Remark after proof of existence of free groups (page-21,Clara loeh)

Question

Context

Definition 2.2.4 (Free groups, universal property). Let $S$ be a set. A group $F$ containing $S$ is freely generated by $S$ if $F$ has the following universal property: For every group $G$ and every map $\varphi: S → G$ there is a unique group homomorphism $\varphi: F\dashrightarrow G$ extending $\varphi$:

Proposition 2.2.6 (Free groups, uniqueness). Let $S$ be a set. Then, up to canonical isomorphism, there is at most one group freely generated by $S$.

The doubt

After the proof of the proposition mentioned in context, the following remark is given:

These isomorphisms are canonical in the following sense: They induce the identity map on $S$, and they are (by the uniqueness part of the universal property) the only isomorphisms between $F$ and $F'$ extending the identity on $S$

In the above, $F$ and $F'$ are two free groups and $S$ is the generating set of them. The isomorphisms are the map from $F$ to $F'$ and back. My questions are the following:

1. How does isomorphism 'induce' an identity map on $S$? In what sense is the word inducing used.

2. How does isomorphism extend the identity on $S$?

Sorry if these are very noob questions! I am still a beginner at this topic. One may find the book here. It's also available from amazon.

Answer 1

If you have a free group $F$ on a generating set $S$, then the elements of $S$ are (or can be seen as) elements of $F$. That is what the inclusion map $S \hookrightarrow F$ indicates.

(Note that in your Definition 2.2.4 it is really given that $S$ is subset of $F$.)

Now suppose that you have two free groups $F$ and $F'$ on the same generating set $S$. Then the point is that there is exactly one isomorphism between $F$ and $F'$ such that the restriction of that isomorphism to $S$ is the identity on $S$.

This is what is meant by 'induces the identity map on $S$'. Another way of saying this is that 'it extends the identity on $S$'. It makes sense because $S$ is a subset of both $F$ and $F'$.

So, just to expand what Proposition 2.2.6 is saying, avoiding the use of the word 'canonical' and being explicit about how you should interpret 'at most one':

Proposition 2.2.6. Let $S$ be a set. Let $F$ and $F'$ both be groups that are freely generated by $S$. Then

1. there is an isomorpism $\phi \colon F \to F'$ with the property that $\phi(s) = s$ for all $s \in S$;
2. if $\psi \colon F \to F'$ is an (other) isomorphism with the property that $\psi(s) = s$ for all $s \in S$, then $\phi = \psi$.