ for all $s \in S$. Consider the composition $\varphi' \circ \varphi: F(S) \rightarrow F(S)$, this is a homomorphism as the composition of homomoorphism and must also satisfy $(\varphi' \circ \varphi)(s) = s$ for every $s \in S$. But note that by the uniqueness of the component homomorphisms, that $\varphi' \circ \varphi = id_{F(S)}$ where this denotes the identity map. Similarly $\varphi \circ \varphi' = id_{F'}$. From this we conclude that $\varphi:F(S) \rightarrow F'$ is an isomorphism and it is the unique.


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*I understand that if $f:S \rightarrow F'$ is any map then the universal property gives a unique homomorphism $\varphi:F(S) \rightarrow F'$ so I see where the first homomorphism comes from, but what part of the universal property is giving us the second homomorphism $\varphi':F' \rightarrow F(S)$? The way I see it the universal property gives us a way to assert the existence of maps from $F(S)$ into some group $G$, even if we take $G = F(S)$ this would still only give us a map from $F(S) \rightarrow F(S)$ not $F' \rightarrow F(S)$ as is stated.

*Why is this called a universal property? I know that at a high level it seems to have something to do with category theory and adjoint pairs, is there any gentle explanation for what the theorem is really getting at here?

*What do we mean by unique up to unique isomorphism? I know that if two groups are isomorphic there may be many isomorphisms between them, so does this mean there is only one isomorphism between $F(S)$ and any other group it may be isomorphic to?

 A: $1.$ The goal is to prove that $F(S)$ is the only group which satisfies the universal property up to isomorphism. So we assume $F'$ is another group which contains $S$ and satisfies the same universal property, and we want to show it is isomorphic to $F(S)$. The universal property states that every function $f:S\to G$ (where $G$ is a group) can be uniquely extended to a homomorphism $F'\to G$. For instance, we have the inclusion $i:S\to F(S)$. Then by this universal property it can be extended to a homomorphism $\varphi':F'\to F(S)$.
$2.$ A universal property is a property which determines the object up to isomorphism. The proposition tells us that up to a natural isomorphism, the group $F(S)$ is the only group which satisfies the given property. So if we wish, we can define the free group on $S$ as a group which satisfies this given property. (of course to do so we first need to know that a group with such property exists, but now we already know that)
$3.$ That's a bad terminology. There might be more than one isomorphism between the two groups. However, they have one natural isomorphism, which comes from the universal property satisfied by the groups, and preserves the set $S$. (as you can see in the proof) If we had to choose an isomorphism between these groups, this specific isomorphism is obviously the most natural choice. So while it might not be the only isomorphism, it is clearly the best one.
