Concrete categories and the concept of to be free Let $F$ be free on the set $S$, let $F′$ be free on the set $S′$, and
assume that $|S| = |S′|$. Prove that there is an isomorphism
$g : F → F′$.
By definition we have: $f:  S → U(G)$, $i:  S → U(F)$ s.t. $f=U(f*) \circ i $ for some $f*:F \to G$. Same for $f', f*', F', G'$. How do you define the isomorphism of $F$ and $F'$ from this diagrams?
thanks!
 A: Let $\mathcal{C}$ be a concrete category with forgetful functor $U : \mathcal{C} \to \mathbf{Set}$.
Suppose that $S, S^\prime$ are sets with $|S|=|S^\prime|$ and $F, F^\prime$ are objects of $\mathcal{C}$ with $F$ free on $S$ and $F^\prime$ free on $S^\prime$. Let $i : S \to U(F)$ and $i^\prime : S^\prime \to U(F^\prime)$ be the canonical injections.
Recall the universal property of $F$: for every object $A$ of $\mathcal{C}$ and every function $f : S \to U(A)$, there exists a unique morphism $g : F \to A$ in $\mathcal{C}$ satisfying $f = U(g) \circ i$. That is, for every object $A$ of $\mathcal{C}$, there is an isomorphism
$$
\operatorname{Hom}_{\mathcal{C}}(F, A)
\overset{\sim}{\to} \operatorname{Hom}_{\mathbf{Set}}(S, U(A)),
$$
given by $g \mapsto U(g) \circ i$.
$F^\prime$ satisfies a similar universal property.
Since $|S|=|S^\prime|$, there is an isomorphism (in $\mathbf{Set}$) $h : S \to S^\prime$. Let $f = i^\prime \circ h : S \to U(F^\prime)$. By the universal property of $F$, there exists a unique morphism $g : F \to F^\prime$ such that $f = U(g) \circ i$.
Similarly, let $f^\prime = i \circ h^{-1} : S^\prime \to F$. By the universal property of $F^\prime$, there exists a unique morphism $g^\prime : F^\prime \to F$ such that $f^\prime = U(g^\prime) \circ i^\prime$. Then $g \circ g^\prime : F \to F$ satisfies
$$
\begin{aligned}
U(g^\prime \circ g) \circ i
&= U(g^\prime) \circ U(g) \circ i \\
&= U(g^\prime) \circ f \\
&= U(g^\prime) \circ i^\prime \circ h \\
&= f^\prime \circ h \\
&= i \circ h \circ h^{-1} \\
&= i.
\end{aligned}
$$
Since $U(\operatorname{id}_F) \circ i = i$, the universal property of $F$ implies that $g^\prime \circ g = \operatorname{id}_F$. An analogous argument shows that $g \circ g^\prime = \operatorname{id}_{F^\prime}$. Therefore $g^\prime = g^{-1}$, and hence $F$ and $F^\prime$ are isomorphic.
